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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

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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

available is proportional to operating

frequency. ∆𝑓 = 𝑄.𝑓

Q = Quality factor

1. Higher Radiating losses in

transmission lines and connecting

wires

2. The dimensions of the antenna gets

minimized to a great extent for a given

directive gain.

2. Transit time effects make conventional

devices unusable at microwave

frequencies.

3. Satellite communications was possible

due to usage of microwaves as

antenna size became practicable.

3. Lumped elements such as Resistors,

Capacitors, and Inductors cannot be

used.

4. Fading effect is less compared to lower

frequencies.

4. Inter electrode capacitances, lead

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

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MICROWAVE FREQUENCY BANDS

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

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

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(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.

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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

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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

7

0

0.5

1

1.5

2

2.5

1 10 100 1000 2000 3000 5000 10000

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

Attenuation dB/m

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

Material Dielectric constant

Relative Velocity 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.

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:

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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

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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

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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

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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 Domain)

TIME-HARMONIC (Frequency Domain)

Faraday‟s Law ∇ × 𝑬 = −𝜕𝑩/𝜕𝑡 ∇ × 𝑬 = −𝑗𝜔𝑩

Ampere‟s Law ∇ × 𝑯 = 𝐽 + 𝜕𝑫/𝜕𝑡 ∇ × 𝑯 = 𝜍 + 𝑗𝜔휀 𝑬

Gauss‟ Law ∇ × 𝑫 = 𝜌 ∇ × 𝑫 = 𝜌

∇ × 𝑩 = 0 ∇ × 𝑩 = 0

∇ × 𝑱 = −𝜕𝜌/𝜕𝑡 ∇ × 𝑱 = −𝑗𝜔𝜌

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]

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

εr = Relative Dielectric Constant

𝜇 = Permeability [ H/m], μ = μr μo

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

μr = Relative Permeability

𝜍 = conductivity [Siemens/m],𝜌 = Volume Charge density [ C/m3]

The relations between the fields with respect to characteristics of the media D = 𝜖 E ; B = 𝜇 H ; J = 𝜍 E

Table 1.4: Maxwell’s equations

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Wave equations for em waves:

𝛁𝟐H = -𝝎𝟐𝜇휀H

𝛁𝟐E = -𝝎𝟐𝜇휀E

The above two equations are called wave equations or Helmholtz wave equations

whose derivation is given below:

From Maxwell‟s equations, We have

∇ × 𝑬 = −𝜕𝑩/𝜕𝑡

B = 𝜇H & 𝜕

𝜕𝑡 = j𝜔

∇ X E = - j𝜔𝜇H, similarly ∇ X H = j𝜔휀E

Now ∇ X (∇ X E) = ∇ 𝑋 – 𝑗𝜔𝜇𝐇

= - j𝜔𝜇 (∇ X H)

= - j𝜔𝜇 (j𝜔휀E)

= 𝜔2𝜇휀E

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

So (∇. E) - ∇2E = ω2𝜇휀E

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

therefore ∇. D = ∇. 휀E = ε(∇. E) = 0

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

−∇2E = ω2𝜇휀E or ∇2E = −ω2𝜇휀E

Similarly for magnetic field we get the equation as ∇2H=−ω2𝜇휀H

Therefore, the wave equations or Helmholtz wave equations are,

𝛻2 Ez = - 𝛚2µεEz (1.1)

𝛻2 Hz = - 𝛚2µεHz (1.2)

Equation 1.1 is used for analysis of TM mode waves

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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),

𝜕2

𝜕𝑥2Ez +

𝜕2

𝜕𝑦2Ez +

𝜕2

𝜕𝑧2Ez = - ω

2µεEz (1.3)

For Propagation along Z-axis 𝜕

𝜕𝑧 = - γ or

𝜕2

𝜕𝑧2= γ 2 (1.4)

Therefore,

𝜕2

𝜕𝑥2Ez + 𝜕2

𝜕𝑦2Ez + γ2Ez = - ω2µεEz (1.5)

𝜕2

𝜕𝑥2Ez + 𝜕2

𝜕𝑦2Ez + (γ2 + ω2µε) Ez =0 (1.6)

Let γ2 + w2µε = h2 (1.7)

𝜕2

𝜕𝑥2Ez + 𝜕2

𝜕𝑦2Ez + h2 Ez =0 (1.8)

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Similarly, for TE mode

𝜕2

𝜕𝑥2Hz + 𝜕2

𝜕𝑦2Hz + h2 Hz =0 (1.9)

Solution to differential equations :-

Let us take first Maxwell equation,

𝛻 x H = j ωεE expanding the equation

𝑎𝑥 𝑎𝑦 𝑎𝑧𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧

Hx Hy Hz

= j ωε [𝑎𝑥 Ex + 𝑎𝑦 Ey+ 𝑎𝑧Ez]

𝑎𝑥 𝑎𝑦 𝑎𝑧𝜕

𝜕𝑥

𝜕

𝜕𝑦−γ

Hx Hy Hz

= j ωε [𝑎𝑥 Ex + 𝑎𝑦 Ey+ 𝑎𝑧Ez]

Equating coefficients of ax, ay and az

𝜕𝐻𝑧

𝜕𝑦 + γ Hy = j ωεEx (1.10)

𝜕𝐻𝑧

𝜕𝑥 + γ Hx = - j ωεEy (1.11)

𝜕𝐻𝑦

𝜕𝑥 -

𝜕𝐻𝑥

𝜕𝑦 = j ωεEz (1.12)

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

𝛻 X E = - j ω𝜇H

𝑎𝑥 𝑎𝑦 𝑎𝑧𝜕

𝜕𝑥

𝜕

𝜕𝑦

𝜕

𝜕𝑧

Ex Ey Ez

= - j ω𝜇 [𝑎𝑥 Hx + 𝑎𝑦 Hy+ 𝑎𝑧Hz]

𝑎𝑥 𝑎𝑦 𝑎𝑧𝜕

𝜕𝑥

𝜕

𝜕𝑦−γ

Ex Ey Ez

= - j ω𝜇 [𝑎𝑥 Hx + 𝑎𝑦 Hy+ 𝑎𝑧Hz]

Equating coefficients of ax, ay and az we get

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𝜕𝐸𝑧

𝜕𝑦 + γEy = - j ωµHx (1.13)

𝜕𝐸𝑧

𝜕𝑥 + γEx = j ωµHy (1.14)

𝜕𝐸𝑦

𝜕𝑥 -

𝜕𝐸𝑥

𝜕𝑦 = - j ωµHz (1.15)

From eqn (1.14) expression for Hy is obtained as

Hy = 1

j ωµ 𝜕𝐸𝑧

𝜕𝑥 +

γ

j ωµ Ex (1.16)

Substituting expression for Hy in eqn (1.10)

𝜕𝐻𝑧

𝜕𝑦 + γ

1

j ωµ 𝜕𝐸𝑧

𝜕𝑥 +

γ2

j ωµ Ex = j ωεEx (1.17)

Ex (j ωε - γ2

j ωµ ) =

𝜕𝐻𝑧

𝜕𝑦 + γ

1

j ωµ 𝜕𝐸𝑧

𝜕𝑥

Ex (−ω2µε - γ2) = j ωµ𝜕𝐻𝑧

𝜕𝑦 + γ

𝜕𝐸𝑧

𝜕𝑥

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

-h2Ex = j ωµ𝜕𝐻𝑧

𝜕𝑦 + γ

𝜕𝐸𝑧

𝜕𝑥

Ex = −𝐣𝛚𝛍

𝐡𝟐

𝝏𝑯𝒛

𝝏𝒚 +

−𝛄

𝐡𝟐 𝝏𝑬𝒛

𝝏𝒙 (1.18)

Similarly

Ey = −𝛄

𝐡𝟐

𝝏𝑬𝒛

𝝏𝒚 +

𝐣 𝛚µ

𝐡𝟐 𝝏𝑯𝒛

𝝏𝒙 (1.19)

Hx = −𝛄

𝐡𝟐

𝝏𝑯𝒛

𝝏𝒙 +

𝐣 𝛚𝛆

𝐡𝟐 𝝏𝑬𝒛

𝝏𝒚 (1.20)

Hy = −𝛄

𝐡𝟐

𝝏𝑯𝒛

𝝏𝒚 -

𝐣 𝛚𝛆

𝐡𝟐 𝝏𝑬𝒛

𝝏𝒙 (1.21)

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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

𝜕2

𝜕𝑥2Ez + 𝜕2

𝜕𝑦2Ez+ h2Ez =0

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

𝜕2Ez

𝜕𝑥2=

𝜕2(𝑋𝑌)

𝜕𝑥2= Y

𝜕2(𝑋)

𝜕𝑥2

𝜕2Ez

𝜕𝑦2=

𝜕2(𝑋𝑌)

𝜕𝑦2= X

𝜕2(𝑌)

𝜕𝑦2

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

𝑌 𝜕2(X)

𝜕𝑥2+ X

𝜕2(𝑌)

𝜕𝑥2+ 𝑕2XY = 0 (1.22)

1

𝑋

𝜕2(X)

𝜕𝑥2+

1

𝑌

𝜕2(𝑌)

𝜕𝑥2+ 𝑕2 = 0 (1.23)

Let 1

𝑋

𝜕2(X)

𝜕𝑥2 = −𝐵2 (1.24)

and 1

𝑌

𝜕2(Y)

𝜕𝑦2 = −𝐴2 (1.25)

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

−𝐴2 − 𝐵2 + 𝑕2 = 0

𝑕2 = 𝐴2 + 𝐵2 (1.26)

The solutions of above equation are

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𝑋 = 𝐶1 cos 𝐵𝑥 +𝐶2sin 𝐵𝑥 (1.27)

𝑌 = 𝐶3 cos 𝐴𝑦 +𝐶4sin 𝐴𝑦 (1.28)

Where 𝐶1,𝐶2,𝐶3,𝐶4 are constants which can be evaluated by applying boundary

conditions.

We have Ez = XY

∴ 𝐸𝑍 = { 𝐶1 cos 𝐵𝑥 +𝐶2sin 𝐵𝑥 }{(𝐶3 cos 𝐴𝑦 +𝐶4sin 𝐴𝑦 (1.29)

,

Fig 1.7: Rectangular waveguide coordinate system

First boundary condition: 𝐸𝑍 = 0 𝑓𝑜𝑟 y= 0 and 0 ≤ 𝑥 ≤ 𝑎

Equation 1.29 becomes

0 = {𝐶1 cos 𝐵𝑥 +𝐶2sin 𝐵𝑥 }𝐶3

𝐶3 = 0 𝑎𝑠 𝑥′ ′ 𝑖𝑠 𝑎 𝑣𝑎𝑟𝑦𝑖𝑛𝑔 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡

Hence Equation 1.29 reduces to

𝐸𝑍 = {𝐶1 cos 𝐵𝑥 +𝐶2sin 𝐵𝑥 } 𝐶4sin 𝐴𝑦 (1.30)

Second boundary condition: 𝐸𝑍 = 0 𝑓𝑜𝑟 x=0 and 0 ≤ 𝑦 ≤ 𝑏

Equation (1.30) becomes

𝐸𝑍 = 𝐶1 𝐶4sin 𝐴𝑦

𝐶1 = 0,𝑎𝑠 ′𝑦′ 𝑖𝑠 𝑎 𝑡𝑖𝑚𝑒 𝑣𝑎𝑟𝑦𝑖𝑛𝑔 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡

Hence Equation 1.29 reduces to

18

𝐸𝑍 = 𝐶2sin 𝐵𝑥 𝐶4sin 𝐴𝑦 (1.31)

Third boundary condition: 𝐸𝑍 = 0 𝑓𝑜𝑟 𝑦 = 𝑏 𝑎𝑛𝑑 0 ≤ 𝑦 ≤ 𝑏 ,

Equation (1.31) becomes

0 = 𝐶2𝐶4sin 𝐵𝑥 sin 𝐴𝑏

𝑓𝑟𝑜𝑚 𝑒𝑎𝑟𝑙𝑖𝑒𝑟 𝑐𝑎𝑠𝑒𝑠 𝐶2 ≠ 0,𝐶4 ≠ 0,

∴ sin 𝐴𝑏 = 0

𝐴𝑏 = 𝑛𝜋

A=𝒏𝝅

𝒃 (1.32)

Hence Equation (1.31) reduces to

𝐸𝑍 = 𝐶2𝐶4sin 𝐵𝑥 sin 𝑛𝜋

𝑏 𝑦

Fourth boundary condition : 𝐸𝑍 = 0 for x = a 𝑎𝑛𝑑 0 ≤ 𝑥 ≤ 𝑏

Equation 1.31 becomes

0 = 𝐶2𝐶4sin 𝐵𝑎 sin 𝑛𝜋

𝑏𝑦

𝑓𝑟𝑜𝑚 𝑒𝑎𝑟𝑙𝑖𝑒𝑟 𝑐𝑎𝑠𝑒𝑠 𝐶2 ≠ 0,𝐶4 ≠ 0, sin 𝑛𝜋

𝑏𝑦 ≠ 0

∴ sin 𝐵𝑎 = 0

𝐵𝑎 = 𝑛𝜋

B = 𝒏𝝅

𝒂 (1.33)

Hence Equation 1.33 reduces to

𝐸𝑍 = 𝐶2𝐶4sin 𝑚𝜋

𝑎𝑥 sin

𝑛𝜋

𝑏𝑦

𝑬𝒁 = 𝑪 𝐬𝐢𝐧 𝒎𝝅

𝒂𝒙 𝐬𝐢𝐧

𝒏𝝅

𝒃𝒚 𝒆𝒋𝝎𝒕−𝜸𝒁 (1.34)

Where C=𝐶2𝐶4 is a constant

19

From equation 1.18 Ex = - γ

h2

𝜕𝐸𝑧

𝜕𝑥 -

j ωµ

h2

𝜕𝐻𝑧

𝜕𝑦

For TM mode Hz = 0 ; Ex= −γ

h2

𝜕𝐸𝑧

𝜕𝑥

𝐸𝑥 = −γ

h2𝐶

𝑚𝜋

𝑎 cos

𝑚𝜋

𝑎𝑥 sin

𝑛𝜋

𝑏𝑦 𝑒𝑗𝜔𝑡 −𝛾𝑍 (1.35)

From equation 1.19,

Ey= −γ

h2

𝜕𝐸𝑧

𝜕𝑦 +

j ωµ

h2

𝜕𝐻𝑧

𝜕𝑥

In TM mode, Hz = 0; Ey= −γ

h2

𝜕𝐸𝑧

𝜕𝑦

𝐸𝑦 =γ

h2𝐶

𝑛𝜋

𝑏 sin

𝑚𝜋

𝑎𝑥 cos

𝑛𝜋

𝑏𝑦 𝑒𝑗𝜔𝑡 −𝛾𝑍 (1.36)

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

Hx= - jὠωε

h2𝐶

𝑛𝜋

𝑏 sin

𝑚𝜋

𝑎𝑥 cos

𝑛𝜋

𝑏𝑦 𝑒𝑗𝜔𝑡−𝛾𝑍 (1.37)

From equation 1.21, with Hz = 0 for TM mode

Hy= jὠωε

h2𝐶

𝑚𝜋

𝑎 cos

𝑚𝜋

𝑎𝑥 sin

𝑛𝜋

𝑏𝑦 𝑒𝑗𝜔𝑡−𝛾𝑍 (1.38)

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.

20

Cut off frequency:

From equations (1.32), (1.33),

h2= 𝛾2+𝜔2𝜖𝜇 = A2 + B2 = 𝑚𝜋

𝑎 2 +

𝑛𝜋

𝑏 2

(1.39)

𝛾 = 𝛼 + 𝑗𝛽 = 𝑚𝜋

𝑎

2 +

𝑛𝜋

𝑏

2 –𝜔2 휀𝜇

When 𝑚𝜋

𝑎 2 +

𝑛𝜋

𝑏 2 > 𝜔2𝜇휀 ; γ will be real and equals α. The wave gets attenuated as

it propagates and over some distance amplitude becomes zero.

When , 𝑚𝜋

𝑎 2 +

𝑛𝜋

𝑏 2 < 𝜔2𝜇휀; γ will be imaginary and γ=jβ , the way travels with no

attenuation but only phase shift. This is condition for propagation. At a particular

frequency called cut off frequency fc ; 𝑚𝜋

𝑎 2 +

𝑛𝜋

𝑏 2= 𝜔c

2𝜇휀

Signals of frequency f<fc will be attenuated and with frequency f>fc will propagate. Thus

wave guide acts like a high pass filter. Now

𝜔c = 2𝜋𝑓c= 1

𝜇휀{

𝑚𝜋

𝑎

2 +

𝑛𝜋

𝑏

2 } (1.40)

fc= 1

2𝜋 𝜖𝜇

𝑚𝜋

𝑎

2 +

𝑛𝜋

𝑏

2

This can be written as,

fc = 𝒄

𝟐

𝒎

𝒂 𝟐 +

𝒏

𝒃 𝟐 (1.41)

where c=1

𝜇𝜖 or 𝜆c =

2𝑎𝑏

𝑚2𝑏2+𝑛2𝑎2 (1.42)

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.

21

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: 𝝀g

𝜆0 ->

𝜆g

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, 𝜆g>𝜆0where 𝜆g = guide wavelength.

𝜆g = 𝜆0

1−𝜆02

𝜆c2

(1.43)

There are two types of velocities in a waveguide,

1) Phase velocity vp

22

2) Group velocity vg

Vp. Vg = c2 where c = 1

𝜇𝜖

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.

vp =𝜆gf =2𝜋𝑓

2𝜋/𝜆𝑔 =

𝜔2𝜋

𝜆𝑔

= 𝜔/𝛽

From equation (1.39) and (1.40)

𝛾2 = 𝑚𝜋

𝑎 2 +

𝑛𝜋

𝑏 2 -𝜔2𝜇휀

(J𝛽)2 = ωc2με - ω2με

𝛽 = 𝜇𝜖 ( 𝜔2 − 𝜔𝑐2) (1.44)

Vp = 𝜔

𝛽 =

𝜔

𝜇𝜖 ( 𝜔− 𝜔𝑐

Vp= 𝒄

𝟏− 𝝎𝒄

𝝎 𝟐

= 𝒄

𝟏− 𝒇𝒄

𝒇 𝟐

(1.45)

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 = 𝑑𝜔

𝑑𝛽

We have 𝛽 = 𝜇𝜖 ( 𝜔2 − 𝜔𝑐2)

𝜕𝛽

𝜕𝜔 =

1

𝜇𝜖 (𝜔2−𝜔𝑐2) . ωμε =

𝜇𝜖

𝟏− 𝝎𝒄

𝝎 𝟐

𝜕𝛽

𝜕𝜔 = 𝜇𝜖

𝟏− 𝒇𝒄

𝒇 𝟐

23

Vg = 𝒅𝝎

𝒅𝜷=

𝟏− 𝒇𝒄

𝒇 𝟐

𝜇𝜖 = c 𝟏 −

𝒇𝒄

𝒇 𝟐 (1.47)

Guide Wave length ‘λg’

From eqn (1.44) we have

vp =𝜆gf = 𝒄

𝟏− 𝝎𝒄

𝝎 𝟐

= 𝒇 𝜆o

𝟏− 𝝎𝒄

𝝎 𝟐

∴ 𝝀g = 𝝀𝐨

𝟏− 𝝎𝒄

𝝎 𝟐

= 𝝀𝐨

𝟏− 𝝀𝐨

𝝀𝒄 𝟐

(1.48)

Analysis of TE Mode propagation in Rectangular Waveguides

For TE mode EZ=0, Therefore expanding eqn 1.2

𝜕2

𝜕𝑥2Hz +

𝜕2

𝜕𝑦2Hz +

𝜕2

𝜕𝑧2Hz = - ω

2µεHz

But ∂2/∂z2 =γ2 and

𝜕2

𝜕𝑥2Hz +

𝜕2

𝜕𝑦2Hz + γ2 Hz = −ω

2µεHz

Using γ2 +ω2 με = h2

𝜕2

𝜕𝑥2Hz +

𝜕2

𝜕𝑦2Hz +h2Hz =0 (1.49)

Let HZ =XY and substituting the same in above equation

𝑌 𝜕2 X

𝜕𝑥2+ X

𝜕2 𝑌

𝜕𝑥2+ 𝑕2XY = 0

1

𝑋

𝜕2 X

𝜕𝑥2+

1

𝑌

𝜕2 𝑌

𝜕𝑥2+ 𝑕2 = 0

𝐿𝑒𝑡

1

𝑋

𝜕2(X)

𝜕𝑥2 = −𝐵2 (1.50)

24

1

𝑌

𝜕2(Y)

𝜕𝑦2 = −𝐴2 (1.51)

𝐴2 + 𝐵2 = 𝑕2 (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 ≤ 𝑥 ≤ a

From eqn 1.18

Ex = −jωµ

h2

𝜕𝐻𝑧

𝜕𝑦 +

−γ2

h2 𝜕𝐸𝑧

𝜕𝑥 since Ez=0

Ex = - j ω µ/h2 (C1cosBx+C2sinBx )(-AC3sinAy+AC4cosAy)

0 = - jωµ/h2(C1cosBx+C2sinBx )AC4

∴ C4 = 0

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

2nd Boundary Condition

Ey = 0 at x=0, 0≤y≤b

From eqn 1.19 Ey = −γ

h2

𝜕𝐸𝑧

𝜕𝑦 +

j ωµ

h2 𝜕𝐻𝑧

𝜕𝑥

Ey= j ωµ

h2

𝜕

𝜕𝑥 [(C1cosBx+C2sinBx )(C3cosAy)]

Ey= j ωµ

h2 [(-BC1sinBx+BC2cosBx )(C3cosAy )]

25

0 = j ωµ

h2 BC2C3Ab

∴ C2= 0

HZ = {C1cosBx }{C3cosAy } (1.55)

3rd Boundary Condition

Ex=0 at y=b, 0 ≤ 𝑥 ≤ a

Ex= - j ωµ

h2 𝜕𝐻𝑧

𝜕𝑦

0 = j ωµ

h2 (C1AcosBx C3sinAb)

SinAb=0

Ab=nπ or A = 𝐧𝛑

𝐛 (1.56)

HZ = {C1 cosBx } {C3 cos(nπ

b)y } (1.57)

4thBoundary Condition

Ey = 0 at x=a, 0≤y≤b

Ey = + j ωµ

h2 𝜕𝐻𝑧

𝜕𝑥

0= j ωµ

h2 (C1C3B sinBa cos(

nπ

b)y)

sinBa = 0 or B = 𝐦𝛑

𝐚 (1.58)

Hz = 𝐶 cos 𝑚𝜋

𝑎𝑥 cos

𝑛𝜋

𝑏𝑦 𝑒𝑗𝜔𝑡 −𝛾𝑍 (1.59)

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

From eqn.1.18 Ex= - j ωµ

h2 𝜕𝐻𝑧

𝜕𝑦

26

Ex= j ωµ

h2 𝐶

𝑛𝜋

𝑏 cos

𝑚𝜋

𝑎𝑥 sin

𝑛𝜋

𝑏𝑦 𝑒𝑗𝜔𝑡−𝛾𝑍 (1.60)

From eqn 1.19 using Ey = + 𝐣 𝛚µ

𝐡𝟐 𝝏𝑯𝒛

𝝏𝒙 we get

Ey= - j ωµ

h2 𝐶

𝑚𝜋

𝑎 sin

𝑚𝜋

𝑎𝑥 cos

𝑛𝜋

𝑏𝑦 𝑒𝑗𝜔𝑡−𝛾𝑍 (1.61)

From eqn 1.20 using Hx = −𝛄

𝐡𝟐

𝝏𝑯𝒛

𝝏𝒙 we get

Hx= 𝛾

h2 𝐶

𝑚𝜋

𝑎 sin

𝑚𝜋

𝑎𝑥 cos

𝑛𝜋

𝑏𝑦 𝑒𝑗𝜔𝑡−𝛾𝑍 (1.62)

From eqn 1.21 using Hy = −𝛄

𝐡𝟐

𝝏𝑯𝒛

𝝏𝒚 we get

Hy= 𝛾

h2 𝐶

𝑛𝜋

𝑏 cos

𝑚𝜋

𝑎𝑥 sin

𝑛𝜋

𝑏𝑦 𝑒𝑗𝜔𝑡−𝛾𝑍 (1.63)

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

27

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

propagating in Z dirección, wave impedance is Zz = 𝐸𝑥

𝐻𝑦 =

−𝐸𝑦

𝐻𝑥 =

𝑬𝒙𝟐+𝑬𝒚𝟐

𝑯𝒙𝟐+𝑯𝒚𝟐

Wave Impedance for TM wave in rectangular waveguides

From equations 1.18 and 1.21

ZTM = 𝐸𝑥

𝐻𝑦 = [−

𝐣𝛚𝛍

𝐡𝟐

𝝏𝑯𝒛

𝝏𝒚 +

−𝛄

𝐡𝟐 𝝏𝑬𝒛

𝝏𝒙] / [−

𝛄

𝐡𝟐

𝝏𝑯𝒛

𝝏𝒚 -

𝐣 𝛚𝛆

𝐡𝟐 𝝏𝑬𝒛

𝝏𝒙 ]

For TM wave Hz = 0

ZTM = [+ −𝛄

𝐡𝟐 𝝏𝑬𝒛

𝝏𝒙] / [ -

𝐣 𝛚𝛆

𝐡𝟐 𝝏𝑬𝒛

𝝏𝒙 ]

ZTM = 𝛾

𝑗𝜔𝜖 =

𝑗𝛽

𝑗𝜔𝜖 =

𝛽

𝜔𝜖

Substituting for β from equation 1.44A

ZTM = 𝝎𝟐𝝁𝝐−𝝎𝒄𝟐𝝁𝝐

𝝎𝝐 =

𝝁

𝝐 𝟏 −

𝒇𝒄

𝒇 𝟐

Or ZTM = 𝝁

𝝐 𝟏 −

𝝀𝒐

𝝀𝒄 𝟐

Since 𝝁

𝝐= 𝜼 impedance of em wave in the given medium

ZTM = 𝜼 𝟏 − 𝒇𝒄

𝒇 𝟐 = 𝜼 𝟏 −

𝝀𝒐

𝝀𝒄 𝟐 (1.64)

The above equation shows that the wave impedance of TM wave is always less than 𝜼 the free

space impedance.

where 𝜼 = 𝜼𝒐 𝝁𝒓

𝝐𝒓 where 𝜼𝒐 = 𝒇𝒓𝒆𝒆𝒔𝒑𝒂𝒄𝒆 𝒊𝒎𝒑𝒆𝒅𝒂𝒏𝒄𝒆= 377 ohms

Wave Impedance for TE wave in rectangular waveguides

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

28

ZTE = 𝐸𝑥

𝐻𝑦 = [−

𝐣𝛚𝛍

𝐡𝟐

𝝏𝑯𝒛

𝝏𝒚 +

−𝛄𝟐

𝐡𝟐 𝝏𝑬𝒛

𝝏𝒙] / [−

𝛄

𝐡𝟐

𝝏𝑯𝒛

𝝏𝒚 -

𝐣 𝛚𝛆

𝐡𝟐 𝝏𝑬𝒛

𝝏𝒙 ]

For TM wave Ez = 0; Z TE = [−𝐣𝛚𝛍

𝐡𝟐

𝝏𝑯𝒛

𝝏𝒚 ] / [−

𝛄

𝐡𝟐

𝝏𝑯𝒛

𝝏𝒚 ]

Z TE = 𝑗𝜔𝜇

𝛾 =

𝑗𝜔𝜇

𝑗𝛽 =

𝜔𝜇

𝛽

Substituting for β from equation 1.44A

Z TE = 𝝎𝝁

𝝎𝟐𝝁𝝐−𝝎𝒄𝟐𝝁𝝐 =

𝝁

𝝐

𝟏

𝟏− 𝒇𝒄

𝒇 𝟐

or Z TE = 𝝁

𝝐

𝟏

𝟏− 𝝀𝒐

𝝀𝒄 𝟐

Since 𝝁

𝝐= 𝜼 ; Z TE =

𝜼

𝟏− 𝝀𝒐

𝝀𝒄 𝟐

= 𝜼

𝟏− 𝒇𝒄

𝒇 𝟐

(1.65)

where 𝜼 = 𝜼𝒐 𝝁𝒓

𝝐𝒓 where 𝜼𝒐 = 𝒇𝒓𝒆𝒆𝒔𝒑𝒂𝒄𝒆 𝒊𝒎𝒑𝒆𝒅𝒂𝒏𝒄𝒆= 377 ohms

The above equation shows that the wave impedance of TE wave is always more than 𝜼

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.

29

1. Reflective attenuation (for f < fc )

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

γ = j 2π/λg = j (2π/λo) 𝟏 − 𝒇𝒄

𝒇 𝟐 = j (2πc/f) 𝟏 −

𝒇𝒄

𝒇 𝟐

𝟏 − 𝒇𝒄

𝒇 𝟐 is imaginary for f< fc; Therefore the propagation constant γ

becomes real and represents attenuation factor given by

∴ α = |(2πc/f) 𝟏 − 𝒇𝒄

𝒇 𝟐 | (1.66)

Where c=3x108 m/s is velocity of e m wave in free space.

2. Dissipative attenuation (for f> fc)

(a) Dielectric loss 𝛼𝑑 =𝜍𝑑 𝜂𝑜

2 1− 𝑓𝑐𝑓

2 Np/ m

Where 𝜍𝑑 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

𝜶𝒄 =𝟐𝑹𝒔

𝒃𝛈× [ {𝟏 +

𝒃

𝒂}

𝒇𝒄

𝒇 𝟐

+ 𝒃𝟐𝒎𝟐+𝒂𝒃𝒏𝟐

𝒃𝟐𝒎𝟐+𝒂𝟐𝒏𝟐{ 𝟏 −

𝒇𝒄

𝒇 𝟐

}] Np / m

Where η= 377 𝜇𝑟

휀𝑟, and Rs =

𝜋𝑓𝜇

σ =

1

𝜍𝛿𝑠

Conducting loss For TM mn

𝜶𝒄 =𝟐𝑹𝒔

𝒃𝛈 𝟏− 𝒇𝒄𝒇 𝟐

× [ 𝒃𝟑𝒎𝟐+𝒂𝟑𝒏𝟐

𝒂𝒃𝟐𝒎𝟐+𝒂𝟑𝒏𝟐] Np/m ; a,b are waveguide dimensions

30

Fig 1.9(a): TE10 Magnetic Field pattern (top view)

Fig 1.9(b): TE10 Field patterns (cross sectional view)

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

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

31

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

Fig 1.12(a): TE20 Magnetic Field pattern cross

view)

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

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

32

Fig 1.13(a): TM11 Field pattern (cross view)

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

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.

We have , β= 2𝜋

𝜆𝑔

and λg= 𝜆

1− 𝑓𝑐𝑓

2

33

𝛽 =2𝜋

𝜆 1 −

𝑓𝑐

𝑓

2

=2𝜋

𝜆 −1

𝑓𝑐𝑓

2

− 1

= 𝑗2𝜋

𝜆

𝑓𝑐𝑓

2

− 1

= 𝑗2𝜋𝑓

𝑐

𝑓𝑐

𝑓

2− 1

∵𝑗2𝜋𝑓

𝑐

𝑓𝑐𝑓

2

− 1 = 𝑗𝑎

Hence the cut of attenuation constant ‘αc’ is given by 𝛂c=𝟐𝝅𝒇

𝝀𝒄 𝟏 −

𝒇𝒄

𝒇 𝟐 Np/m

(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,

𝛼𝑑 =𝜍𝑑 𝜂𝑜

2 1− 𝑓𝑐𝑓

2 Np/ m

Where 𝝈𝒅 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

34

𝜶𝒄 =𝟐𝑹𝒔

𝒃𝛈× [ {𝟏 +

𝒃

𝒂}

𝒇𝒄

𝒇 𝟐

+ 𝒃𝟐𝒎𝟐+𝒂𝒃𝒏𝟐

𝒃𝟐𝒎𝟐+𝒂𝟐𝒏𝟐{ 𝟏 −

𝒇𝒄

𝒇 𝟐

}] Np / m

Where η= 377 𝜇𝑟

휀𝑟, and Rs =

𝜋𝑓𝜇

σ =

1

𝜍𝛿𝑠

Conducting loss For TM mn

𝜶𝒄 =𝟐𝑹𝒔

𝒃𝛈 𝟏− 𝒇𝒄𝒇 𝟐

× [ 𝒃𝟑𝒎𝟐+𝒂𝟑𝒏𝟐

𝒂𝒃𝟐𝒎𝟐+𝒂𝟑𝒏𝟐] Np/m ; a,b are waveguide dimensions

Power Transmission in rectangular waveguides.

Power transmitted = 1

2𝑍 𝐸 2 𝑑𝑎𝑎

= 𝑍

2 𝐻 2 𝑑𝑎𝑎

,

Where Z is the impedance in Z direction

Power transmitted for TE wave = 1

2𝜂 1− 𝑓𝑐𝑓

2

𝑏

0 |𝐸𝑥

2𝑎

0| + |𝐸𝑦

2| dx dy

𝑝𝑚𝑎𝑥 = 27 × 𝐸𝑑

𝑓𝑚𝑎𝑥 × 1 −

𝑓𝑐𝑓

2

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

35

characteristic impedance. The standard sizes of rectangular waveguides are given in

the table below.

Waveguide name

Frequency band name

Recommended frequency

band of operation

(GHz)

Cutoff frequency of lowest

order mode (GHz)

Cutoff frequency

of next mode (GHz)

Inner dimensions of

waveguide opening (inch)

EIA RCSC * IEC

WR2300 WG0.0 R3 0.32 — 0.45 0.257 0.513 23.000 × 11.500

WR2100 WG0 R4 0.35 — 0.50 0.281 0.562 21.000 × 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

WR284 WG10 R32 S band(part) 2.60 — 3.95 2.078 4.156 2.840 × 1.340

†

WR229 WG11A R40 C band(part) 3.30 — 4.90 2.577 5.154 2.290 × 1.145

WR187 WG12 R48 C band(part) 3.95 — 5.85 3.153 6.305 1.872 × 0.872

†

WR159 WG13 R58 C band(part) 4.90 — 7.05 3.712 7.423 1.590 × 0.795

WR137 WG14 R70 C band(part) 5.85 — 8.20 4.301 8.603 1.372 × 0.622

†

WR112 WG15 R84 — 7.05 — 10.00 5.260 10.520 1.122 × 0.497

†

WR90 WG16 R100 X band 8.20 — 12.40 6.557 13.114 0.900 × 0.400

†

WR75 WG17 R120 — 10.00 —

15.00 7.869 15.737 0.750 × 0.375

WR62 WG18 R140 Ku band 12.40 —

18.00 9.488 18.976 0.622 × 0.311

WR51 WG19 R180 — 15.00 —

22.00 11.572 23.143 0.510 × 0.255

WR42 WG20 R220 K band 18.00 —

26.50 14.051 28.102

0.420 × 0.170

†

WR34 WG21 R260 — 22.00 —

33.00 17.357 34.715 0.340 × 0.170

WR28 WG22 R320 Ka band 26.50 —

40.00 21.077 42.154 0.280 × 0.140

WR22 WG23 R400 Q band 33.00 —

50.00 26.346 52.692 0.224 × 0.112

WR19 WG24 R500 U band 40.00 —

60.00 31.391 62.782 0.188 × 0.094

36

WR15 WG25 R620 V band 50.00 —

75.00 39.875 79.750 0.148 × 0.074

WR12 WG26 R740 E band 60.00 —

90.00 48.373 96.746 0.122 × 0.061

WR10 WG27 R900 W band 75.00 — 110.00

59.015 118.030 0.100 × 0.050

WR8 WG28 R1200 F band 90.00 — 140.00

73.768 147.536 0.080 × 0.040

WR6, WR7 WG29 R1400 D band 110.00 —

170.00 90.791 181.583

0.0650 × 0.0325

WR5 WG30 R1800 140.00 —

220.00 115.714 231.429

0.0510 × 0.0255

WR4 WG31 R2200 172.00 —

260.00 137.243 274.485

0.0430 × 0.0215

WR3 WG32 R2600 220.00 —

330.00 173.571 347.143

0.0340 × 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 expres- sion 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.

37

12. Starting with the equation for the propagation constant of a mode in rectan-

gular wave guide, derive the expression for guided wavelength 𝝀g = 𝝀𝐨

𝟏− 𝝀𝐨

𝝀𝒄 𝟐

where 𝜆o is the free space wavelength and 𝜆c 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.

Sol: 𝜆c = 2𝑎𝑏

𝑚2𝑏2+𝑛2𝑎2

Here, a= 2 x 10-2m ; b= 4x 10-2 m

Thus, 𝜆c = 3. 5777cm and fc = 𝑐

𝜆𝑐 = 8.386 GHz.

(b)) In the above case, if sides are increased to 4cm and 8cm, then find the cut off

frequency.

𝜆c = 2𝑎𝑏

𝑚2𝑏2+𝑛2𝑎2

Here, a= 4 x 10-2, b= 8x 10-2 m

Thus, 𝜆c = 7.155 cm and fc = 𝑐

𝜆𝑐 = 4.19 GHz.

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

𝜆c = 2𝑎𝑏

𝑚2𝑏2+𝑛2𝑎2

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

38

Thus, 𝐶𝑢𝑡𝑜𝑓𝑓 𝑊𝑎𝑣𝑒 𝑙𝑒𝑛𝑔𝑡𝑕 𝜆c = 9. 285 cm and fc = 𝑐

𝜆𝑐 = 3.23 GHz.

Wave impedance for TM mode = 𝜼 𝟏 − 𝒇𝒄

𝒇 𝟐

= 𝟑𝟕𝟕 𝟏 − 𝟑.𝟐𝟑

𝟗 𝟐

= 351.8 ohms

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)

Sol: 𝜆c = 2𝑎𝑏

𝑚2𝑏2+𝑛2𝑎2

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

Thus, 𝐶𝑢𝑡𝑜𝑓𝑓 𝑊𝑎𝑣𝑒 𝑙𝑒𝑛𝑔𝑡𝑕 𝜆c = 9. 285 cm and fc = 𝑐

𝜆𝑐 = 3.23 GHz.

Wave impedance for TM mode = 𝜼 𝟏 − 𝒇𝒄

𝒇 𝟐

= 𝟑𝟕𝟕 𝟏 − 𝟑.𝟐𝟑

𝟗 𝟐

= 351.8 ohms

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

(a) fc = 𝒄

𝟐

𝒎

𝒂 𝟐

+ 𝒏

𝒃 𝟐 =

𝒄

𝟐

𝟏

𝟐.𝟐𝟖 𝟐 = 6.5789 G Hz

(b) 𝝀g = = 𝝀𝐨

𝟏− 𝒇𝒄

𝒇 𝟐

= 0.032

𝟏− 𝟔.𝟓𝟕𝟖𝟗

𝟗.𝟐 𝟐

= 0.04664 m

(c) Vp= 𝒄

𝟏− 𝒇𝒄

𝒇 𝟐

= 3x 108 1

𝟏− 𝟔.𝟓𝟕𝟖𝟗

𝟗.𝟐 𝟐

= 4.291 x 108 m/s

(d) Z TE = 𝜼

𝟏− 𝒇𝒄

𝒇 𝟐

= 377

𝟏− 𝟔.𝟓𝟕𝟖𝟗

𝟗.𝟐 𝟐

= 539.3 Ohms

39

5. An aluminum waveguide with a= 4.2 cm, b= 1.5cm, 𝜍c = 3.5x107 mhos/m, filled with

telfon ( μr = 1, ϵr = 2.6, 𝜍 = 10-15 mhos/m) operates at 4 G Hz. Determine

(a) 𝛼𝑐 𝑎𝑛𝑑 𝛼𝑑 for TE10 mode

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

Solution: Cut off frequency for TE10 mode is

f c10 = 𝟏

𝟐𝐚 𝝁𝝐 =

𝟏

𝟐𝐱 𝟎.𝟎𝟒𝟐 𝝁𝒐𝟐.𝟔𝝐 = 2.213 G Hz

δs = 𝟏

𝝅𝒇𝝁𝜍𝑐 = 1.35 x 10-6 m

Rs= 1

𝜍𝛿𝑠=

1

3.5 𝑋 107𝑋 1.35 x 10−6 = 0.0211 Ohms

𝟏 − 𝒇𝒄

𝒇 𝟐

= 1- 𝟐.𝟐𝟏𝟑

𝟒 𝟐

= 0.6939

η = η0 𝝁𝑟

𝜺𝑟 = 233.8 ohms

𝜶𝒄 =𝟐𝑹𝒔

𝒃𝛈× [ {𝟏 +

𝒃

𝒂}

𝒇𝒄

𝒇 𝟐

+ { 𝟏 − 𝒇𝒄

𝒇 𝟐

}] = 0.0133 Np/m

𝛼𝑑 =𝜍𝑑 𝜂𝑜

2 1− 𝑓𝑐𝑓

2 =

10−15 x 377

𝟐 𝒙 𝟏.𝟔𝟔 = 1.4 x 10-13 Np/m Since

𝜶𝒄 ≫ 𝜶𝒅, αd can be neglected The wave propagating in the +z direction in the

rectangular waveguide vary as e-𝜶𝒄𝒛 = e- (0.0133)(1.5) = 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

40

f c10 = 1

2a 𝜇𝜖 =

1

2𝑥0.07𝑥 𝜇09휀0= 7.14x108 Hz=0.714GHz

Phase velocity at 2 GHz =Vp= 𝑐

1− 𝑓𝑐

𝑓 2

=3.211x108 m/s

Guide wave length at 2 GHz=𝜆g = 𝜆0

1−𝜆02

𝜆c2

= 0.16 m or 16 cms

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

Wave impedance for TE mode = Zz = 𝜼/ 𝟏 − 𝒇𝒄

𝒇 𝟐

Cut off frequency for dominant TE10 mode= fc=c/2a=(3x108/2x 0.04)= 3.75 GHz

(since guide is air filled c=1/ 𝜇0휀0=3x108 m/s. For any other dielectric filling with εr

,ε=ε0 εr find c using c=1/ 𝜇0휀)

Zz= 𝜂/ 1 − 𝑓𝑐

𝑓

2

= 408.4 ohms

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

Zznew= 𝜂/ 1 − 𝑓𝑐

1.1𝑓

2

= 402.3 ohms

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

(i) fc = 𝑐

2𝑎 = 2.14 GHz.

41

(ii) Vp = 𝒄

𝟏− 𝒇𝒄

𝒇 𝟐

= 3x 108 1

𝟏− 𝟐.𝟏𝟒

𝟑.𝟓 𝟐

= 3.79 X 108 m/s

(iii) 𝝀g = = λo

1− fc

f 2

= 0.0857

𝟏− 𝟐.𝟏𝟒

𝟑.𝟓 𝟐

= 0.108 m

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.

𝜆c = 2𝑎𝑏

𝑚2𝑏2+𝑛2𝑎2

for TE20, 𝜆c = 2𝑎

2 = 8 cm

for TE30, 𝜆c = 2𝑎

3 = 5.33 cm

for TE40, 𝜆c = 𝑎

2 = 4.0cm

for TM12, 𝜆c = 2𝑎𝑏

𝑚2𝑏2+𝑛2𝑎2 =

2𝑎𝑏

𝑏2+4𝑎2 = 3.88 cm

(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.

42

OBJECTIVE TYPE QUESTIONS

Fill in the blanks

1.Write the microwave bands in the table given below

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

3. Wave guide can carry : TE & TM Modes

4. Cut-off wavelength for TE mode in rectangular waveguide is: 𝟐

𝒎

𝒂 𝟐 +

𝒏

𝒃 𝟐

5. Guide wavelength in rectangular waveguide is: 𝝀𝐨

𝟏− 𝝀𝐨

𝝀𝒄 𝟐

6. Guide wavelength in rectangular waveguide is: 𝝀𝐨

𝟏− 𝒇𝐜𝒇𝒐 𝟐

7. Wave impedance of waveguide in TE mode is: = 𝜼 𝟏 − 𝝀𝒐

𝝀𝒄 𝟐 or 𝜼 𝟏 −

𝒇𝒄

𝒇 𝟐

8. Wave impedance of waveguide in TE mode is: 𝜼 𝟏 − 𝝀𝒐

𝝀𝒄 𝟐

or 𝜼

𝟏− 𝒇𝒄

𝒇 𝟐

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

guides are 𝛻2 Ez = - 𝛚2µεEz and 𝛻2 Hz = - 𝛚2µεHz

12. TEM mode cannot exist in a waveguide.

Frequency

range

Band Frequency

range

Band Frequency

range

Band

1-2 GHz L 4-8 GHz C 12-18 GHz Ku

2-4 GHz S 8-12 GHz X 18-26.5

GHz

K

26.5 GHz-40 GHz Ka

43

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

𝒎

𝒂 𝟐 +

𝒏

𝒃 𝟐

(b) 2𝜋𝑎

𝑃𝑛𝑚 (c)

𝟐𝝅𝒂

𝑷′𝒏𝒎 (d) None of these

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

(a) 𝟐

𝒎

𝒂 𝟐 +

𝒏

𝒃 𝟐

(b) 2𝜋𝑎

𝑃𝑛𝑚 (c)

𝑐

2

𝑚

𝑎

2

+ 𝑛

𝑑

2

(d) Non of these

6. Guide wavelength of rectangular waveguide as

(a) 𝝀𝐨

𝟏− 𝝀𝐨

𝝀𝒄 𝟐

(b) 𝝀𝐨

𝟏− 𝒇𝐜𝒇𝒐 𝟐

(c) Both a and b (d) None of the above

7. Wave impedance of waveguide in TE mode can be

(a) 𝜼

𝟏− 𝒇𝒄

𝒇 𝟐

(b) 𝜂 1 − 𝑓𝑐

𝑓 2 (c) both (a) and (b) (d) None of these

8. Resonant frequency of rectangular cavity resonator is

(a) 𝒄

𝟐

𝒎

𝒂 𝟐

+ 𝒏

𝒃 𝟐

+ 𝒑

𝒅 𝟐

(b) 𝒄

𝟐

𝑚

𝑎

+ 𝑛

𝑏

+ 𝑝

𝑑

(d) 𝒄

𝟐𝝅

𝑚

𝑎

2

+ 𝑛

𝑏

2

+ 𝑝

𝑑

2

(d) None of the above

9. Attenuation constant due to conductor loss is

𝑎 𝑷𝒐𝒘𝒆𝒓 𝒅𝒊𝒔𝒔𝒊𝒑𝒂𝒕𝒆𝒅 / 𝒖𝒏𝒊𝒕 𝒍𝒆𝒏𝒈𝒕𝒉

𝟐 ×𝑷𝒐𝒘𝒆𝒓𝒇𝒍𝒐𝒘 𝑏

2 ×𝑃𝑜𝑤𝑒𝑟𝑓𝑙𝑜𝑤

𝐷𝑖𝑠𝑠𝑖𝑝𝑎𝑡𝑒𝑑 𝑃𝑜𝑤𝑒𝑟 / 𝑢𝑛𝑖𝑡 𝑙𝑒𝑛𝑔𝑡 𝑕

44

𝑐 2 × 𝑃𝑜𝑤𝑒𝑟 𝑑𝑖𝑠𝑠𝑖𝑝𝑎𝑡𝑒𝑑 / 𝑢𝑛𝑖𝑡 𝑙𝑒𝑛𝑔𝑡 𝑕

𝑃𝑜𝑤𝑒𝑟𝑓𝑙𝑜𝑤 (d) None of these

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

(c) Both (a) and (b) (d) None of these

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

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

(c) Both (a) and (b) (d) None of these

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

(c) Cut-off wavelength (d) None of these

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

(c) Travel through the dielectric without touching the walls

(d) Travel along the all the four walls

45

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

46

(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

47

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

48

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 f requencies (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