Parallel Plane Waveguide MCQ || Parallel Plane Waveguide Questions & Answers

Ques.1.  The magnitude of (|E| / |H|) in a uniform plane wave is:

1. √µε
2. Infinity
3. $\sqrt {\frac{\mu }{ \in }}$
4. 1

Answer.3. $\sqrt {\frac{\mu }{ \in }}$

Explanation

Electric field (E) & magnetic field (H) combination called uniform plane wave because E & H has some magnitude through any transverse plane.

Intrinsic Impedance (η) = E/H

And also

$\eta = \sqrt {\frac{{j\omega \mu }}{{\sigma + j\omega \varepsilon }}}$

$\Rightarrow \eta = \frac{E}{H} = \sqrt {\frac{\mu }{\varepsilon }}$

Pointing vector is a vector whose direction is the direction of wave propagation pointing vector =E→×H→ (Hence it is a direction of wave propagation)

Where,

E = Electric filed

B = magnetic field

H = Magnetic field

μ0 = Permeability of free space = 4π x 10-7 H / m

ϵ0 = Permittivity of free space = 8.85 x 10-12 F/m

Ques.2. For transverse electric waves between parallel plates, the lowest value of m, without making all the field components zero, is equal to

1. 3
2. 2
3. 1
4. 0

Explanation

For parallel plate waveguide, the propagation constant is defined as:

β = mπ/a

a = distance between the plates

Where the minimum value of m = 1

Cut-off frequency for the parallel plate waveguide is given by:

fc = m/2a

The wavelength will be:

λc = 2a/m

Ques.3. The dispersion equation of a waveguide, which relates the wavenumber k to the frequency ω, is $k\left( \omega \right) = \left( {1/\:c} \right)\:\sqrt {{\omega ^2} – \omega _o^2}$ where the speed of light c = 3 × 108 m/s, and ωo is a constant. If the group velocity is 2 × 108 m/s, then the phase velocity is

1. 1.5 × 108 m/s
2. 2 × 108 m/s
3. 3 × 108 m/s
4. 4.5 × 108 m/s

Explanation

Group velocity × Phase velocity = c2

2 × 108 × phase velocity = (3 × 108)2

VP = (3 × 1016)/(2 × 108 )

VP = 4.5 × 108 m/s

Ques.4. The phase velocity of waves propagating in a hollow metal waveguide is

1. Equal to the group velocity
2. Equal to the velocity of light in free space
3. Less than the velocity of light in free space
4. Greater than the velocity of light in free space

Answer.4. Greater than the velocity of light in free space

Explanation

The phase velocity of waves propagating in a hollow metal waveguide is Greater than the velocity of light in free space.

Phase velocity is defined as:

Vp = ω/β

β is the phase constant defined as:

$\begin{array}{l} \beta = \sqrt {{\omega ^2}\mu \in – {{\left( {\frac{{m\pi }}{a}} \right)}^2}} \\ \\ {V_p} = \frac{\omega }{{\sqrt {{\omega ^2}\mu \in – {{\left( {\frac{{m\pi }}{a}} \right)}^2}} }}\\ \\ {V_p} = \frac{1}{{\sqrt {\mu \in – {{\left( {\frac{{m\pi }}{{a\omega }}} \right)}^2}} }}\\ \\ {V_p} = \frac{{\frac{1}{{\sqrt {\mu \in } }}}}{{\sqrt {1 – {{\left( {\frac{{m\pi }}{{a\omega \sqrt {\mu \in } }}} \right)}^2}} }} \end{array}$

Using $c = \frac{1}{{\sqrt {\mu C} }}$

where c =  speed of light, the above expression becomes:

${V_p} = \frac{c}{{\sqrt {1 – {{\left( {\frac{{m\pi C}}{{a\omega }}} \right)}^2}} }};$

Also

ωc = mπc/a

${V_p} = \frac{C}{{\sqrt {1 – {{\left( {\frac{{{\omega _c}}}{\omega }} \right)}^2}} }}$

Using Sinθ = ωc

$\begin{array}{l} {V_p} = \frac{c}{{\sqrt {1 – {{\sin }^2}\theta } }}\\ \\ {V_p} = \frac{c}{{\cos \theta }}; \end{array}$

Since -1 ≤ cos θ ≤ 1

Vp > c

Ques.5. A rectangular waveguide of width w and height h has cut-off frequencies for TE10 and TE11 modes in the ratio 1 : 2. The aspect ratio w/h, rounded off to two decimal places, is __________.

1. 2.89
2. 1.73
3. 5.42
4. 0

Explanation

The cutoff frequency of TEmn mode is given by:

${f_c} = \frac{c}{2}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}}$

for TE10 mode

${f_{{c_{10}}}} = \frac{c}{2}\left( {\frac{1}{a}} \right)$ ——- (1)

for TE11 mode

${f_{{c_{11}}}} = \frac{c}{2}\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}}$ —— (2)

Dividing equation 1 by equation 2 we get

$\begin{array}{l} \frac{1}{2} = \frac{{\frac{c}{2}\left( {\frac{1}{a}} \right)}}{{\frac{c}{2}\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}} }}\\ \\ = \frac{1}{2} = \frac{{\frac{c}{{2a}}}}{{\frac{c}{2}\sqrt {\frac{{{a^2} + {b^2}}}{{ab}}} }}\\ \\ = \frac{b}{{\sqrt {{b^2} + {a^2}} }}\\ \\ \frac{1}{2} = \frac{1}{{\sqrt {1 + {{\left( {\frac{a}{b}} \right)}^2}} }}\\ \\ {\left( {\frac{a}{b}} \right)^2} = {\left( {\frac{w}{h}} \right)^2} \end{array}$

= 4 − 1 = 3

ω/h = √3 = 1.732

Ques.6. Transverse Electromagnetic waves are characterized by

1. During wave propagation in Z-direction, the components of H and E are transverse 60° to the direction of propagation of the waves
2. During wave propagation in Z-direction, the components of H and E are transverse to the direction of propagation of the waves.
3. During wave propagation in Z-direction, the components of H and E are transverse 120° to the direction of propagation of the waves
4. None of the above

Answer.2. During wave propagation in Z-direction, the components of H and E are transverse to the direction of propagation of the waves.

Explanation

Transverse Electromagnetic waves:

• In an electromagnetic wave, electric and magnetic field vectors are perpendicular to each other and at the same time are perpendicular to the direction of propagation of the wave.
• This nature of electromagnetic wave is known as Transverse nature.
• Maxwell proved that both the electric(E) and magnetic fields(H) are perpendicular to each other in the direction of wave propagation. He considered an electromagnetic wave propagating along the positive z-axis
• The electric and magnetic fields propagate sinusoidal with the z-axis.

Ques.7. The cut-off frequency of TE01 mode of an air-filled rectangular waveguide having inner dimensions a cm × b cm (a > b) is twice that of the dominant TE10 mode. When the waveguide is operated at a frequency that is 25% higher than the cut-off frequency of the dominant mode, the guide wavelength is found to be 4 cm. The value of b (in cm, correct to two decimal places) is _______.

1. 3
2. 0.5
3. 0.75
4. 2.75

Explanation

Dominant mode of rectangular waveguide is TE10

The cutoff frequency is given as:

${f_c} = \frac{c}{2}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}}$

Given:

(fc)TE01 = 2(fc)TE10 ——-1

Operating frequency:

f = (fc)TE10 + 25% of (fc)TE10

f = 1.25(fc)TE10

For air filled wavelength C, we can write:

${f_c} = \frac{c}{2}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}}$

For TE01 mode

m = 0, n = 1

(fc)TE10 = c/2a

For TE10 mode

m = 1; n = 0

(fc)TE10 = c/2a ——2

From equation (1)

(fc)TE01 = 2(fc)TE10

c/2b = 2c/2a

a = 2b

operating frequency:

f = (1.25) (fc)TE01 = 2(fc)TE10

$f = \left( {1.25} \right)\left[ {\frac{c}{{2a}}} \right]$

=  1.25c/4b

Operating wavelength

λ = c/f = 4b/1.25 = 3.2b

Guide wavelength is given by:

$\begin{array}{l} {\lambda _g} = \frac{\lambda }{{\sqrt {1 – {{\left( {\frac{{{f_c}}}{f}} \right)}^2}} }}\\ \\ 4 = \frac{{3.2b}}{{\sqrt {1 – {{\left( {\frac{1}{{1.25}}} \right)}^2}} }} \end{array}$

4 = 3.2b/0.6

b = 4 × 0.6/3.2

b = 0.75 cm

Ques.8. As the wave frequency approaches to the cut-off frequency of waveguide, the correct statement is:

1. The phase velocity of waves tends to zero.
2. The phase velocity of waves tends to velocity of light.
3. The phase velocity of waves tends to infinite.
4. None of these

Answer.3. The phase velocity of waves tends to infinite.

Explanation

Phase velocity is given as

${v_p} = \frac{c}{{\sqrt {1 – {{\left( {\frac{{{f_c}}}{f}} \right)}^2}{\rm{\;}}} }}$

Where c is the free space velocity, f is the operating frequency and fc is the cut-off frequency.

We can also write phase velocity as:

Vp = c/cosθ

Where θ is the angle with which the wave enters the waveguide.

Given, f = fc

$\begin{array}{l} {v_p} = \frac{c}{{\sqrt {1 – {{\left( {\frac{{{f_c}}}{f}} \right)}^2}} }}\\ \\ {v_p} = \frac{c}{{\sqrt {1 – 1} \:}} = \frac{c}{0} = \infty \end{array}$

vp = ∞ means that,

cos θ = 0, i.e. θ = 90°.

Ques.9. The velocity of electromagnetic waves in a dielectric medium with relative permittivity of 4 will

1. 3 × 108 m/s
2. 6 × 108 m/s
3. 1.5 × 108 m/s
4. 0.75 × 108 m/s

Explanation

The velocity (Vp) of a plane electromagnetic wave is given as-

${v_p} = \frac{c}{{\sqrt {{\mu _r}{\varepsilon _r}} }}$

$c = \frac{1}{{\sqrt {{\mu _o}{\varepsilon _o}} }}$ = 3 × 108m/sec

μ0 = permeability in free space 4π × 10-7 H/m

εo = permittivity in free space 8.854 × 10-12 C2/Nm2

Given:

μr = 4

Vp = 3 × 108/√4

Vp = 1.5 x 108 m/sec

Ques.10. Standard air-filled rectangular waveguides of dimensions a = 2.29 cm and b = 1.02 cm are designed for radar applications. It is desired that these waveguides operate only in the dominant TE10 mode with the operating frequency at least 25% above the cutoff frequency of the TE10 mode but not higher than 95% of the next higher cutoff frequency. The range of the allowable operating frequency f is

1. 8.19 GHz ≤ f ≤ 13.1 GHz
2. 8.19 GHz ≤ f ≤ 12.45 GHz
3. 6.55 GHz ≤ f ≤ 13.1 GHz
4. 1.64 GHz ≤ f ≤ 10.24 GHz

Answer.2. 8.19 GHz ≤ f ≤ 12.45 GHz

Explanation

Cut-off frequency of (rectangular) waveguide

$\Rightarrow {f_c} = \frac{C}{2}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}} \:$ ——1

Dominant mode in rectangular waveguide (a > b) is TE10 and the cutoff frequency is defined as;

${f_{{c_{\left( {10} \right)}}}} = \frac{C}{2}\sqrt {\frac{1}{{{a^2}}}} = \frac{C}{{2a}}$

Calculation:

${f_{{c_{\left( {10} \right)}}}} = \frac{C}{{2a}} = \frac{{3 \times {{10}^{10}}}}{{2 \times 2.29}} = 6.554\:Hz$

Desired operating frequency = 25% more than fc(10)

(1 + 0.25)fc(10) = 1.25fc(10) = 1.25 × 6.55

= 8.199 Hz

Using equation (1), we calculate the different cut-off frequencies for higher modes to find the next dominant mode,

fc for TE01 mode = 14.74 Hz

fc for TE11 mode = 16.4 Hz

and

fc for TE20 mode = 13.144 Hz

So, TE20 is the next higher cut-off frequency,

95% of 13.144 Hz

⇒ 12.484 GHz

Ques.11. For a parallel plate transmission line, let v be the speed of propagation and Z be the characteristic impedance. Neglecting fringe effects, a reduction of spacing between the plates by a factor of two results in

1. Halving of v and no change in Z
2. No change in v and halving of Z
3. No change in both v and Z
4. Halving of both v and Z

Answer.2. No change in v and halving of Z

Explanation

$\begin{array}{l} Z = \sqrt {\frac{L}{C}} \\ \\ Z = \sqrt {\frac{{\frac{{\mu h}}{w}}}{{\frac{{Ew}}{h}}}} \\ \\ Z = \sqrt {\frac{\mu }{E}} \:\frac{h}{w} \end{array}$

We observe that If ‘h’ is halved, z is halved.

V = Ey h

I = Hx ⋅ w

Ques.12. A standard air-filled rectangular waveguide with dimensions a = 8 cm, b = 4cm, operates at 3.4 GHz, For the dominant mode of wave propagation, the phase velocity of the signal is vp. The value (rounded off two decimal places) of vp / C, where C denotes the velocity of light, is _____

1. 1.98
2. 3.5
3. 4.5
4. 0.6

Answer.2. No change in v and halving of Z

Explanation

Given

a = 8 cm

b = 4 cm

f = 3.4 GHz

The cut-off frequency will be:

fc = c/2a = 3 × 1010/ 2 × 8

Sinθ = fc/f

sinθ = 1.875/3.4 = 0.5514

∴ cos θ = 0.83419

Phase velocity is given by:

Vp = c/cosθ

vp/c = 1/cosθ

vp/c = 1.198

Ques.13. A rectangle waveguide of internal dimensions (a = 3 cm and b = 1 cm) is to be operated at TE11 mode. The minimum operating frequency is:

1. 6.25 GHz
2. 10.5 GHz
3. 31.6 GHz
4. 15.8 GHz

Explanation

Given, a = 3 cm

b = 1 cm

The minimum frequency in TE11 is nothing but the cut-off frequency calculated as:

$\begin{array}{l} {f_c} = \frac{{3 \times {{10}^8}}}{2}\sqrt {\frac{1}{{{{\left( {3 \times {{10}^{ – 2}}} \right)}^2}}} + \frac{1}{{{{\left( {1 \times {{10}^{ – 2}}} \right)}^2}}}} \\ \\ {f_c} = 150 \times {10^6}\sqrt {11111.11} \:\: \end{array}$

fc = 150 × 106 × 105.41

fc = 15.81 GHz

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