# Waveguides and Guided Waves MCQ

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 a dominant mode in a rectangular waveguide with breadth 10 cm, the guide wavelength for a signal of 2.5 GHz will be

1. 20 cm
2. 18 cm
3. 15 cm
4. 12 cm

Explanation

The dominant mode of a rectangular waveguide is TE10

The cutoff frequency is given as:

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

(fc)TE10 = c/2a

= (3 × 1010)/20 = 1.5 GHz

Operating frequency:

f = 2.5 GHz

Operating wavelength:

λ = c/f = (3 × 1010)/(2.5 × 109)

= 12 cm

Guide wavelength is given by:

$\begin{array}{l} {\lambda _g} = \frac{\lambda }{{\sqrt {1 – {{\left( {\frac{{{f_c}}}{f}} \right)}^2}} }}\\ \\ {\lambda _g} = \frac{{12}}{{\sqrt {1 – {{\left( {\frac{{1.5}}{{2.5}}} \right)}^2}} }} \end{array}$

λg = 15 cm

Ques.3. In a dispersive medium, the group velocity is:

1. Less than the phase velocity only
2. Equal to the phase velocity only
3. More than the phase velocity, depending on the nature of the dispersive medium
4. More than the phase velocity

Answer.1. Less than the phase velocity only

Explanation

In a dispersive medium, the group velocity is less than the phase velocity only.

Group velocity is given by:

Vg = dω/dβ

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

Vg=c1−(ωCω)2

Vg = c cos θ

Vg < c

Ques.4. Which type of optical fiber is used to eliminate modal dispersion during optical communication?

1. Single-mode step-index fiber
2. Multimode step-index fiber
4. Does not depend on the type of fiber

Explanation

Single-mode step-index fiber is used to eliminate modal dispersion during optical communication.

1) Low signal loss

2) No modal dispersion

3) Does not suffer from modal dispersion

4) Can be used for higher bandwidth applications

5) Long-distance applications

6) Cable TV ends

7) High speed local and wide area network

Ques.5. Optic fibers are used in:

1. CAT scans
2. X-ray photos
3. Ultrasound scans
4. Endoscopy

Explanation

Endoscopes use optical fibers to produce an image of inside the body.

• A doctor can insert a bundle of optical fibers into the body.
• Some carry light into the body, and some carry light reflected off internal body surfaces back out.
• This allows the doctor to see an image of the inside of the body clearly, and help them diagnose diseases like cancer, or see what they are doing during keyhole surgery.

Ques.6. TWT is characterized by

1. Low noise figure, narrow bandwidth, and average gain
2. Gain exceeding 40 dB, wide bandwidth and low noise figure
3. More noise and wide bandwidth
4. More noise, narrow bandwidth, and high Gain

Answer.2. Gain exceeding 40 dB, wide bandwidth and low noise figure

Explanation

Characteristics of a Travelling Wave Tube are as follows:

• The Travelling Wave Tubes (TWT) is a high-gain, low-noise, and wide-bandwidth microwave amplifier.
• It is capable of gains greater than 40dB with bandwidths exceeding an octave (A Bandwidth of 1 octave is one in which the upper frequency is twice the lower frequency).
• The TWT is primarily a voltage amplifier. The wide-bandwidth and low-noise characteristics make the TWT ideal for use as an RF amplifier in microwave equipment.

A comparison of various Microwave tubes used for high power/frequency generation is as shown:

 Parameters Klystron Magnetron TWTs Frequency Few GHz to hundred GHz 1 – 25 GHz 1 to 10 GHz Output Power 10 MW Several kW Order of kW Efficiency 10 % 30 – 60 % 20 % Uses Oscillator and Amplifier Oscillator Oscillator and Amplifier

Ques.7. The waveguide can be considered as

1. High pass filter
2. Low pass filter
3. Bandpass filter
4. None of these

Explanation

Waveguides only allow frequencies above cut-off frequency and do not pass below the cut-off frequencies.

Hence it acts as a high pass filter.

The cut off frequency is given as:

${\lambda _{\rm{C}}} = \frac{2}{{\sqrt {{{\left( {\frac{{\rm{m}}}{{\rm{a}}}} \right)}^2} + {{\left( {\frac{{\rm{n}}}{{\rm{b}}}} \right)}^2}} }}$

Where a and b are the dimensions of the waveguide (a>b)

m and n are mode numbers TEmn

Ques.8. An air-filled rectangular waveguide of dimension 7 × 3.5 cm2 operates in the dominant TE10 mode. The value of phase velocity of the wave in the guide at a frequency of 3.5 GHz is given by:

1. 3.78 × 108 m/s
2. 6.78 × 105 m/s
3. 5.78 × 108 m/s
4. 4.78 × 10m/s

Explanation

Phase velocity is given by:

VP = c/cosθ

Where:

$\cos \theta = \sqrt {1 – {{\left( {\frac{{{f_c}}}{f}} \right)}^2}}$

fc = cut-off frequency

f = frequency of operation

c = speed of light

Calculation:

For the given TE10 mode, the cut-off frequency will be:

fc = c/2a

With a = 7 cm (Given)

fc = (3 × 1010)/(2 × 7) = 2.14 GHz

f = 3.5 GHz

$\cos \theta = \sqrt {1 – {{\left( {\frac{{2.142}}{{3.5}}} \right)}^2}} = 0.79$

VP = (3 × 108)/0.79

= 3.78 × 108 m/s

Ques.9. Which fiber is preferred for long-distance communication?

1. Step index single-mode fiber
3. Step index multimode fiber

Explanation

• Single-mode step indexed fibers are widely used for wideband communications are preferred for long-distance communication.
• Single-mode step-index fiber is used to eliminate modal dispersion during optical communication.
• In this fiber, a light ray can travel on only one path so minimum refraction takes place hence, no pulse spreading permits high pulse repetition rates.

1) Low signal loss

2) No modal dispersion

3) Does not suffer from modal dispersion

4) Can be used for higher bandwidth applications

5) Long-distance applications

6) Cable TV ends

7) High speed local and wide area network

Ques.10. The refractive index of the core is uniform throughout and undergoes an abrupt change at the cladding boundary which is known as ______.

1. Uniform-index fiber
2. Scale-index fiber