CHAPTER-I MICROWAVE TRANSMISSION LINES the Electromagnetic Spectrum Is the Range of All Possible Frequencies of Electromagnetic

CHAPTER-I

MICROWAVE TRANSMISSION LINES

The electromagnetic spectrum is the range of all possible frequencies of electromagnetic radiation emitted or absolved. The electromagnetic spectrum extends from below the low frequencies used for modern radio communication to gamma radiation at the short-wavelength (high-frequency) end, thereby covering wavelengths from thousands of kilometers down to a fraction of the size of an atom as shown in Fig 1.1. The limit for long wavelengths is beyond one‟s imagination. One theory existing depicts that the short wavelength limit is in the vicinity of the Planck length. A few scientists do believe that the spectrum is infinite and continuous.

Microwaves region forms a small part of the entire electromagnetic spectrum as shown in Fig 1.1. Microwaves are electromagnetic waves generally in the frequency range of 1 G Hz to 300 GHz. However with the advent of technology usage of higher end of the frequencies became possible and now the range is extended almost to 1000 G Hz.

Fig:1.1: Electromagnetic Spectrum 1

Brief history of Microwaves

• Modern electromagnetic theory was formulated in 1873 by James Clerk Maxwell, a German scientist solely from mathematical considerations.

• Maxwell‟s formulation was cast in its modern form by Oliver Heaviside, during the period 1885 to 1887.

• Heinrich Hertz, a German professor of physics carried out a set of experiments during 1887-1891 that completely validated Maxwell‟s theory of electromagnetic waves.

• It was only in the 1940‟s (World War II) that microwave theory received substantial interest that led to radar development.

• Communication systems using microwave technology began to develop soon after the birth of radar.

Advantages and disadvantages of Microwaves compared to VHF

Advantages Disadvantages 1. Large bandwidth: The bandwidth 1. Higher Radiating losses in available is proportional to operating transmission lines and connecting frequency. ∆푓 = 푄. 푓 wires Q = Quality factor 2. The dimensions of the antenna gets 2. Transit time effects make conventional minimized to a great extent for a given devices unusable at microwave directive gain. frequencies. 3. Satellite communications was possible 3. Lumped elements such as Resistors, due to usage of microwaves as Capacitors, and Inductors cannot be antenna size became practicable. used. 4. Fading effect is less compared to lower 4. Inter electrode capacitances, lead frequencies. inductors cause severe problems in circuit design 5. As the wavelength is smaller, the attenuation during adverse weather conditions is higher.

Table 1.1: Advantages and disadvantages of Microwaves

MICROWAVE FREQUENCY BANDS

Microwave Band Frequency range L band 1 to 2 GHz S band 2 to 4 GHz C band 4 to 8 GHz X band 8 to 12 GHz

Ku band 12 to 18 GHz K band 18 to 26.5 GHz

Ka band 26.5 to 40 GHz Q band 30 to 50 GHz U band 40 to 60 GHz V band 50 to 75 GHz E band 60 to 90 GHz W band 75 to 110 GHz F band 90 to 140 GHz D band 110 to 170 GHz Table 1.2: Microwave bands

Applications of Microwaves

Microwaves have a broad range of applications in modern technology. They are mostly used in long distance communication systems, radar, radio astronomy, navigation, medical applications etc.

1. Tele Communications (a) Satellite communications (b) Mobile communications (c) Wireless Communications (d) Telemetry links 2. Radars (a) Surveillance Radars (b) Tracking radars (c) Weather radars (d) Terrain mapping radars (e) ATC radars (f) Police radars (g) Sports radars (h) Motion detectors

3. Commercial and Industrial applications

(a) Microwave oven (b) Drying machines(Textile, food, paper, etc.,) (c) Rubber industry, plastics, chemicals etc., (d) Non-destructive Testing (e) Sterilization of instruments (f) Collision avoidance systems (g) Proximity sensors

4. Medical applications (a) Physio-therapy (b) Diagnostics

5. Microwave communication systems handle a large fraction of the world‟s international and other long haul telephone, data and television transmissions. Most of the currently developing wireless telecommunications systems, such as  Direct To Home(DTH) television,  Personal communication systems (PCSs),  Wireless local area networks (WLANS),  Cellular video (CV) systems,  Global positioning satellite (GPS) systems rely heavily on microwave technology.

Transmission lines at Microwave Frequencies

There are generally three types of transmission lines used at microwave frequencies.

1. Coaxial Cables 2. Wave guides 3. Strip lines and micro strip lines

Coaxial Cables

Coax cable, coaxial feeder is normally seen as a thick electrical cable. The cable is made from a number of different elements that when together enable the coaxial cable to carry the radio frequency signals with a low level of loss from one location to another. The main elements within a coaxial cable are:

1. Centre conductor 2. Insulating dielectric 3. Outer conductor 4. Outer protecting jacket or sheath

The overall construction of the coaxial cable can be seen in the Fig 2.1 below and from this it can be seen that it is made up of a number of concentric layers. Although there are many varieties of coaxial cable, the basic overall construction remains the same.

Fig 1.2: Cross section though coaxial cable

1. Centre conductor The centre conductor of the coax is universally made of copper. Sometimes it may be a single conductor whilst in other RF cables it may consist of several strands. 2. Insulating dielectric Between the two conductors of the coaxial cable there is an insulating dielectric. This holds the two conductors apart and in an ideal world would not introduce any loss, although it is one of the chief causes of loss in reality. This coax cable dielectric may be solid or as in the case of many low loss cables it may be semi-air spaced because it is the dielectric that introduces most of the loss. This may be in the form of long "tubes" in the dielectric, or a "foam" construction where air forms a major part of the material. 3. Outer conductor The outer conductor of the RF cable is normally made from a copper braid. This enables the coax cable to be flexible which would not be the case if the outer conductor was solid, although in some varieties made for particular applications it is. To improve the screening double or even triple screened coax cables are sometimes used. Normally this is accomplished by placing one braid directly over another although in some instances a copper foil or tape outer may be used. By using additional layers of screening, the levels of stray pick-up and radiation are considerably reduced. The loss is marginally lower. 4. Outer protecting jacket or sheath Finally there is a final cover or outer sheath to the coaxial cable. This serves no electrical function, but can prevent earth loops forming. It also gives a vital protection needed to prevent dirt and moisture attacking the cable, and prevent the coax cable from being damaged by other mechanical means. How RF coax cable works A coaxial cable carries current in both the inner and the outer conductors. These current are equal and opposite and as a result all the fields are confined within the cable and it neither radiates nor picks up signals. This means that the cable operates by propagating an electromagnetic wave inside the cable. As there are no fields outside the coax cable it is not affected by nearby objects. Coaxial cable attenuation

The power loss caused by a coax cable is referred to as attenuation. It is defined in terms of decibels per unit length, and at a given frequency. Obviously the longer the 5 coaxial cable, the greater is the loss, but it is also found that the loss is frequency dependent, broadly increases with frequency.

For virtually all applications the attenuation or loss is to be minimized. The losses in coaxial cables can be classified into:

(a) Resistive loss (b) Dielectric loss © Radiated loss

Of all these forms of loss, the radiated loss is generally the least important as only a very small amount of power is generally radiated from the cable. Accordingly most of the focus on reducing loss is placed onto the conductive and dielectric losses.

 Resistive loss: Resistive losses within the coax cable arise from the resistance of the conductors and the current flowing in the conductors results in heat being dissipated. The actual area through which the current flows in the conductor is limited by the skin effect, which becomes progressively more apparent as the frequency rises. To help overcome this multi-stranded conductors are often used. To reduce the level of loss due in the coax cable, the conductive area must be increased and this results in low loss coax cables being made larger. However it is found that the resistive losses increase as the square root of the frequency.  Dielectric loss: The dielectric loss represents another of the major losses arising in most coax cables. Again the power lost as dielectric loss is dissipated as heat. It is found that the dielectric loss is independent of the size of the RF cable, but it does increase linearly with frequency. This means that resistive losses normally dominate at lower frequencies. However as resistive losses increase as the square root of frequency, and dielectric losses increase linearly, the dielectric losses dominate at higher frequencies.  Radiated loss: The radiated loss of a coax cable is normally much less than the resistive and dielectric losses. However some very cheap coax cables may have a very poor outer braid and in these cases it may represent a noticeable element of the loss. Power radiated, or picked up by a coax cable is more of a problem in terms of interference. Signal radiated by the coax cable may result in high signal levels being present where they are not wanted. For example leakage from a coax cable carrying a feed from a high power transmitter may give rise to interference in sensitive receivers that may be located close to the coax cable. Alternatively a coax cable being used for receiving may pick up interference if it passes through an electrically noisy environment. It is normally for these reasons that additional measures are taken in ensuring the outer screen or conductor is effective. Double or even triple screened coax cables are available to reduce the levels of leakage to very low levels. Dielectric materials

There are a variety of materials that can be successfully used as dielectrics in coaxial cables. Each has its own dielectric constant, and as a result, coaxial cables that use

6 different dielectric materials will exhibit different velocity factors in relation to velocity in free space as shown in Table 1.3.

Dielectric Relative Velocity Material constant factor Polyethylene 2.3 0.659 Foam polyethylene 1.3 - 1.6 0.88 - 0.79 Solid PTFE 2.07 0.695

Table 1.3: Dielectric constants and velocity factors of some common dielectric materials used in coaxial cables.

Coaxial Cable attenuation dB/m v/s freqeuncy in MHz

2.5

1.5

1 Attenuation dB/m 0.5

0 1 10 100 1000 2000 3000 5000 10000

Fig1.2: Attenuation v/s frequency of a typical coaxial cable

Waveguides

A waveguide is a special form of transmission line consisting of a hollow, metal tube as shown in Fig 1.3. The tube wall provides distributed inductance, while the empty space between the tube walls provides distributed capacitance:

Fig 1.3: Wave guides

Waveguides are practical (considering the size) only for signals of frequencies above 1 GHz. A wave guide has a cutoff frequency which is primarily depends upon waveguide cross sectional dimensions. Below such frequencies (Cut-off frequency), waveguides cannot be used as electrical transmission lines. Therefore waveguide acts as a high pass filter. The cut off frequency increases as dimensions of the waveguide decrease.

Properties of waveguides

 An electromagnetic transmission line generally used within the building.  By construction it is a hallow metal tube  Inner walls of waveguide are coated with gold or silver for smooth finish  Acts as a high pass filter  Used for frequencies above 1 G Hz  The cross sectional dimensions of the waveguide decrease with increase in cutoff frequency value.  Rectangular and circular waveguides are popular  TEM mode does not exist in waveguides.  Mode of propagation is either TE or TM

Some of the issues which are covered in this chapter are

 The propagation rate of information in waveguides  Possible distortions in transmitted information  Attenuation of the EM wave 8

 What frequencies propagate in particular waveguides?  What waveguide dimensions guide EM waves of a particular frequency?  Why would it be important to support only a single mode?  What happens if multiple modes exist within a waveguide?  What determines the excitation of the various modes?  A fundamental understanding of how waveguides work and how to analyze them.

Propagation of em wave in a waveguide

Electromagnetic waves propagate through reflections from the walls of the waveguide. The reflection angle gets acute as the frequency increases as shown in Fig 2.4. The reflection angle is 90o at cutoff, therefore no propagation takes place. The fields of incident ray and reflected ray interact in a waveguide and as a result mode patterns are formed. This phenomenon does not take place in the free space.

Fig 2.4(a): Frequencies just above cut-off

Fig 2.4(b): Increasing effect of frequency

Fig 1.4(c): Frequencies far above cut-off

Modes of propagation:

1. TEM mode ( Does not exist in a waveguide) 2. TE mode 3. TM mode

All electromagnetic waves consist of electric and magnetic fields propagating in the same direction of travel, but perpendicular to each other. Along the length of a normal transmission line, both electric and magnetic fields are perpendicular (transverse) to the direction of wave travel. This is known as the principal mode, or TEM (Transverse Electric and Magnetic) mode. This mode of wave propagation can exist only where there are two conductors, and it is the dominant mode of wave propagation where the cross-sectional dimensions of the transmission line are small compared to the wavelength of the signal.

When an electromagnetic wave propagates down a hollow tube, only one of the fields -- either electric or magnetic -- will actually be transverse to the wave's direction of travel. The other field will “loop” longitudinally to the direction of travel, but still be perpendicular to the other field. Whichever field remains transverse to the direction of travel determines whether the wave propagates in TE mode (Transverse Electric) or TM (Transverse Magnetic) mode as shown in Fig 1.5. In the figure magnetic flux lines appear as continuous loops and electric flux lines appear with beginning and end points.

Fig1.5: Electric and Magnetic field pattern in a rectangular wave guide for TE and TM modes

Maxwell’s equations

The electromagnetic fields of a propagating em wave in a given medium can be solved from the Maxwell‟s equations as given in Table 1.4 below

. MAXWELL'S EQUATIONS Maxwell's equations govern the principles of guiding and propagation of electromagnetic energy and provide the foundations of all electromagnetic phenomena and their applications. The time-harmonic expressions can be used only when the wave is sinusoidal.

STANDARD FORM TIME-HARMONIC (Time Domain) (Frequency Domain)

Faraday‟s Law ∇ × 푬 = −휕푩/휕푡 ∇ × 푬 = −푗휔푩 Ampere‟s Law ∇ × 푯 = 퐽 + 휕푫/휕푡 ∇ × 푯 = 휍 + 푗휔휀 푬 Gauss‟ Law ∇ × 푫 = 휌 ∇ × 푫 = 휌 ∇ × 푩 = 0 ∇ × 푩 = 0 ∇ × 푱 = −휕휌/휕푡 ∇ × 푱 = −푗휔휌

E = Electric Field vector [ V/m] H = Magnetic Field Vector [ A/m] B = Magnetic Flux density vector [ Web/m2 or T] D = Electric Displacement/Flux density vector [ C/ m2] J = Current density [ A/ m2]

휖 = Dielectric Permittivity [ F/m], ε = εr ε0 -12 ε0 = dielectric Permittivity of free space = 8.854 X 10 F/m

εr = Relative Dielectric Constant

휇 = Permeability [ H/m], μ = μr μo

-7 μo = Permeability of free space = 4π X 10 H/m

μr = Relative Permeability 휍 = conductivity [Siemens/m],휌 = Volume Charge density [ C/m3] The relations between the fields with respect to characteristics of the media D = 휖 E ; B = 휇 H ; J = 휍 E

Table 1.4: Maxwell’s equations 11

Wave equations for em waves:

훁ퟐH = -흎ퟐ휇휀H

훁ퟐE = -흎ퟐ휇휀E

The above two equations are called wave equations or Helmholtz wave equations whose derivation is given below:

From Maxwell‟s equations, We have

∇ × 푬 = −휕푩/휕푡

휕 B = 휇H & = j휔 휕푡

∇ X E = - j휔휇H, similarly ∇ X H = j휔휀E

Now ∇ X (∇ X E) = ∇ 푋 – 푗휔휇퐇

= - j휔휇 (∇ X H)

= - j휔휇 (j휔휀E)

= 휔2휇휀E

From Vector analysis ∇ X ∇ X E = ∇ (∇. E) - ∇2E

So (∇. E) - ∇2E = ω2휇휀E

From Maxwell‟s Equation ∇. D = 휌v = 0, and D = 휀E

therefore ∇. D = ∇. 휀E = ε(∇. E) = 0 since 휀 ≠ 0, 훁.E = 0 .Substituting eqn (1) in R.H.S we get

−∇2E = ω2휇휀E or ∇2E = −ω2휇휀E

Similarly for magnetic field we get the equation as ∇2H=−ω2휇휀H

Therefore, the wave equations or Helmholtz wave equations are,

2 2 훻 Ez = - 훚 µεEz (1.1)

2 2 훻 Hz = - 훚 µεHz (1.2)

Equation 1.1 is used for analysis of TM mode waves

Equation 1.2 is used for analysis of TE mode waves

Propagation of em waves in a rectangular waveguide

Fig 1.6: Rectangular coordinates of a rectangular waveguide

Rectangular waveguides are the one of the earliest type of the transmission lines. They are used in many applications. A rectangular waveguide supports TM and TE modes but not TEM waves because we cannot define a unique voltage since there is only one conductor in a rectangular waveguide. The shape of a rectangular waveguide and the rectangular coordinate system representation is as shown fig 1.6. „a‟ and „b‟ are the cross sectional dimensions of the waveguide.

Expanding equation (1.1),

휕2 휕2 휕2 E + E + E = - ω2µεE (1.3) 휕푥 2 z 휕푦 2 z 휕푧 2 z z

휕 휕2 For Propagation along Z-axis = - γ or = γ 2 (1.4) 휕푧 휕푧 2 Therefore,

휕2 휕2 E + E + γ2E = - ω2µεE (1.5) 휕푥 2 z 휕푦2 z z z

휕2 휕2 E + E + (γ2 + ω2µε) E =0 (1.6) 휕푥 2 z 휕푦2 z z

Let γ2 + w2µε = h2 (1.7)

휕2 휕2 E + E + h2 E =0 (1.8) 휕푥 2 z 휕푦2 z z

Similarly, for TE mode

휕2 휕2 H + H + h2 H =0 (1.9) 휕푥 2 z 휕푦2 z z

Solution to differential equations :-

Let us take first Maxwell equation,

훻 x H = j ωεE expanding the equation 푎푥 푎푦 푎푧 휕 휕 휕 휕푥 휕푦 휕푧 = j ωε [푎푥 Ex + 푎푦 Ey+ 푎푧Ez] Hx Hy Hz 푎푥 푎푦 푎푧 휕 휕 −γ 휕푥 휕푦 = j ωε [푎푥 Ex + 푎푦 Ey+ 푎푧Ez] Hx Hy Hz

Equating coefficients of ax, ay and az

휕퐻푧 + γ H = j ωεE (1.10) 휕푦 y x

휕퐻푧 + γ H = - j ωεE (1.11) 휕푥 x y

휕퐻푦 휕퐻푥 - = j ωεE (1.12) 휕푥 휕푦 z

Similarly taking Maxwell’s 2nd equation and expanding in the same way

훻 X E = - j ω휇H 푎푥 푎푦 푎푧 휕 휕 휕 휕푥 휕푦 휕푧 = - j ω휇 [푎푥 Hx + 푎푦 Hy+ 푎푧Hz] Ex Ey Ez 푎푥 푎푦 푎푧 휕 휕 −γ 휕푥 휕푦 = - j ω휇 [푎푥 Hx + 푎푦 Hy+ 푎푧Hz] Ex Ey Ez

Equating coefficients of ax, ay and az we get 14

휕퐸푧 + γE = - j ωµH (1.13) 휕푦 y x

휕퐸푧 + γE = j ωµH (1.14) 휕푥 x y

휕퐸푦 휕퐸푥 - = - j ωµH (1.15) 휕푥 휕푦 z

From eqn (1.14) expression for Hy is obtained as

1 휕퐸푧 γ H = + Ex (1.16) y j ωµ 휕푥 j ωµ

Substituting expression for Hy in eqn (1.10)

휕퐻푧 1 휕퐸푧 γ2 + γ + Ex = j ωεE (1.17) 휕푦 j ωµ 휕푥 j ωµ x

γ2 휕퐻푧 1 휕퐸푧 E (j ωε - ) = + γ x j ωµ 휕푦 j ωµ 휕푥

휕퐻푧 휕퐸푧 E (−ω2µε - γ2) = j ωµ + γ x 휕푦 휕푥

substituting for h2 from (1.7) h2 = γ2 + ω2µε

휕퐻푧 휕퐸푧 -h2E = j ωµ + γ x 휕푦 휕푥

퐣훚훍 흏푯풛 −후 흏푬풛 E = − + (1.18) x 퐡ퟐ 흏풚 퐡ퟐ 흏풙

Similarly

후 흏푬풛 퐣 훚µ 흏푯풛 E = − + (1.19) y 퐡ퟐ 흏풚 퐡ퟐ 흏풙

후 흏푯풛 퐣 훚훆 흏푬풛 H = − + (1.20) x 퐡ퟐ 흏풙 퐡ퟐ 흏풚

후 흏푯풛 퐣 훚훆 흏푬풛 H = − - (1.21) y 퐡ퟐ 흏풚 퐡ퟐ 흏풙

Equations 1.18, to 1.21 give general relationship for field components in a waveguide .

TM (Transverse Magnetic) mode analysis in rectangular wave guide:

From equation (1.8) we have

휕2 휕2 E + E + h2E =0 휕푥 2 z 휕푦2 z z

We use variables and separable method for solving the above equation

Let Ez = XY Where

X is a function of ‘x’ only

And Y is a function of ‘y’ only

휕2Ez 휕2(푋푌) 휕2(푋) = = Y 휕푥2 휕푥2 휕푥2 휕2Ez 휕2(푋푌) 휕2(푌) = = X 휕푦2 휕푦2 휕푦2

By using the above the two equations we can write equation (1.8) as

휕2(X) 휕2(푌) 푌 + X + 푕2XY = 0 (1.22) 휕푥 2 휕푥 2

1 휕2(X) 1 휕2(푌) + + 푕2 = 0 (1.23) 푋 휕푥 2 푌 휕푥 2

1 휕2(X) Let = −퐵2 (1.24) 푋 휕푥 2

1 휕2(Y) and = −퐴2 (1.25) 푌 휕푦2

Substituting equations (1.24) &(1.25) in (1.23) we get

−퐴2 − 퐵2 + 푕2 = 0

푕2 = 퐴2 + 퐵2 (1.26)

The solutions of above equation are

푋 = 퐶1 cos 퐵푥 +퐶2sin 퐵푥 (1.27)

푌 = 퐶3 cos 퐴푦 +퐶4sin 퐴푦 (1.28)

Where 퐶1, 퐶2, 퐶3, 퐶4 are constants which can be evaluated by applying boundary conditions.

We have Ez = XY

∴ 퐸푍 = { 퐶1 cos 퐵푥 +퐶2sin 퐵푥 }{( 퐶3 cos 퐴푦 +퐶4sin 퐴푦 (1.29)

Fig 1.7: Rectangular waveguide coordinate system

First boundary condition: 퐸푍 = 0 푓표푟 y= 0 and 0 ≤ 푥 ≤ 푎 Equation 1.29 becomes

0 = {퐶1 cos 퐵푥 +퐶2sin 퐵푥 } 퐶3

′ ′ 퐶3 = 0 푎푠 푥 푖푠 푎 푣푎푟푦푖푛푔 푐표푚푝표푛푒푛푡 Hence Equation 1.29 reduces to

퐸푍 = {퐶1 cos 퐵푥 +퐶2sin 퐵푥 } 퐶4sin 퐴푦 (1.30)

Second boundary condition: 퐸푍 = 0 푓표푟 x=0 and 0 ≤ 푦 ≤ 푏 Equation (1.30) becomes

퐸푍 = 퐶1 퐶4sin 퐴푦

′ 퐶1 = 0, 푎푠 ′푦 푖푠 푎 푡푖푚푒 푣푎푟푦푖푛푔 푐표푚푝표푛푒푛푡 Hence Equation 1.29 reduces to

퐸푍 = 퐶2sin 퐵푥 퐶4sin 퐴푦 (1.31)

Third boundary condition: 퐸푍 = 0 푓표푟 푦 = 푏 푎푛푑 0 ≤ 푦 ≤ 푏 ,

Equation (1.31) becomes

0 = 퐶2퐶4sin 퐵푥 sin 퐴푏

푓푟표푚 푒푎푟푙푖푒푟 푐푎푠푒푠 퐶2 ≠ 0, 퐶4 ≠ 0, ∴ sin 퐴푏 = 0

퐴푏 = 푛휋 풏흅 A= (1.32) 풃

Hence Equation (1.31) reduces to 푛휋 퐸 = 퐶 퐶 sin 퐵푥 sin 푦 푍 2 4 푏

Fourth boundary condition : 퐸푍 = 0 for x = a 푎푛푑 0 ≤ 푥 ≤ 푏 Equation 1.31 becomes 푛휋 0 = 퐶 퐶 sin 퐵푎 sin 푦 2 4 푏 푛휋 푓푟표푚 푒푎푟푙푖푒푟 푐푎푠푒푠 퐶 ≠ 0, 퐶 ≠ 0, sin 푦 ≠ 0 2 4 푏 ∴ sin 퐵푎 = 0

퐵푎 = 푛휋 풏흅 B = (1.33) 풂 Hence Equation 1.33 reduces to 푚휋 푛휋 퐸 = 퐶 퐶 sin 푥 sin 푦 푍 2 4 푎 푏

풎흅 풏흅 푬 = 푪 퐬퐢퐧 풙 퐬퐢퐧 풚 풆풋흎풕−휸풁 (1.34) 풁 풂 풃

Where C=퐶2퐶4 is a constant

γ 휕퐸푧 j ωµ 휕퐻푧 From equation 1.18 E = - - x h2 휕푥 h2 휕푦

γ 휕퐸푧 For TM mode H = 0 ; E = − z x h2 휕푥

γ 푚휋 푚휋 푛휋 퐸 = − 퐶 cos 푥 sin 푦 푒푗휔푡 −훾푍 (1.35) 푥 h2 푎 푎 푏

From equation 1.19,

γ 휕퐸푧 j ωµ 휕퐻푧 E = − + y h2 휕푦 h2 휕푥

γ 휕퐸푧 In TM mode, Hz = 0; E = − y h2 휕푦

γ 푛휋 푚휋 푛휋 퐸 = 퐶 sin 푥 cos 푦 푒푗휔푡 −훾푍 (1.36) 푦 h2 푏 푎 푏

From equation 1.20, with Hz = 0 for TM mode,;

jὠωε 푛휋 푚휋 푛휋 푗휔푡−훾푍 Hx= - 퐶 sin 푥 cos 푦 푒 (1.37) h2 푏 푎 푏

From equation 1.21, with Hz = 0 for TM mode

jὠωε 푚휋 푚휋 푛휋 푗휔푡−훾푍 Hy= 퐶 cos 푥 sin 푦 푒 (1.38) h2 푎 푎 푏

Consider the following TM mode possibilities:

1) TM0,0 : m=0,n=0 yields Ex = Ey = Ez = Hx= Hy = 0

Thus, TM0.0 can not exist.

2) TM1.0 : Ex = Ey = Ez= Hx=Hy= 0

Thus, TM1,1can not exist.

3) TM0,1: Ex = Ey = Ez= Hx=Hy = 0

Thus, TM0,1can not exist.

4) TM1,1: Ex = Ey = Ez= Hx=Hy≠0

Thus, TM1,1 can exist

Note higher order modes above TM11 also can exist.

Cut off frequency:

From equations (1.32), (1.33),

푚휋 푛휋 h2= 훾2+휔2휖휇 = A2 + B2 = 2 + 2 (1.39) 푎 푏

푚휋 2 푛휋 2 훾 = 훼 + 푗훽 = + – 휔2 휀휇 푎 푏

푚휋 푛휋 When 2 + 2 > 휔2휇휀 ; γ will be real and equals α. The wave gets attenuated as 푎 푏 it propagates and over some distance amplitude becomes zero.

푚휋 푛휋 When , 2 + 2 < 휔2휇휀; γ will be imaginary and γ=jβ , the way travels with no 푎 푏 attenuation but only phase shift. This is condition for propagation. At a particular 푚휋 2 푛휋 2 2 frequency called cut off frequency fc ; + = 휔c 휇휀 푎 푏

Signals of frequency ffc will propagate. Thus wave guide acts like a high pass filter. Now

1 푚휋 2 푛휋 2 휔 = 2휋푓 = { + } (1.40) c c 휇휀 푎 푏

1 푚휋 2 푛휋 2 fc= + 2휋 휖휇 푎 푏 This can be written as,

풄 풎 풏 f = ퟐ + ퟐ (1.41) c ퟐ 풂 풃

1 2푎푏 where c= or 휆c = (1.42) 휇휖 푚2푏2+푛2푎2

To pass lower frequencies in a waveguide we need lower cut off frequencies. But the dimensions increase accordingly. Hence, low frequency signals higher dimension waveguides are used. Similarly, if the mode (m, n) values increase, the cut off frequencies increase for the same dimensions of the waveguide.

In the tables given below we can observe that for a given waveguide dimension, the cutoff frequency increases with the increase of the mode.

Cutoff frequencies for TM Mode (for a =2 cm, b =1 cm)

m /n n=1 n=2 … m=1 1.863 G Hz 3.436 G Hz … m=2 2.357 G Hz 3.727 G Hz … m=3 3.005 G Hz … … … … …

Cutoff frequencies for TE Mode(for a =2 cm, b =1 cm)

m /n n=0 n=1 n=2 … m=0 … 1.667 G Hz 3.333 G Hz … m=1 0.883 G Hz 1.863 G Hz 3.436 G Hz … m=2 1.667 G Hz 2.357 G Hz 3.727 G Hz … m=3 2.5 G Hz 3.005 G Hz … … m=4 3.333 G Hz … … … … … … … …

Guide Wavelength: 흀g

휆0 ->

휆g

Fig 1.8: Guide wavelength in a waveguide

The distance in wave guide over which the phase of wave changes by 2π radians is called guide wavelength.

Here, 휆g>휆0where 휆g = guide wavelength.

휆0 휆g = (1.43) 휆02 1− 휆c2

There are two types of velocities in a waveguide,

1) Phase velocity vp 21

2) Group velocity vg 2 1 Vp. Vg = c where c = 휇휖

Phase Velocity ‘vp’

It is the velocity at which the phase of the wave changes along the length of the guide. It is more than or equal to the velocity of wave in free space.

2휋푓 휔 v =휆 f = = = 휔/훽 p g 2휋/휆푔 2휋 휆푔

From equation (1.39) and (1.40)

푚휋 푛휋 훾2 = 2 + 2 -휔2휇휀 푎 푏

2 2 2 (J훽) = ωc με - ω με

훽 = 휇휖 ( 휔2 − 휔푐2) (1.44) 휔 휔 Vp = = 훽 휇휖 ( 휔− 휔푐 풄 풄 Vp= = (1.45) 흎풄 풇풄 ퟏ− ퟐ ퟏ− ퟐ 흎 풇

Group Velocity

It is defined as the velocity with which the wave (energy) propagates through the wave guide and is always less than the velocity of wave in free space. Vg = 푑휔

푑훽

We have 훽 = 휇휖 ( 휔2 − 휔푐2)

휕훽 1 휇휖 = . ωμε = 휕휔 휇휖 (휔2−휔푐 2) 흎풄 ퟏ− ퟐ 흎

휕훽 휇휖 = 휕휔 풇풄 ퟏ− ퟐ 풇

풇풄 ퟏ− ퟐ 풅흎 풇 풇풄 Vg = = = c ퟏ − ퟐ (1.47) 풅휷 휇휖 풇

Guide Wave length ‘λg’

From eqn (1.44) we have

풄 풇 휆o vp =휆gf = = 흎풄 흎풄 ퟏ− ퟐ ퟏ− ퟐ 흎 흎

흀퐨 흀퐨 ∴ 흀g = = (1.48) 흎풄 흀퐨 ퟏ− ퟐ ퟏ− ퟐ 흎 흀풄

Analysis of TE Mode propagation in Rectangular Waveguides

For TE mode EZ=0, Therefore expanding eqn 1.2

휕2 휕2 휕2 H + H + H = - ω2µεH 휕푥 2 z 휕푦 2 z 휕푧 2 z z

But ∂2/∂z2 =γ2 and

휕2 휕2 H + H + γ2 H = −ω2µεH 휕푥 2 z 휕푦 2 z z z

Using γ2 +ω2 με = h2

휕2 휕2 H + H +h2H =0 (1.49) 휕푥 2 z 휕푦 2 z z

Let HZ =XY and substituting the same in above equation

휕2 X 휕2 푌 푌 + X + 푕2XY = 0 휕푥2 휕푥2

1 휕2 X 1 휕2 푌 + + 푕2 = 0 푋 휕푥 2 푌 휕푥 2 퐿푒푡

1 휕2(X) = −퐵2 (1.50) 푋 휕푥 2

1 휕2(Y) = −퐴2 (1.51) 푌 휕푦 2

퐴2 + 퐵2 = 푕2 (1.52)

The solutions to equation (1.50) and (1.51) are

X= C1 cosBx + C2 sinBx

Y= C3 cosAy+C4 sinAy

Since HZ = XY

HZ = {C1 cosBx + C2 sinBx }{C3 cosAy+C4 sinAy } (1. 53)

1st Boundary Condition

Ex=0 at y=0, 0 ≤ 푥 ≤ a

From eqn 1.18

jωµ 휕퐻푧 −γ2 휕퐸푧 E = − + since Ez=0 x h2 휕푦 h2 휕푥

2 Ex = - j ω µ/h (C1cosBx+C2sinBx )(-AC3sinAy+AC4cosAy)

2 0 = - jωµ/h (C1cosBx+C2sinBx )AC4

∴ C4 = 0

Therefore HZ = {C1cosBx+C2sinBx }{C3cosAy } (1. 54)

2nd Boundary Condition

Ey = 0 at x=0, 0≤y≤b

γ 휕퐸푧 j ωµ 휕퐻푧 From eqn 1.19 Ey = − + h2 휕푦 h2 휕푥

j ωµ 휕 Ey= [(C cosBx+C sinBx )(C cosAy)] h2 휕푥 1 2 3 j ωµ Ey= [(-BC sinBx+BC cosBx )(C cosAy )] h2 1 2 3

j ωµ 0 = BC C Ab h2 2 3

∴ C2= 0

HZ = {C1cosBx }{C3cosAy } (1.55)

3rd Boundary Condition

Ex=0 at y=b, 0 ≤ 푥 ≤ a

j ωµ 휕퐻푧 E - x= h2 휕푦

j ωµ 0 = (C AcosBx C sinAb) h2 1 3 SinAb=0

퐧훑 Ab=nπ or A = (1.56) 퐛 nπ H = {C cosBx } {C cos( )y } (1.57) Z 1 3 b

4thBoundary Condition

Ey = 0 at x=a, 0≤y≤b

j ωµ 휕퐻푧 Ey = + h2 휕푥 j ωµ nπ 0= (C C B sinBa cos( )y) h2 1 3 b 퐦훑 sinBa = 0 or B = (1.58) 퐚

푚휋 푛휋 H = 퐶 cos 푥 cos 푦 푒푗휔푡 −훾푍 (1.59) z 푎 푏

Other components can be written using equations 1.18 to 1.20 with EZ=0

j ωµ 휕퐻푧 From eqn.1.18 E - x= h2 휕푦

j ωµ 푛휋 푚휋 푛휋 푗휔푡−훾푍 Ex= 퐶 cos 푥 sin 푦 푒 (1.60) h2 푏 푎 푏

퐣 훚µ 흏푯풛 From eqn 1.19 using E = + we get y 퐡ퟐ 흏풙

j ωµ 푚휋 푚휋 푛휋 푗휔푡−훾푍 Ey= - 퐶 sin 푥 cos 푦 푒 (1.61) h2 푎 푎 푏

후 흏푯풛 From eqn 1.20 using H = − we get x 퐡ퟐ 흏풙

훾 푚휋 푚휋 푛휋 푗휔푡−훾푍 Hx= 퐶 sin 푥 cos 푦 푒 (1.62) h2 푎 푎 푏

후 흏푯풛 From eqn 1.21 using H = − we get y 퐡ퟐ 흏풚

훾 푛휋 푚휋 푛휋 푗휔푡−훾푍 Hy= 퐶 cos 푥 sin 푦 푒 (1.63) h2 푏 푎 푏 Existance of TE modes in rectangular waveguides

1) TE0,0 : Ex = Ey = Ez = Hx= Hy = 0; Thus, TE0.0 can not exist.

2) TE1.0 : Ex = Hy= 0;Ex = Hy≠ 0; Thus, TE1,0 exist.

3) TE0,1: Ey = Hx = 0;Ex = Hy ≠ 0; Thus, TE0,1 also exists.

4) TE1,1: Ex , Ey , Ez, Hx,Hy≠0;Thus, TE1,1 and all higher order modes can exist. Dominant mode

Dominant mode is that mode among TE/TM modes having lowest value of cutoff frequency. TE10 is the dominant mode for rectangular waveguide.

Degenerative modes

Higher order modes having same cutoff frequency are called degenerative modes.

Eg: TE01 and TE 30 have same cutoff frequency for a/b = 3. Therefore they are degenerative modes.

In a square waveguide where a=b all the TEpq, TEqp, TMpq and TMqp modes are degenerative.

Wave Impedance

Wave Impedance is the ratio of strength of electric field vector in one traverse direction to the strength of magnetic field vector in other traverse direction. While the wave is 퐸푥 −퐸푦 푬풙ퟐ+푬풚ퟐ propagating in Z dirección, wave impedance is Zz = = = 퐻푦 퐻푥 푯풙ퟐ+푯풚ퟐ

Wave Impedance for TM wave in rectangular waveguides

From equations 1.18 and 1.21

퐸푥 퐣훚훍 흏푯풛 −후 흏푬풛 후 흏푯풛 퐣 훚훆 흏푬풛 ZTM = = [− + ] / [− - ] 퐻푦 퐡ퟐ 흏풚 퐡ퟐ 흏풙 퐡ퟐ 흏풚 퐡ퟐ 흏풙

For TM wave Hz = 0

−후 흏푬풛 퐣 훚훆 흏푬풛 Z = [+ ] / [ - ] TM 퐡ퟐ 흏풙 퐡ퟐ 흏풙

훾 푗훽 훽 Z = = = TM 푗휔휖 푗휔휖 휔휖 Substituting for β from equation 1.44A

흎ퟐ흁흐−흎풄ퟐ흁흐 흁 풇풄 ZTM = = ퟏ − ퟐ 흎흐 흐 풇

흁 흀풐 Or ZTM = ퟏ − ퟐ 흐 흀풄

흁 Since = 휼 impedance of em wave in the given medium 흐

풇풄 흀풐 ZTM = 휼 ퟏ − ퟐ = 휼 ퟏ − ퟐ (1.64) 풇 흀풄

The above equation shows that the wave impedance of TM wave is always less than 휼 the free space impedance.

흁풓 where 휼 = 휼풐 where 휼풐 = 풇풓풆풆풔풑풂풄풆 풊풎풑풆풅풂풏풄풆= 377 ohms 흐풓

Wave Impedance for TE wave in rectangular waveguides

Substituting the expression for Ex and Hy from equations1.18 and 1.21

퐸푥 퐣훚훍 흏푯풛 −후ퟐ 흏푬풛 후 흏푯풛 퐣 훚훆 흏푬풛 Z = = [− + ] / [− - ] TE 퐻푦 퐡ퟐ 흏풚 퐡ퟐ 흏풙 퐡ퟐ 흏풚 퐡ퟐ 흏풙

퐣훚훍 흏푯풛 후 흏푯풛 For TM wave E = 0; Z = [− ] / [− ] z TE 퐡ퟐ 흏풚 퐡ퟐ 흏풚

푗휔휇 푗휔휇 휔휇 Z = = = TE 훾 푗훽 훽 Substituting for β from equation 1.44A

흎흁 흁 ퟏ 흁 ퟏ Z TE = = or Z TE = 흎ퟐ흁흐−흎풄ퟐ흁흐 흐 풇풄 흐 흀풐 ퟏ− ퟐ ퟏ− ퟐ 풇 흀풄

흁 휼 휼 Since = 휼 ; Z TE = = (1.65) 흐 흀풐 풇풄 ퟏ− ퟐ ퟏ− ퟐ 흀풄 풇

흁풓 where 휼 = 휼풐 where 휼풐 = 풇풓풆풆풔풑풂풄풆 풊풎풑풆풅풂풏풄풆= 377 ohms 흐풓

The above equation shows that the wave impedance of TE wave is always more than 휼 the free space impedance whose value is 377 Ohms.

Attenuation in rectangular waveguides

In our analysis till now, we assumed loss free conditions. The walls of the guide are assumed of perfect conductor which is loss free. The hollow region is assumed to be free space which is again is free. But in practice, the walls of the guide are made with good but finite conductors and hollow region is a good dielectric but not a perfect one. The attenuation of the wave is divided into two categories. One is reflective attenuation and the other is dissipative attenuation. The first one valid for f < fc is very large. The second one valid for f>fc is very small. The reflective attenuation is due to the frequency of the wave being not high enough to get allowed into the waveguide. The dissipative attenuation has two components, one is dielectric loss and the other is conductor loss.

The attenuation of the wave is divided into two categories.

 Reflective attenuation (Valid for f < fc is very large)

 dissipative attenuation. (Valid for f>fc is very small.)

The dissipative attenuation has two components,  Dielectric loss  Conductor loss.

1. Reflective attenuation (for f < fc )

We have γ = jβ = j 2π/λg , substituting for λg from equation 1.48

풇풄 풇풄 γ = j 2π/λg = j (2π/λo) ퟏ − ퟐ = j (2πc/f) ퟏ − ퟐ 풇 풇 풇풄 ퟏ − ퟐ is imaginary for f< fc; Therefore the propagation constant γ 풇 becomes real and represents attenuation factor given by

풇풄 ∴ α = |(2πc/f) ퟏ − ퟐ | (1.66) 풇 Where c=3x108 m/s is velocity of e m wave in free space.

2. Dissipative attenuation (for f> fc)

휍푑 휂표 (a) Dielectric loss 훼푑 = Np/ m 푓 2 2 1− 푐 푓

Where 휍푑 is conductivity of the dielectric material. ηo = 377 ohms

(b) The attenuation constant due to the imperfect conducting walls for the

TEmn and TMmn modes are given by the following relations.,

Conducting loss For TE mn

ퟐ푹 풃 풇 ퟐ 풃ퟐ풎ퟐ+풂풃풏ퟐ 풇 ퟐ 휶 = 풔 × [ {ퟏ + } 풄 + { ퟏ − 풄 }] Np / m 풄 풃훈 풂 풇 풃ퟐ풎ퟐ+풂ퟐ풏ퟐ 풇

휇푟 휋푓휇 1 Where η= 377 , and Rs = = 휀푟 σ 휍훿푠

Conducting loss For TM mn

ퟐ푹 풃ퟑ풎ퟐ+풂ퟑ풏ퟐ 휶 = 풔 × [ ] Np/m ; a,b are waveguide dimensions 풄 풂풃ퟐ풎ퟐ+풂ퟑ풏ퟐ 풇 ퟐ 풃훈 ퟏ− 풄 풇

Fig 1.9(b): TE Field patterns (cross sectional Fig 1.9(a): TE Magnetic Field pattern (top view) 10 10 view)

Fig 1.9(c): TE10 Electric Field pattern (along the waveguide)

Fig 1.10: TE01 Field pattern ( cross sectional and along the waveguide)

Fig 1.11: TE11 Field pattern ( cross sectional and along the waveguide)

Fig 1.12(b): TE20 Magnetic Field pattern (along the Fig 1.12(a): TE20 Magnetic Field pattern cross line) view)

Fig 1.12(c): TE20 Magnetic Field pattern (along the line top view)

Fig 1.13(b): TM11 Field pattern (along the line) Fig 1.13(a): TM11 Field pattern (cross view)

Fig 1.13(c): TM11 Field pattern (along the line top view)

Electric Field lines are shown in solid line and magnetic field lines are shown in dotted lines. Power losses in a wave guide: Losses in a wave guide can be due to attenuation below cut-off and losses associated with attenuation due to dissipation within the wave guide walls and the dielectric within the Wave guide

„ Case-I: At frequencies below the cut-off frequency ( f < fc) the propagation constant γ' will be real and will have only the attenuation term 'α'. In γ = α + jβ, the phase constant „β‟ itself becomes imaginary implying wave attenuation. The attenuation can be calculated by taking the second part i.e. jβ term.

2휋 We have , β= 휆푔

휆 and λg= 푓 2 1− 푐 푓

2휋 푓 2 훽 = 1 − 푐 휆 푓

2 2휋 푓푐 = −1 − 1 휆 푓

2 2휋 푓 = 푗 푐 − 1 휆 푓

2휋푓 푓 2 = 푗 푐 − 1 푐 푓

푗2휋푓 푓 2 ∵ 푐 − 1 = 푗푎 푐 푓

ퟐ ퟐ흅풇 풇풄 Hence the cut of attenuation constant ‘αc’ is given by 훂c= ퟏ − Np/m 흀풄 풇 (check)

This is the stop band attenuation of the wave guide high pass filter.

Case-II: For f >fc, the wave guide exhibits very low loss.. Attenuation constant due to an imperfect, non- magnetic dielectric in the waveguide is given by,

휍푑 휂표 훼푑 = Np/ m 푓 2 2 1− 푐 푓

Where 흈풅 is conductivity of the dielectric material. ηo = 377 ohms

The attenuation constant due to the imperfect conducting walls for the TEmn and

TMmn modes are given by the following relations.,

Conducting loss For TE mn

ퟐ푹 풃 풇 ퟐ 풃ퟐ풎ퟐ+풂풃풏ퟐ 풇 ퟐ 휶 = 풔 × [ {ퟏ + } 풄 + { ퟏ − 풄 }] Np / m 풄 풃훈 풂 풇 풃ퟐ풎ퟐ+풂ퟐ풏ퟐ 풇

휇푟 휋푓휇 1 Where η= 377 , and Rs = = 휀푟 σ 휍훿푠

Conducting loss For TM mn

ퟐ푹 풃ퟑ풎ퟐ+풂ퟑ풏ퟐ 휶 = 풔 × [ ] Np/m ; a,b are waveguide dimensions 풄 풂풃ퟐ풎ퟐ+풂ퟑ풏ퟐ 풇 ퟐ 풃훈 ퟏ− 풄 풇

Power Transmission in rectangular waveguides.

1 푍 Power transmitted = 퐸 2 푑푎 = 퐻 2 푑푎, 2푍 푎 2 푎

Where Z is the impedance in Z direction

1 푏 푎 Power transmitted for TE wave = |퐸2 | + |퐸2| dx dy 0 0 푥 푦 푓 2 2휂 1− 푐 푓

2 퐸푑 푓푐 푝푚푎푥 = 27 × × 1 − 푓푚푎푥 푓

Where Ed= dielectric strength of the insulating material in me wave guide in V/m

fmax = maximum frequency in M Hz.

f = operating frequency; fc= cutoff frequency

Mode jumping in waveguides

Dominant mode of propagation is generally adopted in waveguides. This is achieved by selecting appropriate excitation arrangement. The waveguide is selected in such a way that the cutoff frequency is just below the operating frequency. Else there is a possibility that mode of em wave in waveguide may jump to possible higher modes and this phenomenon is called mode jumping. We can avoid mode jumping with the appropriate selection of waveguide and thus the cutoff frequency. Mode jumping if allowed could lead to increase in attenuation values and also variations in the

34 characteristic impedance. The standard sizes of rectangular waveguides are given in the table below.

Cutoff Waveguide name Recommended Cutoff frequency Inner frequency frequency Frequency of lowest dimensions of band of of next * band name order waveguide EIA RCSC IEC operation mode mode opening (inch) (GHz) (GHz) (GHz) 23.000 × WR2300 WG0.0 R3 0.32 — 0.45 0.257 0.513 11.500 21.000 × WR2100 WG0 R4 0.35 — 0.50 0.281 0.562 10.500 WR1800 WG1 R5 0.45 — 0.63 0.328 0.656 18.000 × 9.000 WR1500 WG2 R6 0.50 — 0.75 0.393 0.787 15.000 × 7.500 WR1150 WG3 R8 0.63 — 0.97 0.513 1.026 11.500 × 5.750 WR975 WG4 R9 0.75 — 1.15 0.605 1.211 9.750 × 4.875 WR770 WG5 R12 0.97 — 1.45 0.766 1.533 7.700 × 3.850 WR650 WG6 R14 L band(part) 1.15 — 1.72 0.908 1.816 6.500 × 3.250 WR510 WG7 R18 1.45 — 2.20 1.157 2.314 5.100 × 2.550 WR430 WG8 R22 1.72 — 2.60 1.372 2.745 4.300 × 2.150 WR340 WG9A R26 S band(part) 2.20 — 3.30 1.736 3.471 3.400 × 1.700 2.840 × WR284 WG10 R32 S band(part) 2.60 — 3.95 2.078 4.156 † 1.340 WR229 WG11A R40 C band(part) 3.30 — 4.90 2.577 5.154 2.290 × 1.145 1.872 × WR187 WG12 R48 C band(part) 3.95 — 5.85 3.153 6.305 † 0.872 WR159 WG13 R58 C band(part) 4.90 — 7.05 3.712 7.423 1.590 × 0.795 1.372 × WR137 WG14 R70 C band(part) 5.85 — 8.20 4.301 8.603 † 0.622 1.122 × WR112 WG15 R84 — 7.05 — 10.00 5.260 10.520 † 0.497 0.900 × WR90 WG16 R100 X band 8.20 — 12.40 6.557 13.114 † 0.400 10.00 — WR75 WG17 R120 — 7.869 15.737 0.750 × 0.375 15.00 12.40 — WR62 WG18 R140 Ku band 9.488 18.976 0.622 × 0.311 18.00 15.00 — WR51 WG19 R180 — 11.572 23.143 0.510 × 0.255 22.00 18.00 — 0.420 × WR42 WG20 R220 K band 14.051 28.102 † 26.50 0.170 22.00 — WR34 WG21 R260 — 17.357 34.715 0.340 × 0.170 33.00 26.50 — WR28 WG22 R320 Ka band 21.077 42.154 0.280 × 0.140 40.00 33.00 — WR22 WG23 R400 Q band 26.346 52.692 0.224 × 0.112 50.00 40.00 — WR19 WG24 R500 U band 31.391 62.782 0.188 × 0.094 60.00

50.00 — WR15 WG25 R620 V band 39.875 79.750 0.148 × 0.074 75.00 60.00 — WR12 WG26 R740 E band 48.373 96.746 0.122 × 0.061 90.00 75.00 — WR10 WG27 R900 W band 59.015 118.030 0.100 × 0.050 110.00 90.00 — WR8 WG28 R1200 F band 73.768 147.536 0.080 × 0.040 140.00 110.00 — 0.0650 × WR6, WR7 WG29 R1400 D band 90.791 181.583 170.00 0.0325 140.00 — 0.0510 × WR5 WG30 R1800 115.714 231.429 220.00 0.0255 172.00 — 0.0430 × WR4 WG31 R2200 137.243 274.485 260.00 0.0215 220.00 — 0.0340 × WR3 WG32 R2600 173.571 347.143 330.00 0.0170

Table: Standard sizes of rectangular waveguides (source: Radio Components Standardization Committee)

Questions

1. What is a Microwave spectrum bands? 2. What are the applications of microwaves?

3. Discuss the war and peace time applications of microwaves. 4. Derive the wave equation for a TM wave and obtain all the field components in a rectangular waveguide. 5. Derive the wave equation for a TE wave and obtain all the field components in a rectangular waveguide. 6. Derive the expression for guided wave length for a rectangular wave guide. 7. Show that TEM mode cannot exist in a waveguide. 8. What are the advantages of Dominant mode of propagation in Rectangular waveguides? 9. What is mode jumping in a waveguide? How is it avoided? 10. Explain how a waveguide can be used as an attenuator, and obtain an expression for the attenuation constant.

11. Establish the expressions for the phase and group velocities, and Zo of TE and TM modes, and sketch their variation with frequency.

12. Starting with the equation for the propagation constant of a mode in rectan- 흀퐨 gular wave guide, derive the expression for guided wavelength 흀g = 흀퐨 ퟏ− ퟐ 흀풄 where 휆o is the free space wavelength and 휆c is the cutoff wavelength

13. Derive the expressions for cut off frequency, phase constant, group velocity, phase velocity and wave impedance in A rectangular wave guide.

Solved Examples

(1) (a) Find the cut off frequency for TM11 mode, given dimensions of the waveguide 2 cm and 4 cm respectively.

2푎푏 Sol: 휆c = 푚2푏2+푛2푎2 Here, a= 2 x 10-2m ; b= 4x 10-2 m 푐 Thus, 휆 = 3. 5777cm and f = = 8.386 GHz. c c 휆푐 (b)) In the above case, if sides are increased to 4cm and 8cm, then find the cut off frequency.

2푎푏 휆c = 푚2푏2+푛2푎2 Here, a= 4 x 10-2, b= 8x 10-2 m

푐 Thus, 휆 = 7.155 cm and f = = 4.19 GHz. c c 휆푐

NOTE : As the dimensions of waveguide increase, the cut off frequency will reduce

2. A rectangular wave guide of cross section 5 cm X 2 cm is used to propagate TM11 mode at 9 G Hz. Determine the cutoff wave length and wave impedance.(JNTUH Dec 2014) Solution

2푎푏 휆c = 푚2푏2+푛2푎2

Here, a= 2 x 10-2m ; b= 5 x 10-2 m

푐 Thus, 퐶푢푡표푓푓 푊푎푣푒 푙푒푛푔푡푕 휆c = 9. 285 cm and fc = = 3.23 GHz. 휆푐

풇풄 ퟐ ퟑ.ퟐퟑ ퟐ Wave impedance for TM mode = 휼 ퟏ − = ퟑퟕퟕ ퟏ − = 351.8 ohms 풇 ퟗ

3. A rectangular wave guide of cross section 5 cm X 2 cm is used to propagate TM11 mode at 9 G Hz. Determine the cutoff wave length and wave impedance.(JNTUH Dec 2014)

2푎푏 Sol: 휆c = 푚2푏2+푛2푎2 Here, a= 2 x 10-2m ; b= 5 x 10-2 m 푐 Thus, 퐶푢푡표푓푓 푊푎푣푒 푙푒푛푔푡푕 휆 = 9. 285 cm and f = = 3.23 GHz. c c 휆푐

풇풄 ퟐ ퟑ.ퟐퟑ ퟐ Wave impedance for TM mode = 휼 ퟏ − = ퟑퟕퟕ ퟏ − = 351.8 ohms 풇 ퟗ

4. A rectangular waveguide having dimensions a= 2.28 cm, b= 1.01 cm and length =30.48 am operating at 9.2 G Hz with dominant mode. Find out (a) Cutoff frequency (b) Guide Wavelength (c) Phase velocity (d) Wave Impedance.

Solution: m=1, n=0

풄 풎 ퟐ 풏 풄 ퟏ (a) f = + ퟐ = ퟐ = 6.5789 G Hz c ퟐ 풂 풃 ퟐ ퟐ.ퟐퟖ

흀퐨 0.032 (b) 흀g = = = = 0.04664 m 풇풄 ퟔ.ퟓퟕퟖퟗ ퟏ− ퟐ ퟏ− ퟐ 풇 ퟗ.ퟐ

풄 8 1 8 (c) Vp= = 3x 10 = 4.291 x 10 m/s 풇풄 ퟔ.ퟓퟕퟖퟗ ퟏ− ퟐ ퟏ− ퟐ 풇 ퟗ.ퟐ

휼 377 (d) Z TE = = = 539.3 Ohms 풇풄 ퟔ.ퟓퟕퟖퟗ ퟏ− ퟐ ퟏ− ퟐ 풇 ퟗ.ퟐ

7 5. An aluminum waveguide with a= 4.2 cm, b= 1.5cm, 휍c = 3.5x10 mhos/m, filled with -15 telfon ( μr = 1, ϵr = 2.6, 휍 = 10 mhos/m) operates at 4 G Hz. Determine

(a) 훼푐 푎푛푑 훼푑 for TE10 mode

(b) The waveguide loss in dB over a distance of 1.5 m.

Solution: Cut off frequency for TE10 mode is

ퟏ ퟏ f c10 = = = 2.213 G Hz ퟐ퐚 흁흐 ퟐ퐱 ퟎ.ퟎퟒퟐ 흁풐ퟐ.ퟔ흐

ퟏ δs = = 1.35 x 10-6 m 흅풇흁휍푐

1 1 Rs= = = 0.0211 Ohms 휍훿푠 3.5 푋 107푋 1.35 x 10−6

풇 ퟐ ퟐ.ퟐퟏퟑ ퟐ ퟏ − 풄 = 1- = 0.6939 풇 ퟒ

흁푟 η = η0 = 233.8 ohms 휺푟

ퟐ푹 풃 풇 ퟐ 풇 ퟐ 휶 = 풔 × [ {ퟏ + } 풄 + { ퟏ − 풄 }] = 0.0133 Np/m 풄 풃훈 풂 풇 풇

휍푑 휂표 10−15 x 377 훼 = = = 1.4 x 10-13 Np/m Since 푑 ퟐ 풙 ퟏ.ퟔퟔ 푓 2 2 1− 푐 푓

휶풄 ≫ 휶풅, αd can be neglected The wave propagating in the +z direction in the - - (0.0133)(1.5) rectangular waveguide vary as e 휶풄풛 = e = 0.9802 Np or = 20 log10 (0.9868) = - 0.1737 dB

6. An rectangular wave guide is filled by dielectric material of εr= 9 and has dimensions of 7 × 3.5 cm. It operates in the dominant TE mode. [JNTUH 2009] i. Determine the cut off frequency. ii. Find the phase velocity in the guide at a frequency of 2 GHz. iii. Find the guided wave length at 2 GHz.

In this case a=7 cm=0.07m and ε=9ε0,μ=μ0. Dominant mode TE10 .There fore

1 1 8 f c10 = = = 7.14x10 Hz=0.714GHz 2a 휇휖 2푥0.07푥 휇09휀0

푐 8 Phase velocity at 2 GHz =Vp= =3.211x10 m/s 푓푐 1− 2 푓 휆0 Guide wave length at 2 GHz=휆g = = 0.16 m or 16 cms 휆02 1− 휆c2 7. An air filled rectangular wave guide has the dimensions of 4 and 3 cm and is supporting TE10 mode at a frequency of 9800 MHz. Calculate

i. The wave guide impedance ii. The percentage change in the impedance for a 10% increase in the operating frequency.[JNTUH 2010}

operating frequency f = 9800MHz = 9.8 GHz 풇풄 ퟐ Wave impedance for TE mode = Zz = 휼/ ퟏ − 풇 8 Cut off frequency for dominant TE10 mode= fc=c/2a=(3x10 /2x 0.04)= 3.75 GHz 8 (since guide is air filled c=1/ 휇0휀0=3x10 m/s. For any other dielectric filling with εr

,ε=ε0 εr find c using c=1/ 휇0휀)

푓푐 2 Zz= 휂/ 1 − = 408.4 ohms 푓

With increase of 10%in operating frequency f new=1.1f

푓푐 2 Zznew= 휂/ 1 − = 402.3 ohms 1.1푓 Percentage change =[(408.4 -402.3)/ 408.4]x100= 1.51%

8. An air field rectangular wave guide of dimensions (7 X 3.5 cm) operates in the dominant TE10 mode. [ JNTUH May 2010] i. Find the cut off frequency ii. Find the phase velocity of the wave in the guide at the frequency of 3.5 GHz. iii. Determine the guided wave length at the same frequency.

Solution

a = 7 cm, b= 3.5 cm, for dominant mode the cutoff frequency is given by

푐 (i) fc = = 2.14 GHz. 2푎 40

풄 1 (ii) Vp = = 3x 108 = 3.79 X 108 m/s 풇풄 ퟐ.ퟏퟒ ퟏ− ퟐ ퟏ− ퟐ 풇 ퟑ.ퟓ

λo 0.0857 (iii) 흀g = = = = 0.108 m fc ퟐ.ퟏퟒ 1− 2 ퟏ− ퟐ f ퟑ.ퟓ

9. Consider a rectangular wave guide of 8 cm X 4 cm. Given critical wave length of TE10 = 16cm, TM11 = 7.16 cm, TM21 = 5.6 cm. What modes are propagated at a free space wave length of [JNTU May 2010] i. 5 cm and ii. 10 cm.

Solution

For a wave to propagate in a waveguide the wavelength should be smaller than the cutoff of critical wavelength.

2푎푏 휆c = 푚2푏2+푛2푎2 2푎 for TE20, 휆c = = 8 cm 2 2푎 for TE30, 휆c = = 5.33 cm 3 푎 for TE40, 휆c = = 4.0cm 2 2푎푏 2푎푏 for TM12, 휆c = = = 3.88 cm 푚2푏2+푛2푎2 푏2+4푎2

(i) In respect of free space wavelength of 5 cm, following modes are possible.

TE10, TE20, TE30, TM11 and TM21

(ii) In respect of free space wavelength of 10 cm, only TE10 mode is possible.

OBJECTIVE TYPE QUESTIONS Fill in the blanks

1.Write the microwave bands in the table given below

Frequency Band Frequency Band Frequency Band range range range 1-2 GHz L 4-8 GHz C 12-18 GHz Ku 2-4 GHz S 8-12 GHz X 18-26.5 K GHz 26.5 GHz-40 GHz Ka

2. The dominant TE mode in rectangular waveguide is : TE10

3. Wave guide can carry : TE & TM Modes

ퟐ 4. Cut-off wavelength for TE mode in rectangular waveguide is: 풎 풏 ퟐ + ퟐ 풂 풃

흀퐨 5. Guide wavelength in rectangular waveguide is: 흀퐨 ퟏ− ퟐ 흀풄

흀퐨 6. Guide wavelength in rectangular waveguide is: 풇 ퟏ− 퐜 ퟐ 풇풐

흀풐 ퟐ 풇풄 ퟐ 7. Wave impedance of waveguide in TE mode is: = 휼 ퟏ − or 휼 ퟏ − 흀풄 풇

흀풐 ퟐ 휼 8. Wave impedance of waveguide in TE mode is: 휼 ퟏ − or 흀풄 풇풄 ퟏ− ퟐ 풇

9. Degenerative modes are higher order modes having same cut-off frequency

10. Dominant mode is the mode having lowest cut-off frequency 11. The Helmholtz wave equation for TE & TM wave travelling in „Z‟ direction of a wave 2 2 2 2 guides are 훻 Ez = - 훚 µεEz and 훻 Hz = - 훚 µεHz

12. TEM mode cannot exist in a waveguide. 42

Multiple choice questions:

1. The dominant TE mode in rectangular waveguide is

(a) TE01 (b)TE11 (c) TE20 (d) TE10

2. The dominant TE mode in circular waveguide is

(a) TE01 (b)TE11 (c) TE20 (d) TE10

3. Wave guide can carry

(a) TE Mode (b) TM Mode (c) TEM mode (d) Both a and b

4. Cut-off wavelength of circular waveguide in TM mode is

(a) 2 (b) 2휋푎 (c) ퟐ흅풂 (d) None of these 풎 풏 푃 푷′ ퟐ + ퟐ 푛푚 풏풎 풂 풃

5. Cut-off wavelength of rectangular waveguide in TE mode is

ퟐ 2휋푎 푐 푚 2 푛 2 (a) (b) (c) + (d) Non of these 풎 풏 푃 2 푎 푑 ퟐ + ퟐ 푛푚 풂 풃

6. Guide wavelength of rectangular waveguide as

(a) 흀퐨 (b) 흀퐨 (c) Both a and b (d) None of the above 흀퐨 풇 ퟏ− ퟐ ퟏ− 퐜 ퟐ 흀풄 풇풐 7. Wave impedance of waveguide in TE mode can be

휼 푓푐 (a) (b) 휂 1 − 2 (c) both (a) and (b) (d) None of these 풇풄 푓 ퟏ− ퟐ 풇 8. Resonant frequency of rectangular cavity resonator is

풄 풎 ퟐ 풏 ퟐ 풑 ퟐ 풄 푚 푛 푝 (a) + + (b) + + ퟐ 풂 풃 풅 ퟐ 푎 푏 푑

풄 푚 2 푛 2 푝 2 (d) + + (d) None of the above ퟐ흅 푎 푏 푑

9. Attenuation constant due to conductor loss is

푷풐풘풆풓 풅풊풔풔풊풑풂풕풆풅 / 풖풏풊풕 풍풆풏품풕풉 2 ×푃표푤푒푟푓푙표푤 푎 푏 ퟐ ×푷풐풘풆풓풇풍풐풘 퐷푖푠푠푖푝푎푡푒푑 푃표푤푒푟 / 푢푛푖푡 푙푒푛푔푡 푕 43

2 × 푃표푤푒푟 푑푖푠푠푖푝푎푡푒푑 / 푢푛푖푡 푙푒푛푔푡 푕 푐 (d) None of these 푃표푤푒푟푓푙표푤

10. Reflective attenuation comes into being when the frequency of the wave is

(a) Less than cut-off frequency (b) more than cut-off frequency

11. In Rectangular Waveguide (RWG), the mode subscripts m and n indicate

(a) Number of half wave patterns (b) more than cut-off frequency

12. In RWG, for dominant mode, the cut-off wavelength is

(a) 2a (b) 2b (c) a (d) None of these

13. At infinite frequency, the guide wavelength is

(a) Infinite (b) Free space wavelength

14. An air filled rectangular waveguide has dimensions of 6 X 4 cm. its cut-off frequency for TE10 mode is

(a) 2.5 GHz (b) 25 GHz (c) 25 MHz (d) 5 GHz

15. The phase velocity of the guided wave at a frequency of 3.0 GHz in TE10 for the above problem is

(a) 0.1 m/s (b) 5.42 x 108 m/s (c) 5.4 x 106 m/s (d) 3.78 x 108 m/s

16. The group velocity for the above problem is

(a) 1.657 x 108 m/s (b) 5.42 x 108 m/s (c) 0.185 x 108 m/s (d) 3.78 x 108 m/s

17. The waves in a waveguide

(a) Travel along the border walls of the waveguide

(b) Are reflected from side walls but do not travel along them

(d) Travel along the all the four walls

I8. A Joint discontinuity becomes (A) increases at higher frequencies (B) independent of frequency © decreases at higher frequencies (D) increases at lower frequencies

19. The maximum use of microwaves is in (A) Communications B Cooking (C) Industrial heating (D) Research 20. 21. To reduce losses at high frequencies. wave-guide are plated with (A) copper (B) brass (C) gold (D) Aluminum

21. When electromagnetic waves are reflected at an angle front a wall. that wavelength along the wall is (A) greater than in the actual direction of propagation (B) shortened because of the Doppler effect (C) the same as in free space (D) the same as the wavelength perpendicular to the wall

22. A signal propagated in a waveguide has a full wave of electric intensity change between the two further walls, and no component of the electric field in the direction of propagation, the mode is

(A) TE10 (B) TE20 (C) TE22 (D) TE11

23. The cross section of circular waveguide compared to rectangular waveguide (A) Not related (B) bigger (C) smaller (D) equal

24. When electromagnetic waves are propagated in a wave guide (A) They are reflected from the walls but do not travel along than (B) They travel along the broader walls of the guide (C) They travel through the dielectric without touching the walls (D) They travel along the four walls of the waveguide

25. TM01 mode is likely to be preferred to the TE01 mode in circular waveguide because 45

(A) larger diameter (B) dominant anode (C) smaller diameter (D) easy to design

26. Which one of the following modes does not exist in waveguides

(A) TE11 (B) TM11 (C) TEM (D) TM01

27. When microwave signals follow the curvature of the earth this is known as (A) Faraday effect (B) ducting (C) ionospheric reflection (D) tropospheric scatter

28. The ohmic loss in the micro strip line is due to (A) inductance (B) dielectric (C) Capacitance (D) Conductivity

29. TE01 mode in circular waveguide has attenuation which (A) Remains unaltered with increase of frequency (B) Decreased with decrease of frequency (C) Increases with increase of frequency (D) Decreased with increase of frequency

30. Generally the waveguides must be not be preferable below the frequencies or (A) 50 GHz (B) 100 GHz (C) 1 GHz (D) 25 GHz

31. When a wave traveling in an- enters into a waveguide (A) the phase velocity will decrease (B) the group velocity Will decrease © the group velocity will increase (D) the phase velocity will increase 32. A circular waveguide has a cut-off frequency of 9 0Hz in dominant mode, then the inside diameter of the guide slit as air-filled is (A) 0.98 cm (B) 1.2 cm (C) 2.3 cm (D) 0.54 cm 33. The cross section of circular waveguide compared to rectangular waveguide is (A) bigger (B) smaller (C) not related (D) equal 46

34. Waveguide dimensions are decreased if frequency is (A) independent of frequency (B) decreased © unchanged (D) increased

35. Microwave transmission involves propagation of EM waves consisting of 0 (A) changing electric field in medium (B) Changing electric and magnetic field in medium (C) Changing magnetic in medium (D) Does not change both electric and magnetic field in medium

36. Velocity of TE01 wave in a rectangular wave guide is (A) less than free space velocity (B) greater than free space velocity (C) equal to free space velocity (D) twice the free space velocity

37. Power handling ability of circular waveguide compared with the coaxial cable is (A) 10 times less (B) equal C twice (D) 10 times more

38. A joint discontinuity becomes (A) increases at higher frequencies (B) decreases at higher frequencies (C) increases at lower frequencies (D) independent of frequency

39. Two signals with equal magnitude and phase are applied to the two collinear arms of E plane tee the resultant field from the centre arm is (A) zero (B) one (C) sum of two signals (D) infinity

40. The cutoff wavelength of circular waveguide in dominant mode with „d‟ as diameter is (A) 1.7 d (B) 2 d (C) 1.5 et (D) 1.2 d

41. The reflections from the walls of the guide are takes place because the wall of guide are (A) insulators (B) conductors (C) dielectric (D) semiconductor

42. A circular waveguide has a cut-off frequency of 9 GHz in dominant mode, then

47 the inside diameter of the guide if it is air-filled is (A) 0.98 cm (B) 1.2 cm (C) 2.3 cm (D) 0.54 cm

43. Modes in which there is no component of magnetic field in the direction of propagation are (A) EM (B) TE (C) TM (D) TEM

44. A hollow metal pipe of circular cross section of diameter 2 cm can be used as a transmission line at (A) Power frequencies (B) medium frequencies (C) Low frequencies (D) microwave frequencies

45. The higher order modes having the same cut off frequency are called (A) regenerate mode (B) oscillate mode (C) cut-off mode (D) degenerate mode

46. The waveguide behavior is equal to (A) Low Pass Filter (B) Band Pass Filter (C) Band Stop Filter (D) High Pass Filter

47. The main difference between the operation of transmission lines and waveguides is that (A) The former can use stubs and quarter wave transformers unlike the latter (B) The latter are not distributed like transmission lines (C) Transmission lines use the principal mode of propagation and therefore do not suffer from low frequency cutoff (D) Terms such as impedance matching and standing wave ratio cannot be applied to waveguides

48. To reduce losses at high frequencies. waveguide are plated with (A) copper (B) brass (C) aluminum (D) gold

49. The cross section of circular waveguide compared to rectangular waveguide is (A) not related (B) bigger (C) equal (D) smaller

50. The cutoff wavelength for dominant mode in rectangular waveguide is (A) 2/a (B) 2a (C) a+b (D) 2ab 48