Communication Transmission Systems Course

Home , Horn antenna, Isotropic radiator, Phased array, Slot antenna

Part IV: Antennas and Waveguides

1.1 An antenna-Introduction: o A metallic conductor capable of radiating and capturing EM energy. o The interface between the transmission line and the atmosphere o Converts the electrical energy travelling along the TL into an EM wave at the transmit end and vice versa.  Antenna Parameters: o Radiation Pattern: . It is a polar graph representing the field strength/power densities at various angular positions relative to the antenna. . There can be more than one major lobe in the radiation pattern. . Major lobe propagates and receives the most energy (it called front lobe i.e. front of the antenna) . Front-to-Back/Side ratio: it is the ratio of the front lobe power to the back/side lobe power.

o Antenna Types: . Isotropic pattern: The radiation pattern is the same in all directions.  Antennas with isotropic radiation patterns don't exist in practice, but are sometimes discussed as a means of comparison with real antennas. . Omnidirectional Antenna: the radiation pattern of an actual antenna is isotropic in a single plane such as dipole antenna and slot antenna. . Directional Antenna: Do not have symmetry in the radiation pattern. These antennas typically have a single peak direction in the radiation pattern; this is the direction where the bulk of the radiated power travels. Such as dish antenna and slotted waveguide antenna.

o Field Regions:

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Part IV: Antennas and Waveguides

. Near Field (NF): Refers to the field pattern that is close to the antenna.  Reactive NF: Region surrounding the antenna, wherein the reactive field predominates. o Angular Field distribution depends on distance from antenna. 퐷3 > 휆 푅 < 0.62√ 퐷 휆 휆 < 휆 푅 < { 2휋

퐷: Largest antenna dimension 푅: Region Radius

 Radiating NF: Region between reactive near-field and far-field regions (Fresnel zone). o Radiating fields predominate but angular field distribution still depends on distance from antenna. 퐷3 2퐷2 > 휆 0.62√ < 푅 < 퐷 휆 휆 휆 < 휆 < 푅 < 3휆 { 2휋

. Far Field: Refers to the field pattern that is at great distance  Region where angular field distribution is essentially independent of the distance from antenna (Fraunhofer zone) 2퐷2 > 휆 푅 > 퐷 { 휆 < 휆 푅 > 3휆

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Part IV: Antennas and Waveguides

o Radiation Resistance: . Antenna resistance that equals to the ratio of the power radiated by the antenna to the square of the current at its feed input.

푃 푅 = 푟푎푑 Ω 푟 푖2 푃푟푎푑: Radiated power (Watts) 1 푃 = ∯ 푅푒[퐸⃑ × 퐻⃑⃑ ∗]. 휕푆 = ∯ 푊⃑⃑⃑⃑⃑⃑⃑ . 휕푆 푟푎푑 2 푆 푆 푎푣 푊푎푣: The average power density 휕푆: 푛̂휕푎 휕푎: Infinitesimal area of the closed surface =푟2sin (휃)휕휃휕휑 (m2) 푛̂: Unit vector normal to the surface 푖: Antenna current at the feed point (Ampere)

 Example 1: Find the total radiated power for an antenna has an average power sin (휃) density (푊⃑⃑⃑⃑⃑⃑⃑ ) of = 푟̂퐴 푊/푚2. 푎푣 표 푟2

Solution: 1 푃 = ∯ 푅푒[퐸⃑ × 퐻⃑⃑ ∗]. 휕푆 = ∯ 푊⃑⃑⃑⃑⃑⃑⃑ . 휕푆 푟푎푑 2 푆 푆 푎푣 2휋 휋 sin(휃) 푃 = ∯ 푊⃑⃑⃑⃑⃑⃑⃑ . 휕푆 = ∫ ∫ (푟̂퐴 ) . 푟 푟2 sin(휃) 휕휃휕휑 푟푎푑 푆 푎푣 0 0 표 푟2 2 푃푟푎푑 = 휋 퐴표 푊  Example 2: Find the total radiated power for an isotropic antenna has an 2 average power density (푊⃑⃑⃑⃑⃑푎푣⃑⃑ ) of = 푟̂푊표(푟) 푊/푚 .

Solution: 1 푃 = ∯ 푅푒[퐸⃑ × 퐻⃑⃑ ∗]. 휕푆 = ∯ 푊⃑⃑⃑⃑⃑⃑⃑ . 휕푆 푟푎푑 2 푆 푆 푎푣 2휋 휋 푃 = 푊⃑⃑⃑⃑⃑⃑⃑ . 휕푆 = (푟̂푊 (푟)). 푟 푟2 sin(휃) 휕휃휕휑 푟푎푑 ∯푆 푎푣 ∫0 ∫0 표 2 푃푟푎푑 = 4휋푟 푊표 푊 o Radiation Intensity: . The power radiated from an antenna per unit solid angle.

퐸⃑ (푟, 휃, 휑): far zone electric field intensity of the antenna 휂: interstice impedance (120377)

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Part IV: Antennas and Waveguides

 Example 1: Find the total radiated power for an antenna has a radiation 2 intensity 푈(휃, 휑) of 푈(휃, 휑) = 푟 푊푟푎푑 = 퐴표 sin(휃).

Solution:

2휋 휋 푃 = 퐴 sin(휃) . sin(휃) 휕휃휕휑 푟푎푑 ∫0 ∫0 표 2 푃푟푎푑 = 휋 퐴표 푊

 Example 2: Find the total radiated power for an isotropic antenna has a 2 radiation intensity 푈(휃, 휑) of 푈(휃, 휑) = 푟 푊푟푎푑 = 푈표.

Solution:

2휋 휋 푃 = 푈 . sin(휃) 휕휃휕휑 푟푎푑 ∫0 ∫0 표 푃푟푎푑 = 4휋푈표 푊

o Radiation Efficiency: . It is the ratio of the power radiated by an antenna to the sum of the power radiated and the power dissipated.

푃푟푎푑 휉 = × 100% , 푃𝑖푛 = 푃푟푎푑 + 푃푑 푃𝑖푛 푃푟푎푑: Radiated power (W) 푃푑: Dissipated power (W) 푃𝑖푛: Input power (W)

2 푃푟푎푑 = 푅푟푖 . 2 푃푑 = 푅푒푖 . 푅 휉 = 푟 × 100% 푅푟 + 푅푒 푅푟: Radiation resistance 푅푒: Effective antenna resistance

. Takes into account the physical losses of the antenna such as metal and dielectric losses, as well as mismatch between the connection of the antenna and the transmission line.

2 휉 = 푒푟푒푐푒푑 = (1 − |Γ| )푒푐푒푑 푒푟: Reflection efficiency 푒푐: Conductor efficiency 푒푑: Dielectric efficiency Γ: Reflection coefficient

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Part IV: Antennas and Waveguides

o Beamwidth: . It is the angular separation between two identical points on opposite site of the pattern maximum. . Typically, it is between 30o and 60o. . Half-power beamwidth (HPBW): in a plane containing the direction of the maximum of a beam, the angle between the two directions in which the radiation intensity is one-half value of the beam . First-Null beamwidth (FNBW): angular separation between the first nulls of the pattern

 Example 1: For the normalized radiation intensity of an antenna is represented by 푈(휃) = cos2(휃), (0 ≤ 휃 ≤ 휋, 0 ≤ 휑 ≤ 2휋). Find the angle which represents the HPBW; 휃ℎ. Then find the angle which represents the FNBW; 휃푛

Solution:

1- For the HPBW 2 푈(휃)휃=휃 = cos (휃ℎ) = 0.5 ℎ 휋 휃 = cos−1 √0.5 = ℎ 4 Since the pattern is symmetric with respect to the 휋 maximum, then HPBW=2휃 = ℎ 2

2- For the FNBW ( ) 2( ) 푈 휃 휃=휃ℎ = cos 휃ℎ = 0 −1 휃ℎ = cos 0 = 휋 Since the pattern is symmetric with respect to the maximum, 휋 then HPBW=2휃 = 푛 2

o Bandwidth: . It is the frequency range over which antenna operation is satisfactory.

푓 − 푓 퐵푊 = 퐻 퐿 × 100% 푓표 푓퐻: Highest frequency of operation 푓퐿: Lowest frequency of operation 푓표: Optimum frequency of operation

 Example 1: Determine the percent bandwidth for an antenna which has an optimum frequency of 600MHz, and a -3dB frequencies of 570MHz and 630MHz.

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Part IV: Antennas and Waveguides

Solution:

푓 − 푓 630 − 570 퐵푊 = 퐻 퐿 × 100% = × 100% = 10% 푓표 600

o Directivity: . Ratio of radiation intensity in a given direction from the antenna to the average radiation intensity 4휋푈(휃, 휑) 4휋푈(휃, 휑) 퐷(휃, 휑) = = 푃푟푎푑 푃 = 푈(휃, 휑). 휕Ω 푟푎푑 ∯Ω

4휋 퐷(휃, 휑)푚푎푥 = Ω퐴 Ω퐴: Beam Solid angle (all the power of the antenna would flow if its radiation intensity were constant and equals to 푈푚푎푥)

. The average radiation intensity equals to the radiation intensity of an isotropic source. . If the direction is not specified, it implies the directivity of maximum radiation intensity (maximum directivity)

4휋푈푚푎푥 퐷푚푎푥 = 퐷표 = 푃푟푎푑  Example 1: If the radial component of the radiated power density of an infinitesimal linear dipole is given by 푠푖푛2휃 푊 = 푟̂퐴 푊/푚2 푎푣 표 푟2 Then calculate the maximum directivity.

Solution:

2 2 - The radiation intensity is 푈 = 푟 푊푟 = 퐴표푠푖푛 휃 휋  The maximum radiation is directed along 휃 = 2  푈푚푎푥 = 퐴표 - The total radiated power 푃 = 푈(휃, 휑). 휕Ω 푟푎푑 ∯Ω 2휋 휋 2 8휋 푃푟푎푑 = 퐴표 ∫ ∫ 푠푖푛 휃. 푠푖푛휃 휕휃휕휑 = 퐴표 0 0 3 4휋푈푚푎푥 - The maximum directivity 퐷푚푎푥 = 퐷표 = 푃푟푎푑 3 퐷 = 표 2 o Gain: . The ratio of radiation intensity in a given direction to the average radiation intensity that would be obtained if all the power input to the antenna were radiated isotropic. . It takes into account the efficiency of the antenna as well as its directional properties.

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Part IV: Antennas and Waveguides

4휋푈(휃, 휑) 4휋푈(휃, 휑) 퐺(휃, 휑) = = 푒푐푒푑 = 푒푐푒푑퐷(휃, 휑) 푃𝑖푛 푃푟푎푑 2 푃𝑖푛: Input power to antenna = (1 − |Γ| )푃표 = 푃푟푎푑 + 푃푙표푠푠

. Directive/Relative Gain: Ratio of the power density radiated in a particular direction to the power density radiated to the same point by a reference antenna, assuming both antenna are radiating the same amount of power.

. Power Gain (Ap): it is the same as directive gain except that the total power fed to the antenna is used. It is assumed that the given antenna and the reference antenna have the same input power and the reference antenna is lossless. (휉 = 100%, 푃푟푎푑 = 푃𝑖푛, 푃푙표푠푠푒푠 = 0)

퐴푝 = 휉퐷

 Example 1:

A lossless resonant half-wave dipole with an impedance of 73Ωis connected to a transmission line with characteristic impedance of 50Ω. If the radiation pattern is given by 3 푈 = 퐵표푠푖푛 휃

Then calculate the maximum realized gain of the antenna.

Solution:

푈푚푎푥 = 퐵표 - The total radiated power 2휋 휋 2 3 3휋 푃푟푎푑 = 퐵표 ∫ ∫ 푠푖푛 휃. 푠푖푛휃 휕휃휕휑 = 퐵표 ( ) 0 0 4 4휋푈푚푎푥 - The maximum directivity 퐷푚푎푥 = 퐷표 = 푃푟푎푑 퐷표 = 1.697 - Since the antenna is lossless, then 푒푐푒푑 = 1 - 퐺표 = 푒푐푒푑퐷표 = 1.697 - 퐺표(푑퐵) = 10 log(1.697) = 2.297

- With the realized gain, we can take into account the mismatch losses between the TL and the antenna impedance as: 73−50 2 푒 = (1 − |Γ|2) = (1 − | | ) = 0.965 푟 73+50 푒푟(푑퐵) = 10 log(0.965) = −0.155  푒표 = 푒푐푒푑푒푟 = 0.965 = −0.155 푑퐵 퐺표푟푒푎푙𝑖푧푒푑 = 푒푟푒푐푒푑퐷표 = 1.6376 = 2.142 푑퐵

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Part IV: Antennas and Waveguides

o Effective Isotropic Radiation Power (EIRP): . It is equivalent to the equivalent transmit power.

퐸퐼푅푃 = 푃푟푎푑퐷 = 푃𝑖푛퐴푝, 푊푎푡푡 o Power Density: . It is given by

푃𝑖푛퐴푝 푃 퐷 푃 = = 푟푎푑 = 퐸퐼푅푃 − 퐿 , 푊/푚2 4휋푅2 4휋푅2 푝푎푡ℎ  Example 1: Given the free-space radio transmission system with the following specifications: - Transmitter power out = 40 dBm - Transmission line loss = 3 dB - Free-space path loss (퐿푝푎푡ℎ)= 50 dB Determine: a) The antenna input power b) The antenna radiated power c) The EIRP d) The receive power density for an isotropic antenna with a directivity of 1 and efficiency of 100%. e) Repeat the solution if 퐷 = 10 and 휉 = 50%

Solution: a) 푃𝑖푛 = 푃표푢푡 − 푃퐿표푠푠푒푠 = 40푑퐵푚 − 3푑퐵 = 37푑퐵푚 b) 푃푟푎푑(푑퐵) = 푃𝑖푛 − 휉 = 37푑퐵푚 + 0 = 37푑퐵푚 c) 퐸퐼푅푃휉=100% = 푃푟푎푑 × 퐷 = 푃푟푎푑 × 퐴푝 퐸퐼푅푃 = 37푑퐵푚 + 10 log(1) = 37푑퐵푚 휉=100%푑퐵

푃 퐴 푃 퐷 d) 푃 = 𝑖푛 푝 = 푟푎푑 = 퐸퐼푅푃 − 퐿 = 37푑퐵푚 − 50푑퐵 = −13푑퐵푚 4휋푅2 4휋푅2 푝푎푡ℎ e) Homework

1.2 Antenna Types:  Simple Antennas: o The Isotropic Radiator would radiate all the power delivered to it and equally in all directions o The isotropic radiator would also be a point source

 Half-wave Dipole Antennas: o A more practical antenna is the half-wave dipole and the most used for frequencies > 2MHz. o Dipole simply means it is in two parts o A dipole does not have to be one-half wavelength, but that length is handy for impedance matching o A half-wave dipole is sometimes referred to as a Hertz antenna o Typically, the length of a half-wave dipole is 95% of one-half the wavelength measured in free space.

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Part IV: Antennas and Waveguides

Example: What is the Half-wave dipole antenna that will operate on 28 MHz. Solution: 푐 1 푙휆 = × × 0.95 = 5.09푚 2 푓 2 o Each pole of the antenna looks as if it were an open quarter-wavelength section of TL. . The voltage is max., the current is min at the ends. . The voltage is min., the current is max. at the middle. o Assuming the feed-point is in the center of the antenna . The input impedance is 퐸푚𝑖푛 푍푚𝑖푛 = 퐼푚푎푥 . The impedance at the ends of the antenna is 퐸푚푎푥 푍푚푎푥 = 퐼푚𝑖푛 o The impedance varies from the max. value at the ends of ≈ 2.5푘Ω to a min. value at the feed-point of ≈ 72Ω. o For an ideal antenna: . 휁 = 100% . 퐷 = 1.667 = 2.18 푑퐵 . 푅푟 = 푍𝑖푛 = 72Ω . 휃 = 78표 o The free space radiation pattern depends on whether the antenna is placed horizontally or vertically with respect to Earth’s surface.

 The Folded Dipole Antennas: o The folded dipole is the same length as a standard dipole, but is made with two parallel conductors, joined at both ends and separated by a distance that is short compared with the length of the antenna. o The folded dipole differs in that it has wider bandwidth and has approximately four times the feed-point impedance of a standard dipole. This is due to that it has one-half of the input current compared to the half-wave dipole antenna. o For a three-element folded dipole, the input impedance is 32 × 72 = 648Ω

 The Monopole Antennas: o For low- and medium-frequency transmissions, it is necessary to use vertical polarization to take advantage of ground-wave propagation o A vertical dipole would be possible, but similar results are available from a quarter-wavelength monopole antenna o Fed at one end with an unbalanced feedline, with the ground conductor of the feedline taken to earth ground

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Part IV: Antennas and Waveguides

 The Loop Antennas: o Sometimes, smaller antennas are required for certain applications, like AM radio receivers o These antennas are not very efficient but perform adequately o Two types of loop antennas are: . Air-wound loops . Ferrite-core loopsticks

 The Helical Antenna: o Several types of antennas are classified as helical o The antenna in the sketch has its maximum radiation along its long axis o A quarter-wave monopole can be shortened and wound into a helix— common in rubber ducky antenna used with many handheld transceivers

 The Antenna Array: o Simple antenna elements can be combined to form arrays resulting in reinforcement in some directions and cancellations in others to give better gain and directional characteristics o Arrays can be classified as broadside or end-fire . Examples of arrays are: . The Yagi-Uda Array  Widely used, which uses folded dipole as driven element.  The reflector is a straight rod approximately 5% longer than the dipole, which is 5% shorter than the driven element.  The spacing between elements is generally 0.1 and 0.2 wavelengths.  Typically, the directivity is between 7dB and 9dB.  It is a linear array consists of a dipole and two or more parasitic elements (one reflector and one or more directors)  Its Bandwidth can be increased by using more than one folded dipole.  It is used as a HF transmitting antenna, UHF, TV reception because of its wid bandwidth. . The Log-Periodic Dipole Array . The Turnstile Array . The Monopole Phased Array . Other Phased Arrays

 The Reflectors: o It is possible to construct a conductive surface that reflects antenna power in the desired direction o The surface may consist of one or more planes or may be parabolic o Typical reflectors are:

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Part IV: Antennas and Waveguides

. Plane and corner Reflectors . The Parabolic Reflector

 The Cell-Site Antenna: o For cellular radio systems, there is a need for omnidirectional antennas and for antennas with beamwidths of 120º, and less for sectorized cells o Cellular and PCS base-station receiving antennas are usually mounted in such a way as to obtain space diversity o For an omnidirectional pattern, typically three antennas are mounted on a tower with a triangular cross section and the antennas are mounted at 120º intervals

 The Cell-Site Antenna: o Mobile and portable antennas used with cellular and PCS systems have to be omnidirectional and small o The simplest antenna is the quarter-wavelength monopole are these are usually the ones supplied with portable phones o For mobile phones, and common configuration is the quarter-wave antenna with a half-wave antenna mounted collinearly above it

1.3 Antenna-System Noise Power Model:

 Brightness Temperature: o Energy radiated by an object can be represented by an equivalent temperature known as Brightness Temperature; 푇퐵(휃, 휑).

2 푇퐵(휃, 휑) = 휀(휃, 휑)푇푚 = (1 − |Γ| )푇푚 휀: Emissivity (0 ≤ 휀 ≤ 1) 푇푚: Molecular temperature (퐾) Γ(휃, 휑): Reflection coefficient of the surface for the polarization of wave.

 Antenna Temperature: o Energy radiated by various sources appears at antenna terminal as antenna temperature; 푇퐴.

2휋 휋 푇 (휃, 휑)퐺(휃, 휑)sin (휃)휕휃휕휑 ∫0 ∫0 퐵 푇퐴 = 2휋 휋 퐺(휃, 휑)sin (휃)휕휃휕휑 ∫0 ∫0 푇퐴: Antenna temperature (effective noise temperature of the antenna radiation resistance (푘) 퐺(휃, 휑): Gain (power) pattern of the antenna

 Noise Power: o Assuming no losses or other contributions between antenna & receiver, noise power transferred to receiver; 푃푟.

푃푟 = 푘푇퐴∆푓 푃푟: Antenna noise power (Watt)

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Part IV: Antennas and Waveguides

푘: Boltzmann’s constant (1.38 × 10−23 퐽/퐾) 푇퐴: Antenna Temperature (K) ∆푓: Bandwidth (Hz)

 Antenna Temperature at the Receiver: o Effective antenna temperature at receiver terminals; 푇푎.

−2훼푙 −2훼푙 −2훼푙 푇푎 = 푇퐴푒 + 푇퐴푃푒 + 푇표(1 − 푒 ) −1 푇퐴푃 = 푇퐵(푒퐴 − 1) 푇푎: Antenna temperature at the receiver (K) 푇퐴: Antenna temperature at the terminal (K) 푇퐴푃: Antenna temperature at the terminal due to physical temperature (K) 푇표: Physical temperature of the TL (K) 훼: Attenuation constant (Np/m) 푙: Transmission line length (m) 푒퐴: thermal efficiency of the antenna

 Noise Power at the Receiver: o Noise power transferred to receiver ; 푃푟. 푃푟 = 푘푇푎∆푓

o If there’s thermal noise in receiver:

푃푠 = 푘(푇푎 + 푇푟)∆푓 = 푘푇푠∆푓

푃푠: System noise power (Watt) 푇푎: Antenna temperature at the receiver (K) 푇푟: Antenna noise temperature at the receiver (K) 푇푠: Effective system noise temperature at the receiver (K)

 Example 1: Effective antenna temp = 150 K. Antenna is maintained at 300 K and has thermal efficiency 99%. It is connected to a receiver through 10-m waveguide (loss = 0.13 dB/m, temp = 300 K) Find effective antenna temperature at receiver terminals.

Solution: 푇퐴 = 150 퐾 푑퐵 0.13푑퐵 푁푝 훼 ( ) ( )×10푚 훼 ( ) = 푚 = 푚 푚 8.68 8.68 훼푙 = 0.149 푁푝 −1 −1 푇퐴푃 = 푇퐵(푒퐴 − 1) = 300(0.99 − 1) = 3.03

−2훼푙 −2훼푙 −2훼푙 푇푎 = 푇퐴푒 + 푇퐴푃푒 + 푇표(1 − 푒 ) −2×0.149 −2×0.149 −2×0.149 푇푎 = 150 푒 + 3.03푒 + 300(1 − 푒 ) 푇푎 = 190.904 퐾

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Part IV: Antennas and Waveguides

1.4 Waveguides-Introduction:

o Previously, a pair of conductors was used to guide electromagnetic wave propagation. This propagation was via the transverse electromagnetic (TEM) mode, meaning both the electric and magnetic field components were transverse, or perpendicular, to the direction of propagation. o Wave-guiding structures that support propagation in non-TEM modes, namely in the transverse electric (TE) and transverse magnetic (TM) modes. . In TE modes, the electric field is transverse to the direction of propagation. . In TM modes, the magnetic field that is transverse and an electric field component is in the propagation direction. o The term waveguide refers to constructs that only support non-TEM mode propagation. o They are unable to support wave propagation below a certain frequency, termed the cutoff frequency.

 Types of Waveguides: o Rectangular Waveguide: . Waveguide can support TE and TM modes. . The order of the mode refers to the field configuration in the guide, and is given by m and n integer subscripts, TEmn and TM  The m subscript corresponds to the number of half- wave variations of the field in the x direction, and  The n subscript is the number of half-wave variations in the y direction. . Cutoff Frequency: A particular mode is only supported above its cutoff frequency. The cutoff frequency is given by

2 2 2 2 1 m   n  c  m   n  fc           mn 22a   b     a   b  rr 1 1 1 1 c u      o r o r  o o  r r  r r

. Some Standard Rectangular Waveguide

Page 13 of 17 Dr. Omar R. Daoud

Part IV: Antennas and Waveguides

 Example 1: Let us calculate the cutoff frequency for the first four modes of WR284 waveguide. From the Table, the guide dimensions are a = 2.840 mils and b = 1.340 mils. Converting to metric units we have a = 7.214 cm and b = 3.404 cm.

Solution:

o Waveguide Specifications:

. The time tAC it takes for the wavefront to move from A to C (a distance lAC) is

l m 2 t AD AD uuppcos u u u p  cos 2a u  f  u sin c c mf  f c c l m 2 m 2 Distance from A to C AC l  tAC    AD Wavefront Velocity uu cos uu

. The time tAD it takes for the wavefront to move from A to D (a distance lAD) is

m 2 sin  a

2a uu . Since the times t and t must be equal, we have sin AD AC mf . The Wave velocity (푢푢) is given by

1 1 1 1 c uu      o r o r  o o  r r  r r

Page 14 of 17 Dr. Omar R. Daoud

Part IV: Antennas and Waveguides

. The Phase velocity (푢푝) is given by

푢푢 푢푢 푢푢 푢푢 푢푝 = = = = cos (휃) √cos2 (휃) √1 − sin2 (휃) 푓 √1 − ( 푐)2 푓 . The Group velocity (푢퐺) is given by

푓 푢 = 푢 × cos (휃) = 푢 × √1 − ( 푐)2 퐺 푢 푢 푓 . The phase constant (훽) is given by 푓 훽 = 훽 × √1 − ( 푐)2 푢 푓 . The guide wavelength (휆) is given by 휆 휆 = 푢 푓 √1 − ( 푐)2 푓 . The waveguide impedance (휆) is given by

푇퐸 휂푢 푍푚푛 = 푓 2 √1 − ( 푐) 푓

푓 푍푇푀 = 휂 × √1 − ( 푐)2 푚푛 푢 푓  Example 2: For the waveguide WR975 that has been filled with polyethylene, at 600MHz calculate: a) The Wave velocity b) The Phase velocity c) The Group velocity

Solution: From the Table, the guide dimensions are a = 9.75 in. and b = 4.875 in. Then, 푐 300000000 8 a) 푢푢 = = = 2 × 10 푚/푠 √휀푟 √2.26 푢 b) 푢 = 푢 = 2.7 × 108푚/푠 푝 푓 √1−( 푐)2 푓

푐 푓푐01 = = 403 푀퐻푧 2푎√휀푟

푓 c) 푢 = 푢 × √1 − ( 푐)2 = 1.48 × 108푚/푠 퐺 푢 푓

Page 15 of 17 Dr. Omar R. Daoud

Part IV: Antennas and Waveguides

o Circular Waveguide: . A Hollow metallic tube of uniform circular cross section for transmitting electromagnetic waves by successive reflections from the inner walls of the tube is called Circular waveguide. . The circular waveguide is used in many special applications in microwave techniques. . It has the advantage of  greater power – handling capacity and  lower attenuation for a given cutoff wavelength. . However, the disadvantage of somewhat  greater size and weight. . The polarization of the transmitted wave can be altered due to the minor irregularities of the wall surface of the circular guide, whereas the rectangular wave guide the polarization is fixed . The wave of lowest frequency or the dominant mode in the circular waveguide is the TE11 mode.  The first subscript m indicates the number of full – wave variations of the radial component of the electric field around the circumference of the waveguide.  The second subscript n indicates the number of half – wave variations across the diameter. . Cut off wavelength:  The cutoff wavelength for dominant mode of propagation TE11 in circular waveguide of radius ‘a’ is given by 2휋푎 휆 = 푐 1.814

 The cutoff wavelength for dominant mode of propagation TM01 in circular waveguide of radius ‘a’ is given by

2휋푎 휆 = 푐 2.405 . Applications of circular waveguide  Rotating joints in radars to connect the horn antenna feeding a parabolic reflector (which must rotate for tracking)  TE01 mode suitable for long distance waveguide transmission above 10 GHz.  Short and medium distance broad band communication (could replace / share coaxial and microwave links)

Page 16 of 17 Dr. Omar R. Daoud

Part IV: Antennas and Waveguides

 Example 1: For the dominant mode propagated in an air filled circular waveguide, the cut – off wavelength is 10 cm. Find a) the required size or cross sectional area of the guide b) the frequencies that can be used for this mode of propagation

Solution:

a) 2휋푎 휆 = 푐 1.814  휆 × 1.814 푎 = 푐 = 2.93 푐푚 2휋 Area= 휋푎2 = 26.97 푐푚2

b) 푐 푓푐 = = 3퐺퐻푧 휆푐 Therefore the frequency above 3 GHz can be propagated through the waveguide.

 Example 2: For the dominant mode propagated in an air filled circular waveguide, the inner diameter is 4 cm. Find a) the cut-off wavelength b) the cut-off frequency

Solution:

Page 17 of 17 Dr. Omar R. Daoud