Chapter 6

Guided Waves

The main objective of this chapter is to provide the conceptual behaviour of EM waves in guided structures. The topics include:

• fields and propagation characteristics between parallel plates
• fields and propagation characteristics in hollow rectangular and circular waveguides
• TE, TM and TEM waves
• cavity resonators
• solved problems, points/formulae to remember, objective and multiple choice questions and exercise problems.

6.1 INTRODUCTION

In the previous chapter, propagation characteristics of uniform plane waves in free space were considered. These characteristics are very important when waves propagate between transmitter and receiver in free space. However, there are practical situations where propagation is by means of guided waves, that is, the EM waves may be guided between conducting planes or along a pair of wires or coaxial transmission lines. They may also be guided in wave-guide structures. In these cases, the medium of propagation is not free space.

6.2 WAVES BETWEEN PARALLEL PLATES

In these structures, the waves are no more uniform plane waves. Hence their propagation characteristics are described in terms of Transverse Electric (TE) and Transverse Magnetic (TM) waves.

6.3 DERIVATION OF FIELD EQUATIONS BETWEEN PARALLEL PLATES AND PROPAGATION PARAMETERS

Consider Fig. 6.1.

Assume that:

1. the plates are extended to infinity in y and z-directions
2. the plates are perfectly conducting
3. the plates are separated by a distance of ‘a’ metres in x-direction
4. the space between the plates is air or free space
5. the direction of power flow is z
6. all field components in z-direction vary as . Here γ_p is the propagation constant, α_p is the attenuation constant and β_p is the phase constant. In time varying form the fields vary as
   
7. the field is uniform in y-direction as the plates are extended to infinity in y-direction

As the medium between the plates is air, the first and second Maxwell’s equations are given by

 

and

 

The corresponding wave equations in air are:

 

and

 

Expanding Equation (6.1), we get

Equating the respective components on either side, we get

Similarly, expanding Equation (6.2) and equating the respective components on both sides, we get

If H = H⁰, at z = 0, the field components can be written as

and similarly

and

Substituting Equation (6.7) in Equations (6.5) and (6.6), we get

But all terms containing derivatives with respect to y vanish as the fields do not vary with y. Hence Equation (6.8) becomes

As E can be written as

and similarly

The wave equations

and

and become

or,

Solving Equation (6.9) simultaneously, we get the field components as

where

It is seen from Equation (6.12) that if E_z = 0, H_z = 0 as in the case of uniform plane wave, all the components will vanish. Therefore, there should be either E or H in the direction of propagation. In view of this, the solution is divided into two parts:

1. Transverse Electric waves (TE waves) In this E_z = 0 and H_z ≠ 0.
2. Transverse Magnetic waves (TM waves) In this H_z = 0 and E_z ≠ 0.

6.4 FIELD COMPONENTS FOR TE WAVES (E_Z = 0)

TE wave means transverse electric wave for which there is no component of E in the direction of propagation, or, E_z = 0.

For this case, as E_z = 0, it is obvious from Equation (6.12) that H_y = 0 and E_x = 0. Substituting these values in Equation (6.11), we get the component of y as,

that is,

or,

or,

where

This equation has the solution in the form of

 

where A₁ and A₂ are constants.

Using the boundary condition, E_y = 0 at x = 0, A₂ is zero.

So,

At x = a, E_y = 0 requires that

Here if m = 0, all the field components vanish.

Thus,

Substituting Equation (6.19) in Equation (6.18), the other components are obtained. The resultant expressions for the field components are:

These represent the fields for TE_m.

The field components in complete form are:

The instantaneous electric and magnetic fields for TE₁ wave at some instant of time are shown in Fig. 6.2.

TE waves are represented in the form of TE_m where m represents variation along x.

6.5 FIELD COMPONENTS OF TM WAVES (H_Z = 0)

TM wave means transverse magnetic wave for which there is no component of magnetic field in the direction of propagation, or, H_z = 0.

The wave equation for y-component of H from Equation (6.11) can be written

or,

where

The solution of Equation (6.21) appears in the form of

 

As the tangential component of H is not zero at the surface of a conductor, the boundary condition cannot be applied directly to H_y to obtain A₃ and A₄. From Equation (6.9) we have

Putting Equation (6.22) in Equation (6.23), and writing the equation for E_z, we get

The boundary conditions are:

 

 

Applying the first boundary condition, A₃ becomes zero. For the second boundary condition, E_z = 0 at x = a. E_z becomes zero if

Here, m= 0 is also possible as there still exist some more components.

From Equations (6.9) and (6.25) we get the remaining components H_y and E_x. As a whole, the expressions of field components for TM waves are:

The field components in complete form for TM_m waves are:

The field variations for TM₁ wave between parallel plates are shown in Fig. 6.3.

Dominant mode or wave     It is defined as a mode TE₁ which has the lowest cut-off frequency.

6.6 PROPAGATION PARAMETERS OF TE AND TM WAVES

We have the expression for propagation of TE and TM waves as

where             

If           , γ_p will be purely imaginary, or,

 

β_p is the phase constant

So,

If , γ_p will be purely real, or,

 

α_p is the attenuation constant

So,

For all frequencies less than ω_c,

where          

γ_p is real and β_p = 0.. This means that fields will be attenuated exponentially in z-direction and there is no wave motion as β_p = 0.

At f > f_c, γ_p = jβ_p, wave motion takes place and α_p = 0. This f_c is known as cut-off frequency.

Cut-off frequency (f_c)

Definition     Cut-off frequency, f_c is defined as a frequency below which there exists only attenuation and βp = 0 and above which α_p = 0 and β_p exists.

6.7 GUIDE WAVELENGTH

Wavelength between the plates is

But           free space wavelength

where

Now

where              λ_c = cut-off wavelength

The propagation parameters are:

For TE₁

6.8 TRANSVERSE ELECTROMAGNETIC WAVE (TEM WAVE)

 

The field components of TEM wave are obtained from m = 0. They are:

The propagation parameters for TEM wave are

TEM wave between parallel plates are given in Fig. 6.4.

6.9 VELOCITIES OF PROPAGATION

The phase velocity in a waveguide is always greater than free space velocity.

In free space, velocity of propagation is equal to the phase velocity.

Group velocity in waveguide is less than free space velocity.

The velocities of propagation are related by

where

6.10 ATTENUATION IN PARALLEL PLATE GUIDES

Attenuation factor in general

Attenuation factor for TEM wave,

where	a = plate separation, m
	σc = conductor conductivity, mho/m
	μc = conductor permeability, H/m
	ω = 2π f
	ηp = intrinsic impedance, Ω

Attenuation factor is also given by

where	R = resistance, Ω/m
	z0 = characteristic impedance, Ω

Attenuation of TE waves

 

where	m = 1, 2, 3, …
	ω = 2π f
	μ = permeability of medium between plates
	∈ = permittivity of medium between plates

6.11 WAVE IMPEDANCES

Wave impedances at any point between parallel plates are defined by the following relations:

These wave impedances can be evaluated from field expressions. For example,

Problem 6.1     If a wave of 6 GHz is propagating between two parallel conducting plates separated by 30 mm, find the cut-off wavelength, guide wavelength for TE₁ mode.

Solution     Frequency of the wave,

 

Plate separation,

 

Cut-off wavelength for TE₁ is

Free space wavelength,

The guide wavelength

Problem 6.2     When a wave of 6 GHz propagates in parallel conducting plates separated by 3 cm, find the phase velocity, group velocity of the wave for the dominant wave.

Solution     Frequency of the wave,

 

	f = 6 GHz
Plate separation,	a = 3 cm
Free space wavelength,
	λ = 5 cm
Cut-off wavelength,	λc = 2 × a = 2 × 3 = 6 cm

Phase velocity,

Group velocity,

Problem 6.3     When a wave of 6 GHz is to be propagated between two parallel conducting plates separated by 60 mm, find the modes that will propagate through the guide.

Solution     The modes which have their cut-off frequencies less than the frequency of the wave will propagate.

 

Here, plate separation,	a = 60 mm = 6.0 cm
	f = 6 GHz = 6 × 109 Hz
For TEm
For TE1
As fc < f, TE1 propagates.
For TE2
It propagates.
For TE3
As fc > f, it does not propagate.
For TE4
As fc > f, this will not propagate.

6.12 WAVES IN RECTANGULAR WAVEGUIDES

 

6.13 DERIVATION OF FIELD EQUATIONS IN RECTANGULAR HOLLOW WAVEGUIDES

Field expressions can be obtained from the solutions of Maxwell’s equations and wave equations.

Assumptions:

 

As the medium inside the waveguide is air, the first and second Maxwell’s equations are given by

 

 

Expanding these equations, we get

Equating the respective components, we get

and

As the fields are assumed to be varying in the form of , combining time variation, we get

Similarly,

and             

Substituting Equation (6.32) in Equations (6.30) and (6.31), we get

The wave equations are:

 

 

These can be written as

The wave equations for E_z and H_z are given by

Equation (6.33) can be mathematically manipulated to get the following. Consider

and

Equation (6.35) becomes

But E_y from Equation (6.36) is

From Equations (6.37) and (6.38), we get

or,

or,

where             

Similarly,

In the above equations, if E_z = 0 and H_z = 0, all the field components vanish. Hence, the wave cannot satisfy TEM wave characteristics. They are transverse magnetic (TM) and transverse electric (TE) waves. A typical rectangular waveguide is shown in Fig. 6.5.

Transverse Magnetic (TM) Waves in Rectangular Waveguide

TM waves are EM waves for which there is no component of H in the direction of propagation, that is, H_z = 0.

The wave equations given by Equation (6.34) can be easily solved using the method of product solution. In this method, two ordinary differential equations with known solutions are obtained. We know that,

If

 

where X is only a function of x and Y is only a function of y.

From Equations (6.34) and (6.35), we can write

or,

This expression equates a function of x to another function of y. This is possible when each of these functions is equal to some constant. Let the constant be B

and

The general solution of these equations are

 

	X = A1 cos Cx + A2 sin Cx
where
and	Y = A3 cos By + A4 sin By

= XY

       = A₁A₃ cosCx cos By + A₁ A₄ cos Cx sin By

                + A₂ A₃ sin Cx cos By + A₂ A₄ sin Cx sin By

The constants A₁, A₂, A₃, A₄ are evaluated using the boundary conditions. Using the boundary condition

         = 0 at x = 0

         = A₁ A₃ cos By + A₁A₄ sin By

This is zero if A₁ = 0

 

At y = 0, Equation (6.44) becomes

 

For this to vanish, A₂ or A₃ can be zero, while assuming C ≠ 0. Keeping A₂ = 0 in Equation (6.44), becomes zero. Hence instead of A₂ = 0 assume A₃ = 0. Then Equation (6.44) becomes

         = A₂ A₄ sin Cx sin By

               = K sin Cx sin By

 

[K = A₂ A₄]

At x = a

         = K sin Ca sin By

For this to vanish for all values of y (assuming B ≠ 0) the constant C must be

At y = b,

For this to vanish for all values of x, B must be

Hence, the final expression for is

From Equations (6.44) and (6.45), we get

where             

Therefore, the field components for a TM wave are:

TM_mn wavefield components

Field variations of TM₁₁ mode/wave are shown in Fig. 6.6.

Transverse Electric Waves

TE waves are EM waves for which there is no component of E in the direction of propagation, that is, E_z = 0.

The expressions for TE waves are derived in the same manner as in the case of TM waves. From Equation (6.39), we have

But E_z = 0

Hence,       

The first boundary condition is

 

that is,       

But

     = (A₁ cos Cx + A₂ sin Cx).(A₃ cos By + A₄ sin By)

     = (A₁ cos Cx + A₂ sin Cx).(−A₃ B sin By + A₄ B cos By)

As          

The secondary boundary condition is E_x = 0 at y = b

A is             

B should be

Moreover,             

and          

Substituting Equation (6.47) in Equation (6.39), we get

The field expressions for TE waves in complete form are

where

As TE₁₀ is popular, its field equations are

The field patterns are shown in Fig. 6.7.

6.14 PROPAGATION PARAMETERS OF TE AND TM WAVES IN RECTANGULAR WAVEGUIDES

 

We have
and
or,

If

If

If

This particular ω corresponds to ω_c, the angular cut-off frequency.

or,

And cut-off wavelength is

The velocity of wave propagation is

Derivation of guide wavelength in terms of free space and cut-off wavelengths

The expression for guide wavelength is

As

But

or,

The summary of propagation parameters of TE and TM waves are:

 

In TE_mn or TM_mn waves, m represents the number of half-period variations of the field along x-axis and n represents the number of half-period variations of the field along y-axis. Here, the broad wall is along the x-axis and the narrow wall is along the y-axis.

Propagation and field equation of dominant mode, TE₁₀

1. Cut-off frequency for dominant mode, TE₁₀
   
2. Cut-off wavelength for dominant mode, TE₁₀

    

   λ_c of TE₁₀ = 2a

Here	λc = cut-off wavelength (m)
	a = broad wall dimension (m)
	m, n = integers = 1,2,3, …

Transverse Electromagnetic Waves

In TEM wave, both E and H are entirely transverse to the direction of propagation, that is, if the direction of propagation is along z, E_z = 0, and H_z = 0.

TEM wave is called principal wave. Its cut-off frequency is zero and it exists in two conductor transmission lines or in free space.

Characteristics of TEM waves

 

6.15 TEM WAVE DOES NOT EXIST IN HOLLOW WAVEGUIDES

Proof     Method 1     Consider TM wavefield equations given by

As TEM = TM₀₀, that is, when m = 0, n = 0, all the above field components vanish. This itself indicates that there exists no TEM waves in hollow waveguides.

Method 2     Assume that TEM wave exists within a hollow waveguide. Then, the magnetic field lines must be in the transverse plane. Also we know that

 

or,

 

This requires that the lines of H be closed loops. Hence, if a TEM wave exists (by hypothesis) inside the waveguide, the lines of H are closed loops in a plane perpendicular to the axis of the guide. It may be noted that the direction of propagation is along the axis.

By Maxwell’s first equation, we have

that is, the magnetomotive force around the closed loop of H lines is equal to the sum of axial displacement and conduction currents. As the space inside the guide is air or free space,

 

 

[as σ = 0]

that is, conduction current is zero. Hence the axial current must be a displacement current. If there exists displacement current in the axial direction which is the direction of propagation of EM energy, there should be a component of E in the axial direction. The presence of E along the axial or direction of propagation indicates the absence of TEM wave. Therefore, we conclude that there exists no TEM in hollow waveguides of any shape.

6.16 EXCITATION METHODS FOR DIFFERENT TE AND TM WAVES/MODES

The excitation of TE₁₀ by a probe is shown in Fig. 6.8.

The excitation of TE₁₀ by a loop is shown in Fig. 6.9.

The excitation method of TM₁₁ by a probe and a loop are shown in Figs. 6.10 and 6.11.

6.17 EVANESCENT WAVE OR MODE

This is defined as a wave TE_mn or TM_mn in which the operating frequency is less than the cut-off frequency and wave propagation does not take place. For evanescent wave, the TM_mn wave impedance is purely capacitive and this causes only reactive power or energy storage.

6.18 WAVE IMPEDANCE IN WAVEGUIDE

For TE_mn wave, wave impedance is defined as

or,

where η = intrinsic impedance of the unbounded medium.

The wave impedance is purely resistive and average power flow occurs in the waveguide when f > f_c.

For non-propagating TM_mn wave, the wave impedance

when f < f_c for a particular TM_mn mode.

The variation of magnitude of wave impedances of TE_mn and TM_mn for f < f_c is shown in Fig. 6.12.

The wave impedance for the propagating modes is found to be

or,

From the above equations, we get

6.19 POWER TRANSMITTED IN A LOSSLESS WAVEGUIDE

The average power transmitted in z-direction is found by integration of the z component of the complex Poynting vector over a transverse cross-section of the waveguide, that is,

where           H* = complex conjugate of H

Power transmission takes place through TE₁₀ wave. Using the corresponding component values of E and H*, P_av is given by

In an ideal guide, P_av is independent of z-direction.

Power Dissipation in a Lossy Waveguide

When the conductivity of the dielectric (σ_d) in the waveguide is non-zero and the conductivity (σ_c) of the walls is not infinite, wave in the propagating mode will be attenuated and the transmitted power will decrease exponentially with z.

The attenuation factor due to dielectric loss is the indication of power loss for TE₁₀ mode, which is given by

The attenuation factor due to wall loss indicates power loss and is given by

where P_loss = power flow into the first 1m of the inner surface of the wall

and simplified expression for α _w of TE₁₀ mode is given by

where           R_SC = surface resistance at cut-off frequency of TE₁₀, Ω

            

The total attenuation factor is

 

Problem 6.4     Find the cut-off frequencies for TE₁₂ mode in a hollow rectangular waveguide whose dimensions are:

1. a = 2.286 cm, b = 1.016 cm
2. a = 1.016 cm, b = 2.286 cm
3. a = 1 cm, b = 1 cm
4. a = 10 cm, b = 10 cm

Solution     The cut-off frequency for TE₁₂ mode is given by

1. a = 2.286 cm, b = 1.016 cm
   
2. a = 1.016 cm, b = 2.286 cm
   
3. a = 1 cm, b = 1 cm
   
4. a = 10 cm, b = 10 cm
   

Problem 6.5     A rectangular waveguide with dimensions 3 × 2 cm operates at 10 GHz. Find f_c, λ_c, λ, λ_g, β_g, v_p of TE₁₀ mode.

Solution     a = 3 cm, b = 2 cm, f = 10 GHz

For TE₁₀ mode,

Problem 6.6     Find the broad wall dimension of a rectangular waveguide when the cut-off frequency for TE₁₀ mode is (a) 3 GHz, (b) 30 GHz.

Solution

1. f_c = 3 GHz for TE₁₀ mode
   

   or,

   
2. f_c = 30 GHz for TE₁₀ mode
   

Problem 6.7     A hollow rectangular waveguide operates at f = 1 GHz and it has the dimensions of 5 × 2 cm. Check whether TE₂₁ mode propagates or not.

Solution     The propagation constant is given by

Here	a = 5 cm = 0.05m
	b = 2 cm = 0.02m
	f = 1GHz = 109 Hz
For TE21	m = 2, n = 1
	μ0 = 4 π × 10−7 H/m
	∈0 = 8.854 × 10−12 F/m
γg for TE21 mode is

As γ_g is purely real, there is no propagation of TE₂₁ mode.

6.20 WAVEGUIDE RESONATORS

It is used for energy storage. As there is no propagation through the shorted ports, standing waves exist inside the cavity. These resonators are used for various applications, particularly in klystrons and wave metres.

Features of Resonators

 

TM Mode (H_z = 0)

 

If	Ez = (x, y, z) = X(x) Y(y) Z(z),
We can write,	X(x) = C1 cos Ax + C2 sin Ax
	Y(y) = C3 cos By + C4 sin By
	Z(z) = C5 cos Cz + C6 sin Cz

where

and

Here, we have three boundary conditions to solve the constants, C₁, C₂,…etc.

       E_z = 0 at x = 0 and at x = a

       E_z = 0 at y = 0 and at y = b

       E_x = 0, E_y = 0 at z = 0 and at z = c

After simplification, E_zr becomes

where              E_m = C₂ C₄ C₅

The resonant frequency is given by

or,

and

TE Mode (E_z = 0)

As in the case of TM mode, H_z is expressed

 

	Hz (x, y, z) = X(x) Y(y) Z(z)
where	X(x) = p1 cos Bx + p2 sin Bx
	Y(y) = p3 cos Ay + p4 sin Ay
	Z(z) = p5 cos Cz + p6 sin Cz

The set of boundary conditions are

Using the boundary conditions, and simplifying, we get

where	m = 0,1,2,3,…
	n = 0,1,2,3,…
	l = 0,1,2,3,…

Dominant Mode

 

The waves are represented by TE_mnl, TM_mnl.

Degenerate Mode

Modes having the same resonant frequency are called degenerate modes. Ideally, the walls of the resonant cavity have infinite conductivity. But practically, cavity walls have finite conductivity. As a result, some stored energy is lost.

Quality Factor, Q

Quality factor is defined as

where	ω = 2πf
	WL = average power loss in a cycle
	Wav = average stored energy

Quality factor, for dominant mode, TE₁₀₁ is

where           δ = depth of penetration in cavity walls

6.21 SALIENT FEATURES OF CAVITY RESONATORS

1. A completely closed metallic structure forms a cavity and it is called cavity resonator.
2. It stores energy.
3. TE and TM modes exist in the cavity.
4. In TE mode, E_z = 0 (z is propagation direction) and E_x, E_y, H_x, H_y and H_z are present.
5. In TM mode, H_z = 0 and H_x, H_y, E_x, E_y and E_z are present.
6. In cavities, the electric and magnetic fields do not propagate along z-axis but they oscillate with time at a specified location.
7. The lowest order of TM_mnl mode is TM₁₁₀.
8. The resonant frequency of the lowest order TM mode is
   
9. The lowest order for TE_mnl is TE₁₀₁.
10. The resonant frequency of the lowest order TE mode is
    

Problem 6.8     A copper rectangular cavity resonator is structured by 3 × 1 × 4 cm. Find its resonant frequency for TM₁₁₀ mode.

Solution     The dimensions of resonator are:

 

For TM₁₁₀, the resonant frequency is

Problem 6.9     A copper walled rectangular cavity resonator is structured by 3 × 1 × 4 cm and operates at the dominant modes of TE and TM. Find its resonant frequency and quality factor. The conductivity of copper is 5.8 × 10⁷ mho/m. There is air inside the cavity.

Solution     For TM mode, the dominant mode is TM₁₀₀. Its resonant frequency is

Here

For TE mode, the resonant frequency of the dominant, TE₁₀₁ is

The quality factor, Q for TE₁₀₁

where

Problem 6.10     A copper walled resonant cavity is dielectric (∈_r = 4) filled and its dimensions are 5 × 4 × 10 cm. Determine the resonant frequency of TE₁₀₁ and its quality factor.

Solution     The conductivity of copper,

           σ_c = 5.8 × 10⁷ mho/m

             a = 5 cm

             b = 4 cm

             c = 10 cm

The resonant frequency of TE₁₀₁ is

The quality factor, Q for this mode is

and

6.22 CIRCULAR WAVEGUIDES

The waveguides of circular cross-section (Fig. 6.14) are used to transmit EM waves from one point to another. Unlike rectangular waveguides, the circular waveguides do not have unique orientation as it is perfectly symmetrical around the axis.

6.23 SALIENT FEATURES OF CIRCULAR WAVEGUIDES

1. It is easy to manufacture.
2. They are used in rotational coupling.
3. Rotation of polarisation exists and this can be overcome by rotating modes symmetrically.
4. TM01 mode is preferred to TE₀₁ as it requires a smaller diameter for the same cut-off wavelength.
5. TE₀₁ does not have practical application.
6. For f > 10 GHz, TE₀₁ has the lowest attenuation per unit length of the waveguide.
7. The main disadvantage is that its cross-section is larger than that of a rectangular waveguide for carrying the same signal.
8. The space occupied by circular waveguides is more than that of a rectangular waveguide.
9. The determination of fields here consists of differential equations of certain type. Their solutions involve Bessel functions.
10. Here also TE and TM modes exist.

    For TM wave, the solution of axial component

     

    E_z,nm = J_n (K_c r) (A cos nϕ + B sin ϕ)

    and for TE wave, it is

     

    H_z,nm = J_n (K_c r) (A′ cos nϕ + B′ sin nϕ)

where	Jn(Kc r) = Bessel function of the first kind
	r = the radius of the guide
	Kc = the cut-off wave number
	A, B, A′, B′ = constants

The solutions for the Bessel function are obtained for certain values of K_c where these values of K_c are known as eigen values. If K_c is to produce solution of the Bessel function, (K_cr) must be the roots of the Bessel function. Then

 

The propagation parameters for nm^th mode TM waves are:

Phase constant,

 

where
and
where	pnm = the roots of the Bessel function
	K = free space wave number =

The cut-off wavelength for TM wave,

The roots of the Bessel function for TM mode are shown in Table 6.1.

 

The roots of the Bessel function for TE mode are shown in Table 6.2.

 

For circular waveguides, TE₁₁ is the dominant mode. The propagation parameters for TE_nm mode are:

where

Guide wavelength,

Problem 6.11     If the radius of a circular waveguide r = 1.27 cm, f = 10 GHz, find the cut-off wavelength for the dominant mode and phase constant. Assume that the waveguide is air-filled. Take p′₁₁ = 1.841.

Solution     For the dominant mode,

 

For TE₁₁, we have

The phase constant,

Here

Problem 6.12     Determine the size of the circular waveguide required to propagate TE₁₁ mode if λ_c = 8 cm (ρ′₁₁ = 1.841).

Solution     We have

or,

Radius of the guide

 

1. TEM mode exists in parallel plate structure.	(Yes/No)
2. TEM mode exists in a hollow rectangular waveguide.	(Yes/No)
3. The quality factor is nothing but figure of merit of a cavity resonator.	(Yes/No)
4. If quality factor is high, the power loss in the walls of a waveguide is less.	(Yes/No)
5. The phase constant, βmn of TE wave is the same as that of TM wave in a rectangular waveguide.	(Yes/No)
6. The wavelength, λmn of TM wave is the same as that of TE wave in a rectangular waveguide.	(Yes/No)
7. Phase velocity, vp in TE wave is the same as that of TM wave in a rectangular waveguide.	(Yes/No)
8. Group velocity, vg in TE wave is the same as that of TM wave in a rectangular waveguide.	(Yes/No)
9. A rectangular waveguide can be excited for TE10 mode by a loop.	(Yes/No)
10. The presence of modes depends on the shape of the waveguide.	(Yes/No)
11. The presence of modes depends on the dimensions of the waveguide.	(Yes/No)
12. The presence of modes depends on the medium inside the waveguide.	(Yes/No)
13. The presence of modes depends on the operating frequency.	(Yes/No)
14. A rectangular waveguide can be used as an antenna.	(Yes/No)
15. A rectangular waveguide can be used as a transmission line.	(Yes/No)
16. A calibrated cavity resonator is nothing but a frequency meter at microwave frequency.	(Yes/No)
17. In cavity resonator, standing waves are absent.	(Yes/No)
18. A cavity resonator at microwave frequency is similar to a low frequency resonant circuit.	(Yes/No)
19. Each mode of TE wave in a rectangular waveguide has a different cut-off frequency.	(Yes/No)
20. In a hollow rectangular waveguide, TE mode has all the components of E and H.	(Yes/No)
21. In a hollow rectangular waveguide, TM mode has all the components of E and H.	(Yes/No)
22. The phase velocity in a rectangular waveguide is greater than that in free space.	(Yes/No)
23. The group velocity for each TEmn mode in rectangular waveguide is different.	(Yes/No)
24. The phase velocity for each TEmn mode is the same as in a rectangular waveguide.	(Yes/No)
25. Guide wavelength varies with each TE and TM mode.	(Yes/No)
26. Guide wavelength depends on waveguide dimensions.	(Yes/No)
27. Evanescent mode is a mode which does not propagate.	(Yes/No)
28. For evanescent mode, the propagation constant is equal to attenuation constant.	(Yes/No)
29. Wave impedance in a waveguide is smaller than that of free space.	(Yes/No)
30. If the impedance is resistive in a waveguide, the average power flow exists.	(Yes/No)
31.	(Yes/No)
32. The cut-off frequencies of TE10 and TE01 are the same.	(Yes/No)
33. If the operating frequency is 10 GHz and cut-off frequency for a TE10 in a rectangular waveguide is 12 GHz, wave propagation takes place.	(Yes/No)
34. In cavity resonators, there exists no reflected waves.	(Yes/No)
35. In rectangular waveguides, if the propagation constant is purely real, propagation takes place.	(Yes/No)
36. Phase velocity is nothing but the velocity of propagating energy in a waveguide.	(Yes/No)
37. Phase velocity is the speed of the constant phase points on the travelling wave.	(Yes/No)
38. Propagation constant is the same for TE and TM modes in a waveguide.	(Yes/No)
39. If the conductivity of the wall of a waveguide is high, the skin depth is small.	(Yes/No)
40. The resonant frequency depends on the length of the cavity resonator.	(Yes/No)
41. TEM wave means _____.
42. TE wave means _____.
43. TM wave means _____.
44. In TE mode, the electric field in the direction of propagation, z is equal to _____.
45. When an EM wave is propagating in z-direction, the magnetic field in z-direction for TM wave is _____.
46. The cut-off frequency of TEM mode is _____.
47. The cut-off wavelength of TEM mode is _____.
48. Cavity resonator is obtained from a rectangular waveguide by _____.
49. The cut-off frequency for the dominant mode in a rectangular waveguide is _____.
50. The quality factor of a resonating system is _____.
51. The propagation constant, γnm for TEmn at operating frequency less than cut-off frequency is _____.
52. At an operating frequency greater than the cut-off frequency of TEmn wave, the propagation constant is _____.
53. The rectangular waveguide can be excited for TE10 mode by _____.
54. The power flow in z-direction in a waveguide requires the presence of the components of _____.
55. The number of modes in rectangular waveguides are _____.
56. The modes in a waveguide are _____.
57. A waveguide is usually made of _____.
58. The field components in a waveguide are obtained using _____.
59. In TEM mode in a hollow rectangular waveguide, __________ components of E and H are present. _____.
60. The phase velocity in a rectangular waveguide is __________ when f = fc. _____.
61. The relation between phase, group velocities and free space velocity is _____.
62. If the wave impedance is capacitive in a waveguide, the average power flow is _____.
63. For the best cavity resonator, the quality factor is _____.

1. When a wave is propagating in x-direction, TE wave has E_x equal to
   1. zero
   2. E_y
   3. E_z
   4. E_y + E_z
2. For a wave propagating in z-direction in a hollow rectangular waveguide, TEM wave has
   1. E_z = o
   2. H_z = o
   3. all components of E and H are zero
   4. E_x = o, H_z = o
3. The cut-off frequency of TEM wave is
   1. zero
   2. f
   3. infinity
   4. that of TE₁₀
4. If a = 2 cm, b = 1 cm for a waveguide, the cut-off frequency for TE₁₀ mode is
   1. 0.75 GHz
   2. 7.5 GHz
   3. 1.5 GHz
   4. 1.5 MHz
5. If the cut-off frequency for the dominant mode in a rectangular waveguide is 4 GHz, the dimension of the broad wall along x-axis is
   1. 4 cm
   2. 3.75 cm
   3. 5 cm
   4. 37.5 cm
6. If f =10 GHz, the broad dimension of a rectangular waveguide, a = 2 cm, the cut-off frequency for TE₁₀ is
   1. 0.75 GHz
   2. 7.5 GHz
   3. 10 GHz
   4. 5 GHz
7. At f = 10 GHz, a = 2 cm, b = 1 cm, the cut-off frequency for TE₀₁ is
   1. 10 GHz
   2. 1.5 GHz
   3. 15 GHz
   4. 7.5 GHz
8. If in a rectangular waveguide the attenuation constant is 2 dB/m and the phase constant is 2 rad/m, the propagation constant is
   1. 2 + j2 dB/m
   2. 2 + j2 np/m
   3. 2 dB/m
   4. 2 np/m
9. If β = 3oo rad/m for TE₂₁ mode in a rectangular waveguide, the guide wavelength for TE₂₁ is
   1. 0.209 cm
   2. 0.209 m
   3. 2.09 m
   4. 20.9 cm
10. The cut-off frequency for TE₁₀ is 5 GHz at an operating frequency of 6 GHz. The phase constant for TE₁₀ is
    1. 69.5 rad/m
    2. 6.95 rad/m
    3. 0.833 rad/m
    4. 1.2 rad/m
11. If the resonant frequency is 6 GHz, the time averaged stored energy is 10 Joules and the power dissipation is 20 watts in a resonator, the quality factor is
    1. 188.49
    2. 188.49 × 10⁹
    3. 200
    4. 188
12. The ideal value of quality factor is
    1. ∞
    2. 0
    3. 100
    4. very small
13. If , the intrinsic impedance is
    1. 122.47Ω
    2. 1.5Ω
    3. 0.333Ω
    4. 12.247Ω
14. If the phase constant of TM₂₁ is 320rad/m, α = 0 at an operating frequency of 10 GHz, the characteristic impedance of TM₂₁ is
    1. 32Ω
    2. 539Ω
    3. 300Ω
    4. 53.9Ω
15. The cut-off wavelength for the dominant mode in a rectangular waveguide having dimensions 4 × 3 cm is
    1. 12 cm
    2. 8 cm
    3. 1.33 cm
    4. 0.75 cm
16. The cut-off wavelength for TE₃₂ in a waveguide is 10 GHz. The guide wavelength for TE₃₂ mode is
    1. 3 cm
    2. 3.2 cm
    3. 6 cm
    4. 3 m
17. Skin depth in the copper wall of a waveguide operating at f = 6.25 GHz is
    1. 8.39 mm
    2. 83.9 mμ
    3. 83.9 mm
    4. 839 mm
18. The mode of propagation in an air-filled waveguide of 2 × 1 cm dimensions operating at 10 GHz is
    1. TE₀₁
    2. TE₂₁
    3. TE₁₀
    4. TE₁₁
19. If the propagation constant in a waveguide is 105+ j10, np/m the attenuation constant is
    1. 105 np/m
    2. 105 dB/m
    3. 10.5 np/m
    4. 115 np/m
20. If γ_mn = α_mn, in a waveguide
    1. propagation takes place
    2. no propagation takes place
    3. energy is increased
    4. energy is decreased

1. A rectangular waveguide operates at 1 GHz and has dimensions of 5 × 2 cm. Find the distance from the source end at which the electric field of TE₂₁ wave becomes 0.5% of its starting amplitude at z = 0. The amplitude of the electric field in z-direction at z =0 is 10kV/m.
2. A rectangular waveguide with dimensions of 2.5 × 1.0 cm operates below 15.0 GHz. What are the TE modes that can be propagated if the waveguide is filled with material whose relative permittivity, ∈_r = 4 and μ_r = 1, σ = 0? Find their cut-off frequencies and cut-off wavelengths.
3. A hollow rectangular waveguide operates at 10 GHz. If the cut-off frequency of TM₂₁ mode is 6 GHz, find its phase velocity, wavelength, phase constant.
4. A hollow rectangular waveguide has the dimensions of a = 2.286 cm and b = 1.016 cm. Find the cut-off frequencies of TM₁₁ and TM₂₀.
5. Find the cut-off wavelength of TE₁₀ and TE₂₀ mode in the hollow rectangular waveguide of dimensions a = 3 cm, b = 1.5 cm.
6. In a hollow square waveguide, the cut-off frequency for TE₁₀ mode is (a) 10 GHz (b) 20 GHz (c) 30 GHz. Find the dimensions of the waveguide.
7. Find the radius of the circular waveguide required to propagate TE₁₁ mode if λ_c = 10 cm. (ρ′₁₁ = 1.84).
8. Find the cut-off frequency of TE₂₁ mode in a circular waveguide of radius 4 cm. (Take ρ′₂₁ = 3.054).