Waveguides and Guided Waves MCQ

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

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

But for free space

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

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

Answer.3. 15 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:

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

λ_g = 15 cm

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

Answer.1. Single-mode step-index fiber

Explanation

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

Advantages of single-mode fiber:

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

Answer.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.

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:

Answer.1. High pass filter

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 TE_mn

Answer.1. 3.78 × 10⁸ m/s

Explanation

Phase velocity is given by:

V_P = c/cosθ

Where:

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

f_c = cut-off frequency

f = frequency of operation

c = speed of light

Calculation:

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

f_c = 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

Answer.1. Step index single-mode 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.

Advantages of single-mode fiber:

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

Answer.4. Step index  fiber

Explanation

Step-index fiber:

1. The refractive index of the core is uniform throughout and undergoes on abrupt change at the core-cladding boundary.

2. The path of light propagation is zig-zag in a manner