Naval Postgraduate School Antennas & Propagation Distance Learning

TABLE OF CONTENTS (ver 1.2) I-1 Antennas & Propagation (1) I-48 Transmission Line Circuits (2) I-2 Antennas & Propagation (2) I-49 Transmission Line Circuits (3) I-3 Electromagnetic Fields and Waves (1) I-50 Transmission Line Circuits (4) I-4 Electromagnetic Fields and Waves (2) I-51 Transmission Line Impedance (1) I-5 Derivation of the Wave Equation (1) I-52 Transmission Line Impedance (2) I-6 Derivation of the Wave Equation (2) I-53 Transmission Line Impedance (3) I-7 Derivation of the Wave Equation (3) I-54 Transmission Line Impedance (4) I-8 Derivation of the Wave Equation (4) I-55 Impedance Matching I-9 Plane Wave Amplitude I-56 Quarter-Wave Transformers (1) I-10 Poynting's Theorem I-57 Quarter-Wave Transformers (2) I-11 Debye Model (1) I-58 Stub Tuning (1) I-12 Debye Model (2) I-59 Stub Tuning (2) I-13 Permittivity of a Dielectric With Loss I-60 Stub Tuning (3) I-14 Propagation in Lossy Media (1) I-61 Lumped Element Tuning (1) I-15 Propagation in Lossy Media (2) I-62 Lumped Element Tuning (2) I-16 Surface Current and Resistivtiy (1) I-63 Example: Lossless Power Divider I-17 Surface Current and Resistivtiy (2) I-64 Transmission Line Loss (1) I-18 Surface Current and Resistivtiy (3) I-65 Transmission Line Loss (2) I-19 Surface Current and Resistivtiy (4) I-66 Waveguides (1) I-20 Circular Polarization (1) I-67 Waveguides (2) I-21 Circular Polarization (2) I-68 Waveguides (3) I-22 Example I-69 Waveguides (4) I-23 Spherical Waves (1) I-70 Waveguides (5) I-24 Spherical Waves (2) I-71 Waveguides (6) I-25 Spherical Waves (3) I-72 Waveguides (7) I-26 Spherical Wave Amplitude I-73 Waveguides (8) I-27 Ray Representation for Waves I-74 Waveguides (9) I-28 Wave Reflection (1) I-75 Waveguides (10) I-29 Wave Reflection (2) I-76 Mode Patterns in Rectangular Waveguide I-30 Wave Reflection (3) I-77 Table of Waveguide Formulas I-31 Wave Reflection (4) I-78 Waveguide Magic Tee I-32 Wave Reflection (5) I-79 Circulators I-33 Wave Reflection (6) I-80 Microwave Switches I-34 Wave Reflection (7) I-81 Bandwidth (1) I-35 Wave Reflection (8) I-82 Bandwidth (2) I-36 Example (1) I-83 Filter Characteristics I-37 Example (2) I-84 Multiplexers I-38 Example (3) I-85 Waveguide Filters I-39 Example (4) I-86 Decibel Refresher I-40 Example (5) I-87 Coordinate Transform Tables I-41 Common Two-Wire Transmission Lines I-88 Coordinate Systems I-42 Transmission Line Equations (1) I-89 Radar and ECM Frequency Bands I-43 Transmission Line Equations (2) I-90 Electromagnetic Spectrum I-44 Transmission Line Equations (3) II-1 Antennas: Introductory Comments I-45 Transmission Line Equations (4) II-2 Radiation Integrals (1) I-46 Transmission Line Equations (5) II-3 Radiation Integrals (2) I-47 Transmission Line Circuits (1) II-4 Radiation Integrals (3)

1 II-5 Radiation Integrals (4) II-57 Ground Planes and Images (1) II-6 Hertzian Dipole (1) II-58 Ground Planes and Images (2) II-7 Hertzian Dipole (2) II-59 Ground Planes and Images (3) II-8 Hertzian Dipole (3) II-60 Crossed Dipoles (1) II-9 Solid Angles and Steradians II-61 Crossed Dipoles (2) II-10 Directivity and Gain (1) II-62 Crossed Dipoles (3) II-11 Dipole Polar Radiation Plots II-63 Crossed Dipoles (4) II-12 Dipole Radiation Pattern II-64 Polarization Loss (1) II-13 Directivity and Gain (2) II-65 Polarization Loss (2) II-14 Directivity and Gain (3) II-66 Polarization Loss (3) II-15 Examples II-67 Polarization Loss (4) II-16 Beam Solid Angle and Radiated Power II-68 Antenna Polarization Loss II-17 Gain vs. Directivity (1) II-69 Aircraft Blade Antenna II-18 Gain vs. Directivity (2) II-70 Blade Antennas Installed on an Aircraft II-19 Azimuth/Elevation Coordinate System II-71 Small Loop Antenna (1) II-20 Approximate Directivity Formula (1) II-72 Small Loop Antenna (2) II-21 Approximate Directivity Formula (2) II-73 Small Loop Antenna (3) II-22 Thin Wire Antennas (1) II-74 Helix Antenna (1) II-23 Thin Wire Antennas (2) II-75 Helix Antenna (2) II-24 Thin Wire Antennas (3) III-1 Array Antennas (1) II-25 Thin Wire Antennas (4) III-2 Array Antennas (2) II-26 Numerical Integration (1) III-3 Array Antennas (3) II-27 Numerical Integration (2) III-4 Array Antennas (4) II-28 Thin Wires of Arbitrary Length III-5 Array Antennas (5) II-29 Feeding and Tuning Wire Antennas (1) III-6 Array Antennas (6) II-30 Feeding and Tuning Wire Antennas (2) III-7 Principal Planes II-31 Feeding and Tuning Wire Antennas (3) III-8 Polarization Reference II-32 Feeding and Tuning Wire Antennas (4) III-9 Huygen's Principle (1) II-33 Feeding and Tuning Wire Antennas (5) III-10 Huygen's Principle (2) II-34 Calculation of Antenna Impedance (1) III-11 Huygen's Principle (3) II-35 Calculation of Antenna Impedance (2) III-12 Printed Circuit Dipole Array II-36 Calculation of Antenna Impedance (3) III-13 Scanned Arrays (1) II-37 Calculation of Antenna Impedance (4) III-14 Scanned Arrays (2) II-38 Calculation of Antenna Impedance (5) III-15 Grating Lobe Example (1) II-39 Calculation of Antenna Impedance (6) III-16 Grating Lobe Example (2) II-40 Self-Impedance of a Wire Antenna III-17 Sample Linear Array Patterns (1) II-41 The Fourier Series Analog to MM III-18 Sample Linear Array Patterns (2) II-42 Reciprocity (1) III-19 Array Beamwidth (1) II-43 Reciprocity (2) III-20 Array Beamwidth (2) II-44 Mutual Impedance (1) III-21 Linear Array Directivity (1) II-45 Mutual Impedance (2) III-22 Linear Array Directivity (2) II-46 Mutual Impedance (3) III-23 Antenna Sidelobe Control (1) II-47 Mutual Impedance (4) III-24 Antenna Sidelobe Control (2) II-48 Mutual Impedance of Parallel Dipoles III-25 Antenna Sidelobe Control (3) II-49 Mutual Impedance of Colinear Dipoles III-26 Aperture Efficiency II-50 Mutual-Impedance Example III-27 Some Common Amplitude Distributions II-51 Broadband Antennas (1) III-28 Summary of Array Characteristics II-52 Broadband Antennas (2) III-29 Finite Arrays and the "Edge Effect" (1) II-53 Broadband Antennas (3) III-30 Finite Arrays and the "Edge Effect" (2) II-54 Circular Spiral in Low Observable Fixture III-31 Finite Arrays and the "Edge Effect" (3) II-55 Broadband Antennas (4) III-32 Curved vs. Flat Arrays II-56 Yagi-Uda Antenna III-33 Two-dimensional Arrays (1)

2 III-34 Two-dimensional Arrays (2) III-84 Smart Antennas (1) III-35 Two-dimensional Arrays (3) III-85 Smart Antennas (2) III-36 Two-dimensional Arrays (4) III-86 Digital Beamforming III-37 Two-Dimensional Array of Dipoles IV-1 Equivalence Principle (1) III-38 Microwave Beamforming Networks (1) IV-2 Equivalence Principle (2) III-39 Microwave Beamforming Networks (2) IV-3 Equivalence Principle (3) III-40 Corporate Fed Waveguide Array IV-4 Apertures (1) III-41 Series Fed Waveguide Slot Array IV-5 Apertures (2) III-42 Array Example (1) IV-6 Rectangular Aperture (1) III-43 Array Example (2) IV-7 Rectangular Aperture (2) III-44 Array Example (3) IV-8 Rectangular Aperture (3) III-45 Array Example (4) IV-9 Rectangular Aperture (4) III-46 Digital Phase Shifters IV-10 Tapered Aperture (1) III-47 Effect of Phase Shifter Roundoff Errors IV-11 Tapered Aperture (2) III-48 Examples of Time Delay Networks IV-12 Tapered Aperture (3) III-49 Digital Phase Shifter IV-13 Tapered Aperture (4) III-50 Discrete Fourier Transform (1) IV-14 Summary of Aperture Distributions III-51 Discrete Fourier Transform (2) IV-15 Radiation Patterns From Apertures III-52 Discrete Fourier Transform (3) IV-16 Scanned Aperture III-53 Discrete Fourier Transform (4) IV-17 Aperture Example III-54 Discrete Fourier Transform (5) IV-18 Horn Antennas (1) III-55 Matlab Program to Compute the Array Factor IV-19 Horn Antennas (2) III-56 Sample Output of the Matlab Program IV-20 Horn Antennas (3) III-57 Receiving Antennas (1) IV-21 Horn Antennas (4) III-58 Receiving Antennas (2) IV-22 Horn Antennas (5) III-59 Receiving Antennas (3) IV-23 Horn Antennas (6) III-60 Friis Transmission Equation (1) IV-24 Horn Antennas (7) III-61 Friis Transmission Equation (2) IV-25 Horn Example III-62 Radar Range Equation (1) IV-26 Several Types of Horn Antennas III-63 Radar Range Equation (2) IV-27 Microstrip Patch Antennas (1) III-64 Radar Range Equation (3) IV-28 Microstrip Patch Antennas (2) III-65 Radar Range Equation (4) IV-29 Microstrip Patch Antennas (3) III-66 Noise in Systems (1) IV-30 Microstrip Patch Antennas (4) III-67 Noise in Systems (2) IV-31 Microstrip Patch Antennas (5) III-68 Noise in Systems (3) IV-32 Microstrip Patch Antennas (6) III-69 Noise in Systems (4) IV-33 Microstrip Patch Antennas (7) III-70 Calculation of Antenna Temperature IV-34 Reflector Antennas III-71 Noise in Systems (5) IV-35 Singly and Doubly Curved Reflector III-72 Noise Figure & Effective Temperature (1) IV-36 Classical Reflecting Systems III-73 Noise Figure & Effective Temperature (2) IV-37 "Deep Space" Cassegrain Reflector Antenna III-74 SNR of Active and Passive Antennas (1) IV-38 Multiple Reflector Antennas III-75 SNR of Active and Passive Antennas (2) IV-39 Geometrical Optics III-76 Active GPS Antenna IV-40 Parabolic Reflector Antenna (1) III-77 Comparison of SNR: Active vs. Passive IV-41 Parabolic Reflector Antenna (2) III-78 Multiple Beam Antennas IV-42 Reflector Antenna Losses (1) III-79 Radiation Pattern of a Multiple Beam Array IV-43 Reflector Antenna Losses (2) III-80 Active vs. Passive Multipbeam Antennas IV-44 Example (1) III-81 Beam Coupling Losses for a 20 Element IV-45 Example (2) Array IV-46 Example (3) III-82 Active Array Radar Transmit/Receive IV-47 Calculation of Efficiencies (1) Module IV-48 Calculation of Efficiencies (2) III-83 Adaptive Antennas IV-49 Cosine Feed Efficiency Factors

3 IV-50 Feed Example V-21 Wave Reflection at the Earth's Surface (4) IV-51 Calculation of Efficiencies (3) V-22 Atmospheric Refraction (1) IV-52 Reflector Design Using RASCAL (1) V-23 Atmospheric Refraction (2) IV-53 Reflector Design Using RASCAL (2) V-24 Atmospheric Refraction (3) IV-54 Reflector Design Using RASCAL (3) V-25 Atmospheric Refraction (4) IV-55 Reflector Design Using RASCAL (4) V-26 Atmospheric Refraction (5) IV-56 Microwave Reflectors V-27 Atmospheric Refraction (6) IV-57 Reflector Antenna Analysis Methods V-28 Fresnel Zones (1) IV-58 Dipole Fed Parabola (1) V-29 Fresnel Zones (2) IV-59 Dipole Fed Parabola (2) V-30 Fresnel Zones (3) IV-60 Dipole Fed Parabola (3) V-31 Diffraction (1) IV-61 Dipole Fed Parabola (4) V-32 Diffraction (2) IV-62 Crossed Dipole Radiation V-33 Path Clearance Example IV-63 Lens Antennas (1) V-34 Example of Link Design (1) IV-64 Lens Antennas (2) V-35 Example of Link Design (2) IV-65 Lens Antennas (3) V-36 Antennas Over a Spherical Earth IV-66 Lens Antennas (4) V-37 Interference Region Formulas (1) IV-67 Lens Antenna V-38 Interference Region Formulas (2) IV-68 Radomes (1) V-39 Diffraction Region Formulas (1) IV-69 Radomes (2) V-40 Diffraction Region Formulas (2) IV-70 Radomes (3) V-41 Ground Waves (1) IV-71 Radiation Pattern Effects of a Radome V-42 Ground Waves (2) IV-72 Hawkeye V-43 Ground Waves (3) IV-73 JSTARS V-44 Ground Waves (4) IV-74 Carrier Bridge V-45 Ground Waves (5) IV-77 Antenna Measurements (1) V-46 Urban Propagation (1) IV-76 Antenna Measurements (2) V-47 Urban Propagation (2) IV-77 Antenna Measurements (3) V-48 Urban Propagation (3) IV-78 Antenna Measurements (4) V-49 Urban Propagation Simulation IV-79 NRAD Model Range at Point Loma V-50 Attenuation Due to Rain and Gases (1) IV-80 Near-field Probe Pattern Measurement V-51 Attenuation Due to Rain and Gases (2) V-1 Propagation of Electromagnetic Waves V-52 Attenuation Due to Rain and Gases (3) V-2 Survey of Propagation Mechanisms (1) V-53 Ionospheric Radiowave Propagation (1) V-3 Survey of Propagation Mechanisms (2) V-54 Ionospheric Radiowave Propagation (2) V-4 Survey of Propagation Mechanisms (3) V-55 Electron Density of the Ionosphere V-5 Illustration of Propagation Phenomena V-56 The Earth's Magnetosphere V-6 Propagation Mechansims by Frequency V-57 Ionospheric Radiowave Propagation (3) Bands V-58 Ionospheric Radiowave Propagation (4) V-7 Applications of Propagation Phenomena V-59 Ionospheric Radiowave Propagation (5) V-8 Multipath From a Flat Ground (1) V-60 Ionospheric Radiowave Propagation (6) V-9 Multipath From a Flat Ground (2) V-61 Ionospheric Radiowave Propagation (7) V-10 Multipath From a Flat Ground (3) V-62 Ionospheric Radiowave Propagation (8) V-11 Multipath From a Flat Ground (4) V-63 Maximum Usable Frequency V-12 Multipath From a Flat Ground (5) V-64 Ionospheric Radiowave Propagation (9) V-13 Multipath From a Flat Ground (6) V-65 Ionospheric Radiowave Propagation (10) V-14 Multipath From a Flat Ground (7) V-66 Ionospheric Radiowave Propagation (11) V-15 Multipath From a Flat Ground (8) V-67 Ionospheric Radiowave Propagation (12) V-16 Multipath Example V-68 Ionospheric Radiowave Propagation (13) V-17 Field Intensity From the ERP V-69 Ducts and Nonstandard Refraction (1) V-18 Wave Reflection at the Earth's Surface (1) V-70 Ducts and Nonstandard Refraction (2) V-19 Wave Reflection at the Earth's Surface (2) V-71 Ducts and Nonstandard Refraction (3) V-20 Wave Reflection at the Earth's Surface (3)

4 5 Naval Postgraduate School Distance Learning

Antennas & Propagation

LECTURE NOTES VOLUME I REVIEW OF PLANE WAVES, TRANSMISSION LINES AND WAVEGUIDES

by Professor David Jenn

DIRECTION OF CONDUCTOR PROPAGATION SUBSTRATE GROUND (DIELECTRIC) PLANE Er

(ver 1.3) Naval Postgraduate School Antennas & Propagation Distance Learning Antennas & Propagation (1)

Antennas and propagation play a key role in the performance of communication, radar and electronic warfare systems. Antennas serve as transitions from guided wave structures to free space.

Example of a free-space communication channel:

SATELLITE

IONOSPHERE

RAIN, CLOUDS & FOG TROPOSPHERE

TRANSMITTER EARTH RECEIVER

1 Naval Postgraduate School Antennas & Propagation Distance Learning Antennas & Propagation (2)

Example of a monostatic radar:

SATELLITE

IONOSPHERE

RAIN, CLOUDS & FOG TROPOSPHERE

TARGET

TRANSMITTER/RECEIVER EARTH Topics of interest: general EM wave propagation transmission lines antennas propagation system issues involving antennas & propagation

2 Naval Postgraduate School Antennas & Propagation Distance Learning Electromagnetic Fields and Waves (1)

Electrical properties of a medium are specified by its constitutive parameters: -7 · permeability, m = momr (for free space, m º mo = 4p ´10 H/m) -12 · permittivity, e = eoer (for free space, e º eo = 8.85 ´10 F/m) · conductivity, s (for a metal, s ~ 107 S/m) Electric and magnetic field intensities are Er( x,y,z,t) V/m and Hr (x,y,z,t) A/m

· they are vector functions space and time, e.g., in cartesian coordinates r E( x,y,z,t) = xˆE x (x,y,z,t) + yˆE y (x,y,z,t) + zˆ E z (x,y,z,t)

· similar expressions for other coordinates systems r r · fields arise from currents J and charges rv on the source ( J is the volume 2 3 current density in A/m and rv is volume charge density in C/m ) Electromagnetic fields are completely described by Maxwell’s equations: ¶Hr (1) Ñ ´ Er = -m (3) Ñ× Hr = 0 ¶t r r r ¶E r (2) Ñ ´ H = J + e (4) Ñ×E = rv / e ¶t 3 Naval Postgraduate School Antennas & Propagation Distance Learning Electromagnetic Fields and Waves (2)

Most sources of electromagnetic fields have a sinusoidal variation in time (time-harmonic sources). All of the field quantities associated with the sources will have the same sinusoidal time variation. Therefore, we suppress the time dependence for convenience, and work with a time independent quantity called a phasor. The two are related by Er(z,t) = Â{Er(z)e jwt } · Er( z) is the phasor representation; Er( z,t) is the instantaneous quantity

· Â{×} is the real operator (i.e., “take the real part of”) · j = -1

Since the time dependence varies as e jwt , the time derivatives in Maxwell’s equations can be replaced by ¶ / ¶t º jw in the time-harmonic case: (1) Ñ ´ Er = - jwmHr (3) Ñ×Hr = 0 (2) Ñ ´ Hr = Jr + jweEr (4) Ñ× Er = r /e v Any fields or waves that exist in a particular region of space must satisfy Maxwell’s equations and the appropriate boundary conditions.

4 Naval Postgraduate School Antennas & Propagation Distance Learning Derivation of the Wave Equation (1)

r The wave equation in a source free region of space (J = 0, rv = 0) is derived by taking the curl of Maxwell’s first equation:

æ ¶Hr ö ¶ ¶ æ ¶Er ö ¶ 2 Er Ñ´ (Ñ´ Er)= Ñ´ç - m ÷ = -m (Ñ´ Hr )= -m çe ÷ = -me è ¶ t ø ¶ t ¶ t è ¶ t ø ¶ t2 where it is assumed that the medium is time invariant (m and e not time dependent). Now use the vector identity Ñ ´ Ñ ´ Er = Ñ(Ñ · Er)- Ñ2 Er = -Ñ2 Er 123 = rv =0 to obtain ¶ 2Er ¶ 2 Er - Ñ2Er = -me Þ Ñ2 Er - me = 0 ¶ t2 ¶ t2 In the frequency domain, using phasors, and noting that ¶ / ¶t º jw yields

2 r 2 r 2 r 2 r 2 r 2 r Ñ E +w mce c E = Ñ E + kc E = Ñ E -g E = 0

5 Naval Postgraduate School Antennas & Propagation Distance Learning Derivation of the Wave Equation (2)

The subscript “c” denotes the possibility of a complex quantity: ec = e¢ - je¢¢ and mc = m¢ - jm¢¢. The imaginary terms are nonzero if the medium is lossy. Also, we have defined g º a + jb = jkc = jw mcec where a = attenuation constant (Np/m) and b = 2p /l = phase constant (rad/m). In free space, which is a lossless medium, the subscripts “o” are often used

ec = eo ,mc = mo Þ a = 0, b = w eo mo Frequently k is used in place of b when the medium is lossless and unbounded. There is a similar wave equation that can be derived for the magnetic field intensity Ñ2 Hr -g 2Hr = 0 The simplest solutions to the wave equations are plane waves. An example of a plane wave propagating in the z direction is: r -gz E(z) = xˆEoe In general, Eo is a complex constant that depends on the strength of the source and its distance from the observer at z.

6 Naval Postgraduate School Antennas & Propagation Distance Learning Derivation of the Wave Equation (3)

The instantaneous value of the electric field is r r jwt -az E(z,t) = Â{E(z)e }= xˆ Eoe cos(wt - bz) Time snapshots of the field are shown below

· wave propagates in the +z direction E x l · l = wavelength E o DIRECTION OF · w = 2p f (rad/sec) PROPAGATION u p · f = = frequency (Hz) t1 t2 l z w 1 · phase velocity is u = = (in p b me 8 free space up = c = 2.998´10 m/s) · x polarized (direction of the electric - Eo field vector is xˆ ) · maximum amplitude of the wave is Eo

7 Naval Postgraduate School Antennas & Propagation Distance Learning Derivation of the Wave Equation (4)

The magnetic field vector is obtained from Maxwell’s first equation Ñ´ Er = - jwmHr r - jbz Ñ ´ E Ñ ´(xˆE e ) j ¶ - jbz b - jbz Hr = = o = (E e )yˆ = yˆ E e - jwm - jwm we ¶z o we o 123 º Ho wm m E The intrinsic impedance of the medium is h º = = o . Plane waves are b e Ho transverse electromagnetic (TEM) waves, and obey the simple relationship l r kˆ´ Er E Hr = where kˆ is a unit vector in h the direction of propagation (zˆ in this Hr case). The vectors (kˆ,Er,Hr ) are mutually orthogonal and form a right- kˆ PROPAGATION handed system. DIRECTION

8 Naval Postgraduate School Antennas & Propagation Distance Learning Plane Wave Amplitude

r Snapshot of a plane wave propagating in the +y direction E( y,t) = zˆEo cos(wt - by) at time t = 0

9 Naval Postgraduate School Antennas & Propagation Distance Learning Poynting’s Theorem

Poynting’s theorem is a statement of conservation of energy. For a volume of space, V, bounded by a closed surface, S, and filled with a medium (s ,m,e )

r r ¶ 1 2 1 2 2 (E ´H ) · dsr = - eE + mH dv - sE dv ò ò(2 2 ) ò S ¶t V V 1442443 14444244443 14243 POWER FLOWING POWER STORED IN THE POWER THROUGH S FIELDS INSIDE OF S LOSS IN S

The quantity Wr = Er ´ Hr (W/m2 ) is known as the Poynting vector. The instantaneous value of the Poynting vector is Wr (x, y, z,t) = Er(x, y, z,t)´ Hr (x, y, z,t) = Â{Er(x, y, z)e jw t }´Â{Hr (x, y, z)e jw t } and the time-averaged Poynting vector is T Wr = 1 Wr (x, y,z,t)dt = 1 Â Er(x, y,z)´ Hr (x, y,z)* av T ò 2 { } 0 The time-averaged value can be found directly from the phasor fields quantities.

10 Naval Postgraduate School Antennas & Propagation Distance Learning Debye Model (1)

The Debye model has been used to predict the interaction of EM waves with materials since about 1910. Molecules are represented by positive and negative charge centers. ELECTRON CLOUD (-) NUCLEUS (+)

THE CHARGE CENTERS ARE COINCIDENT IN THE ABSENCE OF AN EXTERNAL FIELD

The response of a molecule to an external electric field is expressed in terms of a r polarization, P( t)

r r P = cE ext ELECTRON CLOUD NUCLEUS (+) CHARGE CENTER (-) r E ext This is the simplest form of a dipole: two equal and opposite charges that are slightly displaced. The separation that arises due to the external field is referred to as the electronic polarization and the quantity c is the susceptibility. 11 Naval Postgraduate School Antennas & Propagation Distance Learning Debye Model (2)

It takes a finite amount of time for the molecules to respond to an applied external field. The response is of the form

-t /t P(t) = Po e { c(0)Eext

where t is the relaxation constant (about 10-15 second). Assumptions are that all dipoles are identical, independent, and all relaxation times are the same. In fact, dipoles are spatially and temporally coupled, relaxation times vary, and other types of polarization exist. The Debye model is never seen in real materials, but it can be approached for single particle non-interacting systems such as gases. Other types of polarization: Ionic: mutual displacement of the charge centers (10-13 second) Orientational: rotation of the molecules (10-11 second) Media have a far more complex EM relaxation behavior than previously realized. A new theory (Dissado-Hill) takes all of these factors into account.

12 Naval Postgraduate School Antennas & Propagation Distance Learning Permittivity of a Dielectric With Loss

· Example of a material with resonances in the millimeter wave frequency region · Complex dielectric constant: ec = e¢ - je¢¢ · Below millimeter wave frequencies, e ¢ ³1 and approximately constant and e ¢¢ » 0

(From Bohren and Huffman, Absorption and Scattering of Light by Small Particles)

· Phase velocity (e ¢ is the real part)

1 w u = = p me¢ b

High frequencies travel faster than low frequencies

13 Naval Postgraduate School Antennas & Propagation Distance Learning Propagation in Lossy Media (1)

As waves propagate through a lossy medium, energy is extracted from the wave and absorbed by the medium. There are three general sources of loss: 1. ohmic loss, which is due to the collision of free charges in a conductor, and is accounted for by a finite conductivity, s < ¥ (s = ¥ is a perfect electric conductor, PEC) 2. dielectric loss, due to polarization of molecules caused by an external electric field, and it is accounted for in the imaginary part of ec 3. magnetic loss, due to magnetization of the molecules caused by an external magnetic field, and it is accounted for in the imaginary part of mc

Most materials are non-magnetic (m = mo ) and therefore magnetic losses can be neglected. For all other materials, either ohmic loss or dielectric loss dominates. For an imperfect conductor, an equivalent complex dielectric constant can be derived by introducing the conduction current into Maxwell’s second equation Ñ ´ Hr = Jr + sEr + jweEr æ s ö = Jr + jwçe + ÷Er è jw ø 14243 e c 14 Naval Postgraduate School Antennas & Propagation Distance Learning Propagation in Lossy Media (2)

The attenuation constant determines the rate of decay of the wave. General formulas for the attenuation and phase constants of a conductor are: 1/ 2 1/ 2 ì 2 ü ì 2 ü ïme é æ s ö ùï ïme é æ s ö ùï a = w í ê 1+ ç ÷ -1úý b = wí ê 1+ ç ÷ +1úý 2 èweø 2 èweø îï ëê ûúþï îï ëê ûúþï

For lossless media s = 0 Þ a = 0. 1 0.8

Traditionally, for lossless cases, k is 0.6

used rather than b . For good 0.4 conductors (s /we >> 1), a » p m fs , 0.2 and the wave decays rapidly with 0 distance into the material. A sample -0.2 -0.4

plot of field vs. distance is shown. -0.6

-0.8 ELECTRIC FIELD STRENGTH (V/m) (To apply the formulas to a dielectic -1 0 1 2 3 4 5 with losses, substitute e ® e ¢ and DEPTH INTO MATERIAL (m) s /w ® e ¢¢.)

15 Naval Postgraduate School Antennas & Propagation Distance Learning Surface Current and Resistivity (1)

For good conductors the current is concentrated near the surface. The current can be r approximated by an infinitely thin current sheet, with surface current density, Js A/m and 2 surface charge density, rs C/m Current in a good conductor Surface current approximation r E Er i i ˆ ˆ k i BOUNDARY ki BOUNDARY r r J s J At an interface between two media the boundary conditions must be satisfied: r r r r (1) nˆ21 ´(E1 - E2 ) = 0 (3) nˆ21 ×(E1 - E2 ) = rs / e (2) nˆ ´(Hr - Hr ) = Jr (4) nˆ ×(Hr - Hr ) = 0 21 1 2 s 21 1 2

REGION 1 nˆ21 r J s rs INTERFACE REGION 2

16 Naval Postgraduate School Antennas & Propagation Distance Learning Surface Current and Resistivity (2)

The field in a good conductor is significant only within the first skin depth from the surface. The skin depth is the distance into the material at which the amplitude has decayed by a factor of 1/e. r Hr E kˆ SMALL RECTANGULAR i BRICK AT THE SURFACE nˆ w l r t x J y z

The resistance of the block is R = l = l , where A is the cross sectional area transverse sA stw to the direction of current flow. If we choose a square of surface area, l = w, and the thickness a skin depth ds , then the result is the surface resistivity, which is defined as 1 Rs = . It has units of “ohms per square” (W/ ) sds 17 Naval Postgraduate School Antennas & Propagation Distance Learning Surface Current and Resistivity (3)

For a plane wave normally incident on a metal surface, the time-averaged power density in the material is 1 1 ìE2 ü E 2 zˆRE2 Wr =  Er ´ Hr * =  o e-2az z = o e-2az h = z o e-2az av { } í * ýˆ 2 { } ˆ 2 2 2 îh þ 2h 2h It is assumed that Eo is real for convenience. For a good conductor the intrinsic impedance is approximately 1 + j h º R + jX » sds (Note that the real part is equal to the surface resistivity previously defined.) We can replace the original infinitely thick medium with an infinitesimally thin sheet that satisfies the same boundary condition: r r r r r r r nˆ ´ H = J s ® nˆ ´ nˆ ´ H = nˆ ´ Js ® nˆ (nˆ · H )- H (nˆ · nˆ) = nˆ ´ J s 123 =0 ˆ r r r ki ´ E - nˆ ´ E where H = = and hs is the surface impedance of the thin sheet. hs hs

18 Naval Postgraduate School Antennas & Propagation Distance Learning Surface Current and Resistivity (4)

The boundary condition can be written as r r r r hsnˆ ´ J s = nˆ ´ E ® hs J s = Etan

and the power dissipated by the current flowing on the boundary is

ì ü 1 r r * 1 ï r r * ï Ploss = - Â{E ´ H }· nˆ = - Âí nˆ ´ E · H ý 2 2 123r îïhsnˆ´J s þï ì ü r 2 r 2 1 ï ï Js Js = Â h Jr · nˆ ´ Hr * = Â{h }= R í s s 123ý s s 2 ï r* ï 2 2 î J s þ

The surface impedance concept gives a convenient means of computing the ohmic loss of conductors. We can avoid integrating the volume current inside of the conductor (a volume integral), and need only integrate the surface current (a surface integral). This is only an approximation, but it is very accurate for good conductors. These calculations are necessary in order to determine transmission line loss. 19 Naval Postgraduate School Antennas & Propagation Distance Learning Circular Polarization (1)

A circularly polarized plane wave can be obtained by superimposing two equal amplitude linearly polarized plane waves that are in space and time quadrature (quadrature implies 90 degrees): r r 1. space quadrature, E1^ E2 (for example, Ex vs. E y ) 2. phase quadrature, a e± jp / 2 factor between the two fields Example: Two linearly polarized plane waves propagating in the z direction

Er = xˆE e- jbz and Er = yˆE e- jbz e± jp / 2 1 xo 2 yo Equal amplitudes, E = E º E xo yo o Er(z) = Er (z) + Er (z) = xˆE e- jbz + yˆE e- jbz e± jp / 2 = E e- jbz (xˆ ± j yˆ) 1 2 o o 123 o ± j The instantaneous value at z = 0 is r r jw t E(z,t) = Â{E(z)e }= xˆEo cos(w t) ∓ yˆEo sin(w t) The vector rotates about the z axis. The tip of the electric field vector traces out a circle of radius Eo . The direction of rotation depends on the sign of j.

20 Naval Postgraduate School Antennas & Propagation Distance Learning Circular Polarization (2)

The designation of RHCP is determined by the right-hand rule: the thumb of the right hand is pointed in the direction of propagation, and the fingertips give the direction of rotation of the electric field vector. Similarly, LHCP satisfies the left-hand rule. x x p p + - 2 2

r Er E y y z z

LEFT-HAND CIRCULAR RIGHT-HAND CIRCULAR POLARIZATION (LHCP) POLARIZATION (LHCP)

The above signs hold for e- jbz . If E ¹ E then the tip of the electric field vector xo yo traces an ellipse. The resulting polarization is referred to as elliptical polarization.

21 Naval Postgraduate School Antennas & Propagation Distance Learning Example

We want to find the reflected field when a RHCP plane wave is normally incident on a flat perfectly conducting surface, r - jbz - jbz Ei (z) = xˆEoe - jyˆEoe Assume that the reflected field is of the form r + jbz + jbz Er (z) = xˆErxe + yˆEry e The total tangential field at the boundary (z = 0) must be zero r r Ei (z) + Er (z) = xˆ(Eo + Erx ) + yˆ(Ery - jEo )º 0 Equate x and y components to obtain E = -E rx o r Ery = jEo Ei

The total field is z Er (z) = -xˆE e+ jbz + jyˆE e+ jbz r o o r Er which is a LHCP wave. z = 0

22 Naval Postgraduate School Antennas & Propagation Distance Learning Spherical Waves (1)

An ideal point source for electromagnetic waves has no volume. It radiates a spherical wave (i.e., the equiphase planes are spherical surfaces). An arbitrarily polarized spherical wave can written as e- jbR Er(R) = (E qˆ + E fˆ) R qo fo z · R= distance from the source (Note that if the source is at the origin of the spherical coordinate system then R=r. Thus we will move the source to the kˆ = rˆ origin and use r in the next few charts.) Ef r E · h =impedance of the medium, assumed R = r Eq to be real

SOURCE 2p · b = y l

· Eqo ,Efo are complex constants x

23 Naval Postgraduate School Antennas & Propagation Distance Learning Spherical Waves (2)

Spherical waves are TEM, so the magnetic field intensity is

r ˆ ˆ - jbr kˆ ´ E(r) rˆ ´ (Eqoq + Efof) e Hr (r) = = e- jbr = (- E qˆ + E fˆ) h hr hr fo qo and the time-averaged Poynting vector (assuming Eqo ,Efo are real) r 1 r r * 1 ˆ ˆ ˆ ˆ * Wav = E(r) ´ H (r) = (Eqoq + Efof)´ (- Efoq + Eqof) 2 2r 2h 1 2 2 = ((Eqo ) + (Efo ) )rˆ 2r 2h The power flowing through a spherical surface of radius r is

1 2p p 1 P = Wr · ds = (E )2 + E 2 rˆ· rˆ r2 sin q dq df av òò av ( qo ( fo ) ) ò ò 2 S 2h 0 0 r p 2p 2 2 2p 2 2 = ((Eqo ) + (Efo ) )ò sinq dq = ((Eqo ) + (Efo ) ) 2h 0 h 14243 =2

24 Naval Postgraduate School Antennas & Propagation Distance Learning Spherical Waves (3)

1 Note that the power spreads as (the “inverse square law”). We will see that a far field r2 region can be defined for any antenna. It is the region beyond a minimum distance, rff , where the wave becomes spherical with the following properties: 1. the wave propagates radially outward 2. it is TEM (there are only q and f field components) 1 3. the field components vary as r At a large distance from the source of a spherical wave, the phase front becomes locally plane. FAR FIELD LOCALLY A GOOD APPROXIMATION TO A PLANE WAVE rff

SOURCE SOURCE r

25 Naval Postgraduate School Antennas & Propagation Distance Learning Spherical Wave Amplitude

Snapshot of a spherical wave propagating outward from the origin. The amplitude of the E wave Er(r,t) =qˆ o cos(wt - br) in the x-y plane is plotted at time t = 0 r

26 Naval Postgraduate School Antennas & Propagation Distance Learning Ray Representation for Waves

· Rays are often used to represent a propagating wave. They are arrows in the direction of propagation (kˆ) and are everywhere perpendicular to the equiphase planes (wavefronts) · The behavior of rays upon reflection or refraction is given by a set of rules which form the basis of geometrical optics (the classical theory of ray tracing) · We will see that if an observer gets far enough from a finite source of radiation, then the wavefronts become spherical · At even larger distances the wavefronts become approximately planar on a local scale

SPHERICAL WAVE FRONTS PLANE WAVE FRONTS

RAYS

27 Naval Postgraduate School Antennas & Propagation Distance Learning Wave Reflection (1)

For the purposes of applying boundary conditions, the electric field vector is decomposed into parallel and perpendicular components Er = Er + Er ^ || Er is perpendicular to the plane of incidence ^ Er lies in the plane of incidence || ˆ The plane of incidence is defined by the vectors k i and nˆ DECOMPOSITON OF AN ELECTRIC FIELD PLANE WAVE INCIDENT ON AN VECTOR INTO PARALLEL AND INTERFACE BETWEEN TWO DIELECTRICS PERPENDICULAR COMPONENTS

TRANSMITTED z qt r MEDIUM 2 r E E|| e2 ,m2 r E INTERFACE ^ e1,m1 nˆ MEDIUM 1 q q q i r i ˆ k i NORMAL y

INCIDENT REFLECTED x

28 Naval Postgraduate School Antennas & Propagation Distance Learning Wave Reflection (2)

e m e m r o o r o o E H i i e m e m r r r r ˆ r r E kˆ H i ki i i DIELECTRIC FREE SPACE q DIELECTRIC FREE SPACE qi i nˆ nˆ z q z qt r q t r q H t r r Et r r H r r E r ˆ H kˆ r E kt r t r t ˆ r kˆ H kr Er r r

PERPENDICULAR POLARIZATION PARALLEL POLARIZATION r r The incident fields (Ei ,Hi ) are known in each case. We can write expressions for the r r r r reflected and transmitted fields (Er ,H r ) and (Et ,H t ), and then apply the boundary conditions at z = 0: Er + Er = Er and Hr + Hr = Hr ( i r ) tan ( t )tan ( i r )tan ( t ) tan There is enough information to solve for the coefficients of the reflected and transmitted waves.

29 Naval Postgraduate School Antennas & Propagation Distance Learning Wave Reflection (3)

Summary of results: Reflection and transmission coefficients: Perpendicular polarization: REGION 1 REGION 2 h cosq -h cosq FREE SPACE G = i 0 t e m ^ o o DIELECTRIC h cosqi +h0 cosqt er mr kˆ 2h cosqi i t ^ = h cosqi +h0 cosqt qi nˆ Er^ = G^ Ei^ and Et^ =t ^ Ei^ qr qt Parallel polarization:

ˆ kr INTERFACE h cosqt -h0 cosqi G|| = h cosqt +h0 cosqi sin q = sin q = e m sin q i r r r t 2h cosqi t|| = h cosqt + h0 cosqi mo mrmo mr ho = and h = = ho eo ereo er Er|| = G||Ei|| and Et|| =t||Ei||

30 Naval Postgraduate School Antennas & Propagation Distance Learning Wave Reflection (4)

Example: A boundary between air (ho = 377 W) and glass (er = 4,h =188.5 W). See the following charts for plots of reflection coefficients vs. incidence angle.

Two special cases:

1. Brewster’s angle is the incidence angle at which the reflection coefficient is zero. For parallel polarization this requires hcosqt -ho cosqi = 0, or

h h h 1 2 2 -1 o cosqi = cosqt = 1- sin qt = 1- sin qi Þ q B = tan e r = 63.4 ho ho ho e r

2. Total internal reflection (qt = p /2) occurs at the critical angle of incidence, when the wave is impinging on the boundary from the more dense medium

sinq me sin p /2 me 1 t ( ) o = Þ = = e r Þ sinq c = Þ q c = 30 sinqi moeo sinqc moeo 2

This is the basis of fiber optic transmission lines. 31 Naval Postgraduate School Antennas & Propagation Distance Learning Wave Reflection (5)

Boundary between air (er = 1) and glass (er = 4)

AIR-GLASS INTERFACE,WAVE INCIDENT FROM AIR AIR-GLASS INTERFACE,WAVE INCIDENT FROM GLASS 1 1

0.9 0.9

0.8 PERPENDICULAR 0.8 PERPENDICULAR POLARIZATION POLARIZATION 0.7 0.7

0.6 0.6

0.5 0.5 |gamma| |gamma| 0.4 BREWSTER'S 0.4 ANGLE 0.3 0.3 PARALLEL POLARIZATION 0.2 PARALLEL 0.2 POLARIZATION 0.1 0.1

0 0 0 10 20 30 40 50 60 70 80 90 0 10 20 30 40 50 60 70 80 90 theta, degrees theta, degrees

32 Naval Postgraduate School Antennas & Propagation Distance Learning Wave Reflection (6)

Example of a plane wave incident on a boundary between air and glass (e = 4,q = 45o) r i INCIDENT WAVE 10

8 TRANSMITTED 6 q t 4 GLASS 2 BOUNDARY

z 0

AIR -2 qi qr NORMAL -4 -6

INCIDENT REFLECTED -8

-10 -10 -5 0 5 10 x

33 Naval Postgraduate School Antennas & Propagation Distance Learning Wave Reflection (7)

Example of a plane wave reflection: reflected and transmitted waves (e = 4,q = 45o) r i

REFLECTED WAVE TRANSMITTED WAVE 10 10

8 8

6 6

4 4

2 2 BOUNDARY z 0 z 0 BOUNDARY -2 -2

-4 -4

-6 -6

-8 -8

-10 -10 -10 -5 0 5 10 -10 -5 0 5 10 x x

34 Naval Postgraduate School Antennas & Propagation Distance Learning Wave Reflection (8)

Example of a plane wave reflection: total field

10

8

6 · The total field in region 1 is the sum of the

4 incident and reflected fields

2 BOUNDARY · If region 2 is more dense than region 1 z 0 (i.e., er2 > er1) the transmitted wave is -2 refracted towards the normal -4 -6 · If region 1 is more dense than region 2 -8 (i.e., er1 > er2) the transmitted wave is

-10 refracted away from the normal -10 -5 0 5 10 x

35 Naval Postgraduate School Antennas & Propagation Distance Learning Example (1)

An aircraft is attempting to communicate with a submerged submarine. The frequency is 0.5 MHz and the power density of the normally incident wave at the ocean surface is 12 kW/m2 . The receiver on the submarine requires 0.1 mV/m to establish a reliable link. What is the maximum depth for communication? Hr r i Ei AIR

kˆ mo,eo,s = 0 i x z = 0 r Er Ht t OCEAN z m ,e = 72,s = 4 S/m ˆ o r kt r - jbo z The phasor expression for the incident plane wave is Ei (z) = xˆEoe where 2p c 3´108 b = w m e = , l = = = 600 m. The time-averaged power density is o o o o 6 lo f 0.5 ´10 1 given by the Poynting vector, Wr (z) = Er (z) ´ Hr *(z) avi 2 i i 36 Naval Postgraduate School Antennas & Propagation Distance Learning Example (2)

At the ocean surface, z=0, and from the information provided we can solve for Eo E 2 Wr (z) = o º12 ´103 W/m2 Þ E 2 = (12´103 )(2)(377) Þ E = 3008 V/m avi o o 2ho r -g z Below the ocean surface the electric field is given by Et (z) = xˆEot e , where the transmission coefficient is determined from the Fresnel formulas h -h 2h G = 0 , and t =1 + G = . h +h0 h +ho To evaluate this we need the impedance of seawater m m m h = = o = o e s æ s ö c e - j e e ç1 - j ÷ w o r ç ÷ è eoerw ø s 4 Note that = = 2000 >> 1 which is typical of a good 5 -12 eoerw 2p (5 ´10 )(8.85 ´10 )(72) conductor. Thus we drop the 1 in the denominator for good conductors.

37 Naval Postgraduate School Antennas & Propagation Distance Learning Example (3)

wm 1+ j wm o jF h » j o = o = 0.7(1+ j) = 0.9899e j45 º h e h s 2 s Now the transmission coefficient is

2h 2(1 + j)(0.7) o t = = = 5.25 ´10-3 e j44.89 º t e jFt h +ho (1 + j)(0.7) + 377 At depth z the magnitude of the electric field intensity is

* 1/ 2 Er (z) = Er (z) · Er (z)* = é(xˆE t e-(a + jb ) z )· (xˆE t e-(a + jb ) z ) ù = E t e-a z t t t ëê o o ûú o where the attenuation constant is 1/ 2 ì é 2 ùü ïme æ s ö ï a = w í ê 1+ ç ÷ -1úý 2 èweø îï ëê ûúþï wm s a » o = p fm s = p (5´105 )(4p ´10-7 )(4) = 2.81 Np/m 2 o 38 Naval Postgraduate School Antennas & Propagation Distance Learning Example (4)

Similarly, for a good conductor the phase constant is 1/ 2 ì é 2 ùü ïme æ s ö ï wmos b = w í ê 1+ ç ÷ +1úý » = a 2 èweø 2 îï ëê ûúþï r At what depth is Et (z) = 0.1m V/m?

-6 -2.81z -3 -2.81z 0.1´10 = Eo t e = (3008)(5.24 ´10 )e - 2.81z = -18.88 z = 6.7 m

A common measure of the depth of penetration of the wave in a conductor is the skin depth,d s . It is the distance that the wave travels into the material at which its magnitude is 1/e of its value at the surface

r -1 -1 -a d s Et (0) e = Eo t e = Eo t e Þ ad s =1 Þ d s = 1/a =1/ p fms

(Note that for a nonmagnetic conductor m = mo .)

39 Naval Postgraduate School Antennas & Propagation Distance Learning Example (5)

The instantaneous (time-dependent) expression for the field is r - jb z jw t Et (z,t) = Â{xˆtEoe e } jFo Note that in general Eo can be complex and written in polar form Eo = Eo e . The phase depends on the altitude of the transmitter and the phase of the wave upon leaving the aircraft antenna. We can not determine Fo from the information provided, and furthermore, it is not important in determining whether the link is established. Thus,

r jFt jFo -(a + jb ) z jw t -a z Et (z,t) = Â{xˆt e Eo e e e }= xˆt Eo e cos(w t - bz + Fo + Ft ) kˆ ´ Er (z) yˆE t The magnetic field intensity phasor is Hr (z) = t t = o e-g z and the t h h instantaneous value ì jFt jFo ü r ïxˆt e Eo e -(a + jb ) z jw t ï Ht (z,t) = Âí e e ý jFh îï h e þï xˆt E = o e-a z cos(w t - bz + F + F - F ) h o t h

40 Naval Postgraduate School Antennas & Propagation Distance Learning Common Two-Wire Transmission Lines

Twin lead or two-wire DIRECTION OF PROPAGATION

CONDUCTOR r CONDUCTORS SEPARATION d E (RADIUS a)

Coaxial (“coax”) DIRECTION OF INNER CONDUCTOR PROPAGATION (RADIUS a) Er OUTER CONDUCTOR (RADIUS b)

Microstrip CONDUCTOR DIRECTION OF SUBSTRATE GROUND PROPAGATION (DIELECTRIC) PLANE er Er d

41 Naval Postgraduate School Antennas & Propagation Distance Learning Transmission Line Equations (1)

A short length (Dz) of a two-wire transmission line has the equivalent circuit shown below: i(z,t) i(z + Dz,t) A B + R¢ L¢ + v(z,t) ¢ ¢ v(z + Dz,t) - G C -

Dz R¢ is the total resistance of the conductors (W /m) L¢ is the inductance due to the magnetic field around the conductors (H/m) C¢ is the series capacitance due to the electric field between the conductors (V/m) G¢ is the is the conductance due to loss in the material between the conductors (S/m) Special case: lossless transmission line 1. perfect conductors, s = ¥ and thereforeR¢ = 0 2. perfect dielectric filling the region between the conductors, e ¢¢ = 0 and therefore G¢ = 0

42 Naval Postgraduate School Antennas & Propagation Distance Learning Transmission Line Equations (2)

Use Kirchhoff’s voltage law at node A and take lim Dz®0 ¶v(z,t) ¶i(z,t) - = R¢i(z,t) + L¢ ¶z ¶t Use Kirchhoff’s current law at node B and take lim Dz®0 ¶i(z,t) ¶v(z,t) - = G¢v(z,t) + C¢ ¶z ¶t For the time-harmonic case ¶/¶t ® jw dV(z) - = (R¢+ jwL¢)I(z) (1) dz dI(z) - = (G¢ + jwC¢)V (z) (2) dz This is a set of coupled integral equations. Take d/dz of (1) and substitute it in (2) to get a second order differential equation for V(z) d 2V(z) - (R¢+ jwL¢)(G¢+ jwC¢)V (z) = 0 dz 2 14444244443 ºg 2

43 Naval Postgraduate School Antennas & Propagation Distance Learning Transmission Line Equations (3)

The propagation constant is determined from the transmission line parameters g = (R¢+ jwL¢)(G¢ + jwC¢) º a + jb

The phase velocity is u p = w / b .

In a similar manner a differential equation can be derived for the current. Together they are the transmission line equations (wave equations specialized to transmission lines)

d 2V(z) d 2 I(z) -g 2V (z) = 0 and -g 2 I(z) = 0 dz 2 dz 2 A solution for the voltage is + -gz - +gz V(z) = Vo e +Vo e The first term is a wave traveling in the +z direction and the second a wave traveling in the -z direction. If this is inserted into (1) on the previous page then the result is g I(z) = (V +e-gz -V -e+gz ) R¢ + jwL¢ o o

44 Naval Postgraduate School Antennas & Propagation Distance Learning Transmission Line Equations (4)

The corresponding solution of the differential equation for the current is + -gz - +gz I(z) = Io e + Io e Comparing the coefficients of the terms in the two equations gives the characteristic impedance V + V - R¢ + jwL¢ g R¢ + jwL¢ Z º o = - o = = = o + - ¢ ¢ ¢ ¢ Io Io g G + jwC G + jwC

Example: Air line (e = eo ,m = mo ,s = 0) with perfect conductors (s cond = ¥) operates at 700 MHz and has a characteristic impedance of 50 ohms and a phase constant of 20 rad/m. Find L¢, C¢ and the phase velocity. Since R¢ = G¢ = 0 Þ a = 0, and g = jwL¢jwC¢ = jw L¢C¢ º jb = j20 R¢ + jwL¢ L¢ Z = = = 50. Solve the two equations to obtain C¢ = 90.9 pF/m and o G¢ + jwC¢ C¢ 8 L¢ = 227 nH/m. The phase velocity is u p = w / b = 1/ L¢C¢ = 2.2´10 m/s = 0.733c.

45 Naval Postgraduate School Antennas & Propagation Distance Learning Transmission Line Equations (5)

Formulas are available for computing the transmission line parameters of various configurations. For example, a coax with inner radius a and outer radius b: 2ps 2pe m R G¢ = , C¢ = , L¢ = ln(b/a), R¢ = s (1/a +1/b) ln(b/a) ln(b/a) 2p 2p

p fmcond whereRs = is the surface resistance of the conductors, mcond is its permeability s cond and s cond its conductivity. Note that m, e and s are the constitutive parameters of the material filling the medium between the conductors.

For transmission lines that support transverse electromagnetic (TEM) waves the following relationships hold: G¢ s L¢C¢ = me and = C¢ e

An important characteristic of TEM waves is that Er , Hr and the direction of propagation zˆ are mutually orthogonal. That is, the electric and magnetic field vectors lie in a plane transverse to the direction of propagation. 46 Naval Postgraduate School Antennas & Propagation Distance Learning Transmission Line Circuits (1)

A transmission line circuit is shown below. The source (generator) and receiver are connected by a length l of transmission line. Assume a lossless line (g = jb ) Z g Iin IL + + + Vg Vin Zo VL ZL - - - z = -l z = 0 The current and voltage on the line are given by + - + - jbz - + jbz Vo - jbz Vo + jbz V(z) =Vo e +Vo e and I(z) = e - e Zo Zo The boundary condition at the load (z = 0) can be used to derive a reflection coefficient V (0) V + +V - V + Z - Z Z = = o o Þ G = o = L o º G e jFG L + - - I(0) V V V Z L + Zo o - o o Zo Zo

47 Naval Postgraduate School Antennas & Propagation Distance Learning Transmission Line Circuits (2)

Three special load conditions are:

1. If the load is matched to the characteristic impedance of the line then Z L = Zo and G = 0 2. If the line is open circuited then Z L = ¥ and G =1 ( G =1,FG = 0) 3. If the line is short circuited then ZL = 0 and G = -1 ( G =1,FG = p ) The total voltage at a point on the line is given by æ V - ö V (z) =V +e- jbz +V -e jbz =V + çe- jbz + o e jbz ÷ o o o ç + ÷ è Vo ø + - jbz jbz + - jbz jF G jbz =Vo (e + Ge )= Vo (e + G e e ) * + 2 1/ 2 V(z) = V (z)V(z) = Vo {1+ G + 2G cos(2bz + F G )} The maximum and minimum values of the voltage are + + Vmax = V max = Vo (1+ G) and Vmin = V min = Vo (1- G) If G ¹ 0 there is a standing wave component to the voltage and current.

48 Naval Postgraduate School Antennas & Propagation Distance Learning Transmission Line Circuits (3)

Voltage plots for three load conditions (l = 1m): G = -1 (SHORT) 2

1.5 Zo ZL 1 |V| z = -2.5 z = 0 0.5

0 -2 -1 0 G = 1 (OPEN) z 2 2 jp/ 3 G = 0.2 e 1.5 1.5

1 1 |V| |V|

0.5 0.5

0 0 -2 -1 0 -2 -1 0 z z The load impedance for the last case can be computed from the reflection coefficient

Z L - Zo Z L¢ -1 jp / 3 Z L 1+ G G = = = 0.2e Þ Z L¢ = = =1.14 + j0.41 Z L - Zo Z L¢ -1 Zo 1- G 49 Naval Postgraduate School Antennas & Propagation Distance Learning Transmission Line Circuits (4)

· Voltage maxima occur when cos(2bz + FG ) = 1 Þ 2bz + FG = -2np . (Note that increasing n is in the –z direction.) Maxima are spaced l /2. · Voltage minima occur when cos(2bz + FG ) = -1 Þ 2bz + FG = -(2n + 1)p . Minima are spaced l /2. V (1 + G) · The voltage standing wave ratio (VSWR) is defined as s = max = . Note that V min (1- G) 1< s < ¥.

Plot of voltage and current for Z L¢ = 0.1 (l = 1 m). 2

1.8 VOLTAGE LOAD 1.6 AT z=0

1.4

1.2

1 |V| or |I| 0.8 CURRENT 0.6

0.4

0.2

0 -2.5 -2 -1.5 -1 -0.5 0 z 50 Naval Postgraduate School Antennas & Propagation Distance Learning Transmission Line Impedance (1)

The impedance at any point on the line is the ratio of the voltage to current at that point

+ - jbz + jbz j2bz V (z) Vo [e + Ge ] 1+ Ge Z(z) = = = Zo I(z) V + 1- Ge j2bz o [e- jbz - Ge+ jbz ] Zo At the input of the line z = -l 1+ Ge- j2bl éZ + jZ tan(b )ù Z = Z = Z L o l in o - j2b o ê ú 1- Ge l ëZo + jZL tan(bl)û

For the purpose of computing the power delivered to the load, the load and transmission line can be replaced by an equivalent impedance Zin Z Vg Vg Zin g Iin I = and V = I Z = in Z + Z in in in Z + Z + + g in g in The power delivered to the load and line combination Vg Vin Zin - - * is Pin =Vin Iin

51 Naval Postgraduate School Antennas & Propagation Distance Learning Transmission Line Impedance (2)

Power on a lossless line is computed from the voltage and current

+ - jbz + + jbz V(z) = Vo e + GVo e =Vinc +Vref + + Vo - jbz Vo + jbz I(z) = e - G e = Iinc + I ref Zo Zo 14243 14243 INCIDENT REFLECTED WAVE WAVE The incident instantaneous power in the incident wave is

jwt jwt Pi = Â{Vi e }Â{Ii e } ì + jFo ü + 2 + jFo jwt ï|Vo | e jwt ï |Vo | 2 = Â{|Vo |e e }Âí e ý = cos (wt + Fo ) îï Zo þï Zo

where it has been assumed that Zo is real. A similar analysis of the reflected wave yields + 2 |Vo | 2 Pr = - G cos (wt + Fo + FG ) Zo

52 Naval Postgraduate School Antennas & Propagation Distance Learning Transmission Line Impedance (3)

The time-averaged power is obtained by integrating the instantaneous value over a period

1 T |V + |2 1/ f w p |V + |2 |V + |2 P = P (t)dt = o cos2 wt + F dt = o = o avi ò i ò ( o ) T 0 Zo (1/ f ) 0 2p w Zo 2Zo

Similarly for the reflected power

P = - G 2 P av r avi

and the average power delivered to the load is

|V + |2 P = P + P = o 1- G 2 av L avi avr ( ) Zo

In order to deliver all power to the load we must have G 2 ® 0.

53 Naval Postgraduate School Antennas & Propagation Distance Learning Transmission Line Impedance (4)

Input impedances for the special load conditions

sc é0 + jZo tan(bl)ù 1. Short circuit: Zin = Zo ê ú = jZo tan(bl) » jZobl which is inductive ëZo + j0tan(bl)û bl®0 oc é¥ + jZo tan(bl)ù Zo 2. Open circuit: Zin = Zo ê ú = » - jZo / bl which is ëZo + j¥ tan(bl)û j tan(bl) bl®0 capacitive éZo + jZo tan(bl)ù 3. Matched line: Zin = Zo ê ú = Zo (Note it is independent of the line ëZo + jZo tan(bl)û length.) Input impedances for some special line lengths:

éZ L + jZo tan(p )ù 1. Half-wavelength line: Zin = Zo ê ú = Z L ëZo + jZL tan(p )û éZ L + jZo tan(p /2)ù 2 2. Quarter-wavelength line: Zin = Zo ê ú = Zo / Z L ëZo + jZL tan(p /2)û 54 Naval Postgraduate School Antennas & Propagation Distance Learning Impedance Matching

For “off-the-shelf” components that must be used in a system, fixed values of Zo are used. Common values are 50, 75 and 300 ohms. Most devices (antennas, couplers, phase shifters, etc.) are not “naturally” 50 ohms. An impedance matching circuit must be inserted between the 50 line and the device. The impedance matching circuit is usually incorporated into the device and sold as a single package as illustrated below. MISMATCHED MATCHED JUNCTION JUNCTION

Z Z IMPEDANCE Z o L Zo MATCHING L NETWORK

ZL Zo Three common matching techniques: 1. quarter-wave transformers 2. stub tuners 3. series and parallel lumped elements In general, the imaginary component of the load impedance must be cancelled and the real part shifted to Zo

55 Naval Postgraduate School Antennas & Propagation Distance Learning Quarter-Wave Transformers (1)

MATCHED l / 4 JUNCTION

Z Zo Zo¢ L

Zin = Zo If a quarter-wavelength section is inserted between the transmission line and load, the input impedance is

éZL + jZo¢ tan(p /2)ù 2 Zin = Zo¢ ê ú = Zo¢ / ZL º Zo Þ Zo¢ = ZoZ L ëZo¢ + jZL tan(p /2)û Note that all of the impedances involved must be real. Example: What is the characteristic impedance of a quarter-wave section if it is to match a 100 ohm load to a 50 ohm line?

Zo¢ = Zo ZL = (50)(100) = 70.7 W

56 Naval Postgraduate School Antennas & Propagation Distance Learning Quarter-Wave Transformers (2)

l / 4 A quarter-wave 100 W 200 W 400 W 400 W transformer is designed so that reflections from the two junctions cancel (destructive interference). G = 1/ 3 G = -1/ 3 G = 1/ 3 t = 4 / 3 t = 2 / 3 t = 4 / 3 If the frequency is changed 1 4/3 from its design value, then j4/3 1 1/3 V 16/9 1/3 the cancellation is no j4/9 -4/9 longer complete. -8/27 4/27 j4/27 SUMS 16/81 TO 2 V j4/81 -4/81 -8/243 SUMS 4/243 j4/243 TO -1/3 V 17/729 j4/729 -4/729 -8/2187 M M M 4/2187

ROUND TRIP GIVES p PHASE SHIFT (- SIGN)

57 Naval Postgraduate School Antennas & Propagation Distance Learning Stub Tuning (1)

We have seen that short sections of transmission line can be capacitive or inductive. By adjusting the location and length of a shorted or open stub, any complex impedance can be tuned. d Z ¢ l o d Zo Zo Z L Z ¢ Zo Zo Z L o

l SHORTED SERIES STUB SHORTED PARALLEL STUB · The stub impedance is usually chosen to be the same as that of the transmission line (Zo¢ = Zo ). · Both shorts and opens can be used. Shorts are generally preferred because there is less fringing of the field (and therefore less coupling to external objects) · When dealing with devices in series we work with impedance, because the impedances of devices in series add: Z = R + jX (R =resistance, X =reactance) · When dealing with devices in parallel we work with admittance, because the admittances of devices in parallel add: Y = G + jB (G =conductance, B =susceptance)

58 Naval Postgraduate School Antennas & Propagation Distance Learning Stub Tuning (2)

Example of the parallel stub tuning process for a load with admittance 1 1 YL = GL + jBL = = Z L RL + jX L

YL¢ = GL¢ + jBL¢ = the admittance at distance d from the load, which can be found from the Zin formula and then computing Yin =1/ Zin Ys¢ = jBs¢ = the admittance at a distance l from the short (note that a shorted or open stub can provide only an imaginary part to the impedance or admittance) The total admittance at the junction looking into the parallel combination of stub and load YL¢ is Yin = YL¢ + Ys¢ = GL¢ + jBL¢ + jBs¢ º Yo . Thus d the two conditions are: Yo Yo YL = 1/ ZL 1. Choose d so that GL¢ = Yo Yo 2. Choose l so that BL¢ + Bs¢ = 0 Y in l Ys¢

59 Naval Postgraduate School Antennas & Propagation Distance Learning Stub Tuning (3)

The disadvantages of stub tuning: · the required location and length of the stub may not be practical or convenient · it is designed for a single frequency, and hence is narrow band Example of short-circuited stub tuner on microstrip:

DEVICE BEING TUNED

SHORTED STUB

LOAD LOCATION SHORTING PIN STUB d GROUND PLANE

e r l TRANSMISSION LINE

60 Naval Postgraduate School Antennas & Propagation Distance Learning Lumped Element Tuning (1)

At low frequencies (about 1 GHz and below) lumped inductors, capacitors and resistors are effective and more compact that stubs.

Diode phase shifter with discrete components Printed circuit inductor METALLIZED VIA THICK FILM CAPACITOR

THICK FILM CONDUCTOR

PIN DIODE

61 Naval Postgraduate School Antennas & Propagation Distance Learning Lumped Element Tuning (2)

Example: An antenna with impedance ZL = 90 - j25 ohms must be matched to a 50 ohm transmission line using an inductor or capacitor (no resistor). Design an impedance matching circuit. The tuning element adds only a reactance, and therefore d must be chosen so that the impedance of the combination of load plus transmission line has a real part of 50.

MATCHED JUNCTION Z + jZ tan(bd) jX Z = Z L o in o Z + jZ tan(bd) Z =90- j25W o L Zo L d = 50 + jX L¢

Zo Zin =50+ jX L¢ This equation can be separated into real and imaginary parts the real part solved for d. After much work one finds that the result is d = 0.068l . Using this in the above equation, the impedance is Zin = 50 - j35W. Therefore a series reactance of + j35 W must be added, which is an inductor whose value is determined from jwL = j35.

62 Naval Postgraduate School Antennas & Propagation Distance Learning Example: Lossless Power Divider

Design a lossless power divider that splits the power in the ratio of 2:1 between the two output arms.

2 Vo P1 = 2Z1 2 Vo Z1 Po = 2Zo Zo =50 W Vo

Z 2

2 Vo P2 = 2Z2

Because the device is lossless Po = P1 + P2 . We want 2 2 1 Vo 1 Vo P1 = Po Þ = Þ Z1 = 3Zo = 150 W 3 2Z1 3 2Zo and, similarly, 2 2 2 Vo 2 Vo 3 P2 = Po Þ = Þ Z2 = Zo = 75 W 3 2Z2 3 2Zo 2

63 Naval Postgraduate School Antennas & Propagation Distance Learning Transmission Line Loss (1)

Transmission lines between the antenna and receiver or transmitter in a systems can have significant losses. Traditionally they have been called plumbing loss because the primary contributor was long sections of wavguide. Sources of loss include:

1. cables and waveguide runs (0.25 to 1 dB per meter) 2. devices have insertion loss duplexer, rotary joints, filters, switches, etc. 3. devices and connectors have mismatch loss (VSWR ¹ 1)

DUPLEXER/ CIRCULATOR LOSS TRANSMITTER

TRANSMISSION LINE LOSS

RECEIVER

64 Naval Postgraduate School Antennas & Propagation Distance Learning Transmission Line Loss (2)

Consider a length l of transmission line

Z 2 2 o P ~ E Pin ~ Ein out out

l If the incident wave is TEM, then the field at the output can be expressed as -al Eout = Ein e and the transmission coefficient of the section is E æ E ö out -al out -al t = = e Þ t dB = 20logç ÷ = 20log(e ) Ein è Ein ø Example: A shorted 5m section of transmission line has 8 dB of loss. What is the attenuation coefficient? Because the line is shorted the wave travels 10m and therefore, -0.4 -10a -10a -(8 / 20) ln(10 ) 20log(e )= -8 Þ e = 10 Þ a = = 0.092 Np/m -10

65 Naval Postgraduate School Antennas & Propagation Distance Learning Waveguides (1)

Waveguides are an efficient means of transmitting microwaves. They can be hollow or filled with dielectric or other material. The cross section can be of any shape, but rectangular and circular are most common. First, we examine propagation in a rectangular waveguide of dimension a by b. y

b

z a x Waves propagate in the ± z direction: Er(z), Hr (z) ~ e± jbz . First separate Maxwell’s equations into cartesian components (m,e refer to the material inside of the waveguide) ¶E ü z + jbE = - jwmH ¶y y x ï ï ¶Ez ï r r - jbEx - = - jwmH y ý Ñ ´ E = - jwmH ¶x ï ¶E y ¶Ex ï - = - jwmH z ¶x ¶y þï

66 Naval Postgraduate School Antennas & Propagation Distance Learning Waveguides (2)

¶H ü z + jbH = jweE ¶y y x ï ï ¶H z ï r r - jbH x - = jweEy ý Ñ ´ H = jweE ¶x ï ¶H y ¶H x ï - = jweEz ¶x ¶y þï Rearranging - j æ ¶Ez ¶H z ö Ex = ç b + wm ÷ w 2me - b 2 è ¶x ¶y ø j æ ¶Ez ¶H z ö E y = ç- b + wm ÷ w 2me - b 2 è ¶y ¶x ø j æ ¶Ez ¶H z ö H x = çwe - b ÷ w 2me - b 2 è ¶y ¶x ø - j æ ¶Ez ¶H z ö H y = çwe + b ÷ w 2me - b 2 è ¶x ¶y ø

67 Naval Postgraduate School Antennas & Propagation Distance Learning Waveguides (3)

The wave equations are: Ñ2 Er = -w 2meEr Ñ2 Hr = -w 2meHr ¶2 ¶2 ¶2 ¶2 Note that Ñ2 = + + and = (- jb )2 = -b 2 and the wave equations for the ¶x2 ¶y2 ¶z 2 ¶z2 z components of the fields are

æ ¶2 ¶2 ö ç + ÷E = b 2 - w 2me E ç 2 2 ÷ z ( ) z è ¶x ¶y ø æ ¶2 ¶2 ö ç + ÷H = b 2 - w 2me H ç 2 2 ÷ z ( ) z è ¶x ¶y ø TEM waves do not exist in hollow rectangular waveguides. The wave equations must be solved subject to the boundary conditions at the waveguide walls. We consider two types of solutions for the wave equations: (1) transverse electric (TE) and (2) transverse magnetic (TM).

68 Naval Postgraduate School Antennas & Propagation Distance Learning Waveguides (4)

r Transverse magnetic (TM) waves: H z = 0 and thus H is transverse to the z axis. All field components can be determined from Ez . The general solution to the wave equation is ± jbz ± jbz Ez (x, y, z) = Ez (x, y)e = Ez (x)Ez (y)e ± jbz = (Acos(b x x)+ Bsin(b x x))(Ccos(b y y)+ Dsin(b y y))e

where A, B, C, and D are constants. The boundary conditions must be satisfied:

ìx = 0 ® A = 0 E = 0 at í z îy = 0 ® C = 0

Choose b x and b y to satisfy the remaining conditions. mp E = 0 at x = a: sin(b a) = 0 Þ b a = mp Þ b = (m = 1,2, ) z x x x a … np E = 0 at y = b: sin(b b) = 0 Þ b b = np Þ b = (n = 1,2, ) z x y y b …

69 Naval Postgraduate School Antennas & Propagation Distance Learning Waveguides (5)

For TM waves the longitudinal component of the electric field for a +z traveling wave is given by æ mp ö æ np ö - jbz Ez (x, y, z) = U sinç x÷sinç y÷e è a ø è b ø where the product of the constants AB has been replaced by a new constant U . Each solution (i.e., combination of m and n) is called a mode. Now insert Ez back in the wave equation to obtain a separation equation: 2 2 2 2 æ mp ö æ np ö b = w me - ç ÷ - ç ÷ è a ø è b ø If b 2 > 0 then propagation occurs; b 2 = 0 defines a cuttoff frequency, f , cmn 2 2 1 æ m ö æ n ö fc = ç ÷ + ç ÷ mn 2 me è a ø è b ø Waves whose frequencies are above the cutoff frequency for a mode will propagate, but those below the cutoff frequency are attenuated.

70 Naval Postgraduate School Antennas & Propagation Distance Learning Waveguides (6)

r Transverse electric (TE) waves: Ez = 0 and thus E is transverse to the z axis. All field components can be determined from H z . The general solution to the wave equation is ± jbz ± jbz H z (x, y, z) = H z (x, y)e = H z (x)H z ( y)e ± jbz = (Acos(b x x)+ Bsin(b x x))(Ccos(b y y)+ Dsin(b y y))e

¶H z æ np ö ¶H z æ mp ö But, from Maxwell’s equations, Ex µ ~ cosç y÷ and E y µ ~ cosç x÷. ¶y è b ø ¶x è a ø Boundary conditions: Ex = 0 at y = 0 ® D = 0 Ey = 0 at x = 0 ® B = 0 np E = 0 at y = b ® b = , n = 0,1, x y b … mp E = 0 at x = a ® b = , m = 0,1, y x a … æ mp ö æ np ö - jbz Therefore, H z (x, y, z) =V cosç x÷cosç y÷e (m = n = 0 not allowed) è a ø è b ø The same equation for cutoff frequency holds for both TE and TM waves.

71 Naval Postgraduate School Antennas & Propagation Distance Learning Waveguides (7)

Other important relationships: u · Phase velocity for mode (m,n), u p = where u =1/ me is the phase 1- f / f 2 ( cmn ) velocity in an unbounded medium of the material which fills the waveguide. Note the the phase velocity in the waveguide is larger than in the unbounded medium (and can be greater than c). · Group velocity for mode (m,n), u = u 1 - f / f 2 . This is the velocity of energy g ( cmn ) (information) transport and is less than the velocity in the unbounded medium. · Wave impedance for mode (m,n), h Z = TEmn 1- f / f 2 ( cmn ) Z =h 1- f / f 2 TMmn ( cmn ) where h = m /e is the wave impedance in the unbounded medium. w w · Phase constant for mode (m,n), b = = 1- f / f 2 mn ( cmn ) u p u

72 Naval Postgraduate School Antennas & Propagation Distance Learning Waveguides (8)

l · Guide wavelength for mode (m,n), l = where l is the wavelength gmn 1- f / f 2 ( cmn ) in the unbounded medium.

The dominant mode is the one with the lowest cutoff frequency. For rectangular waveguides with a > b the TE10 mode is dominant. If a mode shares a cutoff frequency with another mode(s), then it is degenerate. For example, TE11 and TM11 are degenerate modes.

Example: If the following field exists in a rectangular waveguide what mode is propagating? æ 2p ö æp ö - j2z Ez = 5sinç x÷sinç y÷e è a ø è b ø

Since Ez ¹ 0 it must be a TM mode. Compare it with the general form of a TM mode field and deduce that m=2 and n=1. Therefore, it is the TM 21 mode.

73 Naval Postgraduate School Antennas & Propagation Distance Learning Waveguides (9)

Example: What is the lowest frequency that will readily propagate through a tunnel with a rectangular cross section of dimension 10m by 5m? If the walls are good conductors, we can consider the tunnel to be a waveguide. The lowest frequency will be that of the dominant mode, which is the TE10 mode. Assume that the tunnel is filled with air 1 æ 1 ö c f = = = 15 MHz c10 ç ÷ 2 moeo è a ø 2(10) Example: Find the five lowest cutoff frequencies for an air-filled waveguide with a=2.29 cm and b=1.02 cm. 2 2 1 æ m ö æ n ö fc = ç ÷ + ç ÷ mn 2 me è 0.029ø è 0.0102ø Use Matlab to generate cutoff frequencies by looping through m and n. Choose the five lowest. Note that when both m,n > 1 then both TE and TM modes must be listed. (The frequencies are listed in GHz.)

TE 01(14.71),TE10 (6.55),TE11 and TM11(16.10),TE 20 (13.10)

74 Naval Postgraduate School Antennas & Propagation Distance Learning Waveguides (10)

Example: Find the field parameters for a TE10 mode, f=10 GHz, a=1.5 cm, b=0.6 cm, filled with dielectric, er = 2.25. Phase velocity in the unbounded medium, u = c/ 2.25 = 3´108 /1.5 = 2´108 m/s Wavelength in the unbounded medium, l = u/ f = 2 ´108 /1´1010 = 0.02 m c/ 2.25 Cutoff frequency, f = u/(2a) = = 0.67 ´1010 Hz c10 (2)(0.015) w 2 2pf 2 Phase constant, b10 = 1- (fc / f ) = 1- (0.067/1) = 74.5p radians u mn c / 2.25 1442443 0.745 l 0.02 Guide wavelength, lg = = = 0.0268 m 1- f / f 2 0.745 ( cmn ) 8 8 Phase velocity, u p = u /0.745 = 2´10 /0.745 = 2.68´10 m/s h h / 2.25 (377) Wave impedance, Z = = o = = 337.4 ohms TE10 1- f / f 2 0.745 (0.745)(1.5) ( cmn ) 8 8 Group velocity, ug = 0.745u = (2´10 )(0.745) =1.49´10 m/s

75 Naval Postgraduate School Antennas & Propagation Distance Learning Mode Patterns in Rectangular Waveguide

From C.S. Lee, S. W. Lee, and L. L. Chuang, “Plot of Modal Field Distribution in Rectangular and Circular Waveguides,” IEEE Trans. on MTT, 1985.

76 Naval Postgraduate School Antennas & Propagation Distance Learning Table of Waveguide Formulas

QUANTITY TEM ( Ez = Hz = 0 ) TM ( Hz = 0) TE ( Ez = 0) WAVE m g jwm GENERAL : ZTM= GENERAL : ZTE = IMPEDANCE, Z ZTEM = h = jwe g e 2 h f > f : h 1 - ( f / f ) f > f : c c c 2 - jh 2 1 - ( fc / f ) f < fc: 1 - ( f / fc ) jwm we f < f : c 2 h 1- ( f / fc ) PROPAGATION 2 2 jk = jw me GENERAL : h 1 - ( fc / f ) GENERAL : h 1 - ( fc / f ) CONSTANT, g 2 2 f > fc: jb = jk 1 - ( f c / f ) f > fc: jb = jk 1 - ( f c / f ) 2 2 f < fc: a = h 1 - ( f / f c) f < fc: a = h 1 - ( f / f c) PHASE 1 GENERAL : w / b GENERAL : w / b u = u u VELOCITY, up f > f : f > f : me c 2 c 2 1 - ( f c / f ) 1 - ( f c / f ) f < fc: NO PROPAGATION f < fc: NO PROPAGATION VECTOR FIELD 1 r g r g r ˆ r E = - Ñ E H = - Ñ H RELATIONSHIP H = k ´ E T 2 T z T 2 T z ZTEM h h r 1 r r r H = zˆ ´ E E = -ZTEzˆ ´ H Z TM h ¶ 2 ¶ 2 Cutoff frequency: f = Propagation constant: g = h2 - k2 Transverse Laplacian: Ñ2 = + c 2p me T ¶x2 ¶y2 æ mp ö 2 æ np ö 2 l For a rectangular waveguide (a by b ): h = ç ÷ + ç ÷ Guide wavelength: l = è a ø è b ø g 2 1 - ( fc / f )

77 Naval Postgraduate School Antennas & Propagation Distance Learning Waveguide Magic Tee

Ports 1 and 2 are the "sidearms." Port 4 is the "sum" port and 3 the "difference" port.

Sidearm excitation Port 3 Port 4 jf jf A1 = ae , A2 = ae A3 = 0 A4 = 2a jf jf+p A1 = ae , A2 = ae A3 = 2a A4 = 0

PORT 3

PORT 2 "Magic" originates from the fact that it is the only 4-port device that can be simultaneously matched at all ports. PORT 4 PORT 1

78 Naval Postgraduate School Antennas & Propagation Distance Learning Circulators

Circulators "circulate" the signal from port to port in the direction indicated by the arrow 1

2

3 Ideally: Signal into port 1 emerges out port 2; signal out port 3 is zero. Signal into port 2 emerges out port 3; signal out port 1 is zero. Signal into port 3 emerges out port 1; signal out port 2 is zero. In practice: 1. There is some insertion loss in the forward (arrow) direction. Values depend on the type of circulator. They range from 0.5 dB to several dB. 2. There is leakage in the reverse (opposite arrow) direction. Typical values of isolation are 20 to 60 dB. That is, the leakage signal is 20 to 60 dB below the signal in the forward direction. 3. Increasing the isolation comes at the expense of size and weight Uses: 1. Allow a transmitter and receiver to share a common antenna without switching 2. Attenuate reflected signals (load the third port)

79 Naval Postgraduate School Antennas & Propagation Distance Learning Microwave Switches

Microwave switches are used to control signal transmission between circuit devices. A general representation of a switch is given in terms of "poles" and "throws" SINGLE POLE - DOUBLE THROW (SPDT) DOUBLE POLE - DOUBLE THROW (DPDT) · POLES THROWS TRANSMISSION ROTATING LINE VANE · · · INPUT · · OUTPUTS · · · · · · ·

· Switches can be constructed in any type of transmission line or waveguide. Common types: Type Principle Applied to: Mechanical Rotating or moving parts All types Diode Forward/backward bias Stripline, microstrip, waveguide yields low/high impedance Gas discharge Confined gas is ionized Waveguide Circulator Magnetized ferrite switches Stripline, microstrip, waveguide circulation direction

80 Naval Postgraduate School Antennas & Propagation Distance Learning Bandwidth (1)

The equivalent circuits of transmission lines and antennas are comprised of combinations of resistors, capacitors and inductors. The transmission coefficient, or gain in the case of an antenna, is frequency dependent. The range of frequencies over which the device has “acceptable performance” is called the bandwidth of the device. For example, the gain of a typical antenna has the following general frequency characteristic:

Gmin G ~ t 2

f

fL fo fH Note that gain can be viewed as a scaled value of the antenna’s transmission coefficient. We will see that other performance measures, not just gain will determine its bandwidth. Specifying frequencies where the gain exceeds the minimum value as in the operating band, the bandwidth is fH - fL . The center of the band is fo = ( fH + fL )/2.

81 Naval Postgraduate School Antennas & Propagation Distance Learning Bandwidth (2)

Information transmission systems, such as radar and communications, require a finite (non- zero) bandwidth. Consider the following waveform as an approximation to a modulated carrier that a radar would employ. In the time domain the signal is

æ e jwot + e- jwot öæ e jwmt + e- jwmt ö s(t) cos( t)cos( t) ç ÷ç ÷ = wo wm = ç ÷ç ÷ è 2 øè 2 ø

The spectrum of this signal has two spikes centered about the carrier frequency, ± wo t B t 1 w o - wm wo +w m w -wo wo Therefore, in order to pass this signal without removing any frequency components, the required bandwidth is B = Df = 2wm . This is an example of a bandpass device. Ideally we would like the amplitude of the transmission coefficient to be constant over the passband. It is usually “bell-shaped” as depicted in the previous chart. Common cutoff choices for the edges of the band are the –3dB, –6dB, and –10dB points. 82 Naval Postgraduate School Antennas & Propagation Distance Learning Filter Characteristics

2 Filters are characterized by their transfer functions H ( f ) = t = 1- G , where G is reflection coefficient. It is usually plotted as return loss in dB, 20log10(G ), or transmission loss in dB, 20log10(t ). Note that in many cases the phase of the characteristic function is also important.

0 0 LOW PASS HIGH PASS PASSBAND PASSBAND FILTER Df Df f f TRANSMISSION LOSS, dB

TRANSMISSION LOSS, dB 0 -¥ -¥ f H fL fH = ¥ BAND STOP FILTER 0 0 BAND PASS FILTER PASSBAND PASSBAND STOPBAND PASSBAND Df f f Df TRANSMISSION LOSS, dB -¥ TRANSMISSION LOSS, dB -¥ f f L H fL fH

83 Naval Postgraduate School Antennas & Propagation Distance Learning Multiplexers

Multiplexers are frequency selective circuits used to separate signals by frequency spectrum. They are comprised of filter networks. The example illustrated is a waveguide manifold multiplexer that separates signals into four subbands: EFFECTIVE SHORT LOCATIONS Df1 + Df2 + Df 3 + Df4

1 2 3 A B C D ì ï Df Df Df Df FILTER í 1 2 3 4 NETWORK îï

The plane at 1 appears as a short in the band Df1, but matched at other frequencies . The waveguide junction at A appears matched at Df1, but shorted at other frequencies. Similarly for planes 2, 3, 4 and junctions B, C, D. Frequency characteristic:

Df1 + Df 2 +Df 3 + Df 4 f Df1 Df2 Df3 Df 4 f

INPUT OUTPUT

84 Naval Postgraduate School Antennas & Propagation Distance Learning Waveguide Filters

From Gamma-F Corporation advertisement 85 Naval Postgraduate School Antennas & Propagation Distance Learning Decibel Refresher

In general, a dimensionless quantity Q in decibels (denoted QdB) is defined by

QdB = 10log10(Q) Q usually represents a ratio of quantities, where the denominator is the reference. Characters are added to the "dB" to denote the reference quantity. For example: P · Power referenced to 1 watt: = P 1w dBw P · Power referenced to 1 milliwatt: = P (= P + 30) 1mw dBm dBw G · Antenna gain referenced to an isotropic source: = G 1 dBi Recall that the gain of an ideal isotropic source is 1. This notation is not usually used because the definition of gain implies an isotropic reference. However, a dipole’s gain (= 1.5) is sometimes is used as a reference, and the notation is dBd. Note: 1. 10 dB represents an order magnitude change in the quantity Q 2. the dB unit does not depend on the reference that is used to define it 3. when quantities are multiplied their dB values add:

ERPdBw = (PG)dBw = PdBw + GdB

86 Naval Postgraduate School Antennas & Propagation Distance Learning Coordinate Transform Tables

xˆ yˆ zˆ Example: from top table, reading across, rˆ cosf sin f 0 rˆ = xˆ cosf + yˆ sinf fˆ -sin f cosf 0 zˆ 0 0 1 and reading down, Rectangular and cylindrical xˆ = rˆ cosf -fˆ sin f xˆ yˆ zˆ rˆ sin q cosf sin qsin f cosq The tables also can be used to transform vectors. The unit vectors in the table qˆ cosqcosf cosqsinf -sin q ˆ -sin f cosf 0 headings are replaced by the corresponding f vector components. For example, given Rectangular and spherical r A = Ax xˆ + Ay yˆ + Azzˆ rˆ fˆ zˆ rˆ sin q 0 cosq in cartesian coordinates, the vector can be qˆ cosq 0 -sin q expressed in cylindrical coordinates as fˆ 0 1 0 A = A cosf + A sinf + A × 0 Cylindrical and spherical r x y z

87 Naval Postgraduate School Antennas & Propagation Distance Learning Coordinate Systems

r = x 2 + y2 + z2 = r2 + z2 æ yö f = tan-1 è xø

z,w -1æ rö r rˆ q = tan è z ø fˆ ® ds = rˆsinq dq df rˆ r r qˆ Direction cosines are the projections of q points on the unit sphere onto the xy y,v plane. They are the x,y, and z f components of rˆ: x,u u = sinq cosf v = sinq sinf w = cosq

88 Naval Postgraduate School Antennas & Propagation Distance Learning Radar and ECM Frequency Bands

89 Naval Postgraduate School Antennas & Propagation Distance Learning Electromagnetic Spectrum

90 Naval Postgraduate School Distance Learning

Antennas & Propagation

LECTURE NOTES VOLUME II BASIC ANTENNA PARAMETERS AND WIRE ANTENNAS

by Professor David Jenn

l Er

ANTENNA Hr kˆ PROPAGATION DIRECTION

(ver 1.3) Naval Postgraduate School Antennas & Propagation Distance Learning Antennas: Introductory Comments

Classification of antennas by size:

Let l be the antenna dimension:

1. electrically small, l << l : primarily used at low frequencies where the wavelength is long 2. resonant antennas, » l / 2: most efficient; examples are slots, dipoles, patches l 3. electrically large, l >> l : can be composed of many individual resonant antennas; good for radar applications (high gain, narrow beam, low sidelobes)

Classification of antennas by type:

1. reflectors 2. lenses 3. arrays

Other designations: wire antennas, aperture antennas, broadband antennas

1 Naval Postgraduate School Antennas & Propagation Distance Learning Radiation Integrals (1)

Consider a perfect electric conductor (PEC) with an electric surface current flowing on S. In the case where the conductor is part of an antenna (a dipole), the current may be caused by an applied voltage, or by an incident field from another source (a reflector). The observation point is denoted by P and is given in terms of unprimed coordinate variables. Quantities associated with source points are designated by primes. We can use any coordinate system that is convenient for the particular problem at hand.

OBSERVATION z POINT P(x, y, z) or P(r,q ,f ) r

Rr = r - r¢ r ¢ 2 ¢ 2 ¢ 2 O R = R = (x - x ) + (y - y ) + (z - z ) r¢ y S r x J s (x¢, y¢,z¢) PEC WITH SURFACE CURRENT

The medium is almost always free space (mo ,eo ), but we continue to use (m,e ) to cover more general problems. If the currents are known, then the field due to the currents can be determined by integration over the surface. 2 Naval Postgraduate School Antennas & Propagation Distance Learning Radiation Integrals (2)

The vector wave equation for the electric field can be obtained by taking the curl of Maxwell’s first equation: r 2 r r Ñ´ Ñ´ E = k E - jwmJs A solution for Er in terms of the magnetic vector potential Ar(rr) is given by Ñ(Ñ · Ar(r)) Er(rr) = - jwAr(r) + (1) jwme r r m J - jkR where (r) is a shorthand notation for (x,y,z) and A(r ) = òò s e ds¢ 4p S R We are particularly interested in the z OBSERVATION case were the observation point is in the POINT far zone of the antenna (P ® ¥). As P r recedes to infinity, the vectors rr and Rr r r r become parallel. rr¢ · rˆ R » r - rˆ(r¢ · rˆ)

r¢ y x

3 Naval Postgraduate School Antennas & Propagation Distance Learning Radiation Integrals (3)

In the expression for Ar(rr) we use the approximation 1/ R »1/ r in the denominator and r rˆ · R » rˆ ·[r - rˆ(r ¢· rˆ)] in the exponent. Equation (2) becomes

r r m r - jkrˆ·[r -rˆ(r ¢·rˆ)] m - jkr r jk (r ¢·rˆ) A(r ) » òò Js e ds¢ = e òò J s e ds¢ 4pr S 4pr S When this is inserted into equation (1), the del operations on the second term lead to 1/ r2 and 1/r 3 terms, which can be neglected in comparison to the - jwAr term, which depends only on 1/ r. Therefore, in the far field,

r r - jwm - jkr r jk (r¢·rˆ) E(r ) » e òò J s e ds¢ (discard the Er component) (3) 4pr S Explicitly removing the r component gives,

r r - jkh - jkr r r jk (r¢·rˆ) E(r ) » e òò[Js - rˆ(J s · rˆ)]e ds¢ 4p r S The radial component of current does not contribute to the field in the far zone.

4 Naval Postgraduate School Antennas & Propagation Distance Learning Radiation Integrals (4)

e- jkr Notice that the fields have a spherical wave behavior in the far zone: Er ~ . The r spherical components of the field can be found by the appropriate dot products with Er . r More general forms of the radiation integrals that include magnetic surface currents (J ms ) are: r ˆ - jkh - jkr é r ˆ Jms ·f ù jkr¢×rˆ Eq (r,q,f) = e òòêJ s ·q + ú e ds¢ 4p r S ë h û r ˆ - jkh - jkr é r ˆ Jms ·q ù jkr¢×rˆ Ef (r,q,f) = e òò êJ s ·f - úe ds¢ 4p r S ë h û The radiation integrals apply to an unbounded medium. For antenna problems the following process is used: 1. find the current on the antenna surface, S, 2. remove the antenna materials and assume that the currents are suspended in the unbounded medium, and 3. apply the radiation integrals.

5 Naval Postgraduate School Antennas & Propagation Distance Learning Hertzian Dipole (1)

Perhaps the simplest application of the radiation integral is the calculation of the fields of an infinitesimally short dipole (also called a Hertzian dipole). Note that the criterion for short means much less than a wavelength, which is not necessarily physically short.

z · For a thin dipole (radius, a << l ) the surface current distribution is independent of f ¢. The current crossing a ring around 2a r the antenna is I = J s 2pa { A/m l · For a thin short dipole (l << l ) we assume that the current is constant and flows along z¢ r the center of the wire; it is a filament of Js (r¢,f¢,z¢) = Jozˆ zero diameter. The two-dimensional f¢ y integral over S becomes a one-dimensional r¢ = a integral over the length, x r r òò J sds¢® 2pa ò I dl¢ S L

6 Naval Postgraduate School Antennas & Propagation Distance Learning Hertzian Dipole (2)

Using rr¢ = zˆz¢ and rˆ = xˆsinq cosf + yˆ sinq sinf + zˆcosq gives rr¢ · rˆ = z¢cosq . The radiation integral for the electric field becomes

l l r - jkh - jkr r¢ - jkhI zˆ - jkr ¢ E(r,q,f) » e I e jk(r ·rˆ) zˆdz¢ = e e jkz cosq dz¢ ò {r ò 4pr 0 dl ¢ 4pr 0 jkz¢cosq However, because l is very short, kz¢ ® 0 and e » 1. Therefore, l r - jkhI zˆ - jkr - jkhI zˆ - jkr E(r,q,f) » e ò (1)dz¢ = l e 4pr 0 4pr leading to the spherical field components z

r - jkhI qˆ · zˆ - jkr Eq = qˆ · E » l e 4pr l jkhI sin q z¢ = l e- jkr 4pr y r x Ef = fˆ · E = 0 SHORT CURRENT FILAMENT

7 Naval Postgraduate School Antennas & Propagation Distance Learning Hertzian Dipole (3)

Note that the electric field has only a 1/r dependence. The absence of higher order terms is due to the fact that the dipole is infinitesimal, and therefore rff ® 0. The field is a spherical wave and hence the TEM relationship can be used to find the magnetic field intensity r kˆ ´ Eˆ rˆ ´ Eqqˆ jkI sin q - jkr H = = = fˆ l e h h 4pr The time-averaged Poynting vector is

2 2 2 2 r 1 ˆ r * 1 * hk I l sin q Wav = Â{E ´ H }= Â{Eq Hf }rˆ = rˆ 2 2 32p 2r2 The power flow is outward from the source, as expected for a spherical wave. The average power flowing through the surface of a sphere of radius r surrounding the source is 2p p 2 2 2 2p p 2 2 2 r hk I 2 2 hk I P = W ·nˆ ds = l sin q rˆ ·rˆr sin q dq df = l W rad ò ò av 2 2 ò ò 12p 0 0 32p r 0 0 1444442444443 =8p / 3

8 Naval Postgraduate School Antennas & Propagation Distance Learning Solid Angles and Steradians

Plane angles: s = Rq , if s = R then q =1 radian

ARC LENGTH R s q

Solid angles: W = A / R2 , if A = R2, then W =1 steradian

W = A / R2 SURFACE AREA R A

9 Naval Postgraduate School Antennas & Propagation Distance Learning Directivity and Gain (1)

The radiation intensity is defined as dP 2 r 2 r U(q,f) = rad = r rˆ ·W = r W dW av av and has units of Watts/steradian (W/sr). The directivity function or directive gain is defined as 2 r power radiated per unit solid angle dP / dW r Wav D(q,f) = = rad = 4p average power radiated per unit solid angle Prad /(4p ) Prad For the Hertzian dipole, hk 2 I 2 2 sin 2 q r2 l 2 r 2 2 r Wav 32p r 3 D(q,f) = 4p = 4p = sin 2 q P 2 2 2 2 rad hk I l 12p The directivity is the maximum value of the directive gain 3 D = D (q ,f ) = D(q ,f ) = o max max max 2

10 Naval Postgraduate School Antennas & Propagation Distance Learning Dipole Polar Radiation Plots

Half of the radiation pattern of the dipole is plotted below for a fixed value of f. The half- power beamwidth (HPBW) is the angular width between the half power points (1/ 2 below the maximum on the voltage plot, or –3dB below the maximum on the decibel plot). FIELD (VOLTAGE) PLOT DECIBEL PLOT

90 1.5 90 10 120 60 120 60 0 1 150 30 150 -10 30 0.5 -20

180 q0 180 q 0

210 330 210 330

240 300 240 300 270 270

The half power beamwidth of the Hertzian dipole, qB :

o o Enorm ~ sinq Þ sin(qHP ) = 0.707 Þ qHP = 45 Þ qB = 2qHP = 90

11 Naval Postgraduate School Antennas & Propagation Distance Learning Dipole Radiation Pattern

Radiation pattern of a Hertzian dipole aligned with the z axis. Dn is the normalized directivity. The directivity value is proportional to the distance from the center.

12 Naval Postgraduate School Antennas & Propagation Distance Learning Directivity and Gain (2)

Another formula for directive gain is 4p r 2 D(q,f) = Enorm (q,f) W A

where WA is the beam solid angle 2p p r 2 W A = ò ò Enorm (q,f) sinq dq df 0 0 r and Enorm (q,f) is the normalized magnitude of the electric field pattern (i.e., the normalized radiation pattern) r r E(r,q,f) Enorm (q,f) = r Emax (r,q,f) Note that both the numerator and denominator have the same 1/r dependence, and hence the ratio is independent of r. This approach is often more convenient because most of our calculations will be conducted directly with the electric field. Normalization removes all of the cumbersome constants.

13 Naval Postgraduate School Antennas & Propagation Distance Learning Directivity and Gain (3)

As an illustration, we re-compute the directivity of a Hertzian dipole. Noting that the maximum magnitude of the electric field is occurs when q = p /2, the normalized electric field intensity is simply r Enorm (q,f) = sinq The beam solid angle is 2p p r 2 W A = ò ò Enorm (q,f) sin q dq df 0 0 p 8p = 2p òsin 3 q dq = 0 3 14243 =4 / 3 and from the definition of directivity,

4p r 2 4p 2 3 2 D(q,f) = Enorm (q,f) = sinq = sin q W A 8p /3 2 which agrees with the previous result.

14 Naval Postgraduate School Antennas & Propagation Distance Learning Example

Find the directivity of an antenna whose far-electric field is given by

90 10 120 60 8 6 jkr 150 30 - 4 ì10e o o 2 cosq , 0 £ q £ 90 ï 180 0 r Eq (r,q,f ) = í - jkr ïe 210 330 cosq , 90o £ q £ 180o îï r 240 300 270 r The maximum electric field occurs when cosq = 1 ® Emax = 10/r . The normalized electric field intensity is ïì cosq , 0o £q £ 90o E (q,f) = qnorm í îï0.1 cosq , 90o £q £180o which gives a beam solid angle of 2p p / 2 2p p 2 2 2p W A = ò ò cos q sin q dq df + 0.01 ò ò cos q sinq dq df = (1.1) 0 0 0 p / 2 3

and a directivity of Do = 5.45 = 7.37 dB.

15 Naval Postgraduate School Antennas & Propagation Distance Learning Beam Solid Angle and Radiated Power

In the far field the radiated power is

2p p 2p p 1 ì r r * ü 1 r 2 2 P = ÂíE ´ H ý · rˆds = E r sinq dq df Þ F = 2hP rad 2 ò ò î þ 2h ò ò rad rad 0 0 1442443 0 0 2 Er /h 14444244443 ºFrad From the definition of beam solid angle

2p p r 2 W A = ò ò Enorm (q,f) sin q dq df 0 0 2p p 1 r 2 2 r 2 2 = E r sinq dq df = Þ Frad = W A Emax r r 2 2 ò ò Emax r 0 0 14444244443 ºFrad Equate the expressions for Frad r 2 2 W A Emax r Prad = 2h

16 Naval Postgraduate School Antennas & Propagation Distance Learning Gain vs. Directivity (1)

Directivity is defined with respect to the radiated power, Prad . This could be less than the power into the antenna if the antenna has losses. The gain is referenced to the power into the antenna, Pin . ANTENNA I P inc R R Pref Pin l a

Define the following:

Pinc = power incident on the antenna terminals Pref = power reflected at the antenna input Pin = power into the antenna P = power loss in the antenna (dissipated in resistor R , P = 1 I 2 R ) loss l loss 2 l P = power radiated (delivered to resistor R , P = 1 I 2 R , R is the radiation rad a rad 2 a a resistance)

The antenna efficiency, e, is Prad = ePin where 0 £ e £ 1.

17 Naval Postgraduate School Antennas & Propagation Distance Learning Gain vs. Directivity (2)

Gain is defined as dP /dW dP /dW G(q,f) = rad = 4p rad = eD(q ,f) Pin /(4p ) Prad / e

Most often the use of the term gain refers to the maximum value of G(q,f).

Example: The antenna input resistance is 50 ohms, of which 40 ohms is radiation resistance and 10 ohms is ohmic loss. The input current is 0.1 A and the directivity of the antenna is 2.

The input power is P = 1 I 2 R = 1 0.12 (50) = 0.25 W in 2 in 2 The power dissipated in the antenna is P = 1 I 2 R = 1 0.12 (10) = 0.05 W loss 2 l 2 The power radiated into space is P = 1 I 2 R = 1 0.1 2 (40) = 0.2 W rad 2 a 2 æ Prad ö æ 0.2 ö If the directivity is Do = 2 then the gain is G = eD = ç ÷D = ç ÷(2) = 1.6 è Pin ø è0.25ø

18 Naval Postgraduate School Antennas & Propagation Distance Learning Azimuth/Elevation Coordinate System

Radars frequently use the azimuth/elevation coordinate system: (Az,El) or (a,g ) or (qe ,fa ). The antenna is located at the origin of the coordinate system; the earth's surface lies in the x-y plane. Azimuth is generally measured clockwise from a reference (like a compass) but the spherical system azimuth angle f is measured counterclockwise from the x axis. Therefore a = 360- f and g = 90 -q degrees.

ZENITH

z

CONSTANT ELEVATION

P q r g y f a

x HORIZON

19 Naval Postgraduate School Antennas & Propagation Distance Learning Approximate Directivity Formula (1)

Assume the antenna radiation pattern is a “pencil beam” on the horizon. The pattern is constant inside of the elevation and azimuth half power beamwidths (qe ,fa ) respectively: z q = 0

y

q x e f = 0 fa q = p / 2

20 Naval Postgraduate School Antennas & Propagation Distance Learning Approximate Directivity Formula (2)

Approximate antenna pattern - jkr ìEoe r ï qˆ, (p / 2 -qe / 2) £ q £ (p / 2 +qe / 2) and - fa / 2 £ f £ fa / 2 E(q,f) = í r îï0, else The beam solid angle is p qe fa + 2 2 2 W = sin q dq df A ò ò { p qe fa »1 - - 2 2 2 = fa [sin(qe / 2)- sin(-qe / 2)] » fa [qe / 2 - (-qe / 2)] = faqe 4p 4p This leads to an approximation for the directivity of Do = = . Note that the W A qefa angles are in radians. This formula is often used to estimate the directivity of an omni- directional antenna with negligible sidelobes.

21 Naval Postgraduate School Antennas & Propagation Distance Learning Thin Wire Antennas (1)

Thin wire antennas satisfy the condition a << l . If the length of the wire (l) is an integer multiple of a half wavelength, we can make an “educated guess” at the current based on an open circuited two-wire transmission line FEED POINTS

l / 4 l / 2 z

I(z) OPEN CIRCUIT For other multiples of a half wavelength the current distribution has the following features FEED POINT LOCATED AT MAXIMUM

= l / 2 = l l l CURRENT GOES TO ZERO AT END

= 3l / 2 l

22 Naval Postgraduate School Antennas & Propagation Distance Learning Thin Wire Antennas (2)

On a half-wave dipole the current can be approximated by

I(z) = Io cos(kz) for - l / 4 < z < l / 4 Using this current in the radiation integral l / 4 r - jkh - jkr jkz¢cosq E(r,q,f) = e zˆ ò Io cos(kz¢)e dz¢ 4pr -l / 4 - jkhI l / 4 = o e- jkr zˆ òcos(kz¢)e jkz¢cosq dz¢ 4pr -l / 4 From a table of integrals we find that =0 =±1 Az¢ 64748 64748 e [Acos(Bz¢) + B sin(Bz¢)] ò cos(Bz¢) eAz¢dz¢ = A2 + B2 where A = jk cosq and B = k , so that A2 + B2 = -k 2 cos2 q + k2 = k 2 sin 2 q . The q component requires the dot product zˆ ·qˆ = -sin q .

23 Naval Postgraduate School Antennas & Propagation Distance Learning Thin Wire Antennas (3)

Evaluating the limits gives p 2 cosçæ cosq ÷ö è 2 ø 6444447444448 æp ö jp cosq / 2 - jp cosq / 2 cosç cosq ÷ jkhIo - jkr k[e - (-1)e ] jhIo - jkr è 2 ø Eq = e sin q = e 4pr k2 sin 2 q 2pr sinq The magnetic field intensity in the far field is æp ö r cosç cosq ÷ kˆ ´ E E jI 2 Hr = = fˆ q = o e- jkr è øfˆ h Hf 2pr sin q The directivity is computed from the beam solid angle, which requires the normalized electric field intensity 2 2 cos p cosq r 2 Eq (2 ) Enorm = = 2 sinq Eq max

24 Naval Postgraduate School Antennas & Propagation Distance Learning Thin Wire Antennas (4)

p p cos2 (p cosq / 2) cos2 (p cosq / 2) W A = 2p sin q dq = 2p dq = (2p )(1.218) ò sin 2 q ò sin 2 q 0 0 14444244443 Integrate numerically The directive gain is 2 p 2 p 4p 2 4p cos ( cosq ) cos ( cosq ) D = Er (q ,f) = 2 = 1.64 2 norm 2 2 W A (2p )(1.218) sin q sin q The radiated power is 2 W E r2 2 2 2 A qmax 2p(1.218)h Io r 2 1 2 Prad = = = 36.57 Io º Io Ra 2h 2h(2pr)2 2

where Ra is the radiation resistance of the dipole. The radiated power can be viewed as the power delivered to resistor that represents “free space.” For the half-wave dipole the radiation resistance is 2P R = rad = (2)(36.57) = 73.13 ohms a 2 Io 25 Naval Postgraduate School Antennas & Propagation Distance Learning Numerical Integration (1)

The rectangular rule is a simple way of evaluating an integral numerically. The area under the curve of f (x) is approximated by a sum of rectangular areas of width D and height f (x ), where x = D + (n -1) + a is the center of the interval nth interval. Therefore, if n n 2 all of the rectangles are of equal width b b ò f (x)dx » Då f (xn ) a a Clearly the approximation can be made as close to the exact value as desired by reducing the width of the triangles as necessary. However, to keep computation time to a minimum, only the smallest number of rectangles that provides a converged solution should be used.

2 3 f ( x) 1 4 N D L · · · x a b x1 x2 x N

26 Naval Postgraduate School Antennas & Propagation Distance Learning Numerical Integration (2)

p cos2 (p cosq /2) Example: Matlab programs to integrate dq ò sinq 0 Sample Matlab code for the rectangular rule % integrate dipole pattern using the rectangular rule clear rad=pi/180; % avoid 0 by changing the limits slightly a=.001; b=pi-.001; N=5 delta=(b-a)/N; sum=0; for n=1:N theta=delta/2+(n-1)*delta; sum=sum+cos(pi*cos(theta)/2)^2/sin(theta); end I=sum*delta Convergence: N=5, 1.2175; N=10, 1.2187; N=50, 1.2188. Sample Matlab code using the quad8 function % integrate to find half wave dipole solid angle clear I=quad8('cint',0.0001,pi-.0001,.00001); disp(['cint integral, I: ',num2str(I)])

function P=cint(T) % function to be integrated P=(cos(pi*cos(T)/2).^2)./sin(T);

27 Naval Postgraduate School Antennas & Propagation Distance Learning Thin Wires of Arbitrary Length

For a thin-wire antenna of length l along the z axis, the electric field intensity is é æ kl ö æ kl öù êcosç cosq ÷ - cosç ÷ú jhI - jkr è 2 ø è 2 ø Eq = e ê ú 2pr ê sinq ú ëê ûú Example: l = 1.5l (left: voltage plot; right: decibel plot) 90 2 90 10 120 60 120 60 0

150 1 30 150 -10 30 -20

180 0 180 0

210 330 210 330

240 300 240 300 270 270

28 Naval Postgraduate School Antennas & Propagation Distance Learning Feeding and Tuning Wire Antennas (1)

When an antenna terminates a transmission line, as shown below, the antenna impedance (Za ) should be matched to the transmission line impedance (Zo ) to maximize the power delivered to the antenna

Za - Zo 1+ G Z Z G = and VSWR = o a Za + Zo 1- G

The antenna’s input impedance is generally a complex quantity, Z = R + R + jX . a ( a l ) a The approach for matching the antenna and increasing its efficiency is

1. minimize the ohmic loss, R ® 0 l 2. “tune out” the reactance by adjusting the antenna geometry or adding lumped elements, X a ® 0 (resonance occurs when Za is real) 3. match the radiation resistance to the characteristic impedance of the line by adjusting the antenna parameters or using a transformer section, Ra ® Zo

29 Naval Postgraduate School Antennas & Propagation Distance Learning Feeding and Tuning Wire Antennas (2)

Example: A half-wave dipole is fed by a 50 ohm line

Z - Z 73- 50 G = a o = = 0.1870 Za + Zo 73+ 50 1+ G VSWR = = 1.46 1- G

The loss due to reflection at the antenna terminals is

t 2 = 1- G 2 = 0.965 10log(t 2 )= -0.155 dB

which is stated as “0.155 dB of reflection loss” (the negative sign is implied by using the word “loss”).

30 Naval Postgraduate School Antennas & Propagation Distance Learning Feeding and Tuning Wire Antennas (3)

The antenna impedance is affected by 1. length 2. thickness 3. shape 4. feed point (location and method of feeding) 5. end loading Although all of these parameters affect both the real and imaginary parts of Za , they are generally used to remove the reactive part. The remaining real part can be matched using a transformer section. Another problem is encountered when matching a balanced radiating structure like a dipole to an unbalanced transmission line structure like a coax. UNBALANCED FEED BALANCED FEED

COAX TWO-WIRE LINE + + l / 2 Vg l /2 Vg - -

DIPOLE DIPOLE

31 Naval Postgraduate School Antennas & Propagation Distance Learning Feeding and Tuning Wire Antennas (4)

If the two structures are not balanced, a return current can flow on the outside of the coaxial cable. These currents will radiate and modify the pattern of the antenna. The unbalanced currents can be eliminated using a balun (balanced-to-unbalanced transformer) UNBALANCED FEED BALANCED FEED

I(z) a. b. a .b .

Va ¹Vb Va =Vb

Baluns frequently incorporate chokes, which are circuits designed to “choke off” current by presenting an open circuit to current waves.

32 Naval Postgraduate School Antennas & Propagation Distance Learning Feeding and Tuning Wire Antennas (5)

An example of a balun employing a choke is the sleeve or bazooka balun

I2 I1

HIGH IMPEDANCE (OPEN CIRCUIT) PREVENTS CURRENT FROM FLOWING l / 4 ON OUTSIDE Zo¢ I2 I1

SHORT CIRCUIT Zo

The choke prevents current from flowing on the exterior of the coax. All current is confined to the inside of surfaces of the coax, and therefore the current flow in the two directions is equal (balanced) and does not radiate. The integrity of a short circuit is easier to control than that of an open circuit, thus short circuits are used whenever possible. Originally balun referred exclusively to these types of wire feeding circuits, but the term has evolved to refer to any feed point matching circuit.

33 Naval Postgraduate School Antennas & Propagation Distance Learning Calculation of Antenna Impedance (1)

The antenna impedance must be matched to that of the feed line. The impedance of an antenna can be measured or computed. Usually measurements are more time consuming (and therefore expensive) relative to computer simulations. However, for a simulation to accurately include the effect of all of the antenna’s geometrical and electrical parameters on Za , a fairly complicated analytical model must be used. The resulting equations must be solved numerically in most cases. One popular technique is the method of moments (MM) solution of an integral equation (IE) for the current. z 1. I(z¢) is the unknown current distribution on the / 2 wire l 2. Find the z component of the electric field in 2a terms of I(z¢) from the radiation integral + b / 2 3. Apply the boundary condition Eg Vg -b / 2 - ì 0, b /2 £ z £ l /2 Ez (r = a) = í î Eg , b /2 ³ z

in order to obtain an integral equation for I(z¢) - l /2

34 Naval Postgraduate School Antennas & Propagation Distance Learning Calculation of Antenna Impedance (2)

One special form of the integral equation for thin wires is Pocklington’s equation

2 l / 2 - jkR æ 2 ¶ ö e ì0 , b/2 £ z £ /2 ç k + ÷ I(z ¢ ) dz ¢ = í l è ¶z2 ø ò 4pR î- jweEg, b / 2 ³ z -l / 2 where R = a2 + (z - z¢)2 . This is called an integral equation because the unknown quantity I(z¢) appears in the integrand. 4. Solve the integral equation using the method of moments (MM). First approximate the current by a series with unknown expansion coefficients {In } N I(z ¢ ) = å InFn(z ¢ ) n=1

The basis functions or expansion functions {Fn } are known and selected to suit the particular problem. We would like to use as few basis functions as possible for computational efficiency, yet enough must be used to insure convergence.

35 Naval Postgraduate School Antennas & Propagation Distance Learning Calculation of Antenna Impedance (3)

Example: a step approximation to the current using a series of pulses. Each segment is called a subdomain. Problem: there will be discontinuities between steps. I(z ¢ ) CURRENT STEP APPROXIMATION I2F2

I1F1 IN FN D L L · · · · z ¢ - l z1 z2 0 zN l 2 2 A better basis function is the overlapping piecewise sinusoid I(z¢ ) CURRENT PIECEWISE SINUSOID I2F2

D INFN I1F1 L L · · · z ¢ - l z1 z2 0 zN l 2 2 36 Naval Postgraduate School Antennas & Propagation Distance Learning Calculation of Antenna Impedance (4)

A piecewise sinusoid extends over two segments (each of length D) and has a maximum at the point between the two segment ì z ¢ - z n¢ -1 , z n¢ -1 £ z ¢ £ z n¢ Fn(z ¢ ) ï z n¢ - z n¢ -1 ï ï z n¢ +1 - z ¢ Fn(z ¢ ) = í , z n¢ £ z ¢ £ z n¢ +1 z n¢ +1 - z n¢ z ¢ ï ï 0, elsewhere zn- D z zn+ D n ï (= zn-1) (= zn+1) î

Fn(z ¢ )

Entire domain functions are also F1 possible. Each entire domain basis function extends over the z ¢ entire wire. Examples are - F l F3 2 l sinusoids. 2 2

37 Naval Postgraduate School Antennas & Propagation Distance Learning Calculation of Antenna Impedance (5)

Solving the integral equation: (1) insert the series back into the integral equation 2 l / 2 æ N ö - jkR æ 2 ¶ ö e ì0 , b / 2 £ z £ /2 ç k + ÷ ç InFn (z ¢ )÷ dz ¢ = í l 2 ò å 4pR î- jweEg, b/ 2 ³ z è ¶z ø è n=1 ø -l / 2 Note that the derivative is with respect to z (not z ¢ ) and therefore the differential operates only on R. For convenience we define new functions f and g: N ì / 2 2 - jkR(z,z¢) ü ï l æ 2 ¶ ö e ï ì0, b/ 2 £ z £ l / 2 å I í ò çk + ÷F (z¢) dz¢ý = í n ç 2 ÷ n ¢ - jweE , b / 2 ³ z n=1 îï- / 2 è ¶z ø 4pR(z,z ) þï î g l 1444442444443 14444444244444443 ºg º f (Fn )® fn (z) Once Fn is defined, the integral can be evaluated numerically. The result will still be a function of z, hence the notation fn (z). (2) Choose a set of N testing (or weighting) functions {Cm}. Multiply both sides of the equation by each testing function and integrate over the domain of each function D m to obtain N equations of the form

N Cm (z) å In fn (z)dz = Cm (z)g(z)dz, m =1,2,..., N ò n=1 ò D m D m

38 Naval Postgraduate School Antennas & Propagation Distance Learning Calculation of Antenna Impedance (6)

Interchange the summation and integration operations and define new impedance and voltage quantities N ì ü ï ï å In í ò Cm (z) fmn(z)dzý = ò Cm (z)g(z)dz n =1 ï ï îDm þ D m 144424443 1442443 V Zmn m This can be cast into the form of a matrix equation and solved using standard matrix methods N -1 å InZmn = Vm ® [Z][I]= [V] ® [I]= [Z] [V] n=1 [Z] is a square impedance matrix that depends only on the geometry and material characteristics of the dipole. Physically, it is a measure of the interaction between the currents on segments m and n. [V ] is the excitation vector. It depends on the field in the gap and the chosen basis functions. [I] is the unknown current coefficient vector. After [I] has been determined, the resulting current series can be inserted in the radiation integral, and the far fields computed by integration of the current.

39 Naval Postgraduate School Antennas & Propagation Distance Learning Self-Impedance of a Wire Antenna

The method of moments current allows calculation of the self impedance of the antenna by taking the ratio Zself = Vg / Io

RESISTANCE REACTANCE

40 Naval Postgraduate School Antennas & Propagation Distance Learning The Fourier Series Analog to MM

The method of moments is a general solution method that is widely used in all of engineering. A Fourier series approximation to a periodic time function has the same solution process as the MM solution for current. Let f (t) be the time waveform ¥ ao 2 f (t) = + å[an cos(wnt)+ bn sin(wnt)] T T n=1 For simplicity, assume that there is no DC component and that only cosines are necessary to represent f (t) (true if the waveform has the right symmetry characteristics) 2 ¥ f (t) = åan cos(wnt) T n =1 The constants are obtained by multiplying each side by the testing function cos(wmt) and integrating over a period T / 2 2 T / 2 æ ¥ ö ì0, m ¹ n ò f (t)cos(wmt)dt = ò ç åan cos(wnt)÷cos(wmt)dt = í -T / 2 T -T / 2 è n=1 ø îan , m = n

This is analogous to MM when f (t) ® I(z¢), an ® In , Fn ® cos(wnt), and Cm ® cos(wmt). (Since f (t) is not in an integral equation, a second variable t ¢ is not required.) The selection of the testing functions to be the complex conjugates of the expansion functions is referred to as Galerkin’s method.

41 Naval Postgraduate School Antennas & Propagation Distance Learning Reciprocity (1)

When two antennas are in close proximity to each other, there is a strong interaction between them. The radiation from one affects the current distribution of the other, which in turn modifies the current distribution of the first one.

ANTENNA 1 ANTENNA 2 (SOURCE) (RECEIVER) I 1 I2

+ V V1 Eg 21 -

Z1 Z2 Consider two situations (depicted on the following page) where the geometrical relationship between two antennas does not change. 1. A voltage is applied to antenna 1 and the current induced at the terminals of antenna 2 is measured. 2. The situation is reversed: a voltage is applied to antenna 2 and the current induced at the terminals of antenna 1 is measured.

42 Naval Postgraduate School Antennas & Propagation Distance Learning Reciprocity (2)

Case 1:

ANTENNA 1 I2

CURRENT METER + V1 - ENERGY FLOW ANTENNA 2

Case 2:

ANTENNA 1 + V2 -

I1 ENERGY FLOW

ANTENNA 2 CURRENT METER

43 Naval Postgraduate School Antennas & Propagation Distance Learning Mutual Impedance (1)

Define the mutual impedance or transfer impedance as

V2 V1 Z12 = and Z21 = I1 I2 The first index on Z refers to the receiving antenna (observer) and the second index to the source antenna. Reciprocity Theorem: If the antennas and medium are linear, passive and isotropic, then the response of a system to a source is unchanged if the source and observer (measurer) are interchanged. With regard to mutual impedance: · This implies that Z21 = Z12 · The receiving and transmitting patterns of an antenna are the same if it is constructed of linear, passive, and isotropic materials and devices. In general, the input impedance of antenna number n in the presence of other antennas is obtained by integrating the total field in the gap (gap width bn ) V 1 r r Z = n = E · d n I I ò n l n n b n 44 Naval Postgraduate School Antennas & Propagation Distance Learning Mutual Impedance (2)

Er is the total field in the gap of antenna n (due to its own voltage plus the incident fields n from all other antennas).

Vn En

In

45 Naval Postgraduate School Antennas & Propagation Distance Learning Mutual Impedance (3)

If there are a total of N antennas Er = Er + Er + + Er n n1 n2 L nN Therefore, 1 N r r Z = å E · d n I ò nm l n n b m=1 n Define r r Vnm = ò Enm· dln bn The impedance becomes N N N 1 Vnm Zn = åVnm = å = å Znm In m=1 m=1 In m=1 123 { ºVn ºZnm For example, the impedance of dipole n=1 is written explicitly as Z = Z + Z + + Z 1 11 12 L 1N When m = n the impedance is the self impedance. This is approximately the impedance that we have already computed for an isolated dipole using the method of moments.

46 Naval Postgraduate School Antennas & Propagation Distance Learning Mutual Impedance (4)

The method of moments can also be used to compute the mutual coupling between antennas. The mutual impedance is obtained from the definition

Vm mutual impedance at port m due to a Zmn = = I current in port n, with port m open ciruited n Im =0 Vn By reciprocity this is the same as Zmn = Znm = . Say that we have two dipoles, one I m In =0 distant (n) and one near (m). To determine Zmn a voltage can be applied to the distant dipole and the open-circuited current computed on the near dipole using the method of moments. The ratio of the distant dipole’s voltage to the current induced on the near one gives the mutual impedance between the two dipoles. Plots of mutual impedance are shown on the following pages: 1. high mutual impedance implies strong coupling between the antennas 2. mutual impedance decreases with increasing separation between antennas 3. mutual impedance for wire antennas placed end to end is not as strong as when they are placed parallel

47 Naval Postgraduate School Antennas & Propagation Distance Learning Mutual Impedance of Parallel Dipoles

80

60 l /2 d 40 R ) 12 ohms

( 20 X or R 0 X12

-20

-40 0 0.5 1 1.5 2 Separation, d (wavelengths)

48 Naval Postgraduate School Antennas & Propagation Distance Learning Mutual Impedance of Colinear Dipoles

25 l / 2 20 s 15

10

(ohms) R21 X 5 or R 0 X 21

-5

-10 0 0.5 1 1.5 2 Separation, s (wavelengths)

49 Naval Postgraduate School Antennas & Propagation Distance Learning Mutual-Impedance Example

Example: Assume that a half wave dipole has been tuned so that it is resonant (Xa = 0). We found that a resonant half wave dipole fed by a 50 ohm transmission line has a VSWR of 1.46 (a reflection coefficient of 0.1870). If a second half wave dipole is placed parallel to the first one and 0.65 wavelength away, what is the input impedance of the first dipole? From the plots, the mutual impedance between two dipoles spaced 0.65 wavelength is

Z21 = R21 + jX21 = -24.98- j7.7W

Noting that Z21 = Z12 the total input impedance is

Zin = Z11 + Z21 = 73 + (-24.98 - j7.7) = 48.02 - j7.7 W The reflection coefficient is

Z - Z (48.02 - j7.7)- 50 o G = in o = = -0.0139- j0.0797 = 0.081e- j99.93 Z + Z (48.02 - j7.7) + 50 in o which corresponds to a VSWR of 1.176. In this case the presence of the second dipole has improved the match at the input terminals of the first antenna.

50 Naval Postgraduate School Antennas & Propagation Distance Learning Broadband Antennas (1)

The required frequency bandwidth increases with the rate of information transfer (i.e., high data rates require wide frequency bandwidths). Designing an efficient wideband antenna is difficult. The most efficient antennas are designed to operate at a resonant frequency, which is inherently narrow band. Two approaches to operating over wide frequency bands: 1. Split the entire band into sub-bands and use a separate resonant antenna in each band Advantage: The individual antennas are easy to design (potentially inexpensive) Disadvantage: Many antennas are required (may take a lot of space, weight, etc.)

TOTAL SYSTEM BANDWIDTH OF AN 1 BANDWIDTH Df1 INDIVIDUAL ANTENNA BROADBAND 2 INPUT SIGNAL Df2 t M L FREQUENCY MULTIPLEXER f N Df N Df1 Df2 DfN

51 Naval Postgraduate School Antennas & Propagation Distance Learning Broadband Antennas (2)

2. Use a single antenna that operates over the entire frequency band

Advantage: Less aperture area required by a single antenna Disadvantage: A wideband antenna is more difficult to design than a narrowband antenna Broadbanding of antennas can be accomplished by:

1. interlacing narrowband elements having non-overlapping sub-bands (stepped band approach) Example: multi-feed point dipole TANK CIRCUIT USING LUMPED ELEMENTS

dmin OPEN CIRCUIT AT HIGH FREQUENCIES

dmax SHORT CIRCUIT AT LOW FREQUENCIES

52 Naval Postgraduate School Antennas & Propagation Distance Learning Broadband Antennas (3)

2. design elements that have smooth geometrical transitions and avoid abrupt discontinuities Example: biconical antenna dmin

dmax spiral antenna

dmax FEED POINT

dmin

WIRES

53 Naval Postgraduate School Antennas & Propagation Distance Learning Circular Spiral in Low Observable Fixture

54 Naval Postgraduate School Antennas & Propagation Distance Learning Broadband Antennas (4)

Another example of a broadband antenna is the log-periodic array. It can be classified as a single element with a gradual geometric transitions or as discrete elements that are resonant in sub-bands. All of the elements of the log periodic antenna are fed.

DIRECTION dmin dmax OF MAXIMUM RADIATION

The range of frequencies over which ana antenna operates is determined approximately by the maximum and minimum antenna dimensions

lH lL dmin » and dmax » 2 2

55 Naval Postgraduate School Antennas & Propagation Distance Learning Yagi-Uda Antenna

A Yagi-Uda (or simply Yagi) is used at high frequencies (HF) to obtain a directional azimuth pattern. They are frequently employed as TV/FM antennas. A Yagi consists of a fed element and at least two parasitic (non-excited) elements. The shorter elements in the front are directors. The longer element in the back is a reflector. The conventional design has only one reflector, but may have up to 10 directors. POLAR PLOT OF RADIATION PATTERN

REFLECTOR DIRECTORS BACK LOBE MAIN LOBE FED MAXIMUM MAXIMUM ELEMENT DIRECTIVITY DIRECTIVITY The front-to-back ratio is the ratio of the maximum directivity in the forward direction to that in the back direction: Dmax / Dback

56 Naval Postgraduate School Antennas & Propagation Distance Learning Ground Planes and Images (1)

In some cases the method of images allows construction of an equivalent problem that is easier to solve than the original problem. When a source is located over a PEC ground plane, the ground plane can be removed and the effects of the ground plane on the fields outside of the medium accounted for by an image located below the surface. ORIGINAL PROBLEM EQUIVALENT PROBLEM

ORIGINAL ® ® SOURCES ® ® h I dl I dl I dl h I dl REGION 1 GROUND PLANE PEC REMOVED REGION 2 h IMAGE SOURCES ® ® I dl I dl The equivalent problem holds only for computing the fields in region 1. It is exact for an infinite PEC ground plane, but is often used for finite, imperfectly conducting ground planes (such as the Earth’s surface).

57 Naval Postgraduate School Antennas & Propagation Distance Learning Ground Planes and Images (2)

The equivalent problem satisfies Maxwell’s equations and the same boundary conditions as the original problem. The uniqueness theorem of electromagnetics assures us that the solution to the equivalent problem is the same as that for the original problem. Boundary conditions at the surface of a PEC: the tangential component of the electric field is zero. ORIGINAL PROBLEM EQUIVALENT PROBLEM ® ® Idz TANGENTIAL Idz q1 COMPONENTS CANCEL h h E|| =0 PEC E^ h q2 Eq Eq1 Eq2

® Idz A similar result can be shown if the current element is oriented horizontal to the ground plane and the image is reversed from the source. (A reversal of the image current direction implies a negative sign in the image’s field relative to the source field.) 58 Naval Postgraduate School Antennas & Propagation Distance Learning Ground Planes and Images (3)

Half of a symmetric conducting structure can be removed if an infinite PEC is placed on the symmetry plane. This is the basis of a quarter-wave monopole antenna.

I l / 4 I l / 4 + + V g b Vg 2b z = 0 - PEC -

ORIGINAL MONOPOLE DIPOLE · The radiation pattern is the same for the monopole as it is for the half wave dipole above the plane z = 0 · The field in the monopole gap is twice the field in the gap of the dipole · Since the voltage is the same but the gap is half of the dipole’s gap

1 73.12 R = R = = 36.56 ohms a monopole 2 a dipole 2

59 Naval Postgraduate School Antennas & Propagation Distance Learning Crossed Dipoles (1)

y Crossed dipoles (also known as a turnstile) consists of two orthogonal x dipoles excited 90 degrees out of phase. Iy I = I x o I x jp / 2 z Iy = Ioe = jIo

The radiation integral gives two terms æ ö ç / 2 / 2 ÷ - jkho - jkr l jx¢k sin q cosf l jy¢k sin q sin f Eq = e ç ò Io cosq cosfe dx¢+ j ò Io cosq sin fe dy¢÷ 4pr ç - / 2 14243 - / 2 14243 ÷ è l qˆ· xˆ l qˆ· yˆ ø l/ 2 jx¢k sin q cosf If kl << 1 then ò e dx¢»l and similarly for the y integral. Therefore, -l/ 2 - jkr - jkr - jkhIo e e E = l cosq (cosf + j sin f) = E cosq (cosf + j sin f ) q 4p r o r 14243 ºEo

60 Naval Postgraduate School Antennas & Propagation Distance Learning Crossed Dipoles (2)

A similar result is obtained for Ef jkhI e- jkr e- jkr E = ol (sin f - j cosf) = -E (sin f - j cosf ) f 4p r o r 123 º-Eo Consider the components of the wave propagating toward an observer on the z axis q = f = 0 : Eq = Eo , Ef = jEo, or e- jkr Er = E (xˆ + jyˆ ) o r which is a circularly polarized wave. If the observer is not on the z axis, the projected lengths of the two dipoles are not equal, and therefore the wave is elliptically polarized. The axial ratio (AR) is a measure of the wave’s ellipticity at the specified q,f: E AR = max , 1 £ AR £ ¥ Emin For the crossed dipoles Ef 1 1 AR = = = 2 2 2 Eq cos q(cos f +sin f) cosq

61 Naval Postgraduate School Antennas & Propagation Distance Learning Crossed Dipoles (3)

The rotating linear pattern is shown. A linear receive antenna rotates like a propeller blade as it measures the far field at range r. The envelope of the oscillations at any particular angle gives the axial ratio at that angle. For example, at 50 degrees the AR is about 1/ 0.64 =1.56 = 1.93 dB. Ef =1 1

0.9

0.8

0.7

0.6

0.5 Eq 0.4

0.3

NORMALIZED FIELD ROTATING LINEAR 0.2 POLARIZATION

0.1

0 0 50 100 150 200 250 300 350 PATTERN ANGLE, q (DEGREES)

62 Naval Postgraduate School Antennas & Propagation Distance Learning Crossed Dipoles (4)

Examples of rotating linear patterns on crossed dipoles that are not equal in length

90 1 90 0 120 60 120 60 -10

150 0.5 30 150 -20 30 -30

180 0 180 0

210 330 210 330

240 300 240 300 270 270

VOLTAGE PLOT DECIBEL PLOT

63 Naval Postgraduate School Antennas & Propagation Distance Learning Polarization Loss (1)

r For linear antennas an effective height (he ) can be defined r h e r r r V = E · h Voc Einc oc inc e

The open circuit voltage is a maximum when the antenna is aligned with the incident electric field vector. The effective height of an arbitrary antenna can be determined by casting its far field in the following form of three factors - jkr r ée ù r E(r,q,f) = [Eo ]ê ú[he (q,f)] ë r û The effective height accounts for the incident electric field projected onto the antenna element. The polarization loss factor (PLF) between the antenna and incident field is r r 2 Einc · he PLF, p = r 2 r 2 Einc he

64 Naval Postgraduate School Antennas & Propagation Distance Learning Polarization Loss (2)

Example: The Hertzian dipole’s far field is é - jkr ù r é jhkIo ù e ˆ E(r,q ,f) = ê ú ê ú [lsinqq ] ë 4p û ë r û 14r 243 he (q ,f ) If we have a second dipole that is rotated by an angle d in a plane parallel to the plane containing the first dipole, we can calculate the PLF as follows. First, r r r r r r r r Voc = Einc · he = Einc zˆ · he zˆ ¢ = Einc he zˆ ·zˆ ¢ = Einc he cosd z ¢ z z r d h Er e y inc y x Voc

TRANSMIT RECEIVE

65 Naval Postgraduate School Antennas & Propagation Distance Learning Polarization Loss (3)

The PLF is r r 2 r 2 r 2 2 Einc · he Einc he cos d p = = = cos2 d r 2 r 2 r 2 r 2 Einc he Einc he

When the dipoles are parallel, p=1, and there is no loss due to polarization mismatch. However, when the dipoles are at right angles, p=0 and there is a complete loss of signal. A more general case occurs when the incident field has both q and f components r ˆ ˆ Einc = Ei q + Ei f q f ˆ ˆ r 2 (Eiq q + Eif f )· he p = ˆ ˆ 2 r 2 Eiq q + Eif f he Example: The effective height of a RHCP antenna which radiates in the +z direction is r ˆ ˆ given by the vector he = ho (q - jf). A LHCP field is incident on this antenna (i.e., the incident wave propagates in the –z direction): r ˆ ˆ jkz Einc = Eo (q - jf )e 66 Naval Postgraduate School Antennas & Propagation Distance Learning Polarization Loss (4)

The PLF is 2 ˆ ˆ ˆ ˆ jkz Eoho (q - jf )· (q - jf )e p = = 0 2 2 2Eo 2ho If a RHCP wave is incident on the same antenna, again propagating along the z axis in the r ˆ ˆ jkz negative direction, Einc = Eo (q + jf )e . Now the PLF is 2 ˆ ˆ ˆ ˆ jkz Eoho (q - jf )· (q + jf )e p = = 1 2 2 2Eo 2ho r ˆ jkz Finally, if a linearly polarized plane wave is incident on the antenna, Einc = qEoe 2 ˆ ˆ ˆ jkz Eoho (q - jf )·qe p = =1/ 2 2 2 Eo 2ho If a linearly polarized antenna is used to receive a circularly polarized wave (or the reverse situation), there is a 3 dB loss in signal.

67 Naval Postgraduate School Antennas & Propagation Distance Learning Antenna Polarization Loss

TRANSMIT RECEIVE ANTENNA ANTENNA

0 dB RHCP Summary of polarization RHCP losses for polarization mismatched antennas > 25 dB

3 dB LHCP

3 dB 0 dB V

V 3 dB

> 25 dB H

68 Naval Postgraduate School Antennas & Propagation Distance Learning Aircraft Blade Antenna

Blade antennas are used for telemetry and communications. They have nearly hemispherical coverage, allowing the aircraft to maneuver without a complete loss of signal. The resulting polarization is often called slant because it contains both horizontal and vertical components. Comparison of measured and calculated Blade antenna patterns for a blade installed on a cylinder l / 4 THIN WIRE 1.58¢ ¢ h (4.01 cm) o 45

3.0¢ ¢ ( 7.62 cm) Method of moments model:

Location of source excitation

69 Naval Postgraduate School Antennas & Propagation Distance Learning Blade Antennas Installed on an Aircraft

The top antenna (A) provides coverage in the upper hemisphere, while the bottom antenna (B) covers the lower hemisphere. The two antennas can be duplexed (switched) or their signals combined using a coupler.

Method of moments patch model: Elevation pattern: (signals combined with 3 dB coupler)

Pit ch y z Yaw x Ro ll

A

B

70 Naval Postgraduate School Antennas & Propagation Distance Learning Small Loop Antenna (1)

By symmetry, we expect that the field of a small wire loop located in the z = 0 plane will depend only on the angle off of the wire axis, q . Because of the azimuthal symmetry cylindrical coordinates are required to solve this problem. If the wire is very thin (a ˆ filament) and has a constant current Iof¢ flowing in it, the radiation integral is 2p r - jkh - jkr jk(r ¢·rˆ) ˆ E(r,q ,f ) » e ò Io e f¢a df¢ 4pr 14243 0 dlr¢

z Using the transformation tables rˆ = xˆ sinq cosf + yˆ sinq sin f + zˆ cosq q rr r Rr r¢ = arˆ¢ = a(xˆ cosf¢ + yˆ sin f¢) rˆ · r¢ = a sinq(cosf cosf¢ + sin f sinf¢)

f¢ y We also need fˆ¢ in terms of the cartesian x r¢ = arˆ¢ unit vectors (fˆ¢ is not a constant that can be moved outside of the integral) fˆ¢ = -xˆsin f¢ + yˆ cosf¢

71 Naval Postgraduate School Antennas & Propagation Distance Learning Small Loop Antenna (2)

The radiation integral becomes 2p r - jkhI a - jkr ¢ ¢ E(r,q,f) » o e ò (- xˆ sin f¢ + yˆ cosf¢)e jkasin q (cosf cosf +sin f sin f )df¢ 4pr 0 For a small loop ka << 1 and the exponential can be represented by the first two terms of a jkasin q (cosf sin f ¢+sin f cosf ¢) Taylor’s series to get e » 1+ jkasin q(cosf cosf¢+ sin f sin f¢) Inserting the approximation in the integral 2p ò (- xˆsin f¢+ yˆ cosf¢)[1+ jkasin q (cosf cosf¢+ sin f sin f¢)]df¢ 0 2p 2p Since ò sinf¢df¢ = ò cosf¢df¢ = 0 the 1 in the square brackets can be dropped. The 0 0 remaining terms in the integrand involve the following factors: sinf sin 2 f¢ sinf ¢cosf cosf¢ ® integrates to zero because sinf ¢cosf¢ is an odd function of f ¢ cosf cos2 f¢ sinf ¢sin f cosf¢ ® integrates to zero because sin f¢cosf¢ is an odd function of f¢

72 Naval Postgraduate School Antennas & Propagation Distance Learning Small Loop Antenna (3)

The only two terms that do not integrate to zero are of the form 2p 2p ò sin 2 f¢df¢ = ò cos2 f¢df¢ = p 0 0 Therefore, 2p 2p r - jkahIo ( jkasin q ) - jkr æ 2 2 ö E(r,q,f) » e ç- xˆsin f ò sin f¢df¢ + yˆ cosf ò cos f¢df¢÷ 4pr è 0 0 ø k2hI a2 k 2hI a2 = o sinq e- jkr (- xˆ sin f + yˆ cosf ) = fˆ o sin q e- jkr 4r 14444244443 4r = fˆ The radiation pattern of the small loop is the same as that of a short dipole aligned with the loop axis. The radiated power is k4hI 2pa4 P = o rad 12 and the radiation resistance 4 4 4 4 4 2P k hp a (2p / l) (120p )p a 4 æ a ö R = rad = = = 320p a 2 ç ÷ Io 6 6 è l ø

73 Naval Postgraduate School Antennas & Propagation Distance Learning Helix Antenna (1)

A helix is described by the following parameters: z D D = diameter A = axial length S S C = circumference (pD) S = center-to-center spacing d C = pD L between turns a L = length of one turn N = number of turns a = tan-1(S /pD) =pitch angle

GROUND PLANE

Conventional helices are constructed with air or low-dielectric cores. A helix is capable of operating in several different radiation modes and polarizations, depending on the combination of parameter values and the frequency.

74 Naval Postgraduate School Antennas & Propagation Distance Learning Helix Antenna (2)

Radiation modes of a helix:

NORMAL MODE AXIAL MODE In both cases the radiation is circularly polarized: 2Sl Normal mode: AR = D2p 2 2n +1 Axial mode: AR = 2n In the axial mode, the beamwidth decreases with increasing helix length, NS SMALL (LOOKS 115o LIKE A LOOP) BWFN = LARGE (C / l) NS / l

C > l for 12o < a <13o.

75 Naval Postgraduate School Distance Learning

Antennas & Propagation

LECTURE NOTES VOLUME III ARRAYS, ANTENNAS IN SYSTEMS AND ACTIVE ANTENNAS

by Professor David Jenn

(ver 1.3) Naval Postgraduate School Antennas & Propagation Distance Learning Array Antennas (1)

Arrays are collections of antennas that are geometrically arranged and excited in such a way as to achieve a desired radiation pattern. In most applications the array elements are identical (e.g., an array of dipoles). The general geometry of an array is shown below. · The observation point is in the far field · The global coordinate system is located at the reference element (#0) · The location of element n is given by the position vector rrn

z TO FAR FIELD

r J r REFERENCE so ELEMENT, o r Rn r q r ¢ y rr ¢ r r n Jsn rn r r ¢ x ELEMENT, n

1 Naval Postgraduate School Antennas & Propagation Distance Learning Array Antennas (2)

The radiation integral for the nth element is

r - jkh - jkr r jkrn¢ ×rˆ E(r,q,f) = e Js e ds¢ 4p r òò n Sn The total field at P is the sum of all of the radiated element fields. Assume that the current distribution on all elements is the same except for a complex constant scaling factor, jFn An = ane (an º An ): r r J sn =An J so Noting that rrn¢ =r¢ + rn , the total field is N r æ - jkh ö - jkr r jk (r¢+rn )×rˆ E(r,q ,f) = åç ÷e òò An J so e ds¢ 4p rn n =0è ø Sn é ù é N ù - jkh - jkr r jkr¢×rˆ jkrn ×rˆ = ê e Js e ds¢ú ê Ane ú 4p r òò o å ëê Sn ûú ëên=0 ûú 1444442444443 1442443 ELEMENT FACTOR, EF ARRAY FACTOR, AF This result is referred to as the principle of pattern multiplication.

2 Naval Postgraduate School Antennas & Propagation Distance Learning Array Antennas (3)

As a simple application of the array formulas, consider a linear array of dipoles along the x axis. z

r rN rr 2 l x r L 1 d 0 1 2 N Assume: 1. equally spaced dipole elements 2. uniformly excited (equal power to all elements, An = 1) 4. neglect "edge effects" (mutual coupling changes near edges)

From the coordinate tables rˆ = xˆsin q cosf + yˆ sin q sin f + zˆcosq . The position vector to element n is rrn = nd xˆ . Therefore

rˆ · rrn = nd sinq cosf

3 Naval Postgraduate School Antennas & Propagation Distance Learning Array Antennas (4)

Now we can write the array factor N N N jy N +1 jkr ×rˆ jknd sin q cosf jy n 1- e AF = A e n = (1)e = e = ( ) å n å å( ) jy n=0 n=0 n=0 1- e 14243 GEOMETRIC SERIES where y = kd sin q cosf . Using trigonometric identities sin[(N +1)y / 2] sin[(N +1)y /2] AF = e jNy / 2 Þ AF = sin(y / 2) sin(y /2) The maximum value of AF is N+1 and it occurs whenever the denominator is zero

sin(y /2) = 0 Þ ym /2 = mp Þ y m = 2mp , m = 0,±1,…

A plot of the array factor is shown on the next page. The maximum at y = 0 is the main beam or main lobe. The other maxima are called grating lobes. Generally grating lobes are undesirable, because most applications such as radar require a single focussed beam. There are N-1 sidelobes between each grating lobe; the notches between lobes are nulls.

4 Naval Postgraduate School Antennas & Propagation Distance Learning Array Antennas (5)

MAXIMUM VALUE 20 log(N+1)

GRATING MAIN ) LOBE BEAM GRATING LOBE dB ( d = 1.4l N+1 = 20 ARRAY FACTOR

-80 -60 -40 -20 0 20 40 60 80 q (DEG) The array factor is defined for - ¥

5 Naval Postgraduate School Antennas & Propagation Distance Learning Array Antennas (6)

We have considered an array along the x axis, as shown. To obtain the array factor for arrays on other axes, simply replace the x direction cosine with the direction cosine of the appropriate axis. z P Direction cosines: z z = q cosz x = sin q cosf = u z z y cosz y = sin q sin f = v x y cosz z = cosq = w f

x

For example, if the array is located along the y axis use y = kdsin q sin f 14243 v

6 Naval Postgraduate School Antennas & Propagation Distance Learning Principal Planes

The principal planes of a linearly polarized antenna are defined by looking directly at the antenna (along the normal nˆ) from the far field. If the electric and magnetic fields in the aperture plane are TEM, then Er and Hr are orthogonal. The two principal planes are · E-plane: defined by the vectors nˆ and Er · H-plane: defined by the vectors nˆ and Hr y For example, if an array of dipoles are aligned with the y axis, then the y-z ANTENNA Er r plane is the E plane ( is parallel to the E x dipoles). The x-z plane is the H plane.

H-PLANE Hr nˆ

E-PLANE

7 Naval Postgraduate School Antennas & Propagation Distance Learning Polarization Reference

In a principal plane, the component of Er parallel to the field in the aperture (i.e., lies in the r principal plane and is parallel to the effective height vector, h ) is the co-polarized e component. The component of Er perpendicular to the field in the aperture is the cross polarized component. · in the E-plane: Ecpol = Eq and Expol = Ef · in the H-plane: Ecpol = Ef and Expol = Eq The crossed polarized radiation for well- y r Ecpol designed array antennas is very low. The

r maximum crossed polarized value is typically Expol 40 to 60 dB below the maximum value of the ANTENNA x co-polarized component. r E xpol When an array of crossed dipoles operate as r Ecpol two orthogonal linear arrays, the two sets of H-PLANE dipoles can operate simultaneously with high nˆ isolation between the two polarizations. This E-PLANE is referred to as polarization diversity (or polarization reuse).

8 Naval Postgraduate School Antennas & Propagation Distance Learning Huygen’s Principle (1)

Huygen’s principle states that any wavefront can be decomposed into a collection of point sources. New wavefronts can be constructed from the combined “spherical wavelets” from the point sources of the old wavefront.

ORIGINAL - jkR WAVE ¥ e n FRONT RECONSTRUCTED E(P) ~ å WAVE FRONT n =-¥ Rn POINT SAMPLES OF WAVE FRONT

where Rn is the distance from the SPHERICAL 4 R4 WAVELET nth source to the observation point, P. Sources closest to P DIRECTION OF 3 R3 PROPAGATION will contribute most to the field

P

2 R2

1 R1

9 Naval Postgraduate School Antennas & Propagation Distance Learning Huygen’s Principle (2)

An array of sources can be used to generate wavefronts in any desired direction. This is easily illustrated in the time domain. Assume that source 1 is turned on first, then source 2, and finally source 3. A plane wave approximation will be generated, and the angle of propagation depends on the time delay increment between sources. z

DIRECTION OF PROPAGATION

WAVE FRONT OF q SUPERIMPOSED s WAVES x 1 2 3

SPHERICAL WAVE SOURCE

Since time-harmonic waves are sinusoidal in time and space, the same result occurs if a phase delay is introduced between adjacent sources.

10 Naval Postgraduate School Antennas & Propagation Distance Learning Huygen’s Principle (3)

Time-harmonic waves vary as ~ cos(wt - kz) where z is the direction of propagation. Increasing z has the same effect as delaying time. Parallel feed arrangement: · Line lengths for time delay: d l1 is arbitrary 1 2 3 x l2 = l1 + dsinq s = + 2dsinq l 3 l1 s Pin / 3 · The associated phases are l 2 Pin / 3 l1 k 1,k 2 and k 3 l 3 l l l 2P /3 in · In phasor form e- jkl1 , e- jkl2 and e- jkl3 · After subtracting out the common Pin phase factor, the residual is

e0, e- jkd sin qs and e- j2kd sin qs

11 Naval Postgraduate School Antennas & Propagation Distance Learning Printed Circuit Dipole Array

12 Naval Postgraduate School Antennas & Propagation Distance Learning Scanned Arrays (1)

An appealing feature of arrays is that the pattern can be scanned electronically by adjusting the array coefficients, An . This approach permits beams to be moved from point to point in space in just microseconds, whereas mechanical scanning of a large antenna could take several seconds. Let the excitation coefficients be an imaginary quantity (a “phase shift”) that has a constant linear progression from element to element

- jnkd sinqs cosfs - jny s An = e = e

where y s = kd sinq s cosfs and (qs ,fs ) is the direction in which the beam is to be pointed. The array factor becomes N N sin[(N +1)(y -y )/ 2] AF = åe jknd(sin q cosf -sin qs cosfs ) = åe jn(y -y s ) = s e jN (y -y s )/ 2 n=0 n=0 sin[(y -y s )/ 2] The magnitude of the array factor is sin[(N +1)(y -y )/ 2] AF = s sin[(y -y s )/ 2]

The main beam is now located in the direction y =y s . The array factor slides through the visible region window until the main beam is located at ys . 13 Naval Postgraduate School Antennas & Propagation Distance Learning Scanned Arrays (2)

The location of the array factor in the visible window is shown for the cases of no scan and with scan. Notice that grating lobes originally outside of the visible region when the beam is not scanned can enter the visible region when the beam is scanned.

Broadside beam (no scan) Scanned beam

BROAD- SCANNED SIDE MAIN MAIN BEAM BEAM

y y

- kd 0 kd - kd 0 kd

14 Naval Postgraduate School Antennas & Propagation Distance Learning Grating Lobe Example (1)

Example: A linear array of elements d = 0.6l . At what scan angle will the first grating lobe appear at the edge of the visible region? The result will be the same whether the beam can be scanned in either the positive or negative direction. The first grating lobe on the positive side is m =1, and it will enter the visible region when the beam is scanned in the negative direction. For the given spacing, 2p kd = 0.6l = 1.2p . The grating lobe and main beam locations are depicted below. l WITH SCAN NO SCAN m = -1 m = 0 m =1 y -kd 0 kd Working with the m =1 lobe, which enters at q = 90o , y -y kd sin 90o - sinq s = ( s ) = p 2 2 sin qs = 1- l / d = -0.667 Þ qs = -41.8o

15 Naval Postgraduate School Antennas & Propagation Distance Learning Grating Lobe Example (2)

Array factor plot for a 20 element array that is scanned to –41.8 degrees. The element spacing is d = 0.6l . Note that the grating lobe location does not depend on the number of elements, but only the spacing in wavelengths. The total pattern is obtained by multiplying by EF, which could be chosen to place a null at the same location as the grating lobe. Note how the main beam broadens when it is scanned away from broadside.

0

-10

-20

-30

-40

normalized array factor (dB) -50

-60 -80 -60 -40 -20 0 20 40 60 80 pattern angle, theta (deg)

16 Naval Postgraduate School Antennas & Propagation Distance Learning Sample Linear Array Patterns (1)

AF for 5 element array, d = 0.4l

17 Naval Postgraduate School Antennas & Propagation Distance Learning Sample Linear Array Patterns (2)

AF for 5 element array, d = 0.4l

18 Naval Postgraduate School Antennas & Propagation Distance Learning Array Beamwidth (1)

The main lobe width is commonly described by either the 3 dB, half-power beamwidth, (HPBW) or the beamwidth between first nulls (BWFN). The beamwidth between first nulls can be determined from the numerator of the array factor, éN +1 ù N +1 sin (y1 -ys ) = 0 Þ (y1 -y s ) = ±p ëê 2 ûú 2 where y1 = kd sinq1. Assume that the scan angle is zero, and that the array length is approximately L » (N +1)d L 2p sinq1 = ±p 2 l l sinq1 = ± L -1æ l ö q1 = ±sin ç ÷ è L ø -1æ l ö Therefore, for a symmetrical beam, BWFN = 2q1 = 2sin ç ÷ è L ø

19 Naval Postgraduate School Antennas & Propagation Distance Learning Array Beamwidth (2)

If the array is very long in terms of wavelength, then L / l is large and the small angle approximation is valid 2l BWFN » L More commonly the HPWB is used. Since power is proportional to field squared:

2 2 sin[(N +1)y HP / 2] AFnorm = = 0.5 (N +1)sin(y HP /2)

where y HP = kd sinqHP and qHP is the half power angle. Generally this must be done on the computer or by interpolating from a table. Once qHP is found, the half power beamwidth is HPBW, qB = 2qHP This assumes that the beam is symmetrical about the peak. If the beam is asymmetrical, which occurs when it is scanned off of broadside, the half power points on the left and right sides of the beam are not equal. Each one must be computed separately and then added to obtain the HPBW.

20 Naval Postgraduate School Antennas & Propagation Distance Learning Linear Array Directivity (1)

From the definition of directivity, 4p Do = W A 2p p r 2 W A = ò ò Enorm (q,f) sinq dq df 0 0 Assume a uniform array. For convenience align the array along the z axis. Then the z direction cosine is used rather than the x direction cosine in the argument of AF

sin[(N +1)(y -y )/2] AF = s sin[(y -y s )/2]

where y = kd cosq and y s = kd cosqs . If the beam is not scanned far from broadside, then the pattern will not deviate significantly from the non-scanned case. Therefore let y s = 0. Now evaluate the integral 2 2p p sin[(N +1)y / 2] W A = ò ò sinq dq df 0 0 (N +1)sin[y / 2]

21 Naval Postgraduate School Antennas & Propagation Distance Learning Linear Array Directivity (2)

There is an identity that can be used if the beam is near broadside:

1 2 N AF 2 = + (N +1- n)cos(nkd cosq ) norm 2 å N +1 (N +1) n=0 Inserting this in the integrand and evaluating the integral gives

N ïì 1 2 é(N +1- n) ùïü W = 4p + sin (nkd ) A í 2 åê úý îïN +1 (N +1) n=0ë nkd ûþï Finally, N +1 Do = 1 N é(N +1- n) ù 1+ åê sin (nkd )ú kd n =1ë (N +1)n û Example: N elements with d = l / 2; kd = p ® sin(nkd) = 0 and

Do = N +1

22 Naval Postgraduate School Antennas & Propagation Distance Learning Antenna Sidelobe Control (1)

Many antenna applications require low sidelobes and scanned beams. Some of the important advantages and disadvantages of sidelobe reduction are:

Advantages Disadvantages reduced clutter return more complicated feed required low probability of intercept (LPI) reduced gain less susceptible to jamming increased beamwidth; nulls move outward RADAR TARGET

ANTENNA MAIN LOBE

SIDELOBE CLUTTER GROUND Classification of sidelobe level (not standardized): · low sidelobes: -25 to – 40 relative to the beam maximum · ultra-low sidelobes: < – 40 relative to the beam maximum 23 Naval Postgraduate School Antennas & Propagation Distance Learning Antenna Sidelobe Control (2)

The shape of the amplitude distribution, determined by an = An , can be used to reduce the sidelobes of the radiation pattern. For a focused beam the amplitude distribution is always symmetric about the center of the array. To scan a focused beam a linear phase is introduced across the array length. AMPLITUDE DISTRIBUTION SAMPLES AT ELEMENT LOCATIONS an = An

x L L END CENTER END x = 0 x = L Common aperture distribution functions: 1. Chebyshev: · yields the minimum beamwidth for a specified sidelobe level · all sidelobes are equal · only practical for a small number of elements

24 Naval Postgraduate School Antennas & Propagation Distance Learning Antenna Sidelobe Control (3)

2. Binomial: · has no sidelobes · only practical for a small number of elements 3. Taylor: · specify the maximum sidelobe level and rate of falloff of sidelobe level 4. Bayliss: · for low sidelobe difference beams (a difference beam has a null where the pattern would normally have a maximum) 5. Cosine-on-a-pedestal: (cosine raised to a power plus a constant) · wide range of sidelobe levels and falloff rates · Hamming window is one of these

1 m æ p n ö 1- c an = c1 + (1 - c1)cos ç ÷ an 1 è 2N ø

c1 x 0 L 25 Naval Postgraduate School EO4612 Microwave Devices & Radar Distance Learning Aperture Efficiency

The reduction in directivity due to non-uniform amplitude is given by the illumination efficiency N 2 ån=0 an ei = N 2 (N +1)ån =0 an For an array this factor is sometimes called the aperture efficiency. However, for other types of antennas such as reflectors, the aperture and illumination efficiencies are not necessarily the same.

Example: For a uniform distribution, an = 1, and therefore 2 2 N N 2 ån=0 an ån=01 (N +1) ei = = = = 1 N 2 N 2 (N +1)(N +1) (N +1)ån=0 an (N +1)ån=0 1

æpxn ö For a cosine distribution, an = cosç ÷, where xn is the location of element n. The è L ø efficiency can be found with a simple computer calculation. Choose an arbitrary value for N, generate the coefficients, and then calculate ei . The result is ei » 0.81 = -0.91 dB.

26 Naval Postgraduate School Antennas & Propagation Distance Learning Some Common Amplitude Distributions

The table summarizes the characteristics of some commonly encountered amplitude distributions. It is taken from Introduction to Radar Systems by Skolnik, (3rd edition) but similar tables can be found in any antenna textbook. The values are usually computed for continuous distributions. We can use the numbers for arrays (which can be viewed as sampled versions of continuous distributions) if the elements are closely spaced enough so that the true shape of the distribution function is approximated.

Table 9.1 in Skolnik

· z is the distance along the array ( x in previous charts) · a is the array length (L in previous charts) · Relative gain is ei

27 Naval Postgraduate School Antennas & Propagation Distance Learning Summary of Array Characteristics

1. The radiation pattern of an array of identical elements is the product of an element factor and an array factor. This implies that the current distribution on every element is the same except for a constant scale factor, An . This assumption is not accurate for small arrays, or for the elements of large arrays that are located near edges. 2. Grating lobes (secondary maxima) occur when the spacing relative to wavelength becomes too large. Grating lobes originally outside of the visible region can move into the visible region when the beam is scanned. 3. The directivity increases with array length if there are no grating lobes. This implies that more elements must be added to increase the length. 4. The HPBW (and BWFN) decrease with increasing array length. 5. Scanning can be achieved by providing a linear phase progression per element. 6. The HPBW (and BWFN) increase as the beam is scanned from broadside. The beam shape between the half power points (or first nulls) becomes asymmetrical when scanned. 7. The sidelobe levels can be reduced by symmetrically tapering the excitation amplitude from the center to the edges. 8. A reduction in sidelobe level is accompanied by an increase in beamwidth and a reduction in gain.

28 Naval Postgraduate School Antennas & Propagation Distance Learning Finite Arrays and the “Edge Effect” (1)

The patterns of finite arrays differ from those of infinite arrays. The difference is referred to as the edge effect. · Mutual coupling variations are significant near the edges of the array. Elements near the edges have fewer neighbors than those in the center of a large array.

CENTER EDGE MUTUAL COUPLING DIPOLE DIPOLE TERMS MISSING · A truncated ground plane affects the wide angle radiation in two ways: 1. The image of an element near the edge of the ground plane is disrupted 2. Diffraction occurs at the edges of the ground plane

DIRECT EDGE DIPOLES DIFFRACTED

IMAGE CONTRIBUTION MISSING BECAUSE GROUND IMAGES PLANE TOO SMALL

29 Naval Postgraduate School Antennas & Propagation Distance Learning Finite Arrays and the “Edge Effect” (2)

Example of H-plane element patterns for a 10 element array (#1 is at an edge; #5 is near the center). The ground plane is infinite (i.e., there is no ground plane edge effect)

H-plane element patterns (infinte ground plane) 10

5

0

-5 ELEMENT 1 ELEMENT 2 -10 ELEMENT 3 ELEMENT 4 -15 ELEMENT 5 RELATIVE POWER, dB -20

-25

-30 -80 -60 -40 -20 0 20 40 60 80 THETA, DEG

30 Naval Postgraduate School Antennas & Propagation Distance Learning Finite Arrays and the “Edge Effect” (3)

These patterns illustrate the effect of a finite ground plane on the pattern of a finite array of 10 dipoles. 0 INFINITE GROUND PLANE FINITE GROUND PLANE -5

-10

-15

-20

-25 RELATIVE POWER, dB -30

-35

-40 -80 -60 -40 -20 0 20 40 60 80 THETA, DEG 31 Naval Postgraduate School Antennas & Propagation Distance Learning Curved vs. Flat Arrays

0 CURVED GP (12.7 dB) FLAT GP (14.4 dB)

-5

FLAT GROUND PLANE

-10

CURVED GROUND PLANE -15 Relative Power (dB)

-20

-25 -80 -60 -40 -20 0 20 40 60 80 Theta (degrees)

32 Naval Postgraduate School Antennas & Propagation Distance Learning Two-dimensional Arrays (1)

A two-dimensional array can be formed by constructing a linear array whose elements are themselves linear arrays. In this case the element lattice (or geometrical arrangement) is rectangular. In general, a two-dimensional array can be constructed on any shaped surface. By far the most common situation is where the elements lie in a plane, i.e., a planar array.

Assume that an array is formed by linear arrays along the x axis, each having N x elements spaced d x . There are N y linear arrays distributed uniformly along the y axis with spacing d y . z

m = 0 m = N y n = 0 y

n = N x dx

x d y

33 Naval Postgraduate School Antennas & Propagation Distance Learning Two-dimensional Arrays (2)

The array factor of the 2-d array is

N x N y jnkd x sinq cosf jmkd y sinq sin f AF = å å Amne e n=0 m=0

From the assumed feeding arrangement, Amn = Am An , which allows separation of the two sums N x N y jnkd x sin q cosf jmkd y sinq sin f AF = AFx AFy = å Ane å Ame n=0 m =0 The sums are the same as in the linear array case éN +1 ù éN x +1 ù y sin kd (sin q cosf - sin q cosf ) sin kd y (sinq sinf - sin qs sin fs ) ê x s s ú ê 2 ú AF = ë 2 û ë û sin[kd x (sinq cosf - sinqs cosfs )/ 2] sin[kd y (sin q sin f - sinqs sinfs )/ 2]

The physical area of the array is approximately A = (N x +1)d x (N y +1)d y . The main beam

direction is given by (qs,fs) 34 Naval Postgraduate School Antennas & Propagation Distance Learning Two-dimensional Arrays (3)

Mesh plot of a two dimensional array factor

N x = 10 N y = 4 d x = d y = 0.5 Uniform illumination ( An = Am = 1)

35 Naval Postgraduate School Antennas & Propagation Distance Learning Two-dimensional Arrays (4)

The directivity of a two-dimensional array can be obtained by considering it to be a linear array with “elements” that are themselves linear arrays. z Directivity of the x and y linear ELEMENT OF THE x m = 0 m = N y LINEAR ARRAY IS A n = 0 y arrays for l /2 spacing: y LINEAR ARRAY Dox = (N x +1) Doy = (N y +1) n = N x dx Beamwidths of the linear arrays: x d y LINEAR ARRAY ALONG qBx » l / Lx = l /(dx (N x +1)) THE x AXIS qBy » l / Ly = l /(d y (N y +1)) Using the approximate equation for directivity that was derived earlier 4p 4p 4p A D = = L L = o 2 x y 2 qBxqBy l l This equation holds for any antenna that has an aperture with area A. It will be derived in a more rigorous manner later. The gain is G = eDo , where e is the antenna efficiency.

36 Naval Postgraduate School Antennas & Propagation Distance Learning Two-Dimensional Array of Dipoles

37 Naval Postgraduate School Antennas & Propagation Distance Learning Microwave Beamforming Networks (1)

Beamforming networks are used to distribute a signal from the input to the individual element (or, on receive, to combine the signals from the elements and deliver it to the receiver). The networks fall into three broad categories: 1. Parallel feeds (or corporate feeds): they have a tree-like structure whereby the signal is split between branches. OUTPUTS TO ELEMENTS

POWER DIVIDER

TRANSMISSION LINES INPUT OR WAVEGUIDES 2. Series feeds: the signal to the individual elements is tapped off of a main line.

POWER DIVIDER OUTPUTS TO ELEMENTS TRANSMISSION LINES INPUT OR WAVEGUIDES 3. Space feeds: a region of space is used to combine or distribute signals. Space feeds generally incorporate reflectors or lenses

38 Naval Postgraduate School Antennas & Propagation Distance Learning Microwave Beamforming Networks (2)

Example: A parallel feed splits the input power uniformly N ways. There is a cable loss of LdB dB between the output of each branch and the element. What is the loss of this feed if the only loss is due to the cables? If 1 W is applied to the input there is 1/N W at the input to each cable. The total power out is of the cables is the sum N N Pout = å (L / N) = L å (1/ N ) = L n=1 n=1

OUTPUTS TO ELEMENTS (1/ N )(10-LdB /10 ) The feed loss is 14243 L L Pout L 1/ N = = L = LdB Pin 1 Note that the total loss is not NLdB because the sources of the loss (resistors) are in parallel. 1 W This result does not hold for a series feed.

39 Naval Postgraduate School Antennas & Propagation Distance Learning Corporate Fed Waveguide Array

40 Naval Postgraduate School Antennas & Propagation Distance Learning Series Fed Waveguide Slot Arrays

41 Naval Postgraduate School Antennas & Propagation Distance Learning Array Example (1)

Design an array to meet the following specifications:

1. Azimuth sidelobe level 30 dB 2. + 45 degree scan in azimuth; no elevation scan; no grating lobes 3. Elevation HPBW of 5 degrees 4. Gain of at least 30 dB over the scan range

AIRCRAFT y BODY

L x x z ARRAY APERTURE L y

Azimuth: f=0o Elevation: f=90o Restrictions: 1. Elements are vertical (yˆ ) dipoles over a ground plane 2. Feed network estimated to have 3 dB of loss 3. Dipole spacings are: 0.4l £ dx £ 0.8l and 0.45l £ dy £ 0.6l 42 Naval Postgraduate School Antennas & Propagation Distance Learning Array Example (2)

Restrictions (continued): 4. errors and imperfections will increase the SLL about 2 dB so start with a -32 dB sidelobe distribution 2 0.2 + 0.8cos (p x¢/2) (ei = 0.81) 5. minimize the number of phase shifters used in the design Step 1: start with the gain to find the required physical area of the aperture 4pA G = De = ecosq ³ 30 dB l2 cosq is the projected area factor, which is a minimum at 45 deg. The efficiency includes tapering efficiency (0.81) and feed loss (0.5). Therefore 4pA G = (0.707)(0.81)(0.5) =103 l2 2 or, A/ l = (Lx / l)(Ly / l) » (N x + 1)(d x / l)( N y + 1)(d y / l) = 278.1 Step 2: uniform illumination in elevation; must have a HPBW of 5 degrees sin((N +1)kd sinq / 2) y y = 0.707 Þ L = (N +1)d = 10l N +1 sin kd sinq / 2 y y y ( y ) ( y )q = 2.5o

43 Naval Postgraduate School Antennas & Propagation Distance Learning Array Example (3)

This leads to Lx = A/ Ly = 28l . To minimize the number of elements choose the largest allowable spacing

d y = 0.6l Þ N y +1 = Ly /d y = 17 Step 3: azimuth spacing must avoid grating lobes which occur when o o sin qn -sinq s = nl / dx (qn £ -90 , n = -1,qs = 45 ) -1- 0.707 = -l /d Þ d £ 0.585l x x Again, to minimize the number of elements, use the maximum allowable spacing (0.585l ) which gives N x + 1 = Lx / d x = 48. Because the beam only scans in azimuth, one phase shifter per column is sufficient.

AZIMUTH FEED · (48 COLUMNS) · · · · · · PHASE SHIFTER · ELEVATION FEED · · (17 DIPOLES PER COLUMN) · ·

44 Naval Postgraduate School Antennas & Propagation Distance Learning Array Example (4)

Step 4: Find the azimuth beamwidth at scan angles of 0 and 45 degrees. Letting qH = qB /2

sin((N +1)kd (±sin q - sin q ) / 2) x x H s = 0.707 (N x +1)sin(kdx (± sinqH - sinqs ) / 2)

Solve this numerically for qs = 0 and 45 degrees. Note that the beam is not symmetrical when it is scanned to 45 degrees. Therefore the half power angles are different on the left and right sides of the maximum

q = 0: q = -q = 0.91 Þ HPBW = q - q = 2(0.91) = 1.82o s B+ B- B+ B- o qs = 45: q + = 46.3, q - = 43.75 Þ HPBW = 46.3 - 43.75 = 2.55 B B We have not included the element factor, which will affect the HPBW at 45 degrees.

45 Naval Postgraduate School Antennas & Propagation Distance Learning

Digital Phase Shifters

Phase shifters introduce a precise phase shift to a wave passing through it. They are used to "tilt" the phase across the array aperture for beam scanning. Diode phase shifters are capable of providing only discrete phase intervals. A n bit phase shifter has 2n phase states. The quantization levels are separated by D = 360o /2n . The fact that the exact phase cannot always be obtained results in: 1. gain loss 2. increase in sidelobe level 3. beam pointing error Example of phase truncation: QUANTIZED PHASE · PHASE REQUIRED · · TO SCAN THE BEAM · ALLOWABLE · PHASE STATES · ARRAY · · ELEMENT · · · · · · · ·

46 Naval Postgraduate School Antennas & Propagation Distance Learning Effect of Phase Shifter Roundoff Errors

0 Truncation causes beam pointing NO PHASE -10 SHIFTER errors. Random roundoff methods ROUNDOFF destroy the periodicity of the -20 quantization errors. The resultant rms error is smaller than the maximum -30 error using truncation. -40 Linear array, 60 elements, d = 0.4l -50 4 bit phase shifters -60 0 -50 0 50 0 TRUNCATION WEIGHTED -10 -10 RANDOM ROUNDOFF -20 -20

-30 -30

-40 -40

-50 -50

-60 -60 -50 0 50 -50 0 50

47 Naval Postgraduate School Antennas & Propagation Distance Learning Examples of Time Delay Networks

4D

3D

INPUT 2D OUTPUT

D

D IS A TIME DELAY BIT TIME DELAY FIBERS SWITCH

4D 3D D 2D INPUT OUTPUT

48 Naval Postgraduate School Antennas & Propagation Distance Learning Digital Phase Shifter

· X-band 5-bit PIN diode phase shifter

From Hughes Aircraft Co.(Raytheon)

49 Naval Postgraduate School Microwave Devices & Radar Distance Learning Discrete Fourier Transform (1)

The discrete Fourier transform (DFT) is a sampled version of the continuous Fourier transform æ 2p ö N -1 - jç ÷mn è N ø F(um ) = å f (xn )e (m = 0,1,…, N -1) n =0 The corresponding inverse discrete Fourier transform (IDFT) is

æ 2p ö N -1 jç ÷mn 1 è N ø f (xn ) = å F(um)e (n = 0,1,…, N -1) N n=0 F(m) and f (n) are sometimes used to denote sampled data. The fast Fourier transform (FFT) is an efficient means of calculating the DFT because it dramatically reduces the number of required arithmetic operations. Clearly the DFT and IDFT calculations are similar, so the FFT can be used to compute the IDFT as well. In general, the array factor for elements along the x axis is N -1 jxnk sin q cosf AF(q,f) = å A(xn )e n=0

50 Naval Postgraduate School Microwave Devices & Radar Distance Learning Discrete Fourier Transform (2)

For simplicity, let the observation point lie in the x-z plane (f = 0o). Evaluating the array factor at a set of observation angles qm , m = 0,1,…, N-1 N -1 N -1 jxnk sin qm jxnkum AF(qm ) = å An e = å An e n=0 n=0

where um = sinqm is the x direction cosine. If Du is the spacing between pattern values in direction cosine space, then the array factor is N -1 N -1 j(nd)(kmDu) jmn(kdDu) AF(um ) = å An e = å An e n=0 n=0 The summation conforms to the definition of the IDFT

æ 2p ö N -1 - jç ÷mn è N ø AF(m) = å A(n)e (m = 0,1,…, N -1) n=0 2p l where kdDu = 2p / N , or Du = = . The FFT algorithm will return N / 2 negative Nkd Nd values of u, N /2 -1 positive values of u, and u = 0 (for a total of N). 51 Naval Postgraduate School Microwave Devices & Radar Distance Learning Discrete Fourier Transform (3)

The FFT computation has the following properties: 1. Most software is based on the original radix 2 algorithm which requires the number of inputs (aperture distribution samples) to satisfy N = 2M where M is an integer. In Matlab when M is not an integer a less efficient calculation is used. 2. The number of outputs (pattern angles) is equal to the number of inputs. 3. A characteristic of the algorithm is a wrap-around of the output data. In Matlab the output data is arranged in the proper order by calling FFTSHIFT on the array returned by FFT. 4. If more pattern angles than aperture samples are desired (usually the case) then the input can be appended with zeros. 5. The FFT returns N samples of half of a period of the array factor. In other words, if the aperture samples are taken at 0.5l intervals (i.e., the array elements are spaced 0.5l ), then the samples will correspond to the range - 90o £q £ 90o or equivalently -1 £ kd £1. 6. If the element spacing is less than 0.5l then some of the output data must be discarded; if the element spacing is greater than 0.5l then more periods of the AF must be generated. This is accomplished by inserting one or more zeros between the aperture distribution samples.

52 Naval Postgraduate School Microwave Devices & Radar Distance Learning Discrete Fourier Transform (4)

8 1 FROM FFT 0.8 6 EXCITED ELEMENT 0.6 PADDED ZERO 4 0.4 |AF|

Amplitude, An 0.2 2 0 0 0 5 10 15 0 5 10 15 Distance, x Angle Number, m 8 FROM 6 FFTSHIFT SAMPLE OUTPUTS FOR UNIFORM DISTRIBUTION

4 |AF| NUMBER OF SAMPLES, N = 16 NUMBER OF ELEMENTS = 4 2

0 0 5 10 15 Angle Number, m

53 Naval Postgraduate School Microwave Devices & Radar Distance Learning Discrete Fourier Transform (5)

8 1 FROM FFT 0.8 6 0.6 EXCITED ELEMENT PADDED ZERO 4 0.4 |AF|

Amplitude, An 0.2 2 0 0 0 5 10 15 0 5 10 15 Distance, x Angle Number, m 8 FROM 6 FFTSHIFT SAMPLE OUTPUTS FOR UNIFORM DISTRIBUTION

4 |AF| NUMBER OF SAMPLES, N = 16 NUMBER OF ELEMENTS = 4 2

0 0 5 10 15 Angle Number, m

54 Naval Postgraduate School Microwave Devices & Radar Distance Learning Matlab Program to Compute the Array Factor

% program to compute the array factor for a % throw away output values that are not in % linear array of elements using the MATLAB % the visible region for this spacing % FFT routine for k=1:N clear,clf arg=(k-N/2)*npack/N/d; rad=pi/180; if arg < 1 M=11; nmax=max(nmax,k); N=2^M; end d=0.5; % spacing in wavelengths if arg > -1 Nel=10; % number of elements nmin=min(nmin,k); thetas=20; % scan angle end % phase for scanning the beam if abs(arg) < 1 psis=2*pi*d*sin(thetas*rad); x(k)=atan(arg/sqrt(1-arg^2))/rad; % number of periods required from the end spacing end npack=0; for k=nmin:nmax while (npack/2/d) < 1 AFdb(k-nmin+1)=ydb(k); npack=npack+1; theta(k-nmin+1)=x(k); end end % assign excitation values and pad the plot(theta,AFdb),grid % distribution with zeros axis([-90,90,-50,0]) A=zeros(N,1); xlabel('Angle, degrees') for k=1:Nel ylabel('Normalized Array Factor, dB') A(npack*k)=1*exp(j*(k-1)*psis); end z=fft(A,N); y=fftshift(z); ydb=20*log10(abs(y)/Nel+1e-2); nmin=N/2; nmax=N/2;

55 Naval Postgraduate School Microwave Devices & Radar Distance Learning Sample Output of the Matlab Program

o Sample calculation for 20 elements, d = 0.8l, N = 2048 (FFT samples),q s = 20 . A grating lobe occurs in the visible region for this spacing.

0

-10

-20

-30

Normalized Array Factor, dB -40

-50 -50 0 50 Angle, degrees 56 Naval Postgraduate School Antennas & Propagation Distance Learning Receiving Antennas (1)

When an antenna is receiving, it is INCIDENT convenient to define an effective WAVE FRONT ANTENNA area (or effective aperture) Ae . The power delivered to a load at r r the antenna terminals is Winc Pr = Winc Ae

r Pr = Winc Ae

where Wr is the incident power inc density. An equivalent circuit for the antenna is shown below. The current is I = Vinc /(Za + ZL). I r Winc Za ZL Vinc ZL

EQUIVALENT ANTENNA CIRCUIT

57 Naval Postgraduate School Antennas & Propagation Distance Learning Receiving Antennas (2)

Let the load be conjugate matched to the antenna impedance (which is the condition for maximum power transfer) and assume there are no losses (R = 0) l * ZL = Za (RL = Ra and X L = -X a ). The equivalent circuit becomes I

Vinc 2Ra

The power delivered to the receiver can be found in terms of the effective area 2 1 2 1 Vinc r Pr = I Ra = º Winc Ae 2 2 4Ra 2 2 h(kl) r 1 Einc For a Hertzian dipole R = , Einc = Vinc / , and Winc = . Now solve for Ae . a 6p l 2 h

58 Naval Postgraduate School Antennas & Propagation Distance Learning Receiving Antennas (3)

1 V 2 1 V 2 3 p l2 A = inc = inc = = 0.119l2 e r 2 2 2 (4Ra )Winc 2 é1 Einc ù 2 (2p ) (4Ra )ê ú ë2 h û For a Hertzian dipole the directivity is 3/2, and therefore the effective area can be written as 2 2 3 æ l ö æ l ö 4p Ae A = ç ÷ = Dç ÷ Þ D = m em ç ÷ ç ÷ 2 2 è 4p ø è 4p ø l The subscript m denotes that it is the maximum effective area because the losses are not included. If losses are included then the gain is substituted for directivity

æ l2 ö 4p A A = Gç ÷ Þ G = e e ç ÷ 2 è 4p ø l The formula holds for any type of antenna that has a well-defined aperture, or surface area through which all of the radiated power flows. From the formula one can deduce that the effective area is related to the physical area A by A = eA. em

59 Naval Postgraduate School Antennas & Propagation Distance Learning Friis Transmission Equation (1)

Consider two antennas that form a communication or data link. The range between the antennas is R. (The pattern can depend on both q and f , but only q is indicated.)

RECEIVE qt Pt Gt ,Gt R A ,G ,G er r r TRANSMIT qr Pr

The power density at the receive antenna is POWER INTO ANTENNA 64748 2 r Pt (1- Gt ) Winc = Gt (qt ) 4pR2 \ r 2 and the received power is Pr = Winc Aer (qr )(1- Gr )p (p is the polarization loss factor, PLF). 60 Naval Postgraduate School Antennas & Propagation Distance Learning Friis Transmission Equation (2)

But A = G (q )l2 /(4p), er r r 2 PtGt (qt )Gr (q r )l 2 2 Pr = (1- Gt )(1- Gr )pL (4pR)2 L is a general loss factor (0 £ L £ 1). This is known as the Friis transmission equation (sometimes called the link equation). Example: (Satcom system) Parameters at the ground station (uplink):

Gt = 54 dB = 251188.6, L = 2 dB = 0.6310, Pt =1250 W R=23,074 miles = 37,132 km, f =14 GHz At the satellite (downlink)

Gr = 36 dB = 3981, Pt = 200 W, f =12 GHz If we assume polarization matched antennas and no reflection at the antenna inputs, (1250)(251188.6)(3981)(0.0214)2 P = (0.6310) r 2 (4p )2 (37132 ´103 ) = 1.66 ´10-9 W = -87.8 dBw = -57.8 dBm

61 Naval Postgraduate School Antennas & Propagation Distance Learning Radar Range Equation (1)

“Quasi-monostatic” geometry:

TX Gt

Pt R

RX Gr s

Pr s = radar cross section (RCS) in square meters Pt = transmitter power, watts Pr = received power, watts Gt = transmit antenna gain in the direction of the target (assumed to be the maximum) Gr = transmit antenna gain in the direction of the target (assumed to be the maximum) PtGt = effective radiated power (ERP) 4pA From antenna theory: G = er r l2 Aer = Ae = effective area of the receive antenna A = physical aperture area of the antenna l = wavelength (c / f ) e = antenna efficiency

62 Naval Postgraduate School Antennas & Propagation Distance Learning Radar Range Equation (2)

r Power density incident on the target, Winc

POWER DENSITY AT RANGE R R PtGt 2 Winc = 2 (W / m ) pR Pt Gt 4

Power collected by the radar target and scattered back towards the radar

TARGET EFFECTIVE INCIDENT WAVE FRONT COLLECTION AREA IS IS APPROXIMATELY PLANAR s AT THE TARGET P G s P = sW = t t c inc 4pR2

63 Naval Postgraduate School Antennas & Propagation Distance Learning Radar Range Equation (3)

The RCS gives the fraction of incident power that is scattered back toward the radar. Therefore, P = P and the scattered power density at the radar, Wr , is obtained s c s by dividing by 4pR2 .

RECEIVER (RX)

TARGET SCATTERED POWER RCS s DENSITY AT RANGE P R FROM THE TARGET W = s s 4pR2

The target scattered power collected by the receive antenna is Ws Aer . Thus the maximum target scattered power that is available to the radar is

P G sA P G G sl2 P = t t er = t t r r (4pR2 )2 (4p)3 R4

This is the classic form of the radar range equation (RRE).

64 Naval Postgraduate School Antennas & Propagation Distance Learning Radar Range Equation (4)

Including the reflections at the antenna terminals 2 PtGtsAer 2 2 PtGtGrsl 2 2 Pr = (1- Gt )(1- Gr )L = (1- Gt )(1- Gr )L (4pR2 )2 (4p )3 R4 For monostatic systems a single antenna is generally used to transmit and receive so Gt = Gr º G and Gr = Gt . The above form of the RRE is too crude to use as a design tool. Factors have been neglected that have a significant impact on radar performance: · noise, · system losses, · propagation behavior, · clutter, · waveform limitations, etc. However, this form of the RRE does give some insight into the tradeoffs involved in radar 4 design. The dominant feature of the RRE is the 1/ R factor. Even for targets with 4 relatively large RCS, high transmit powers must be used to overcome the 1/ R when the range becomes large.

65 Naval Postgraduate School Antennas & Propagation Distance Learning Noise in Systems (1)

One way that noise enters communication and radar systems is from background radiation of the environment. (This refers to emission by the background as opposed to scattering of the system’s signal by the background, which is clutter.) Noise is also generated by the components in the radar’s receive channel. Under most conditions it is the internally generated thermal noise that dominates and limits the system performance.

SINUSOIDAL PULSE WITHOUT NOISE IDEAL TRANSMITTED WAVEFORM NOISE

TRANSMITTER TARGET RETURN

NOISY OUTPUT RECEIVER WAVEFORM ANTENNA

SINUSOIDAL PULSE WITH NOISE THERMAL NOISE NOISE

66 Naval Postgraduate School Antennas & Propagation Distance Learning Noise in Systems (2)

A high noise level will hide a weak signal and possibly cause a loss in communications or, in the case of radar, prevent detection of a target with a low radar cross section. · Thermal noise is generated by charged particles as they conduct. High temperatures result in greater thermal noise because of increased particle agitation. · Noise is a random process and therefore probability and statistics must be invoked to access the impact on system performance. · Thermal noise exists at all frequencies. We will consider the noise voltage to be constant with frequency (so called white noise) and its statistics (average and variance) independent of time (stationary). If the noise voltage generated in a resistor at temperature T Kelvin (K) is measured, it is found to obey Plank’s blackbody radiation law 4hfBR Vn = ehfkT -1 h = 6.546´10-34 J-sec is Plank’s constant, k =1.38 ´10-23 J/oK is Boltzmann’s constant, B is the system bandwidth in Hz, f is the center frequency in Hz, and R is the resistance in ohms.

67 Naval Postgraduate School Antennas & Propagation Distance Learning Noise in Systems (3)

At microwave frequencies hf << kT and the exponential can be approximated by the first two terms of a Taylor’s series hf ehfkT -1 » kT and therefore, Vn = 4kTBR, which is referred to as the Rayleigh-Jeans approximation. If the noisy resistor is used as a generator and connected to a second load resistor, R, the power delivered to the load in a bandwidth B is

2 2 æ Vn ö Vn N º Pn = ç ÷ R = = kTB W è 2Rø 4R

NOISY RESISTOR R NOISELESS AT TEMPERTURE T RESISTOR, R

+ + CONJUGATE MATCHED LOAD FOR MAXIMUM R Vn Þ Vn B R R POWER TRANSFER - - FILTER

68 Naval Postgraduate School Antennas & Propagation Distance Learning Noise in Systems (4)

Limiting cases:

· B ® 0 Þ Pn ® 0: Narrow band systems collect less noise · T ® 0 Þ Pn ® 0: Cooler devices generate less noise · B ® ¥ Þ Pn ® ¥: Referred to as the ultraviolet catastrophe, it does not occur because noise is not really white over a wide band. Any source of noise (for example, a mixer or cable) that has a resistance R and delivers noise power Pn can be described by an equivalent noise temperature, Te . The noise source can be replaced by a noisy resistor of value R at temperature Te , so that the same noise power is delivered to the load P T = n e kB The noise at the antenna terminals due to the background is described by an antenna temperature, TA. The total noise in a bandwidth B is determined from the system noise temperature, Ts = Te + TA:

N = kTsB

69 Naval Postgraduate School Antennas & Propagation Distance Learning Calculation of Antenna Temperature

SKY BACKGROUND MAINBEAM SIDELOBE

Pr EMITTED REFLECTED

EARTH BACKGROUND

The antenna collects noise power from background sources. The noise level can be characterized by the antenna temperature p 2p ò0 ò0 TB(q,f)G(q,f)sinq dq df TA = p 2p ò0 ò0 G(q,f)sin q dq df

T B is the background brightness temperature and G the antenna gain.

70 Naval Postgraduate School Antennas & Propagation Distance Learning Noise in Systems (5)

Returning to the radar range equation, we can calculate the signal-to-noise ratio (SNR) P PG sA PG G sl2L SNR = r = t t er L = t t r 2 2 3 4 N N(4pR ) (4p ) R kTs B Given the minimum SNR that is required for detection (detection threshold) it is possible to determine the maximum range at which a target can be “seen” by the radar PG G sl2L R = 4 t t r max 3 (4p ) (kTs B)SNR min If the return is greater than the detection threshold a target is declared. A is a false alarm (the noise is greater than the threshold level but there is no target); B is a miss (a target is present but the return is not detected). TARGET RETURNS RANDOM NOISE A DETECTION B THRESHOLD (RELATED TO SNR min ) RECEIVED POWER TIME

71 Naval Postgraduate School Antennas & Propagation Distance Learning Noise Figure & Effective Temperature (1)

Noise figure is used as a measure of the noise added by a device. It is defined as:

(S / N)in Sin / Nin Nout Fn = = = (S / N)out Sout / Nout kTo BnG S where G = out . By convention, noise figure is defined at the standard temperature Sin of To = 290 K. The noise out is the amplified noise in plus the noise added by the device GNin + DN DN Fn = =1 + k ToBnG k ToBnG DN can be viewed as originating from an increase in temperature. The effective temperature is k TeBnG Te Fn = 1 + = 1+ k ToBnG To Solve for effective temperature in terms of noise figure yielding the relationship

Te = (Fn -1)To

72 Naval Postgraduate School Antennas & Propagation Distance Learning Noise Figure & Effective Temperature (2)

The overall noise figure for M cascaded devices with noise figures F ,F , , F and 1 2 … M gains G1,G2,…,GM is F -1 F -1 F -1 F = F + 2 + 3 + + M o 1 G G G L G G G 1 1 2 1 2L M-1 The overall effective temperature for M cascaded devices with temperatures T , T , ,T and gains G ,G , ,G is 1 2 … M 1 2 … M T T T T = T + 2 + 3 + + M e 1 G G G L G G G 1 1 2 1 2L M-1 Example: For the receive channel shown below. The antenna gain is 30 dB, T A = 200 K; mixer has 10 dB conversion loss and 3 dB noise figure; IF amplifier: 6 dB noise figure

Nin Nout IF AMP G1, F1 G2 , F2

LO F2 -1 4 -1 The noise figure of the dashed box is Fo = F1 + = 2 + = 32 G1 0.1

Therefore, the system noise temperature is Ts = Te + TA = 31To + TA = 9190 K. 73 Naval Postgraduate School Antennas & Propagation Distance Learning SNR of Active and Passive Antennas (1)

Consider two arrays of isotropic elements: one has the amplifier before the signals are combined and the second has the amplifier after the signals are combined. Note that the latter, amplification after beamforming, is the conventional passive antenna with an amplifier at its output. Passive (left) and active (right) two- s s s s element (isotropic) arrays are n n shown. A uniform plane wave is L L incident. The signal power s is the same at the output of each element. C C 1 2 Noise generators are shown in each n G G branch to signify that a white noise power n is added to the signal. L C1 C2 L G Each branch has a loss L. S S We want to compare the signal-to- noise ratio at the outputs for the two Power coupling coefficients for a lossless power cases. (Lower case signifies a “per divider satisfiy: C1 + C2 = 1 element” quantity. All quantities are power.)

74 Naval Postgraduate School Antennas & Propagation Distance Learning SNR of Active and Passive Antennas (2)

Since the signal is correlated, voltage is added and then squared to get power, whereas the uncorrelated noise powers are simply added. Assume that the couplers are equal power split, C1 = C2 = 1/2. For amplification after beamforming: N = nG 2 2 S = ( sGLC1 + sGLC2 ) = sGL( C1 + C2 ) 1442443 =2 S 2sLG æ s ö æ s ö = = 2Lç ÷ = (AF)ç ÷L N nG è n ø è n ø where AF is the array factor for the two element array and s/n is the SNR for a single channel (element). For amplification before beamforming

N = nGLC1 + nGLC2 = nGL(C1 + C2 ) = nGL 2 S = ( sGLC1 + sGLC2 ) = 2sGL S 2sGL æ s ö æ s ö = = 2ç ÷ = (AF)ç ÷ N nGL è nø è n ø The SNR is not affected by the loss factor L when the amplifier is moved to the element. The gain of an active antenna is no longer a measure of the increase in SNR at its output.

75 Naval Postgraduate School Antennas & Propagation Distance Learning Active GPS Antenna

Active microstrip patch GPS antenna

Top side: radiating patch Bottom side: amplifier circuit

76 Naval Postgraduate School Microwave Devices & Radar Distance Learning Comparison of SNR: Active vs Passive

The previous analysis is approximate because it does not take into account the exact relationships between gain and noise figure for each of the devices in a channel. However, it illustrates the important points with regard to amplifiers and beamforming:

PASSIVE ANTENNAS (Amplification after beamforming): · The antenna gain is the SNR improvement (neglecting noise introduced by the antenna).

ACTIVE ANTENNAS (Amplification before beamforming): · The SNR performance can be significantly better than the gain indicates. · Beam coupling losses can be recovered. · SNR degradation is determined only by the aperture efficiency. All other losses are recovered. · The coupler match looking into the sidearms does not affect the SNR.

77 Naval Postgraduate School Antennas & Propagation Distance Learning Multiple Beam Antennas

Several beams share a common aperture (i.e., use the same radiating elements) Arrays Reflectors Lenses

Advantages: Cover large search volumes quickly Track multiple targets simultaneously Form "synthetic" beams Disadvantages: Beam coupling loss Increased complexity in the feed network Sources of beam coupling loss: (1) leakage and coupling of signals in the feed network, and (2) non-orthogonality of the beam patterns

78 Naval Postgraduate School Antennas & Propagation Distance Learning Radiation Patterns of a Multiple Beam Array

0 25 dB Taylor -5 N =30 -10 d = 0.4l -15 Dq = 2.3o s -20

-25 RELATIVE POWER, dB -30

-35

-40 -30 -20 -10 0 10 20 30 PATTERN ANGLE, DEGREES

79 Naval Postgraduate School Microwave Devices & Radar Distance Learning Active vs Passive Multibeam Antennas

LNA 1 1

2 2 TO DISPLAY 3 3

M M CONVENTIONAL N M (LNA PER BEAM)

APERTURE BEAM FORMER SIGNAL M BEAMS RECEIVER PROCESSOR N RADIATING LNA ELEMENTS 1 1

2 2 TO DISPLAY 3 3

ACTIVE (LNA PER RADIATING ELEMENT) N M N M M

80 Naval Postgraduate School Antennas & Propagation Distance Learning Beam Coupling Losses for a 20 Element Array

2 Beams 4 Beams

p 2p 1 r r* 2 For m and n to be orthogonal beams: ò ò Em (q,f)×En (q,f)R sinqdqdf = 0 h 0 0 Example: If the beams are from a uniformly illuminated line source then Er (q,f), n Er (q,f) µ sinc(×) and adjacent beams are orthogonal if the crossover level is 4 dB. m 81 Naval Postgraduate School Antennas & Propagation Distance Learning Active Array Radar Transmit/Receive Module

82 Naval Postgraduate School Antennas & Propagation Distance Learning Adaptive Antennas

Adaptive antennas are capable of changing their radiation patterns to place nulls in the direction of jammers or other sources of interference. The antenna pattern is controlled by adjusting the relative magnitudes and phases of the signals to/from the radiating elements.

COMPLEX WEIGHTS w1 w2 w3 L wN-1 wN jFn wn =| wn| e

|wn|/|wmax | S SUMMING NETWORK Fn - Fref

In principle, a N element array can null up to N -1 jammers, but if the number of jammers becomes a significant fraction of the total number of elements, then the pattern degradation is unacceptable. The performance of an adaptive array depends on the algorithm (procedure used to determine and set the weights). Under most circumstances the antenna gain is lower for an adaptive antenna than for a conventional antenna.

83 Naval Postgraduate School Antennas & Propagation Distance Learning Smart Antennas (1)

Antennas with built-in multi-function capabilities are often called smart antennas. If they are conformal as well, they are known as smart skins. Functions include: · Self calibrating: adjust for changes in the physical environment (i.e., temperature). · Self-diagnostic (built-in test, BIT): sense when and where faults or failures have occurred. Tests can be run continuously (time scheduled with other radar functions) or run periodically. If problems are diagnosed, actions include: · Limit operation or shutdown the system · Adapt to new conditions/reconfigure

SOURCE WIRE Example: a test signal is used to

GROUND isolate faulty dipoles and PLANE transmission lines DIPOLE

TEST SIGNAL

84 Naval Postgraduate School Antennas & Propagation Distance Learning Smart Antennas (2)

Example of a self-calibrating, self-diagnostic transmit/receive module

TX TX ANTENNA · · ELEMENT PA COUPLER · · · LNA TX A B RX · · RX RX

RX TX C D

TX RX POWER SPLITTER CONTROLLER C on TX B on RX A on TX AMPLITUDE AND PHASE COMPARISON D on RX

85 Naval Postgraduate School Antennas & Propagation Distance Learning Digital Beamforming

RADIATING 1 2 3 N ELEMENT L

SYNCHRONOUS DETECTOR

I Q I Q I Q I Q TWO CHANNEL ANALOG-TO- DIGITAL CONVERTER s1(t) s2 (t) sN (t)

SIGNAL PROCESSOR (COMPUTER)

OUTPUT, y(t) N Assuming a narrowband signal, y(t) = åwnsn(t). The complex signal (I and Q, or n=1 equivalently, amplitude and phase) are measured and fed to the computer. Element responses become array storage locations in the computer. The weights are added and the sums computed to find the array response. In principle any desired beam characteristic can be achieved, including multiple beams.

86 Naval Postgraduate School Distance Learning

Antennas & Propagation

LECTURE NOTES VOLUME IV APERTURES, HORNS AND REFLECTORS

by Professor David Jenn

ELLIPSOID F · PARABOLOID

GREGORIAN

(ver1.3) Naval Postgraduate School Antennas & Propagation Distance Learning Equivalence Principle (1)

There is symmetry between the electric and magnetic quantities that occur in electro- magnetics. This relationship is referred to as duality. However, a major difference between the two views is that there are no magnetic charges and therefore no magnetic current. Fictitious magnetic current Jr and charge r can be introduced m vm r r r r (1)Ñ´ E = - jwmH - Jm (3)Ñ× H = rvm / m r r r r (2)Ñ´ H = J + jweE (4)Ñ× E = rv /e If magnetic current is allowed, then the radiation integrals must be modified. The far field radiation integral becomes

r r - jkh - jkr é r r 1 r ù jk(r ¢·rˆ) E(r ) » e òò êJs - rˆ(Js · rˆ)+ J ms ´ rˆúe ds¢ 4pr S ë h û where Jr is the magnetic surface current density (V/m). The boundary conditions at an ms interface must also be modified to include the magnetic current and charg nˆ r r r - nˆ ´(E1 - E2 )= Jms 1 r r Ñ ×(H1 - H2 )= rvs / m 2

1 Naval Postgraduate School Antennas & Propagation Distance Learning Equivalence Principle (2)

In some cases it will be advantageous to replace an actual current distribution with an equivalent one over a simpler surface. An example is illustrated below. The currents on the antennas inside of an arbitrary surface S set up electric and magnetic fields everywhere. The same external fields will exist if the antennas are removed and replaced with the proper equivalent currents on the surface. ORIGINAL PROBLEM EQUIVALENT PROBLEM r r r r E , H E , H 2 2 2 2 r r Etan J s S r r S E1 , H1 r r r r Htan E1 , H1 J ms ANTENNAS nˆ nˆ

The required surface currents are: r r r r r r Jms = -nˆ ´ (E1 - E2 ) and J s = nˆ ´ (H1 - H 2 )

2 Naval Postgraduate School Antennas & Propagation Distance Learning Equivalence Principle (3)

Important points regarding the equivalence principle: 1. The tangential fields are sufficient to completely define the fields everywhere in space, both inside and outside of S.

2. If the fields inside do not have to be identical to those in the original problem, then the currents to provide the same external fields are not unique.

3. Love’s equivalence principle refers to the case where the interior fields are set to zero. The equivalent currents become r r J ms = nˆ ´ E2 Jr = -nˆ ´ Hr s 2 or in terms of the outward normal r r J ms = -nˆ ´ E2 Jr = nˆ ´ Hr s 2

3 Naval Postgraduate School Antennas & Propagation Distance Learning Apertures (1)

The equivalence principle can be used to determine the radiation from an aperture (opening) in an infinite ground plane. The aperture lies in the z = 0 plane. Region 1 contains the source. In order to apply the radiation integrals, we S need to find the currents in unbounded space (no objects present). PEC · Apply Love’s equivalence principle to find r E tan = 0 the currents on S. The currents are nonzero only in the aperture. Er OPENING · Both electric and magnetic currents exist a z in the aperture. To simplify the integration we would like to eliminate one of the SOURCE OF currents. Since Er is specified, we will PLANE WAVE a INSIDE use the magnetic current. The steps involved in eliminating the electric current are illustrated in the figure on the next page.

4 Naval Postgraduate School Antennas & Propagation Distance Learning Apertures (2)

1. Since Er = Hr = 0 inside, we can place any object 1 1 r inside without affecting the fields. Put a PEC just inside E1 = 0 of region 1. INSERT PEC 2. Now remove the PEC and introduce images of the JUST INSIDE S sources Jr and Jr s ms 3. Allow the images and sources to approach the PEC. Jr The PEC shorts out the electric current. (The image of CURRENT ms an electric current element is opposite the source.) IMAGES Jr Only the magnetic current remains. s r r r ì -2nˆ ´ E2 = -2nˆ ´ Ea , in the aperture J ms = í î0, else

Note: Alternatively a perfect magnetic conductor (PMC) could be placed inside S. The magnetic current would short out and the electric current would double.

5 Naval Postgraduate School Antennas & Propagation Distance Learning Rectangular Aperture (1)

One basic application of the equivalence principle is radiation from a rectangular aperture of width 2b (in y) and height 2a (in x). Assume that the incident plane wave is r - jkz Ei = xˆEoe . Evaluating the incident field at z = 0 gives the aperture field x ì xˆE , x £ a, y £ b INCIDENT r INFINITE r o Ei Ea = í PLANE GROUND î 0, else WAVE PLANE

The equivalent current in the aperture is REGION 1 z < 0 z r r REGIONz > 0 2 Jms = -2nˆ ´ Ea = -2Eo yˆ y All objects are removed so that the APERTURE currents exists alone in free space. Now the radiation integral can be applied. Since the electric current is zero, the far field at observation points in region 2 is r r - jk - jkr r jk(r ¢·rˆ) E(r ) = e òò J ms ´ rˆe ds¢ 4pr S Jr ´ rˆ = -2E yˆ ´(xˆsin q cosf + yˆ sinq sin f + zˆ cosq) where ms o = 2Eo (zˆsin q cosf - xˆ cosq ) 6 Naval Postgraduate School Antennas & Propagation Distance Learning Rectangular Aperture (2)

The position vector to an integration point in the aperture is rr¢ = xˆx¢ + yˆy¢ and therefore the dot product in the exponent is

rˆ · rr¢ = x¢sinq cosf + y¢sinq sin f The integral becomes a b - jk Er(rr) = e- jkr 2E (zˆsin q cosf - xˆ cosq) e jkx¢sin q cosfdx¢ e jky¢sinq sin fdy¢ 4pr o ò ò -a -b 144424443144424443 2asinc(ka sinq cosf) 2bsinc(kb sin q sin f) The dot products with the spherical components, zˆ ·qˆ = -sin q and xˆ ·qˆ = cosq cosf lead to 2 2 qˆ · (zˆsin q cosf - xˆ cosq ) = - sin q cosf - cos q = cosf

Using the fact that the aperture area is A = 4ab gives jkAE E = o e- jkr cosf sinc(kasin q cosf)sinc(kbsinq sin f) q 2pr where r is the distance from the center of the aperture to the observation point.

7 Naval Postgraduate School Antennas & Propagation Distance Learning Rectangular Aperture (3)

Similarly, the dot products zˆ ·fˆ = 0 and xˆ ·fˆ = -sin f lead to

fˆ · (zˆsinq cosf - xˆ cosq ) = sin f cosq and - jkAE E = o e- jkr cosq sin f sinc(kasin q cosf)sinc(kbsinq sin f) f 2pr Example: Contour plots for a = 3l and b = 2l in direction cosines are shown

E-theta E-phi 1 1

0.5 0.5

0 0

-0.5 -0.5 V=sin(theta)*sin(phi) V=sin(theta)*sin(phi)

-1 -1 -1 -0.5 0 0.5 1 -1 -0.5 0 0.5 1 U=sin(theta)*cos(phi) U=sin(theta)*cos(phi)

8 Naval Postgraduate School Antennas & Propagation Distance Learning Rectangular Aperture (4)

Properties of the “sinc” function

Maximum value at x =0 1 ·

0.9 sin( x) 0.8 sin(0) = = 1 0.7 x x=0

0.6

0.5 · First sidelobe level: -13.2 dB below the |sinc(x)| 0.4 maximum

0.3 · Caution: some authors and Matlab define

0.2 sin(px) 0.1 sin(x) = 0 x -30 -20 -10 0 10 20 30 x

9 Naval Postgraduate School Antennas & Propagation Distance Learning Tapered Aperture (1)

Just as in the case of array antennas, the sidelobe level can be reduced and the main beam scanned by controlling the amplitude and phase of the aperture field. As an example, let a rectangular aperture be excited by the TE10 mode from a waveguide. The field in the aperture is given by

r ìyˆEo cos(px¢/ a), x¢ £ a /2 and y¢ £ b/ 2 Ea = í î0, else =-xˆ r r } The equivalent magnetic current is Jms = -2nˆ ´ Ea = -2(zˆ ´ yˆ)Eo cos(px¢/ a) in the aperture. y

b

z a x

10 Naval Postgraduate School Antennas & Propagation Distance Learning Tapered Aperture (2)

The radiation integral is

- jkE r Er(rr) = o e- jkr [xˆ ´ rˆ]òòcos(px¢/ a)e jk (r ¢·rˆ)dx¢dy¢ 2p r S The cross product reduces to xˆ ´ rˆ = zˆsin q sin f - yˆ cosq The integrals are separable. The y integral is the same as the uniformly illuminated case - jk Er(rr) = e- jkr E (zˆsin q sin f - yˆ cosq ) 2p r o a / 2 b / 2 ´ ò cos(px¢/ a)e jkx¢sin q cosf dx¢ ò e jky¢sin q sin fdy¢ -a / 2 -b / 2 1444442444443144424443 æ ka ö kb 2pa cosç sin q cosf ÷ bsincçæ sin q sin f ÷ö è 2 ø è 2 ø p 2 -(ka sin q cosf)2

11 Naval Postgraduate School Antennas & Propagation Distance Learning Tapered Aperture (3)

The q component is obtained from 2 2 qˆ · (zˆ sin q sinf - yˆ cosq)= -sin q sinf - cos qsin f = -sin f or, æ æ ka ö ö ç cosç sin q cosf ÷ ÷ jkEo A - jkr ç è 2 ø ÷æ æ kb öö Eq = e sin f çsincç sin q sin f ÷÷ r çp 2 - (kasin q cosf )2 ÷è è 2 øø ç ÷ è ø The aperture illumination efficiency is 2 r òò nˆ ´ Ea dxdy e = S i r 2 Aòò nˆ ´ E a dxdy S The numerator is a / 2 b / 2 a / 2 éa ù 2abEo zˆ ´ yˆEo cos(px¢/a)dx¢dy¢ = bEo sin(px¢/a) = ò ò ëêp ûú p -a / 2 -b / 2 -a / 2

12 Naval Postgraduate School Antennas & Propagation Distance Learning Tapered Aperture (4)

The denominator is

a / 2 b / 2 a / 2 abE2 zˆ ´ yˆE cos(px¢/ a)2dx¢dy¢ = bE2 cos2 (px¢/ a)dx¢ = o ò ò o o ò 2 -a / 2 -b / 2 -a / 2 The ratio gives the illumination (or taper) efficiency,

2abE /p 2 e = o = 8/p 2 i 2 AabEo / 2

The directivity is 4pA 32Aæ 2p /l ö æ 64 öæ A ö D = ei = ç ÷ = ç ÷ç ÷ l2 l2p è k ø è kl øè l2 ø 14243 =1 Example: WR-90 waveguide (a = 0.9 inch, b = 0.4 inch) and l = 3 cm: D = 2.63.

13 Naval Postgraduate School Antennas & Propagation Distance Learning Summary of Aperture Distributions

This table is similar to Table 7.1 from Skolnik presented previously. This table includes entries for circular apertures. (Note: x and r are normalized aperture variables and a(x) = A(x) , where A(x) is the complex illumination coefficient.

FIRST 3 DB BEAM- LINEAR APERTURE CIRCULAR APERTURE SIDELOBE WIDTH, a(x) ei a(r) ei LEVEL, DB RADIANS 13.2 0.88l /(2a) 1 1 1 1 17.6 1.02l /(2a) 1- x2 0.865 1- r 2 0.75 20.6 1.15l /(2a) 1- x2 0.833 1- r 2 0.64 1.27 /(2a) 2 3 / 2 2 3 / 2 24.6 l (1- x ) 0.75 (1- r ) 1.36 /(2a) 2 2 2 2 28.6 l (1- x ) 0.68 (1- r ) 0.55 1.47 /(2a) 2 5 / 2 30.6 l (1- x ) 0.652

14 Naval Postgraduate School Antennas & Propagation Distance Learning Radiation Patterns From Apertures

Comparison of patterns for different aperture Uniform vs. triangular aperture illumination widths

40 2.5 BY 10 WAVELENGTHS 0 10 BY 10 WAVELENGTHS UNIFORM TRIANGULAR 30 -5

20 -10

-15 10

-20 D (dB) 0 -25

-10 RELATIVE POWER, dB -30

-20 -35

-30 -40 -80 -60 -40 -20 0 20 40 60 80 -80 -60 -40 -20 0 20 40 60 80 THETA, DEG THETA, DEG

15 Naval Postgraduate School Antennas & Propagation Distance Learning Scanned Aperture

A linear phase progression across the aperture causes the beam to scan. z DIRECTION OF BEAM SCAN q -a a x

r GROUND PLANE E i q s The far field has the same form as the non-scanned case, but with the argument modified to include the linear phase jkAE Er = o e- jkr (qˆcosf -fˆsin f cosq ) 2pr ´sinc[ka(sinq cosf - sin qs cosfs )]sinc[kb(sin q sinf - sin qs sin fs )] Example: What phase shift is required to scan the beam of an aperture with 2a = 10l to 30o? 2p(10l) k(2a)sin 30o cos0o = (0.5) = 10p =1800o l

16 Naval Postgraduate School Antennas & Propagation Distance Learning Aperture Example

Example: A radar antenna requires a beamwidth of 25 degrees in elevation and 2 degrees in azimuth. The azimuth sidelobes must be 30 dB and the elevation sidelobes 20 dB. Find a, b and G. Let the x-z plane be azimuth and the y-z plane elevation. Based on the required sidelobe levels, from the table,

æ p ö æ l ö 2a (1.47)(90) Azimuth HPBW: (2o )ç ÷ = 1.47ç ÷ Þ = = 42.1 è180o ø è 2a ø l p

æ p ö æ l ö 2b (1.15)(7.2) Elevation HPBW: (25o )ç ÷ = 1.15ç ÷ Þ = = 2.64 è180o ø è 2bø l p

At 1 GHz the dimensions turn out to be 12.63m and 0.79m. The gain is

4p (42.1l)(2.64l) G = (0.833)(0.6522) = 758 = 28.8 dB l2

17 Naval Postgraduate School Antennas & Propagation Distance Learning Horn Antennas (1)

An earlier example dealt with an open-ended waveguide in an infinite ground plane. This configuration is not practical because the wave impedance in the guide is much different than the impedance of free space, and therefore a large reflection occurs at the opening. Very little energy is radiated; most is reflected back into the waveguide.

THROAT Flares are used to improve the match a r b ¢ E-PLANE and increase the dimensions of the Ea HORN b radiating aperture (to reduce beamwidth and increase gain). The result is a horn INTERSECTION a antenna. An E-plane horn has the top OF FLARE WALLS and bottom walls flared; an H-plane FLARE APERTURE horn has the side walls flared. A REGION pyramidal horn has all four walls flared, r as shown on the next page. a Ea H-PLANE b a ¢ HORN b

18 Naval Postgraduate School Antennas & Propagation Distance Learning Horn Antennas (2)

In most applications, the horn is not installed in a ground plane. Without a ground plane currents can flow on the outside surfaces of the horn, which modifies the radiation pattern slightly (mostly in the back hemisphere). We will neglect the exterior currents and compute the radiation pattern from the currents in the aperture only. The geometry of a H-plane horn is shown below. x

TOP VIEW OF PYRAMIDAL H-PLANE HORN HORN R2 r b ¢ x E D(x) z a a a y R1 b a ¢

a ¢ SPHERICAL D WAVE FRONT max

19 Naval Postgraduate School Antennas & Propagation Distance Learning Horn Antennas (3)

Assume that the waveguide has a TE10 mode at the opening. If the flare is long and gradual the following approximations will hold:

1. The amplitude of the field in the aperture is very close to a TE10 mode distribution. 2. The wave fronts at the aperture are spherical, with the phase center (spherical wave origin) at the intersection of the flare walls. The deviation of the phase from that of a plane wave is given by kD(x), where

é 2 ù 2 2 2 æ x ö x D(x) = R1 + x - R1 = R1ê 1+ ç ÷ -1 ú » ê è R1 ø ú 2R1 ë 14243 û 2 1æ x ö »1+ ç ÷ 2è R1 ø The phase error depends on the square of the distance from the center of the aperture, and therefore is called a quadratic phase error. The electric field distribution in the aperture is approximately

r æpx ö - jkR(x) æpx ö - jk(R1+D(x)) æpx ö - jkD(x) Ea = yˆEo cosç ÷ e = yˆEo cosç ÷e ® yˆEo cosç ÷e è a¢ ø è a¢ ø è a¢ ø

20 Naval Postgraduate School Antennas & Propagation Distance Learning Horn Antennas (4)

The R1 in the exponent has been dropped because it is a common phase that does not affect the far-field pattern. The equivalent magnetic current in the aperture is

r r æpx ö - jkD(x) Jms = -2zˆ ´ Ea = xˆ2Eo cosç ÷e è a¢ ø

If the wave at the aperture is spherical (i.e., TEM) then the magnetic field is easily obtained from the electric field, and the equivalent electric current can be found r r r zˆ ´ Ea 2Eo æpxö - jkD(x) J s = 2zˆ ´ H a = 2zˆ ´ = -yˆ cosç ÷ e h h è a¢ ø These currents are used in the radiation integral. Because of the presence of kD(x) in the exponential, the integrals cannot be reduced to a closed form result. The major tradeoff in the design of a horn: in order to increase the directivity the aperture dimensions must be increased, but increasing the aperture dimensions also increases the quadratic phase error, which in turn decreases the directivity. What is the optimum aperture size?

21 Naval Postgraduate School Antennas & Propagation Distance Learning Horn Antennas (5)

Patterns of a 10l aperture with and without quadratic phase error. The phase error decreases the directivity and increases the beamwidth and sidelobe level.

0

-5 150 DEGREES

-10 75 DEGREES

-15

-20

-25 RELATIVE POWER (dB) -30

-35

-40 0 5 10 15 20 PATTERN ANGLE (DEG)

22 Naval Postgraduate School Antennas & Propagation Distance Learning Horn Antennas (6)

A maximum phase error of 45 degrees is one criterion used to limit the length of the flare: kDmax £ p / 4 2p p (R2 - R1) £ l 4 2p p R2 (1- cos(y / 2)) £ l 4 2p a¢ p [1- cos(y /2)]£ l 2sin(y /2) 4

Use the identities 1- cos(y /2) = 2sin 2 (y / 4) and sin(y / 2) = 2sin(y /4)cos(y / 4)

2sin 2 (y /4) l l £ ® tan(y / 4) £ 2sin(y / 4)cos(y / 4) 4a¢ 4a¢

This is a good guideline for limiting the length of the flare based on pattern degradation, but does not necessarily give the optimum directivity.

23 Naval Postgraduate School Antennas & Propagation Distance Learning Horn Antennas (7)

The optimum aperture width depends on the length of the flare, as shown below for an H-plane horn. A similar plot can be generated for an E-plane horn. (The separate factor on directivity is the reduction due to phase error.) H-plane optimum:

a¢ = 3lR1H a¢b æ 1 ö D = 10.2 opt 2 ç ÷ 100 è1.3ø R1 = 100l l kDmax » 0.75p lD 75 E-plane optimum:

b b¢ = 2lR1E 50 ab¢ æ 1 ö D = 10.2 opt 2 ç ÷ 25 R1 = 10l l è1.25ø kDmax » 0.5p

10 20 a¢ a¢b¢ Pyramidal optimum: a¢ = 3lR1 , b¢ = 2lR1 , Dopt = 6.4 (R1 = R1 = R1) l2 H E

24 Naval Postgraduate School Antennas & Propagation Distance Learning Horn Example

Example: An E-plane horn has R1 = 20l and a = 0.5l .

(a) The optimum aperture dimension for maximum directivity

b¢ = 2lR1E = l 40 = 6.3l

(b) The flare angle for the optimum directivity

b¢/ 2 6.3l /2 tan(y / 2) = = = 0.1575 R1 20l y / 2 = 8.95o y = 17.9o (c) The optimum directivity is

(0.5l)(6.3l)æ 1 ö Dopt = 10.2 ç ÷ = 25.7 = 14.1 dB l2 è1.25ø

25 Naval Postgraduate School Antennas & Propagation Distance Learning Several Types of Horn Antennas

26 Naval Postgraduate School Antennas & Propagation Distance Learning Microstrip Patch Antennas (1)

Microstrip patch antennas (or simply patch antennas) consist of a thin substrate of grounded dielectric this is plated on top with a smaller area of metal that serves as the element. The advantages and disadvantages include: · Lend themselves to printed circuit fabrication techniques · Low profile - ideal for conformal antennas · Circular or linear polarization determined by feed configuration · Difficult to increase bandwidth beyond several percent · Substrates support surface waves · Lossy

SURFACE FEED LINE

RADIATING SUBSTRATE PATCH (DIELECTRIC)

GROUND PLANE

27 Naval Postgraduate School Antennas & Propagation Distance Learning Microstrip Patch Antennas (2)

Several methods of feeding patch antennas are illustrated below: TOP VIEW

PROXIMITY COUPLING

SURFACE LINE

FEED THROUGH LINE

28 Naval Postgraduate School Antennas & Propagation Distance Learning Microstrip Patch Antennas (3)

Several methods of broadbanding patch antennas are illustrated: PARASITIC ELEMENTS SUPERSTRATES

RADIATING PATCH PARASITIC PATCH SUPERSTRATE SUBSTRATE

FEED POINT GROUND PLANE FEED POINT

REACTIVE LOADING VARIABLE SUBSTRATES

RADIATING PATCH

SUBSTRATE SUBSTRATE

SHORTED STUB FEED FEED GROUND PLANE POINT POINT

29 Naval Postgraduate School Antennas & Propagation Distance Learning Microstrip Patch Antennas (4)

FOLDED BOW RECTANGULAR DIPOLE TIE SLOTTED Modification of the basic element geometry can also provide some increased bandwidth.

CIRCULAR CIRCULAR SLOTTED WITH "EARS"

30 Naval Postgraduate School Antennas & Propagation Distance Learning Microstrip Patch Antennas (5)

The rigorous formulas for a rectangular patch can be obtained from the Sommerfeld solution for an x-directed unit strength infinitesimal dipole located at the top of the substrate. The exact radiation patterns are given by z

SUBSTRATE æ jwmo ö - jkr PATCH e , m Ef = ç ÷sinf e F(q ) r r è 4p r ø w l y æ jwmo ö - jkr Eq = -ç ÷cosf e G(q ) h è 4p r ø x where 2tan(k h) F(q) = 1 tan(k1h) - j(n1(q)secq) / mr 2 tan(k h)cosq G(q) = 1 tan(k1h) - j(er cosq) /n1(q )

2 2 and k1 = kn1(q), n1(q ) = n1 - sin q , n1 = ermr

31 Naval Postgraduate School Antennas & Propagation Distance Learning Microstrip Patch Antennas (6)

The power radiated into space (assuming h << l ) is

2 æ khpm ö æ 1 2 ö P = 80 r ç1- + ÷ rad ç ÷ ç 2 4 ÷ è l ø è n1 5n1 ø However, some power may be captured by a surface wave. If the substrate is thin (h << l ): 3 60(khpm )3 æ 1 ö P » r ç1- ÷ surf 2 ç 2 ÷ l è n1 ø Prad The radiation efficiency is erad » . Define a new constant p that is a function Prad + Psurf of the ratio of the patch’s radiated power to a Hertzian dipole’s radiated power

a 3a b æ 1 2 ö p = é1+ 2 (kw)2 + 4 (kw)4 + 2 (k )2 ùç1- + ÷ ê l úç 2 4 ÷ ë 20 560 10 ûè n1 5n1 ø

where a2 = -0.16605, a4 = 0.00761, and b2 = -0.09142.

32 Naval Postgraduate School Antennas & Propagation Distance Learning Microstrip Patch Antennas (7)

The input resistance is 2 90erad æ l ö 2 æp xo ö Rin » mrer ç ÷ sin ç ÷ p è wø è l ø

( xo , yo ) is the location of the feed point. The bandwidth (defined as VSWR < 2) is 16p æ w öæ h ö BW » ç ÷ç ÷ 3 2ererad è l øè l ø

Example: Nonmagnetic substrate with er = 2.2, f = 3.0 GHz, w /l = 1.5, and h / l = 0.025. The length is chosen for resonance, l = 0.0311 m.

From the formulas presented, if the feed location is (xo = 0.0057, yo = 0), then the input resistance is 43 ohms, the bandwidth approximately 0.037 (3.7%), and the radiation efficiency 0.913.

33 Naval Postgraduate School Antennas & Propagation Distance Learning Reflector Antennas

Reflector systems have been used in optical devices (telescopes, microscopes, etc.) for centuries. They are a simple means of generating a large radiating aperture, which results in a high gain and narrow beamwidth. The most common is the “satellite dish,” a single surface parabolic reflector. The advantages are: · Simple · Broadband (provided that the feed antenna is broadband) · Very large apertures possible The disadvantages are: · Slow beam scanning · Mechanical limitations (wind resistance, gravitational deformation, etc.) · Surface roughness must be controlled · Limited control of aperture illumination

34 Naval Postgraduate School Antennas & Propagation Distance Learning Singly and Doubly Curved Reflectors

A singly curved reflector is generated by translating a plane curve (such as a parabola) along an axis. The radius of curvature in one dimension is finite; in the second dimension it is infinite. The focus is a line, and therefore a linear feed antenna is used.

FULL REFLECTOR OFFSET REFLECTOR

SIDE VIEW SIDE VIEW A doubly curved reflector has two finite radii of curvature. The focus is a point. Spherical wave sources are used as feeds.

FOCUS PARENT PARABOLOID

ROTATIONALLY OFFSET SYMMETRIC PARABOLIC REFLECTOR REFLECTOR 35 Naval Postgraduate School Antennas & Propagation Distance Learning Classical Reflecting Systems

NEWTONIAN PFUNDIAN HERSCHELLIAN PLATE

PLATE F PARABOLOID · F PARABOLOID

PARABOLOID ·

· F

PARABOLOID

HYPERBOLOID ELLIPSOID F F F · HYPERBOLOID · ·

PARABOLOID PARABOLOID

GREGORIAN CASSEGRAIN SCHWARTZCHILD

36 Naval Postgraduate School Antennas & Propagation Distance Learning “Deep Space” Cassegrain Reflector Antenna

37 Naval Postgraduate School Antennas & Propagation Distance Learning Multiple Reflector Antennas

Dual reflecting systems like the Cassegrain and Gregorian are not uncommon. Some specialized systems have as many a four or five reflectors.

FEED FEED

TERTIARY

SECONDARY SECONDARY PRIMARY

PRIMARY

TOP OF SCAN TOP OF SCAN

BOTTOM BOTTOM OF SCAN OF SCAN

38 Naval Postgraduate School Antennas & Propagation Distance Learning Geometrical Optics

Geometrical optics (GO) refers to the high-frequency ray tracing methods that have been used for centuries to design systems of lenses and reflectors. The postulates of GO are: · Wavefronts are locally plane and TEM · Wave directions are specified rays, which are vectors normal to the wavefronts (equiphase planes) · Rays travel in straight lines in a homogeneous medium · Polarization is constant along a ray in an isotropic medium · Power contained in a bundle of rays (a flux tube) is conserved

FLUX TUBE (RAY BUNDLE)

POWER THROUGH BOTH CROSS SECTIONS IS EQUAL · Reflection and refraction obeys Snell’s law and is described by the Fresnel formulas · The reflected field is linearly related to the incident field at the reflection point by a reflection coefficient (i.e., Eref = Einc G) 39 Naval Postgraduate School Antennas & Propagation Distance Learning Parabolic Reflector Antenna (1)

What is the required shape of a surface so that it converts a spherical wave to a plane wave on reflection? All paths from O to the plane wave front AB must be equal: FP + PA = FV + VB SPHERICAL A WAVE FROM P PA = FP cosq ¢+ FB SOURCE nˆ r¢ VB = FV + FB V q ¢ F B z¢ z Plug in for VB and PA f FP + (FPcosq ¢ + FB) PLANE WAVE REFLECTED FROM S = FV + (FV + FB) SURFACE FP(1+ cosq ¢) = 2FV F is the focus r¢(1+ cosq ¢) = 2 f V is the vertex r¢ = 2 f /(1+ cosq ¢) f is the focal length This is an equation for a parabola.

40 Naval Postgraduate School Antennas & Propagation Distance Learning Parabolic Reflector Antenna (2)

The feed antenna is located at the focus. The design parameters of the parabolic reflector are the diameter D, and the ratio f / D. The edge angle is given by -1é 1 ù qe = 2 tan ê ú ë4 f / Dû P Ideally, the feed antenna should have the q¢/2 following characteristics: r¢ nˆ 1. Maximize the feed energy intercepted by the reflector (small HPBW ® large feed) D = 2a q ¢ q 2. Provide nearly uniform illumination in the F z focal plane and no spillover (feed pattern z¢ abruptly goes to zero at qe ) f qe 3. Radiate a spherical wave (reflector must be in the feed’s far field ® small feed) 4. Must not significantly block waves reflected off of the surface ® small feed FOCAL PLANE

41 Naval Postgraduate School Antennas & Propagation Distance Learning Reflector Antenna Losses (1)

1. Feed blockage reduces gain and increases sidelobe levels (efficiency factor, eb ). Support struts can also contribute to blockage loss. PARABOLIC FEED ANTENNA SURFACE

BLOCKED RAYS

2. Spillover reduces gain and increases sidelobe levels (efficiency factor, es ) x FEED PATTERN PARABOLIC SURFACE

qe FOCUS

42 Naval Postgraduate School Antennas & Propagation Distance Learning Reflector Antenna Losses (2)

3. Aperture tapering reduces gain (this is the same illumination efficiency that was encountered in arrays; efficiency factor, ei ) 4. Phase error in the aperture field (i.e., due to the roughness of the reflector surface, random phase errors occur in the aperture field, efficiency factor, ep ). Note that there are also random amplitude errors in the aperture field, but they will be accounted for in the illumination efficiency factor.

5. Cross polarization loss (efficiency factor, ex ). The curvature of the reflector surface gives rise to cross polarized currents, which in turn radiate a crossed polarized field. This factor accounts for the energy lost to crossed polarized radiation. 6. Feed efficiency (efficiency factor, e f ). This is the ratio of power radiated by the feed to the power into the feed. This gain of the reflector can be written as 4pA 4pA G = e = e e e e e e 2 a 2 i p x f s b l l 123 ºeA For reflectors, the product denoted as eA is termed the aperture efficiency.

43 Naval Postgraduate School Antennas & Propagation Distance Learning Example (1)

A circular parabolic reflector with f / D = 0.5 has a feed pattern E(q ¢) = cosq ¢ for q ¢ £ p /2. The edge angle is 1 -1æ ö o qe = 2 tan ç ÷ = (2)(26.56) = 53.1 è 4 f / D ø The aperture illumination is e- jkr¢ e- jkr¢ A(q¢) = E(q¢) = cosq¢ r¢ r¢ but r¢ = 2 f /(1+ cosq ¢) cosq¢(1+ cosq¢) A(q¢) = 2 f The edge taper is the ratio of the field at the edge of the reflector to that at the center A(q ) cosq (1+ cosq )/(2 f ) e = e e = 0.4805 = -6.37 dB A(0) 2/(2 f ) The feed pattern required for uniform amplitude distribution is A(q ) E(q ¢)(1+ cosq ¢)/(2 f ) 2 e = º 1 ® E(q ¢) = = sec2 (q ¢/2) A(0) E(0) / f (1+ cosq ¢)

44 Naval Postgraduate School Antennas & Propagation Distance Learning Example (2)

The spillover loss is obtained from fraction of feed radiated power that falls outside of the reflector edge angles. The power intercepted by the reflector is 2p q q e 2 e 2 Pint = ò ò cos q ¢sin q ¢df dq ¢ = 2p ò cos q ¢sinq ¢dq ¢ 0 0 0 q e écos3q ¢ù = -2p ê ú = 0.522p 3 ë û0 The total power radiated by the feed is

2p p / 2 2 p / 2 2 Prad = ò0 ò0 cos q ¢sinq ¢df dq ¢ = 2p ò0 cos q ¢sinq ¢dq ¢ p /2 écos3 q ¢ù = -2p ê ú = 0.667p 3 ë û0 Thus the fraction of power collected by the reflector (spillover efficiency) is

es = Pint / Prad = 0.522 / 0.667 = 0.78

The spillover loss in dB is 10 log(0.78) = -1.08 dB. 45 Naval Postgraduate School Antennas & Propagation Distance Learning Example (3)

A fan beam is generated by a cylindrical paraboloid fed by a line source that provides uniform illumination in azimuth and a cos(py¢/ Dy ) distribution in elevation

Dx

Dy PARABOLIC REFLECTOR

LINE SOURCE

The sidelobe levels (from Table 7.1of Slolnik or equivalent): uniform distribution in azimuth (x), SLL = 13.2 dB, e = 1 ix cosine in elevation (y), SLL = 23 dB, e = 0.81 iy

Find Dx and Dy for azimuth and elevation beamwidths of 2 and 12 degrees o qel = 69l / Dy = 12 Þ Dy = 5.75l q = 51l / D = 2o Þ D = 25.5l az x x The aperture efficiency is ei = (1)(.81) and the gain is 4pAp 4p (5.75l)(25.5l) G = ei = (1)(0.81) =1491.7 = 31.7 l2 l2 46 Naval Postgraduate School Antennas & Propagation Distance Learning Calculation of Efficiencies (1)

Spillover loss can be computed from the feed antenna pattern. If the feed pattern can be e- jkr¢ expressed as Er (r¢,q ¢) = g(q¢) eˆ where g(q ¢) gives the angular dependence and f r¢ f eˆ f denotes the electric field polarization, then the spillover efficiency is 2p qe ò ò g(q¢) sinq¢ dq¢df¢ FEED POWER INTERCEPTED e = = 0 0 s FEED POWER RADIATED 2p p ò ò g(q¢) sinq¢ dq¢df¢ 0 0 Example: What is the spillover loss when a dipole feeds a paraboloid with f / D = 0.4? o 2p 64o 3 64 2 é cos q ¢ù sin q ¢ sinq ¢ dq ¢df¢ êcosq ¢ + ú ò ò 3 0 0 ë û0 -1.3 es = = = = 0.488 = -3.1 dB 2p p o 3 180 - 2.667 sin 2 q ¢ sinq ¢ dq ¢df¢ é cos q ¢ù ò ò êcosq ¢+ ú 3 0 0 ë û0

47 Naval Postgraduate School Antennas & Propagation Distance Learning Calculation of Efficiencies (2)

The illumination efficiency (also known as tapering efficiency) depends on the feed pattern as well 2 2p qe g(q¢) tan(q ¢/ 2) dq¢df¢ 2 ò ò æ f ö 0 0 ei = 32ç ÷ è D ø 2p p ò ò g(q ¢) sinq ¢ dq¢df¢ 0 0 ì2(n +1)cosn q¢, 0 £q ¢ £ p /2 A general feed model is the function g(q¢) = í î0, else The formulas presented yield the following efficiencies for this simple feed model: qe n+1 n é æqe öù es = (n +1) cos q¢sinq¢ dq¢ =1- cosç ÷ ò ê è 2 øú 0 ë û 2 2 éqe / 2 ù æ f ö n / 2 ei = ç ÷ 2(n +1)ê cos q¢tan(q¢/2) dq¢ ú è D ø ê ò ú ë 0 û

48 Naval Postgraduate School Antennas & Propagation Distance Learning Cosine Feed Efficiency Factors

n Efficiencies for a cos q¢ feed: (full aperture angle is 2qe )

Aperture efficiency (eies ) Spillover efficiency es

1 1

0.9 0.9

0.8 0.8 0.7 0.7

0.6 0.6

0.5 0.5 0.4 0.4 n=2 Efficiency Factor 0.3 n=2 Spillover Efficiency 0.3 n=4 n=4 n=6 0.2 n=6 0.2 n=8 n=8 0.1 0.1

0 0 0 50 100 150 0 50 100 150 Full Aperture Angle, Degrees Full Aperture Angle, Degrees 49 Naval Postgraduate School Antennas & Propagation Distance Learning Feed Example

Given a reflector with f / D = 0.5 find n for a cosn q¢ feed that gives optimum efficiency. Estimate the directivity of the feed antenna.

1 -1æ ö o For f / D = 0.5 the edge angle is qe = 2 tan ç ÷ = 53.1 . Therefore the full aperture è 4 f / D ø o angle is 2qe = 106.3 . From the figure on the previous page, the curve with the maximum in the vicinity of 106.3o is n = 4, and therefore the feed exponent should be 4. The HPWB is 4 g(q¢) = cos qHP¢ = 0.5 cosqHP¢ = 0.84 o o qHP¢ = 32.8 Þ HPBW = 65.5 We can use the formula for the directivity of the cosine pattern presented previously D = 2(n +1) = 2(5) = 10 = 10 dB. The approximate directivity formula can also be used 4p 4p D » = = 9.6 = 9.8 dB 2 qefa (1.14)

50 Naval Postgraduate School Antennas & Propagation Distance Learning Calculation of Efficiencies (3)

Feed blockage causes an additional loss in gain. For large reflectors, the null field hypothesis can be used to estimate the loss. Essentially it says that the current in the shadow of the feed projected on the aperture is zero. The shadow area is illustrated below for a rectangular aperture. a a b PROJECTED SHADOW b FEED REFLECTOR For a circular aperture, the area where nonzero currents exist is approximately 2 2 æ D ö æ D f ö p 2 2 Ae » p ç ÷ - p ç ÷ = (D - D f ) è 2 ø è 2 ø 4 This assumes that all of the currents in the illuminated part of the aperture are uniform and in phase, which is not D D f always the case. Both D andD f should be much greater than the wavelength.

51 Naval Postgraduate School Antennas & Propagation Distance Learning Reflector Design Using RASCAL (1)

Axially symmetric parabolic design:

52 Naval Postgraduate School Antennas & Propagation Distance Learning Reflector Design Using RASCAL (2)

Axially symmetric Cassegrain design:

53 Naval Postgraduate School Antennas & Propagation Distance Learning Reflector Design Using RASCAL (3)

Offset single surface paraboloid:

54 Naval Postgraduate School Antennas & Propagation Distance Learning Reflector Design Using RASCAL (4)

Offset Cassegrain configuration:

55 Naval Postgraduate School Antennas & Propagation Distance Learning Microwave Reflectors

Reflectors are used as microwave relays. The antennas with curved tops are horn-fed offset reflectors that are completely enclosed (called hog horns). The enclosures cut down on noise and interference cause by spillover. They also protect the antenna components from the elements (weather, birds, etc.) Axially symmetric reflector systems are also visible. They too are completely enclosed by a transparent radome.

OFFSET REFLECTOR APERTURE

FEED HORN

56 Naval Postgraduate School Antennas & Propagation Distance Learning Reflector Antenna Analysis Methods

Geometrical optics is used to design reflector surfaces, but is usually not accurate enough to use for predicting the secondary pattern (from the reflector). There are two common methods for computing the scattered field from the reflector:

1. Find or estimate the current induced on the reflector and use it in the radiation integrals. a) Rigorously calculate the current using a numerical approach such as the method of moments b) Estimate the current using the physical optics (PO) approximation r r J s = 2nˆ ´ H f (r¢,q ¢,f¢) If the feed field is shadowed from a part of the surface, then the current is assumed to be zero on that part. r 2. Equivalent aperture method. Find Ea in the aperture plane and compute an equivalent magnetic current that can be used in the radiation integral r r Jms = -2nˆ ´ Ea (x¢, y¢, z¢)

57 Naval Postgraduate School Antennas & Propagation Distance Learning Dipole Fed Parabola (1)

A dipole is used to feed a parabola. This is not practical because at least half of the the dipole’s radiated power is directed in the rear hemisphere and misses the reflector (i.e., there is at least 3 dB of spillover loss). However, this example illustrates how crossed polarized currents and fields are generated. For the dipole aligned with the y axis jkhI e- jkr¢ e- jkr¢ E (r¢,q ¢,f¢) = eˆ l siny º E eˆ siny f f 4p r¢ o f r¢

D y¢ There are two coordinate systems x, y, z ® r,q,f ® rˆ,qˆ,fˆ y r¢ x¢, y¢, z¢ ® r¢,q¢,f¢ ® rˆ¢,qˆ¢,fˆ¢ q ¢ z cosy is just the y¢ direction cosine z¢ cosy = rˆ¢ · yˆ¢ = sinq ¢sin f¢ f x, x¢ Because y is the angle from the dipole axis, y the dipole pattern depends on

siny = 1- cos2y = 1- sin 2 q ¢sin 2 f¢

58 Naval Postgraduate School Antennas & Propagation Distance Learning Dipole Fed Parabola (2)

If the reflector is in the far field of the dipole, then the electric field vector will have only qˆ¢,fˆ¢ components ˆ ˆ eˆ f = rˆ¢sinq¢sinf¢+q¢cosq¢sin f¢ +f¢cosf¢ 1442443 DROP THIS Re-normalizing the vector qˆ¢cosq¢sinf¢+fˆ¢cosf¢ qˆ¢cosq¢sin f¢ +fˆ¢cosf¢ qˆ¢cosq¢sinf¢+fˆ¢cosf¢ eˆ f = = = cos2 q¢sin 2 f¢ + cos2 f¢ 1- sin 2 q¢sin 2 f¢ siny which we rearrange to find ˆ ˆ eˆ f siny =q¢cosq¢sin f¢ +f¢cosf¢ After the spherical wave is reflected from the parabola, a plane wave exists and there is no 1/r dependence. The field in the aperture is the field reflected from the parabola. At the reflector, the tangential components cancel and the normal components double r r r r (Ei + Er )norm = 2(Ei )norm = 2(nˆ · Ei )nˆ

59 Naval Postgraduate School Antennas & Propagation Distance Learning Dipole Fed Parabola (3)

Therefore, r r r r Ea = Er = 2(nˆ · E f )nˆ - E f which is nothing more than a vector form of Snell’s Law. The normal at a point on the reflector surface is given by q ¢ q ¢ nˆ = -rˆ¢cos +qˆ¢sin 2 2 After some math, which involves the used of several trig identities, the aperture field is e- jkr¢ Er (r¢,q¢,f¢) = E {xˆ cosf¢sin f¢(1- cosq¢) - yˆ(cosq¢sin 2 f¢ + cos2 f¢)} a o r¢ The magnetic current in the aperture is r r r ˆ¢ J = -2nˆ ´ E = -2zˆ ´ E q q ¢/2 ms a a rˆ¢ This current exists over a circular aperture, nˆ and it is used in the radiation integral to get r¢ q¢/2 the far field. Since the integration is over a q ¢ circular region, it is convenient to use polar coordinates, (r¢,f ¢) z¢

60 Naval Postgraduate School Antennas & Propagation Distance Learning Dipole Fed Parabola (4)

The important characteristic of the aperture field is that there are both x and y components, even though the feed dipole is purely y polarized. Since the radiation integral has the form - jkr 2p D / 2 r jkh e æ Jms ´ rˆö ìqˆü - jkrˆ·r¢ E q (r,q,f) = ç ÷ · í ý e r¢dr¢df¢ { } 4p r ò ò h îfˆþ f 0 0 è ø the x directed currents result in a crossed polarized far field component.

D PROJECTED APERTURE J ms y r¢ r¢ q ¢ Jms z x x z¢ f¢ x CURRENT ON THE PROJECTED APERTURE y z¢ y

61 Naval Postgraduate School Antennas & Propagation Distance Learning Crossed Polarized Radiation

Below is a comparison of the crossed polarized radiation in the principal plane of a axially symmetric parabolic reflector to that of an offset parabolic reflector. In the symmetric case, the radiation from the crossed polarized components cancel when the observation point is in the principal plane. Note that the feed is not a dipole, but a raised cosine.

62 Naval Postgraduate School Antennas & Propagation Distance Learning Lens Antennas (1)

Lens antennas are also based on geometrical optics principles. The major advantage of lenses over reflectors is the elimination of blockage. Lenses can be constructed the same way at microwave frequencies as they are at optical frequencies. A dielectric material is shaped to provide equal path lengths from the focus to the aperture, as illustrated below.

It is important to keep the reflection at the nˆ air/dielectric boundary as small as z possible. The wavelength in the dielectric r is l = lo /n , where n = er is the index q F z of refraction. D The axial path length is f+nt. The path f t length along the ray shown is r+nz. Since er they must be equal, f + t = r cosq + z t = r cosq + z - f Inserting t back in the original equation: f + r cosq + z - f = r cosq + z

63 Naval Postgraduate School Antennas & Propagation Distance Learning Lens Antennas (2)

Solving for r gives f (n -1) r = n cosq -1 which is the equation for a hyperbola. There are several practical problems with “optical type” lenses at microwave frequencies. 1. For a high gain a large D is required, yet the focal length must be small to keep the overall antenna volume small. Lenses can be extremely heavy and bulky. 2. Hyperoloids are difficult to manufacture, so a spherical approximation is often used for the lens shape. The sphere’s deviation from a hyperbola results in phase errors called aberrations. The errors distort the far field pattern similar to quadratic phase errors in horns. 3. Reflection loss occurs at the air/dielectric interface. There are also multiple reflections inside of the lens that cause aperture amplitude and phase errors. 4. As in the case of reflectors, there is spillover, non-uniform amplitude at the aperture, and crossed-polarized far fields. Special design tricks can be employed at microwave frequencies. Since the wavelength is relatively large compared to optical case, a sampled version of the lens is practical.

64 Naval Postgraduate School Antennas & Propagation Distance Learning Lens Antennas (3)

A sampled version of a lens would use two arrays placed back to back. The array of pickup elements receives the feed signal and transmits it to the second array at the output aperture. The cable between the elements provides the same phase shift that a path through a solid dielectric would provide. In fact, there is no need to curve the pickup array aperture. A plane surface can be used and any phase difference between the curved and plane surfaces are then included in the phase of the connecting cable.

end In a dielectric lens, the shortest PICKUP l OUTPUT APERTURE APERTURE electrical path length (klend ) is at the r edge and the longest electrical path D length (klcenter ) is in the center. F q z Therefore the cable in the center must lcenter be longer than the cable at the edge. f Phase shifters could be inserted between the arrays to scan the output aperture beam. This approach is referred to as a constrained lens (i.e., the signal paths are constrained to cables).

65 Naval Postgraduate School Antennas & Propagation Distance Learning Lens Antennas (4)

Constrained lenses still suffer spillover loss. However, the aperture surfaces can be planar (rather than hyperbolic or spherical). Generally they are much lighter weight than a solid lens. Conventional reflectors and lenses must be scanned mechanically; that is, rotated or physically pointed. A limited amount of scanning can be achieved by moving the feed off of the focus. However, the farther the feed is displaced from the focal point, the larger the aperture phase deviation from a plane wave. This type of scanning is limited to just a few degrees. Reflectors and lenses can be designed with multiple focii. Surfaces more complicated that parabolas and hyperbolas are required, and often they are difficult to fabricate.

A Luneberg lens is a spherical structure 2 that has a precisely controlled n(r) = 2 - (r / a) inhomogenous relative dielectric constant (or index of refraction, n(r)). HORN Because of the spherical symmetry the FEED r feed can scan over 4p steradians. It is D = 2a heavy and bulky. a

66 Naval Postgraduate School Antennas & Propagation Distance Learning Lens Antenna

67 Naval Postgraduate School Antennas & Propagation Distance Learning Radomes (1)

Radome, a term that originates from radar dome, refers to a structure that is used to protect the antenna from adverse environmental elements. It must be structurally strong yet transparent to electromagnetic waves in the frequency band of the antenna. Aircraft radomes are subjected to a severe operating environment. The heat generated by high velocities can cause ablation (a wearing away) of the radome material. Testing of a charred space shuttle tile HARM (high-speed anti-radiation missile) radome testing

68 Naval Postgraduate School Antennas & Propagation Distance Learning Radomes (2)

The antenna pattern with a radome will always be different than that without a radome. Undesirable effects include: 1. gain loss due to the dielectric loss in the radome material and multiple reflections 2. beam pointing error from refraction by the radome wall 3. increased sidelobe level from multiple reflections

GIMBAL SCANNED MOUNT ANTENNA TRANSMITTED RAYS REFRACTED

AIRCRAFT BODY LOW LOSS REFLECTIONS DIELECTRIC RADOME

These effects range are small for flat non-scanning antennas with flat radomes, but can be severe for scanning antennas behind doubly curved radomes.

69 Naval Postgraduate School Antennas & Propagation Distance Learning Radomes (3)

Geometrical optics can be used to estimate the effects of radomes on antenna patterns if the following conditions are satisfied: 1. The radome is electrically large and its surfaces are “locally plane” (the radii of curvature of the radome surfaces are large compared to wavelength) 2. The radome is in the far field of the antenna 3. The number of reflections is small, so that the sum of the reflected rays converges quickly to an accurate result Reconstructed aperture method:

PROJECTED RADOME D APERTURE RECONSTRUCTED RECONSTRUCTED AT ANTENNA FROM RAYS nˆ nˆ AMPLITUDE ANTENNA APERTURE DIRECTION OF PHASE REFLECTION LOBE D

70 Naval Postgraduate School Antennas & Propagation Distance Learning Radiation Pattern Effects of a Radome

Comparison of measured horn patterns with and without a radome

0 Method of moments patch H-PLANE model of a HARM radome

-5

-10

-15 Relative Power (dB)

-20

HORN HORN WITH RADOME -25 -80 -60 -40 -20 0 20 40 60 80 Theta (Degrees)

71 Naval Postgraduate School Antennas & Propagation Distance Learning Hawkeye

72 Naval Postgraduate School Antennas & Propagation Distance Learning JSTARS

73 Naval Postgraduate School Antennas & Propagation Distance Learning Carrier Bridge

74 Naval Postgraduate School Antennas & Propagation Distance Learning Antenna Measurements (1)

Purpose of antenna measurements: 1. Verify analytically predicted gain and patterns (design verification) 2. Diagnostic testing (troubleshooting) 3. Quality control (verify assembly methods and tolerances) 4. Investigate installation methods on patterns and gain 5. Determine isolation between antennas

General measurement technique: 1. The measurement system is essentially a communication link with transmit and receive antennas separated by a distance R. 2. The antenna under test (AUT), that is, the antenna with unknown gain, is usually the receive antenna. 3. A calibration is performed by noting the received power level when a standard gain horn is used to receive (the gain of a standard gain horn is known precisely). 4. The AUT is substituted for the reference antenna, and the change in power is equivalent to the change in gain (since all other parameters in the Friis equation are the same for the two measurement conditions).

75 Naval Postgraduate School Antennas & Propagation Distance Learning Antenna Measurements (2)

Conditions on the measurement facility include: 1. R must be large enough so that the spherical wave at the receive antenna is approximately a plane wave. (In other words, the receive antenna must be in the far field of the transmit antenna, and vice versa.)

D The phase error at the edge of the antenna is

TRANSMIT typically limited to p /8 SOURCE 2 2 R kDmax = k R +(L / 2) - R = p / 8 L

TARGET or, 2 SPHERICAL 2L WAVEFRONT r º R = ff min l

2. Reflections from the walls, ceiling and floor must be negligible so that multipath contributions are insignificant. 3. Noise in the instrumentation system must be low enough so that low sidelobe levels can be measured reliably.

76 Naval Postgraduate School Antennas & Propagation Distance Learning Antenna Measurements (3)

Examples of measurement chambers. (AUTs are installed on an aircraft.)

Far field chamber: a communication link in a closed environment.

Tapered chamber: the tapered region behaves like a horn transition

Compact range: a plane wave is reflected from the reflector, which allows very small values of R (mostly used for radar cross section and scattering measurements).

77 Naval Postgraduate School Antennas & Propagation Distance Learning Antenna Measurements (4)

Antenna measurement facility descriptors: SYSTEM DESCRIPTOR CATEGORIES physical configuration indoor/outdoor near field far field compact tapered instrumentation time domain frequency domain continuous wave (CW) pulsed CW data analysis & presentation fixed frequency/variable aspect fixed aspect/frequency sweep two-dimensional frequency aspect time domain trace imaging of currents and fields polar or rectangular plots

78 Naval Postgraduate School Antennas & Propagation Distance Learning NRAD Model Range at Point Loma

SHIP MODEL

79 Naval Postgraduate School Antennas & Propagation Distance Learning Near-field Probe Pattern Measurement

80 Naval Postgraduate School Distance Learning

Antennas & Propagation

LECTURE NOTES VOLUME V ELECTROMAGNETIC WAVE PROPAGATION

by Professor David Jenn

(ver1.3) Naval Postgraduate School Antennas & Propagation Distance Learning Propagation of Electromagnetic Waves

Radiating systems must operate in a complex changing environment that interacts with propagating electromagnetic waves. Commonly observed propagation effects are depicted below.

4 SATELLITE

IONOSPHERE 1 DIRECT 3 5 2 REFLECTED 3 TROPOSCATTER 1 4 IONOSPHERIC HOP 2 5 SATELLITE RELAY 6 GROUND WAVE

6 TRANSMITTER RECEIVER EARTH

Troposphere: lower regions of the atmosphere (less than 10 km) Ionosphere: upper regions of the atmosphere (50 km to 1000 km)

Effects on waves: reflection, refraction, diffraction, attenuation, scattering, and depolarization. 1 Naval Postgraduate School Antennas & Propagation Distance Learning Survey of Propagation Mechanisms (1)

There are may propagation mechanisms by which signals can travel between the radar transmitter and receiver. Except for line-of-sight (LOS) paths, the mechanism’s effectiveness is generally a strong function of the frequency and transmitter-receiver geometry. 1. direct path or "line of sight" (most radars; SHF links from ground to satellites) RX TX o o

SURFACE 2. direct plus earth reflections or "multipath" (UHF broadcast; ground-to-air and air- to-air communications) TX o o RX

SURFACE 3. ground wave (AM broadcast; Loran C navigation at short ranges) TX RX o o SURFACE

2 Naval Postgraduate School Antennas & Propagation Distance Learning Survey of Propagation Mechanisms (2)

4. ionospheric hop (MF and HF broadcast and communications)

F-LAYER OF IONOSPHERE TX E-LAYER OF o o RX IONOSPHERE SURFACE

5. waveguide modes or "ionospheric ducting" (VLF and LF communications)

D-LAYER OF TX IONOSPHERE o o RX SURFACE

Note: The distinction between ionospheric hops and waveguide modes is based more on the mathematical models than on physical processes.

3 Naval Postgraduate School Antennas & Propagation Distance Learning Survey of Propagation Mechanisms (3)

6. tropospheric paths or "troposcatter" (microwave links; over-the-horizon (OTH) radar and communications) TROPOSPHERE

TX o RX o SURFACE 7. terrain diffraction

TX o o RX MOUNTAIN 8. low altitude and surface ducts (radar frequencies) SURFACE DUCT (HIGH DIELECTRIC CONSTANT) TX o o SURFACE RX 9. Other less significant mechanisms: meteor scatter, whistlers

4 Naval Postgraduate School Antennas & Propagation Distance Learning Illustration of Propagation Phenomena

(From Prof. C. A. Levis, Ohio State University)

5 Naval Postgraduate School Antennas & Propagation Distance Learning Propagation Mechanisms by Frequency Bands

VLF and LF Waveguide mode between Earth and D-layer; ground wave at short (10 to 200 kHz) distances LF to MF Transition between ground wave and mode predominance and sky (200 kHz to 2 MHz) wave (ionospheric hops). Sky wave especially pronounced at night. HF Ionospheric hops. Very long distance communications with low power (2 MHz to 30 MHz) and simple antennas. The “short wave” band. VHF With low power and small antennas, primarily for local use using direct (30 MHz to 100 MHz) or direct-plus-Earth-reflected propagation; ducting can greatly increase this range. With large antennas and high power, ionospheric scatter communications. UHF Direct: early-warning radars, aircraft-to satellite and satellite-to-satellite (80 MHz to 500 MHz) communications. Direct-plus-Earth-reflected: air-to-ground communications, local television. Tropospheric scattering: when large highly directional antennas and high power are used. SHF Direct: most radars, satellite communications. Tropospheric refraction (500 MHz to 10 GHz) and terrain diffraction become important in microwave links and in satellite communication, at low altitudes.

6 Naval Postgraduate School Antennas & Propagation Distance Learning Applications of Propagation Phenomena

Direct Most radars; SHF links from ground to satellites Direct plus Earth UHF broadcast TV with high antennas; ground-to-air and air-to- reflections ground communications Ground wave Local Standard Broadcast (AM), Loran C navigation at relatively short ranges Tropospheric paths Microwave links Waveguide modes VLF and LF systems for long-range communication and navigation (Earth and D-layer form the waveguide) Ionospheric hops MF and HF broadcast communications (including most long-distance (E- and F-layers) amateur communications) Tropospheric scatter UHF medium distance communications Ionospheric scatter Medium distance communications in the lower VHF portion of the band Meteor scatter VHF long distance low data rate communications

7 Naval Postgraduate School Antennas & Propagation Distance Learning Multipath From a Flat Ground (1)

When both a transmitter and receiver are operating near the surface of the earth, multipath (multiple reflections) can cause fading of the signal. We examine a single reflection from the ground assuming a flat earth. RECEIVER d q ¢ = 0 · TRANSMITTER C · A D .· Ro q = 0 hr R · 2 ht R1 B y y

EARTH'S SURFACE (FLAT) h t IMAGE REFLECTION POINT jf · re G The reflected wave appears to originate from an image.

8 Naval Postgraduate School Antennas & Propagation Distance Learning Multipath From a Flat Ground (2)

Multipath parameters: 1. Reflection coefficient, G = rejfG . For low grazing angles, y » 0, the approximation G » -1 is valid for both horizontal and vertical polarizations. 2. Transmit antenna gain: Gt (qA) for the direct wave; Gt (qB ) for the reflected wave. 3. Receive antenna gain: Gr (qC ) for the direct wave; Gr (qD ) for the reflected wave. 4. Path difference: DR = (R1 + R2 ) - Ro 14 2 4 3 { REFLECTED DIRECT Gain is proportional to the square of the electric field intensity. For example, if Gto is the gain of the transmit antenna in the direction of the maximum (q = 0), then 2 G (q) = G E (q) º G f (q)2 t to tnorm to t where E is the normalized electric field intensity. Similarly for the receive antenna tnorm with its maximum gain in the direction q ¢ = 0 2 G (q¢) = G E (q¢) º G f (q ¢)2 r ro rnorm ro r

9 Naval Postgraduate School Antennas & Propagation Distance Learning Multipath From a Flat Ground (3)

Total field at the receiver

Etot = Eref + Edir { { REFLECTED DIRECT º F 644444474444448

- jkRo e é ft (qB ) fr (qD ) - jkDR ù = ft (q A) fr (qC ) 1+ G e 4p Ro ëê ft (q A ) fr (qC ) ûú The quantity in the square brackets is the path-gain factor (PGF) or pattern-propagation factor (PPF). It relates the total field at the receiver to that of free space and takes on values 0 £ F £ 2. · If F = 0 then the direct and reflected rays cancel (destructive interference) · If F = 2 the two waves add (constructive interference) Note that if the transmitter and receiver are at approximately the same heights, close to the ground, and the antennas are pointed at each other, then d >> ht ,hr and

Gt (q A) » Gt (qB ) Gr (qC ) » Gr (qD ) 10 Naval Postgraduate School Antennas & Propagation Distance Learning Multipath From a Flat Ground (4)

An approximate expression for the path difference is obtained from a series expansion:

1 (h - h )2 R = d 2 + (h - h )2 » d + r t o r t 2 d 1 (h + h )2 R + R = d 2 + (h + h )2 » d + t r 1 2 t r 2 d Therefore, 2h h DR » r t d and - jk 2hrht / d jkhrht / d - jkhrht / d jkhrht / d | F |=1-e = e (e - e ) = 2sin(khrht /d )

The received power depends on the square of the path gain factor 2 2 2 æ khthr ö æ kht hr ö Pr µ| F | = 4sin ç ÷ » 4ç ÷ è d ø è d ø

The last approximation is based on hr ,ht << d and G » -1.

11 Naval Postgraduate School Antennas & Propagation Distance Learning Multipath From a Flat Ground (5)

Two different forms of the argument are frequently used.

1. Assume that the transmitter is near the ground ht » 0 and use its height as a reference. The elevation angle is y where

h - h Dh h Ro tany = r t º » r Dh = hr - ht d d d y d 2. If the transmit antenna is very close to the ground, then the reflection point is very near to the transmitter and y is also the grazing angle:

DR = b - a = 2h siny a t y b ht y y

If the antenna is pointed at the horizon (i.e., its maximum is parallel to the ground) then y »q A.

12 Naval Postgraduate School Antennas & Propagation Distance Learning Multipath From a Flat Ground (6)

Thus with the given restrictions the PPF can be expressed in terms of y

| F |= 2sin(kht tany )

The PPF has minima at: kht tany = np (n = 0,1,…,¥) 2p h tany = np l t tany = nl /ht

Maxima occur at: kht tany = mp / 2 (m = 1,3,5,…,¥) 2p 2n +1 h tany = p (n = 0,1, ,¥ ) l t 2 … (2n +1)l tany = 4ht Plots | F | are called a coverage diagram. The horizontal axis is usually distance and the vertical axis receiver height. (Note that because d >> hr the angle y is not directly measurable from the plot.)

13 Naval Postgraduate School Antennas & Propagation Distance Learning Multipath From a Flat Ground (7)

Coverage diagram: Contour plots of | F | in dB for variations in hr and d normalized to a reference range do . Note that when d = do then Etot = Edir . æ do ö | F |= 2ç ÷sin(kht tany ) è d ø

60 d o = 2000 m 50 (m)

hr ht = 100l 40

30

20

10 RECEIVER HEIGHT,

0 1000 2000 3000 4000 5000 RANGE, d (m)

14 Naval Postgraduate School Antennas & Propagation Distance Learning Multipath From a Flat Ground (8)

Another means of displaying the received field is a height-gain curve. It is a plot of | F | in dB vs hr at a fixed range. · The constructive and destructive interference as a function of height can be identified. · At low frequencies the periodicity of the curve at low heights can be destroyed by the ground wave. · Usually there are many reflected wave paths between the transmitter and receiver, in which case the peaks and nulls are distorted. · This technique is often used to determine the optimum tower height for a broadcast radio antenna. 10

5

0

-5

-10 PATH GAIN FACTOR (dB) -15

-20 0 10 20 30 40 50 60 RECEIVER HEIGHT, hr (m)

15 Naval Postgraduate School Antennas & Propagation Distance Learning Multipath Example

A radar antenna is mounted on a 5 m mast and tracks a point target at 4 km. The target is 2 m above the surface and the wavelength is 0.2 m. (a) Find the location of the reflection point on the x axis and the grazing angle y . (b) Write an expression for the one way path gain factor F when a reflected wave is present. Assume a reflection coefficient of G » -1. (b) The restrictions on the heights and distance are satisfied for the following 5 m 2 m formula y y x æ kh h ö æ 2p (2)(5) ö x=0 x=4 km F = 2sinç t r ÷ = 2sinç ÷ d (0.2)(4000 Reflection è ø è ø Point = (2)(0.785) = 0.157 (a) Denote the location of the reflection point by xr and use similar triangles The received power varies as F 2 , thus 5 2 tany = = 2 xr 4000 - xr 10log(F )= -16.1 dB xr = 2.86 km The received power is 16.1 dB below the y = tan-1(5/2860) = 0.1o free space value

16 Naval Postgraduate School Antennas & Propagation Distance Learning Field Intensity From the ERP

The product PtGt is called the effective radiated power (ERP, or sometimes the effective isotropic radiated power, EIRP). We can relate the ERP to the electric field intensity as follows:

· The Poynting vector for a TEM wave: r 2 Edir Wr = Â{Er ´ Hr *}= ho · For the direct path: PG Wr = t t 2 4pRo · Equate the two expressions: (note that ho » 120p ) r 2 Edir P G 30PG E = t t Þ Er = t t º o 2 dir ho 4pRo d d

where Eo is called the unattenuated field intensity at unit distance.

17 Naval Postgraduate School Antennas & Propagation Distance Learning Wave Reflection at the Earth’s Surface (1)

Fresnel reflection coefficients hold when: 1. the Earth’s surface is locally flat in the vicinity of the reflection point 2. the surface is smooth (height of irregularities << l ) Traditional notation: o o 1. grazing angle, y = 90 -qi , and the grazing angle is usually very small (y < 1 ) s æ s ö 2. complex dielectric constant, ec = e re o - j = eo çer - j ÷ º eo (e r - jc ), w è e ow ø 14243 erc s where c = weo 3. horizontal and vertical polarization reference is used VERTICAL POL HORIZONTAL POL Also called Also called Er = Er r r transverse || V E^ = EH transverse magnetic nˆ nˆ electric ˆ ˆ (TM) pol y ki y ki (TE) pol

SURFACE SURFACE

18 Naval Postgraduate School Antennas & Propagation Distance Learning Wave Reflection at the Earth’s Surface (2)

Reflection coefficients for horizontal and vertical polarizations:

(e - jc )siny - (e - jc ) - cos2y - G º R = r r || V 2 (er - jc )siny + (er - jc ) - cos y

siny - (e - jc ) - cos2y G º R = r ^ H 2 siny + (er - jc ) - cos y

For vertical polarization the phenomenon of total reflection can occur. This yields a surface guided wave called a ground wave. From Snell’s law, assuming mr = 1 for the Earth,

sinqi sinqi = sinqr = (er - jc )mr sinqt Þ sinqt = mr =1 er - jc p Let q be complex, q = + jq , where q is real. t t 2

19 Naval Postgraduate School Antennas & Propagation Distance Learning Wave Reflection at the Earth’s Surface (3)

p æp ö Using qt = + jq : sinqt = sinç + jq ÷ = cos( jq) = coshq 2 è 2 ø cosqt = - jsin( jq ) = - jsinhq

sinqi Snell’s law becomes sinqt = coshq = er - jc 2 2 cosqt = 1-sin qt = 1- cosh q = sinhq Reflection coefficient for vertical polarization:

jh sinhq +ho cosqi G|| º -RV = jhsinhq -ho cosqi

mo where h = . Note that G|| = 1 and therefore all of the power flow is along the eo (e r - jc) surface. The wave decays exponentially with distance into the Earth.

20 Naval Postgraduate School Antennas & Propagation Distance Learning Wave Reflection at the Earth’s Surface (4)

Example: surface wave propagating along a perfectly conducting plate

· 5l plate · 15 degree grazing angle · TM (vertical) polarization · the total field is plotted (incident plus scattered) · surface waves will follow curved surfaces if the radius of curvature >> l

INCIDENT WAVE (75 DEGREES OFF OF NORMAL)

CONDUCTING PLATE

21 Naval Postgraduate School Antennas & Propagation Distance Learning Atmospheric Refraction (1)

Refraction by the lower atmosphere causes waves to be bent back towards the earth’s surface. The ray trajectory is described by the equation: n Re sinq =CONSTANT Two ways of expressing the index of refraction n (= er ) in the troposphere:

1. n = 1+ cr / rSL +HUMIDITY TERM REFRACTED R = 6378 km = earth radius q RAY q e q c » 0.00029 = Gladstone-Dale constant r, rSL = mass densities at altitude and sea level

R R EARTH'S 77.6 R e e 2. n = (p + 4,810e/T)10-6 - 1 e SURFACE T p = air pressure (millibars) T = temperature (K) e = partial pressure of water vapor (millibars)

22 Naval Postgraduate School Antennas & Propagation Distance Learning Atmospheric Refraction (2)

Refraction of a wave can provide a significant level of transmission over the horizon. A bent refracted ray can be represented by a straight ray if an equivalent earth radius Re¢ is used. For most atmospheric conditions Re¢ = 4Re / 3 = 8500 km

REFRACTED RAY REFRACTED BECOMES A RAY TX STRAIGHT LINE RX

TX RX hr ht LINE OF SIGHT (LOS) ht hr BLOCKED BY EARTH'S BULGE EARTH'S EARTH'S EQUIVALENT EARTH RADIUS, R SURFACE SURFACE e¢

STANDARD EARTH CONDITIONS: RADIUS , R 4 e R¢ » R e 3 e

23 Naval Postgraduate School Antennas & Propagation Distance Learning Atmospheric Refraction (3)

2 2 Distance from the transmit antenna to the horizon is Rt = (Re ¢ + ht ) -(Re ¢ ) but Re¢ >> ht so that Rt » 2Re¢ ht . Similarly Rr » 2Re ¢ hr . The radar horizon is the sum

RRH » 2Re ¢h t + 2Re¢ hr Example: A missile is flying 15 m TX Rt Rr above the ocean towards a ground RX based radar. What is the approximate ht hr range that the missile can be detected assuming standard atmospheric EARTH'S R¢ R ¢ conditions? SURFACE e e Re ¢

Using ht = 0 and hr = 15 gives a radar horizon of

RRH » 2Re¢ hr » (2)(8500´103 )(15) »16 km

24 Naval Postgraduate School Antennas & Propagation Distance Learning Atmospheric Refraction (4)

Derivation of the equivalent Earth radius

h q (h) RAY PATH TANGENT n(h) q o VERTICAL

SURFACE Break up the atmosphere into thin horizontal layers. Snell’s law must hold at the boundary between each layer, e(h)sin[q(h)]= eo sinqo h (h) h3 M q n(h3) q h2 2 n(h2) q h1 1 n(h1) THIN LAYER IN WHICH qo n » CONSTANT

SURFACE

25 Naval Postgraduate School Antennas & Propagation Distance Learning Atmospheric Refraction (5)

In terms of the Earth radius,

Re eo sin qo = (Re + h) e(R) sin[q (R)] 1442443 14444244443 AT THE AT RADIUS SURFACE R=Re +h Using the grazing angle, and assuming that e(h) varies linearly with h ì d ü Re eo cosyo = (Re + h)í eo + h e(h) ýcos[y (h)] î dh þ Expand and rearrange ì d ü 2 d Re eo {cosy o - cos[y(h)]}= í eo + Re e(h) ýhcos[y(h)]+ h e(h) cos[y(h)] î dh þ dh 2 If h << Re then the last term can be dropped, and since y is small, cosy » 1+y /2 2 2 2h é Re d ù [y (h)] »y o + ê1 + e (h)ú Re ë eo dh û The second term is due to the inhomogenity of the index of refraction with altitude.

26 Naval Postgraduate School Antennas & Propagation Distance Learning Atmospheric Refraction (6)

Define a constant k such that -1 2 2 2h 2 2h é Re d ù [y (h)] »y o + =y o + where k = ê1+ e(h)ú kRe Re¢ ë e o dh û

Re¢ = kRe is the effective (equivalent) Earth radius. If Re¢ is used as the Earth radius then rays can be drawn as straight lines. This is the radius that would produce the same geometrical relationship between the source of the ray and the receiver near the Earth’s surface, assuming a constant index of refraction. The restrictions on the model are: 1. Ray paths are nearly horizontal 2. e(h) versus h is linear over the range of heights considered Under standard (normal) atmospheric conditions, k » 4/3. That is, the radius of the Earth æ 4ö is approximately Re¢ = ç ÷6378 km = 8500 km. This is commonly referred to as “the four- è 3ø thirds Earth approximation.”

27 Naval Postgraduate School Antennas & Propagation Distance Learning Fresnel Zones (1)

For the direct path phase to differ from the reflected path phase by an integer multiple of 180o the paths must differ by integer multiples of l /2

DR = nl /2 (n = 0,1,…) The collection of points at which reflection would produce an excess path length of nl /2 is called the nth Fresnel zone. In three dimensions the surfaces are ellipsoids centered on the direct path between the transmitter and receiver

LOCUS OF REFLECTION DIRECT POINTS (SURFACES OF PATH (LOS) REVOLUTION) n = 2 RECEIVER n =1 TRANSMITTER hr ht

REFLECTING SURFACE

28 Naval Postgraduate School Antennas & Propagation Distance Learning Fresnel Zones (2)

A slice of the vertical plane gives the following geometry

d d d r RX t h TX r R2 h R t 1 nth FRESNEL ZONE

REFLECTION POINT

For the reflection coefficient G = r e jp = -r : · If n is even the two paths are out of phase and the received signal is a minimum · If n is odd the two paths are in phase and the received signal is a maximum

Because the LOS is nearly horizontal Ro » d and therefore Ro = dt + d r » d . For the nth Fresnel zone R1 + R2 = d + nl /2.

29 Naval Postgraduate School Antennas & Propagation Distance Learning Fresnel Zones (3)

The radius of the nth Fresnel zone is nld d F = t r n d or, if the distances are in miles, then ndtdr Fn = 72.1 (feet) fGHzd Transmission path design: the objective is to find transmitter and receiver locations and heights that give signal maxima. In general: 1. reflection points should not lie on even Fresnel zones 2. the LOS should clear all obstacles by 0.6F1, which essentially gives free space transmission

The significance of 0.6F1 is illustrated by examining two canonical problems: (1) knife edge diffraction and (2) smooth sphere diffraction.

Conversions: 0.0254 m = 1 in; 12 in = 1 ft; 3.3 ft = 1 m; 5280 ft = 1 mi; 1 km = 0.62 mi

30 Naval Postgraduate School Antennas & Propagation Distance Learning Diffraction (1)

Knife edge diffraction

l = CLEARANCE DISTANCE l > 0 l = 0,SHADOW BOUNDARY l < 0

SHARP ht OBSTACLE hr d Smooth sphere diffraction

l = CLEARANCE DISTANCE = 0,SHADOW l > 0 l BOUNDARY l < 0 BULGE hr h t SMOOTH CONDUCTOR

31 Naval Postgraduate School Antennas & Propagation Distance Learning Diffraction (2)

Etot A plot of shows that at 0.6F1 the free space (direct path) value is obtained. Edir

SHADOW BOUNDARY

0 Er r Edir FREE SPACE in dB FIELD VALUE

-5 -6

l < 0 l > 0

-10 0 0.6F1 CLEARANCE DISTANCE, l > 0 32 Naval Postgraduate School Antennas & Propagation Distance Learning Path Clearance Example

Consider a 30 mile point-to-point communication link over the ocean. The frequency of operation is 5 GHz and the antennas are at the same height. Find the lowest height that provides the same field strength as in free space. Assume standard atmospheric conditions.

The geometry is shown below (distorted The maximum bulge occurs at the midpoint. scale). The bulge factor (in feet) is given d » d + d d d t r approximately by b = t r , where d (15)(15) 1.5k t b = =112.5 ft max (1.5)(4/3) and d r are in miles. nd d 0.6F t r TX 1 RX Fn = 72.1 ft fGHzd d ht b bmax hr 0.6F1 = 53 ft

dt dr Compute the minimum antenna height: h = b + 0.6F Re¢ max 1 =112.5 + 53 =165 ft

33 Naval Postgraduate School Antennas & Propagation Distance Learning Example of Link Design (1)

34 Naval Postgraduate School Antennas & Propagation Distance Learning Example of Link Design (2)

35 Naval Postgraduate School Antennas & Propagation Distance Learning Antennas Over a Spherical Earth

When the transmitter to receiver distance becomes too large the flat Earth approximation is no longer accurate. The curvature of the surface causes: 1. divergence of the power in the reflected wave in the interference region 2. diffracted wave in the shadow region (note that this is not the same as a ground wave)

The distance to the horizon is dt = RRH » 2Re¢ht or, if ht is in feet, dt » 2ht miles. The maximum LOS distance between the transmit and receive antennas is

dmax = dt + dr » 2ht + 2hr (miles)

TANGENT RAY INTERFERENCE (SHADOW BOUNDARY) REGION

hr DIFFRACTION REGION dr ht dt

SMOOTH R¢ e CONDUCTOR

36 Naval Postgraduate School Antennas & Propagation Distance Learning Interference Region Formulas (1)

Interference region formulas

Ro

R2 R1 h y y r ht dr dt

SMOOTH Re¢ CONDUCTOR

The path-gain factor is given by

F = 1 + r e jfG e- jkDR D

where D is the divergence factor (power) and DR = R1 + R2 - Ro .

37 Naval Postgraduate School Antennas & Propagation Distance Learning Interference Region Formulas (2)

Approximate formulas1 for the interference region: 1 ì 2 2 éfG - kDRùü2 F = í(1+ G D) - 4G D sin ê úý î ë 2 ûþ where -1 2h h h + h é 4S S 2T ù DR = 1 2 J(S,T), tany = 1 2 K(S,T ), D = 1+ 1 2 (power) ê 2 ú d d ë S(1- S2 )(1+ T )û d1 d2 S1 = , S2 = where h1 is the smallest of either ht or hr 2Re¢h1 2Re¢h2

d S1T + S2 S = = , T = h1 /h2 (< 1 since h1 < h2) 2Re¢h1 + 2Re¢h2 1+ T 2 2 2 2 2 (1 - S1 ) + T (1- S2 ) J(S,T ) = (1- S1 )(1- S2 ), and K(S,T ) = 1 + T 2

1D. E. Kerr, Propagation of Short Radio Waves, Radiation Laboratory Series, McGraw-Hill, 1951 (the formulas have been reprinted in many other books including R. E. Collin, Antennas and Radiowave Propagation, McGraw-Hill, 1985).

38 Naval Postgraduate School Antennas & Propagation Distance Learning Interference Region Formulas (3)

The distances can be computed from d = d1 + d2 and

d æ F + p ö æ 2R¢ (h - h )d ö d = + p cos , F = cos-1ç e 1 2 ÷, 1 ç ÷ ç 3 ÷ 2 è 3 ø è p ø and 1/ 2 2 é d 2 ù p = êRe¢ (h1 + h2 ) + ú 3 ëê 4 ûú Another form for the phase difference is 2kh h kDR = 1 2 (1- S 2 )(1- S 2 ) =nzp d 1 2 where 3 / 2 3 / 2 4h1 h1 h2 / h1 2 2 n = = , z = (1- S1 )(1- S2 ), l 2Re¢ 1030l d / d RH

and dRH = 2Re¢h1 (distance to the radio horizon).

39 Naval Postgraduate School Antennas & Propagation Distance Learning Diffraction Region Formulas (1)

DIRECT RAY SHADOW TO HORIZON BOUNDARY

h r DIFFRACTED ht d RAYS

Re¢

Approximate formulas for the diffraction region (frequencies > 100 MHz):

F =V1(X )U1(Z1)U1(Z2 )

where U 1 is available from tables or curves, Zi = hi / H (i = 1,2), X = d / L, and 1 1 2 æ (R¢ ) ö3 æ R¢ ö3 V (X ) = 2 pXe-2.02 X , L = 2ç e ÷ = 28.41l1/ 3(km), H = e = 47.55l2 / 3(m) 1 ç ÷ ç 2 ÷ è 4k ø è 2k ø

40 Naval Postgraduate School Antennas & Propagation Distance Learning Diffraction Region Formulas (2)

A plot of U1(Z)

Fig. 6.29 in R. E. Collin, Antennas and Radiowave Propagation, McGraw-Hill, 1985 (axis labels corrected)

41 Naval Postgraduate School Antennas & Propagation Distance Learning Surface Waves (1)

At low frequencies (1 kHz to about 3 MHz) the interface between air and the ground acts like an efficient waveguide at low frequencies for vertical polarization. Collectively the space wave (direct and Earth reflected) and surface wave are called the ground wave.

SURFACE WAVE

ht d hr

Re¢

The power density at the receiver is the free space value times an attenuation factor 2 Pr = Pdir 2As

where the factor of 2 is by convention. Most estimates of As are based calculations for a surface wave along a flat interface. Approximations for a flat surface are good for 1/ 3 d £ 50/( fMHz ) miles. Beyond this distance the received signal attenuates more quickly.

42 Naval Postgraduate School Antennas & Propagation Distance Learning Surface Waves (2)

kd Define a two parameters: p = (numerical distance) 2 2 2 er + (s /weo ) -1æe e w ö b = tan ç r o ÷ è s ø 1.8´104s A convenient formula is s /weo = . The attenuation factor for the ground wave fMHz 2 0.3p + -0.6 p o is approximately As = - p /2 e sin b (b £ 90 ) 2 + p + 0.6p2 Example: A CB link operates at 27 MHz with low gain antennas near the ground. Find the received power at the maximum flat Earth distance. The following parameters hold: -3 Pt = 5 W; Gt = Gr = 1; er =12 and s = 5´10 S/m. The maximum flat Earth range is 1/ 3 dmax = 50/(27) =16.5 miles. pd / l æ16.5 ö d p = = 0.25d / l = 0.0225ç ÷(1000) » 601 ® = 4p 122 + (90 / 27)2 è 0.62ø l

43 Naval Postgraduate School Antennas & Propagation Distance Learning Surface Waves (3)

Check b to see if formula applies (otherwise use the chart on the next page)

æ (12)(8.85´10-12 )(2p )(27 ´106 )ö b = tan-1ç ÷ = 74.5o ç -3 ÷ è 5 ´10 ø

Attenuation constant

2 + 0.3p -0.6 p -4 As = - p / 2 e sin b » 8.33´10 2 + p + 0.6p2

The received power for the ground wave is

2 2 PtGt Aer 2 Pt (1)(l /4p ) 2 Pr = Pdir 2As = 2As = 2As 4pd 2 4p d 2 (5)(8.33´10-4 )2 = = 1.52 ´10-14 W (4p )(4)2 (601)2

44 Naval Postgraduate School Antennas & Propagation Distance Learning Surface Waves (4)

FLAT EARTH

Fig. 6.35 in R. E. Collin, Antennas and Radiowave Propagation, McGraw-Hill, 1985

45 Naval Postgraduate School Antennas & Propagation Distance Learning Ground Waves (5)

-2 SPHERICAL EARTH (er = 15 and s = 10 S/m)

Fig. 6.36 in R. E. Collin, Antennas and Radiowave Propagation, McGraw-Hill, 1985

46 Naval Postgraduate School Antennas & Propagation Distance Learning Urban Propagation (1)

Urban propagation is a unique and relatively new area of study. It is important in the design of cellular and mobile communication systems. A complete theoretical treatment of propagation in an urban environment is practically intractable. Many combinations of propagation mechanisms are possible, each with different paths. The details of the environment change from city to city and from block to block within a city. Statistical models are very effective in predicting propagation in this situation. In an urban or suburban environment there is rarely a direct path between the transmitting and receiving antennas. However there usually are multiple reflection and diffraction paths between a transmitter and receiver.

BASE · Reflections from objects close to the STATION ANTENNA mobile antenna will cause multiple signals to add and cancel as the mobile unit moves. Almost complete cancellation can occur resulting in “deep fades.” These small-scale (on the order of tens of

MOBILE wavelengths) variations in the signal are ANTENNA predicted by Rayleigh statistics.

47 Naval Postgraduate School Antennas & Propagation Distance Learning Urban Propagation (2)

· On a larger scale (hundreds to thousands of wavelengths) the signal behavior, when measured in dB, has been found to be normally distributed (hence referred to a lognormal distribution). The genesis of the lognormal variation is the multiplicative nature of shadowing and diffraction of signals along rooftops and undulating terrain. · The Hata model is used most often for predicting path loss in various types of urban conditions. It is a set of empirically derived formulas that include correction factors for antenna heights and terrain.

Path loss is the 1/ r 2spreading loss in signal between two isotropic antennas. From the 2 Friis equation, with Gt = Gr = 4pAe / l = 1 2 P (1)(1)l2 æ 1 ö L = r = = s 2 ç ÷ Pt (4p r) è 2krø Note that path loss is not a true loss of energy as in the case of attenuation. Path loss as defined here will occur even if the medium between the antennas is lossless. It arises because the transmitted signal propagates as a spherical wave and hence power is flowing in directions other than towards the receiver.

48 Naval Postgraduate School Antennas & Propagation Distance Learning Urban Propagation (3)

* Hata model parameters : d = transmit/receive distance (1 £ d £ 20 km) f = frequency in MHz (100 £ f £ 1500 MHz) hb = base antenna height (30 £ hb £ 200 m) hm = mobile antenna height (1 £ hm £ 10 m) The median path loss is

Lmed = 69.55 + 26.16log( f ) -13.82log(hb ) + [44.9 - 6.55log(hb )]log(d) + a(hm )

In a medium city: a(hm) = [0.7 -1.1log( f )]hm +1.56log( f ) - 0.8 2 ì1.1- 8.29log (1.54hm ), f £ 200 MHz In a large city: a(hm ) = í 2 î4.97 - 3.2log (11.75hm ), f ³ 400 MHz ì- 2log2 ( f /28) - 5.4, suburban areas Correction factors: Lcor = í 2 î- 4.78log ( f ) +18.33log( f ) - 40.94, open areas The total path loss is: Ls = Lmed - Lcor

*Note: Modified formulas have been derived to extend the range of all parameters.

49 Naval Postgraduate School Antennas & Propagation Distance Learning Urban Propagation Simulation

Urban propagation modeling using the wireless toolset Urbana (from Demaco/SAIC). Ray tracing (geometrical optics) is used along with the geometrical theory of diffraction (GTD) Closeup showing antenna placement (below)

Carrier to Interference (C/I) ratio (right)

50 Naval Postgraduate School Antennas & Propagation Distance Learning Measured Data

Two different antenna heights Measured data in h = 2.7 m h = 1.6 m an urban

f = 3.35 GHz environment

f = 8.45 GHz Three different frequencies

f = 15.75 GHz

From Masui, “Microwave Path Loss Modeling in Urban LOS Environ- ments,” IEEE Journ. on Selected Areas in Comms., Vol 20, No. 6, Aug. 2002.

51 Naval Postgraduate School Antennas & Propagation Distance Learning Attenuation Due to Rain and Gases (1)

Sources of signal attenuation in the atmosphere include rain, fog, water vapor and other gases. Most loss is due to absorption of energy by the molecules in the atmosphere. Dust, snow, and rain can also cause a loss in signal by scattering energy out of the beam.

52 Naval Postgraduate School Antennas & Propagation Distance Learning Attenuation Due to Rain and Gases (2)

53 Naval Postgraduate School Antennas & Propagation Distance Learning Attenuation Due to Rain and Gases (3)

There is no complete, comprehensive macroscopic theoretical model to predict loss. A wide range of empirical formulas exist based on measured data. A typical model:

A = aRb , attenuation in dB/km R is the rain rate in mm/hr Ea a = Ga fGHz Eb b = Gb fGHz

where the constants are determined from the following table:

-5 Ga = 6.39 ´10 Ea = 2.03 fGHz < 2.9 -5 = 4.21´10 = 2.42 2.9 £ fGHz < 54 -2 = 4.09 ´10 = 0.699 54 £ fGHz < 180

Gb = 0.851 Eb = 0.158 fGHz < 8.5 = 1.41 = -0.0779 8.5 £ fGHz < 25 = 2.63 = -0.272 25 £ fGHz <164

54 Naval Postgraduate School Antennas & Propagation Distance Learning Ionospheric Radiowave Propagation (1)

The ionosphere refers to the upper regions of the atmosphere (90 to 1000 km). This region is highly ionized, that is, it has a high density of free electrons (negative charges) and positively charged ions. The charges have several important effects on EM propagation:

1. Variations in the electron density (N e) cause waves to bend back towards Earth, but only if specific frequency and angle criteria are satisfied. Some examples are shown below. Multiple skips are common thereby making global communication possible.

Nemax 4 IONOSPHERE 3

2

1 TX SKIP DISTANCE

EARTH’S SURFACE

55 Naval Postgraduate School Antennas & Propagation Distance Learning Ionospheric Radiowave Propagation (2)

2. The Earth’s magnetic field causes the ionosphere to behave like an anisotropic medium. Wave propagation is characterized by two polarizations (“ordinary” and “extra- ordinary” waves). The propagation constants of the two waves are different. An arbitrarily polarized wave can be decomposed into these two polarizations upon entering the ionosphere and recombined on exiting. The recombined wave polarization will be different that the incident wave polarization. This effect is called Faraday rotation.

The electron density distribution has the general characteristics shown on the next page. The detailed features vary with · location on Earth, · time of day, · time of year, and · sunspot activity.

The regions around peaks in the density are referred to as layers. The F layer often splits into the F1 and F2 layers.

56 Naval Postgraduate School Antennas & Propagation Distance Learning Electron Density of the Ionosphere

(Note unit is per cubic centimeter)

57 Naval Postgraduate School Antennas & Propagation Distance Learning The Earth’s Magnetosphere

58 Naval Postgraduate School Antennas & Propagation Distance Learning Ionospheric Radiowave Propagation (3)

Relative dielectric constant of an ionized gas (assume electrons only): w 2 e =1 - p r w(w - jn) where: n = collision frequency (collisions per second) 2 N ee w p = , plasma frequency (radians per second) meo 3 Ne = electron density (/ m ) e = 1.59´10-19 C, electron charge m = 9.0´10-31 kg, electron mass For the special case of no collions, n = 0 and the corresponding propagation constant is

2 w p kc = w moe reo = ko 1- w 2

where ko = w moeo .

59 Naval Postgraduate School Antennas & Propagation Distance Learning Ionospheric Radiowave Propagation (4)

Consider three cases: - jkc z - j kc z 1. w > w p : kc is real and e = e is a propagating wave - jkc z - kc z 2. w < w p : kc is imaginary and e = e is an evanescent wave 3. w = w p : kc = 0 and this value of w is called the critical frequency, wc

At the critical frequency the wave is reflected. Note that wc depends on altitude because the electron density is a function of altitude. For electrons, the highest frequency at which a reflection occurs is

wc REFLECTION fc = » 9 Ne max w = wc Þ e r = 0 POINT 2p Reflection at normal incidence requires IONOSPHERE h¢ the greatest Ne .

TX EARTH’S 1 SURFACE The critical frequency is where the propagation constant is zero. Neglecting the Earth’s magnetic field, this occurs at the plasma frequency, and hence the two terms are often used interchangeably.

60 Naval Postgraduate School Antennas & Propagation Distance Learning Ionospheric Radiowave Propagation (5)

At oblique incidence, at a point of the ionosphere where the critical frequency is fc , the ionosphere can reflect waves of higher frequencies than the critical one. When the wave is incident from a non-normal direction, the reflection appears to occur at a virtual reflection point, h¢, that depends on the frequency and angle of incidence.

VIRTUAL HEIGHT

IONOSPHERE h¢

EARTH’S SURFACE TX SKIP DISTANCE

61 Naval Postgraduate School Antennas & Propagation Distance Learning Ionospheric Radiowave Propagation (6)

To predict the bending of the ray we use a layered approximation to the ionosphere just as we did for the troposphere.

M

LAYERED er(z3) z3 y IONOSPHERE 3 e (z ) z2 y2 r 2 APPROXIMATION ALTITUDE z er (z1) 1 y1

yi er = 1

Snell’s law applies at each layer boundary

siny i = sin(y1 ) e r (z1) =L

The ray is turned back when y (z) = p /2, or siny i = e r (z)

62 Naval Postgraduate School Antennas & Propagation Distance Learning Ionospheric Radiowave Propagation (7)

Note that:

1. For constant yi , N e must increase with frequency if the ray is to return to Earth (because er decreases with w). 2. Similarly, for a given maximum N e (Ne max ), the maximum value of yi that results in the ray returning to Earth increases with increasing w. There is an upper limit on frequency that will result in the wave being returned back to Earth. Given Ne max the required relationship between yi and f can be obtained

siny i = er (z) 2 2 w p sin y i = 1- w 2 2 81Ne max 1- cos y i = 1- f 2 f 2 cos2y 81N N = i Þ f = e max e max max 2 81 cos y i

63 Naval Postgraduate School Antennas & Propagation Distance Learning Ionospheric Radiowave Propagation (8)

Examples:

o 10 3 10 2 1. y i = 45 , N e max = 2´10 / m : fmax = (81)(2 ´10 ) /(0.707) =1.8 MHz o 10 3 10 2 2. y i = 60 , N e max = 2´10 / m : fmax = (81)(2´10 ) /(0.5) = 2.5 MHz

The value of f that makes er = 0 for a given value of Ne max is the critical frequency defined earlier: fc = 9 Ne max

Use the Ne max expression from previous page and solve for f

f = 9 Ne max secy i = fc secy i

This is called the secant law or Martyn’s law. When secy i has its maximum value, the frequency is called the maximum usable frequency (MUF). A typical value is less than 40 MHz. It can drop as low as 25 MHz during periods of low solar activity. The optimum usable frequency (OUF) is 50% to 80% of the MUF.

64 Naval Postgraduate School Antennas & Propagation Distance Learning Maximum Usable Frequency

The maximum usable frequency (MUF) in wintertime for different skip distances. The MUF is lower in the summertime.

Fig. 6.43 in R. E. Collin, Antennas and Radiowave Propagation, McGraw-Hill, 1985

65 Naval Postgraduate School Antennas & Propagation Distance Learning Ionospheric Radiowave Propagation (9)

Multiple hops allow for very long range communication links (transcontinental). Using a simple flat Earth model, the virtual height (h¢), incidence angle (yi ), and skip distance (d ) d are related by tany = . This implies that the wave is launched well above the horizon. i 2h¢ However, if a spherical Earth model is used and the wave is launched on the horizon then d = 2 2Re¢h¢.

EFFECTIVE SPECULAR REFLECTION POINT

IONOSPHERE IONOSPHERE

h ¢

yi TX EARTH’S SURFACE d

Single ionospheric hop Multiple ionospheric hops (flat Earth) (curved Earth)

66 Naval Postgraduate School Antennas & Propagation Distance Learning Ionospheric Radiowave Propagation (10)

Approximate virtual heights for layers of the ionosphere

Layer Range for h¢ (km) F2 250 to 400 (day) F1 200 to 250 (day) F 300 (night) E 110

Example: Based on geometry, a rule of thumb for the maximum incidence angle on the ionosphere is about 74o. The MUF is

o MUF = fc sec(74 ) = 3.6 fc 12 3 For Ne max =10 / m , fc » 9 MHz and the MUF = 32.4 MHz. For reflection from the F2 layer, h¢ » 300 km. The maximum skip distance will be about

3 3 d max » 2 2Re¢h¢ = 2 2(8500 ´10 )(300 ´10 ) = 4516 km

67 Naval Postgraduate School Department of Electrical & Computer Engineering Monterey, California Ionospheric Radiowave Propagation (7)

1 + h¢/R¢ - cosq 1 For a curved Earth, using the law of sines for a triangle e = sinq tany i where d q = y 2Re¢ R/2 i R/2 h¢ and the launch angle (antenna pointing angle above the horizon) D is o o D = f -90 = 90 -q -y i d / 2 f LAUNCH ANGLE: The great circle path via the o o D = 90 -q -yi = f - 90 reflection point is R, which can be obtained from Re¢ 2R¢ sin q R = e q siny i

68 Naval Postgraduate School Department of Electrical & Computer Engineering Monterey, California Ionospheric Radiowave Propagation (8)

Example: Ohio to Europe skip (4200 miles = 6760 km). Can it be done in one hop? To estimate the hop, assume that the antenna is pointed on the horizon. The virtual height required for the total distance is

d / 2 = Re¢q ® q = d /(2Re¢ ) = 0.3976 rad = 22.8 degrees (Re¢ + h¢)cosq = Re¢ ® h¢ = Re¢ /cosq - Re¢ = 720 km This is above the F layer and therefore two skips must be used. Each skip will be half of the total distance:. Repeating the calculation for d / 2 =1690 km gives

q = d /(2Re¢ ) = 0.1988 rad = 11.39 degrees h¢ = Re¢ /cosq - Re¢ =171 km This value lies somewhere in the F layer. We will use 300 km (a more typical value) in computing the launch angle. That is, still keep d / 2 =1690 km and q = 11.39 degrees, but point the antenna above the horizon to the virtual reflection point at 300 km 300 -1 o é o ù o tany i = sin(11.39 ) 1 + - cos(11.39 ) ® y i = 74.4 ëê 8500 ûú

69 Naval Postgraduate School Department of Electrical & Computer Engineering Monterey, California Ionospheric Radiowave Propagation (9)

The actual launch angle required (the angle that the antenna beam should be pointed above the horizon) is

o o o o o launch angle, D = 90 -q -y i = 90 -11.39 - 74.4 = 4.21 11 3 The electron density at this height (see chart, p.3) is N e max » 5´10 /m which corresponds to the critical frequency

fc » 9 Ne max = 6.36 MHz and a MUF of

MUF » 6.36sec74.4o = 23.7 MHz Operation in the international short wave 16-m band would work. This example is oversimplified in that more detailed knowledge of the state of the ionosphere would be necessary: time of day, time of year, time within the solar cycle, etc. These data are available from published charts.

70 Naval Postgraduate School Department of Electrical & Computer Engineering Monterey, California Ionospheric Radiowave Propagation (10)

Generally, to predict the received signal a modified Friis equation is used:

Pt GtGr Pr = Lx La (4pR / l)2 where the losses, in dB, are negative: Lx = Lpol + Lrefl - Giono Lrefl = reflection loss if there are multiple hops Lpol = polarization loss due to Faraday rotation and earth reflections Giono = gain due to focussing by the curvature of the ionosphere La = absorption loss R = great circle path via the virtual reflection point

Example: For Pt = 30 dBW, f = 10 MHz, Gt = Gr = 10 dB, d = 2000 km, h¢ = 300 km, Lx = 9.5 dB and La = 30 dB (data obtained from charts).

o From geometry compute: y i = 70.3 , R = 2117.8 km, and thus Pr = -108.5 dBw

71 Naval Postgraduate School Antennas & Propagation Distance Learning Ducts and Nonstandard Refraction (1)

Ducts in the atmosphere are caused by index of refraction rates of decrease with height over short distances that cause rays to bend back towards the surface.

TOP OF DUCT

EARTH’S SURFACE TX · The formation of ducts is due primarily to water vapor, and therefore they tend to occur over bodies of water (but not land-locked bodies of water) · They can occur at the surface or up to 5000 ft (elevated ducts) · Thickness ranges from a meter to several hundred meters · The trade wind belts have a more or less permanent duct of about 1 to 5 m thickness · Efficient propagation occurs for UHF frequencies and above if both the transmitter and receiver are located in the duct · If the transmitter and receiver are not in the duct, significant loss can occur before coupling into the duct 72 Naval Postgraduate School Antennas & Propagation Distance Learning Ducts and Nonstandard Refraction (2)

Because variations in the index of refraction are so small, a quantity called the refractivity is used 6 N(h) = [n(h) - 1]10 n(h) = er (h)

In the normal (standard) atmosphere the gradient of the vertical refractive index is linear with height, dN/dh » -39 N units/km. If dN/dh < -157 then rays will return to the surface. Rays in the three Earth models are shown below.

True Earth Equivalent Earth Flat Earth (Standard Atmosphere)

Re Re¢ ¥

From Radiowave Propagation, Lucien Boithias, McGraw-Hill

73 Naval Postgraduate School Department of Electrical & Computer Engineering Monterey, California Ducts and Nonstandard Refraction (3)

Another quantity used to solve ducting problems is the modified refractivity 6 M (h) = N(h) +10 (h / Re¢ ) In terms of M, the condition for ducting is dM /dh = dN/dh +157. Other values of dN / dh (or dM /dh) lead to several types of refraction as summarized in the following figure and table. They are: 1. Super refraction: The index of refraction decrease is more rapid than normal and the ray curves downward at a greater rate 2. Substandard refraction (subrefraction): The index of refraction decreases less rapidly than normal and there is less downward curvature than normal

From Radiowave Propagation, Lucien Boithias, McGraw-Hill 74 Naval Postgraduate School Department of Electrical & Computer Engineering Monterey, California Ducts and Nonstandard Refraction (4)

Summary of refractivity and ducting conditions

Ray Atmospheric Virtual Horizontally dN / dh Curvature k Refraction Earth Launched Ray > 0 up < 1 more convex 0 none 1 below actual dN normal moves 0 > > -39 dh > 1 away -39 4/3 normal less from Earth dN > 4/3 convex - 39 > > -157 down dh above normal plane parallel to -157 Earth super-refraction concave draws closer < -157 to Earth

75 Naval Postgraduate School Antennas & Propagation Distance Learning

INDEX TO LECTURE NOTES (ver 1.2) active antenna III-74 constrained lens IV-65 adaptive antenna III-83 coordinate systems I-87 antenna temperature III-70 co-polarized III-8 aperture distributions III-27, IV-14 cosine feed directivity IV-49 aperture efficiency III-26 cosine feed efficiency IV-49 aperture field IV-4 cosine-on-a-pedestal III-25 aperture in a ground plane IV-4 coverage diagram V-13 array directivity III-21 critical frequency V-57 array edge effect III-29 cross polarization IV-62 array factor, AF III-2 cross polarization loss IV-43 arrays III-1 crossed dipoles II-60 atmospheric attenuation V-50 cross-polarized III-8 attenuation constant I-6, I-15 cylindrical coordinate system I-87 axial ratio II-61 cylindrical paraboloid IV-46 azimuth II-19 Debye model I-11 balun II-31 decibel (dB) I-86 bandwidth I-81 detection threshold III-71 basis function II-35 diffraction region V-36, V-39 Bayliss III-25 digital beamforming III-86 bazooka balun II-33 dipole feed IV-58 beam coupling loss III-81 direct V-2 beam solid angle II-13 direction cosines III-6 beamforming networks III-38 directive gain II-10 beamwidth between first nulls III-19 directivity II-10 (BWFN) biconical antenna II-53 discrete Fourier transform (DFT) III-50 binomial distribution III-25 dispersive dielectric I-13 blade antenna II-69 doubly curved surface IV-35 blockage IV-42, IV-51 duct V-4, V-69 bootlace lens IV-65 earth radius V-22 Brewster's angle I-31 edge diffraction III-29 broadband antennas II-51 edge taper, edge illumination IV-44 bulge factor V-33 effective aperture III-57 capacitance I-45 effective area III-57 cascaded devices III-73 effective height II-64 Cassegrain IV-36, IV-53 effective isotropic radiated power V-17 chamber configurations IV-77 effective radiated power (ERP) V-17 characteristic impedance I-45, I-51 effective temperature III-69 Chebyshev III-24 efficiency II-17 choke II-32 electron charge V-57 circular polarization I-20 electron density V-54, V-57 circulator I-79 electron mass V-57 classification of antennas II-1 element factor, EF III-2 clutter III-65 elevation II-19 coaxial cable I-41 elliptical polarization II-61 collision frequency V-57 entire domain II-37 conductance I-45 E-plane III-7 conductivity I-3 E-plane horn IV-18 conjugate matched antenna III-58 equivalence principle IV-1 1 equivalent currents IV-1 illumination efficiency III-26 equivalent earth radius V-23, V-27 image II-57 equivalent noise temperature III-69 impedance I-8, I-45, I-37 excitation vector II-39 impedance matching I-55 expansion function II-35 impedance matrix II-39 false alarm III-71 incident fields IV-2 far field, far zone II-3, II-4 inductance I-45 fast Fourier transform (FFT) III-51 integral equation (IE) II-35 feeding antennas II-29 interference region V-36, V-37 filament of current II-7 ionosphere V-1 filter I-83 ionospheric hop V-3, V-53 flat earth approximation limit V-36, V-41 knife edge diffraction V-31 flux tube, ray bundle IV-39 launch angle V-66 focused beam III-24 layers V-54 Fourier series II-41 lens antenna IV-63 frequency bands I-89, V-6 linear array III-3 frequency spectrum I-90 linear phase progression III-24 Fresnel reflection I-28 link equation III-60 Fresnel zone V-28 log periodic antenna II-55 Friis transmission equation III-60 lognormal distribution V-47 front-to-back ratio II-56 loop antenna II-71 gain II-17 loss I-64 geometric series III-4 lossy medium I-14 geometrical optics I-27, IV-40 Love's equivalence principle IV-3 Gladstone-Dale constant V-22 low sidelobe III-23 GPS antenna III-76 lumped elements I-61 grating lobe III-15 Luneberg lens IV-66 grazing angle V-9 magic tee I-78 ground plane II-58 magnetic current II-5 ground wave V-2, V-41 magnetic vector potential II-3 group velocity I-72, I-77 magnetosphere V-56 half wave dipole II-23 main lobe, main beam III-4 half wave dipole, directivity II-25 Martyn's law V-62 half wave dipole, radiated power II-25 matched load I-48,I-54 HARM IV-71 maximum usable frequency (MUF) V-62 Hata model V-47 Maxwell's equations I-3 height-gain curve V-15 measurement IV-75 helix antenna II-74 method of images II-57 Hertzian (short) dipole II-6 method of moments (MM) II-34 Hertzian dipole, directivity II-10 microstrip I-41 Hertzian dipole, gain II-17 microstrip antenna IV-27 Hertzian dipole, HPBW II-11 miss III-71 Hertzian dipole, radiated power II-8 modes I-70 Hertzian dipole, radiation pattern II-11 modulation I-82 Hertzian dipole, radiation resistance II-17 monopole II-59 hog-horn antenna IV-56 monostatic radar I-2, III-62 horizon, distance to V-24 multipath V-2 horizontal polarization V-18 multipath, flat earth V-8 horn antenna IV-18 multiple beam antennas III-78 H-plane III-7 multiplexer I-84 H-plane horn IV-18 mutual impedance II-44, II-47 Huygen's principle III-9 noise III-65 2 noise bandwidth III-67 radiation integral II-2 noise figure III-72 radiation intensity II-10 noise power III-68 radio (radar) horizon V-24 nonstandard refraction V-69 radome IV-68 null-field hypothesis IV-51 rain attenuation V-50 nulls III-4 random roundoff III-46 numerical distance V-42 Rascal IV-53 numerical integration II-26 ray tracing IV-39 offset reflector IV-37, IV-55 Rayleigh-Jeans approximation III-68 ohmic loss II-17 receiving antennas III-57 open circuit I-48,I-54 reciprocity II-42 open-ended waveguide IV-10 reciprocity theorem II-44 optimum horn IV-24 rectangular aperture IV-6 optimum usable frequency (OUF) V-62 reflected V-2 orthogonal beams III-81 reflecting systems IV-36 parabolic reflector IV-40 reflection coefficient I-30, I-47 parallel feed III-38 reflection loss II-30 parasitic II-56 reflector antenna IV-34 passive antenna III-74 refraction V-22 patch antenna IV-27 refractive modulus V-71 path loss V-47 refractivity V-70 path-gain factor (PGF) V-10 resistance I-45 pattern multiplication III-2 rotating linear polarization II-62 pattern propagation factor (PPF) V-10 rough surface V-18 perfect electric conductor (PEC) II-2 roundoff error III-46 perfect magnetic conductor (PMC) IV-5 scanned aperture IV-16 permeability I-3 scanned array III-13 permittivity I-3 scattered fields IV-2 phase constant I-6, I-15, secant law V-62 I-73, I-77 phase delay III-11 self impedance II-40 phase shifter III-48 series feed III-37 phase velocity I-73,I-77 shadow boundary V-31 phasor I-4 short circuit I-48,I-54 piecewise sinusoid II-36 sidelobe control III-23 planar array III-33 sidelobes III-4 plane waves I-6 signal-to-noise ratio (SNR) III-71 Plank's blackbody law III-67 sinc function, sin(x)/x IV-9 plasma frequency V-57 singly curved surface IV-35 polarization diversity (reuse) III-8 skin depth I-17, I-39 polarization loss factor, PLF II-64, II-68 skip distance V-59 power density I-10 smart antenna III-84 power divider I-63 smooth surface V-18 Poynting theorem I-10 space feed III-38 principal planes III-7 sphere diffraction V-32 propagation V-1 spherical coordinate system I-87 propagation constant I-6,I-15,I-44 spherical wave I-23 propagation mechanisms V-1 spillover IV-42 pyramidal horn IV-19 spiral antenna II-53 quadratic phase error IV-20 standard atmospheric conditions V-27 quarter wave transformer I-56 standing wave I-48 radar cross section III-64 steradian II-9 radar range equation (RRE) III-62 Stratton-Chu (radiation) integral II-2

3 stub tuning I-58 subdomain II-36 super refraction V-70 surface current I-16 surface duct V-69 surface resistivity I-17 surface wave V-21 switch I-80 system noise temperature III-69 taper III-24 Taylor III-25 TE10 mode IV-10 TEM wave I-8, I-24 terrain diffraction V-4 thermal noise III-67 thin wire antennas, patterns II-28 thin wires, current distribution II-22 thin-wire approximation II-6 time delay III-11, III-48 total internal reflection I-31 transfer impedance II-44 transformations I-87 transmission line I-41 transmission line circuit I-47 transmission line equation I-42 transverse electric (TE) I-68, I-71 transverse magnetic (TM) I-68, I-69 troposcatter V-3 troposphere V-1 turnstile antenna II-60 ultra-low sidelobe III-23 uniqueness theorem II-58 urban propagation V-46 vertical polarization V-18 virtual height V-58 visible region III-5 voltage minimum/maximum I-50 VSWR II-30 wave equation I-4 wave equation, vector II-3 wave impedance I-72, I-77 waveguide radiation IV-10 waveguides I-66 wavelength in waveguide I-73, I-77 white noise III-67 yagi-uda II-56

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