Waveguides

Question 1
A waveguide consists of two infinite parallel plates (perfect conductors) at a separation of 10^{-4} cm, with air as the dielectric. Assume the speed of light in air to be 3 \times 10^{8} m/s. The frequency/frequencies of TM waves which can propagate in this waveguide is/are _______.
A
6 \times 10^{15}Hz
B
0.5 \times 10^{12}Hz
C
8 \times 10^{14}Hz
D
1 \times 10^{13}Hz
GATE EC 2022   Electromagnetics
Question 1 Explanation: 
Cut-off frequency.
\begin{aligned} f_c&=\frac{c}{2a}\;\;\;(m=1) &=\frac{3 \times 10^8}{2 \times 10^{-4} \times 10^{-2}}\\ &=1.5 \times 10^{14}Hz \end{aligned}
f \gt f_c will only propagate
A and C will propage.
Question 2
A standard air-filled rectangular waveguide with dimensions \text{a=8 cm, b=4 cm}, operates at \text{3.4 GHz}. For the dominant mode of wave propagation, the phase velocity of the signal in v_{p}. The value (rounded off to two decimal places) of v_{p}/c, where c denotes the velocity of light, is _____
A
2.24
B
3.82
C
1.2
D
4.6
GATE EC 2021   Electromagnetics
Question 2 Explanation: 
f_{c 10}=\frac{c}{2 a}=\frac{3 \times 10^{8}}{2\left(8 \times 10^{-2}\right)}=1.875 \mathrm{GHz}
Guide phase velocity, V_{p}=\frac{C}{\sqrt{1-\left(\frac{f_{C 10}}{f}\right)^{2}}}
\frac{V_{p}}{C}=\frac{1}{\sqrt{1-\left(\frac{f_{C 10}}{f}\right)^{2}}}=\frac{1}{\sqrt{1-\left(\frac{1.875}{3.4}\right)^{2}}}=1.198
Question 3
The refractive indices of the core and cladding of an optical fiber are 1.50 and 1.48, respectively. The critical propagation angle, which is defined as the maximum angle that the light beam makes with the axis of the optical fiber to achieve the total internal reflection, (rounded off to two decimal places) is _____ degree.
A
18.3
B
3.45
C
9.37
D
5.82
GATE EC 2021   Electromagnetics
Question 3 Explanation: 
Given that
Refractive index of core \eta_{1}=1.50
Refractive index of clad \eta_{2}=1.48
Critical propagation angle \left(\theta_{P}\right)
\theta_{P}=\sin ^{-1}\left[ \frac{ \sqrt{\eta_{1}^{2}-\eta_{2}^{2}}}{\eta_{1}} \right]=\sin ^{-1} \left[ \frac{ \sqrt{1.5^{2}-1.48^{2}} }{1.5} \right] =9.37
Question 4
Consider a rectangular coordinate system (x,y,z) with unit vectors a_{x}\:a_{y} and a_{z}. A plane wave traveling in the region z\geq 0 with electric field vector E=10\cos\left ( 2\times 10^{8}t\:+\:\beta z\right )a_{y} is incident normally on the plane at z=0, where \beta is the phase constant. The region z\geq 0 is in free space and the region z \lt 0 is filled with a lossless medium (permittivity \varepsilon \:=\:\varepsilon _{0}, permeability \mu \:=\:4\mu _{0} , where \varepsilon _{0}\:=\:8.85\times 10^{-12}\:\text{F/m} and \mu _{0}\:=\:4\pi \times 10^{-7}\:\text{H/m}). The value of the reflection coefficient is
A
\frac{1}{3}
B
\frac{3}{5}
C
\frac{2}{5}
D
\frac{2}{3}
GATE EC 2021   Electromagnetics
Question 4 Explanation: 
Given \vec{E}=10 \cos \left(2 \pi \times 10^{8} t+\beta z\right) \hat{a}_{y} for z \geq 0 having free space. For z \lt 0 medium has
\epsilon_{r 2}=1 ; \mu_{r 2}=4


\Gamma=\frac{\eta_{2}-\eta_{1}}{\eta_{2}+\eta_{1}}=\frac{120 \pi(2)-120 \pi}{120 \pi(2)+120 \pi}=\frac{1}{3}
Question 5
A rectangular waveguide of width w and height h has cut-off frequencies for TE_{10} \; and \;TE_{11} modes in the ratio 1:2. The aspect ratio w/h, rounded off to two decimal places, is ___________
A
0.85
B
1.25
C
1.73
D
1.92
GATE EC 2019   Electromagnetics
Question 5 Explanation: 
f_{c m n}=\frac{c}{2} \sqrt{\left(\frac{m}{a}\right)^{2}+\left(\frac{n}{b}\right)^{2}}
For TE mode
f_{c10}=\frac{c}{2 w} \quad\ldots(i)
and For TE_{11} mode.
\begin{aligned} f_{c11} &=\frac{c}{2} \sqrt{\frac{1}{w}^{2}+\left(\frac{1}{h}\right)^{2}} &\ldots(i)\\ &=\frac{c}{2 w} \sqrt{1+\left(\frac{w}{h}\right)^{2}} &\ldots(ii)\\ \text{given}\quad\frac{f_{c10}}{f_{11}} &=\frac{1}{2} &\ldots(iii) \end{aligned}
put (i). (in) in (ii)
\Rightarrow \frac{\frac{c}{2 w}}{\frac{c}{2 w} \sqrt{1+\left(\frac{w}{h}\right)^{2}}}=\frac{1}{2} \Rightarrow \sqrt{1+\left(\frac{w}{h}\right)^{2}}=2
On solving above equation, we get,
\frac{w}{h}=\sqrt{3}=1.732
Question 6
The dispersion equation of a waveguide,which relates the wavenumber kto the frequency \omega, is

k(\omega) =(1/c)\sqrt{\omega ^2 -{\omega _0}^2}

where the speed of light c=3 \times 10^8 m/s, and \omega_0 is a constant. If the group velocity is 2 \times 10^8 m/s, then the phase velocity is
A
1.5 \times 10^8 m/s
B
2 \times 10^8 m/s
C
3 \times 10^8 m/s
D
4.5 \times 10^8 m/s
GATE EC 2019   Electromagnetics
Question 6 Explanation: 
By definition v_{p}=\frac{\omega}{\beta}=\frac{\omega}{k}
where, k(\omega)=\left(\frac{1}{c}\right) \sqrt{\omega^{2}-\omega_{0}^{2}} \quad (given)
\therefore v_{p}=\frac{c}{\sqrt{1-\left(\frac{\omega_{0}}{\omega}\right)^{2}}}
by definition,
\begin{aligned} v_{g} &=\frac{d \omega}{d \beta}=\frac{d \omega}{d k} \\ &=\frac{d k}{d \omega}=\frac{1}{c} \frac{1}{2 \sqrt{\omega^{2}-\omega_{0}^{2}}} \times 2 \omega \\ \text{or}\quad v_{g} &=c \sqrt{1-\left(\frac{\omega_{0}}{\omega}\right)^{2}}\\ \because \quad v_{p} \cdot v_{g} &=c^{2} \\ \therefore \quad v_{p} &=\frac{c^{2}}{v_{g}}=\frac{\left(3 \times 10^{8}\right)^{2}}{2 \times 10^{8}} \\ &=4.5 \times 10^{8} \mathrm{m} / \mathrm{sec} \end{aligned}
Question 7
The cutoff frequency of TE_{01} mode of an air filled rectangular waveguide having inner dimensions a cm \times b cm ( a \gt b) is twice that of the dominant TE_{10} mode. When the waveguide is operated at a frequency which is 25% higher than the cutoff 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 _______.
A
1.10
B
0.55
C
0.65
D
0.75
GATE EC 2018   Electromagnetics
Question 7 Explanation: 
\begin{array}{l} f_{c(01)}=2 f_{c(10)}=\frac{2 c}{2 a}=\frac{c}{a} \\ \frac{c}{2 b}=\frac{c}{a} \Rightarrow a=2 b \Rightarrow b=\frac{a}{2} \end{array}
Operating frequency,
f=1.25 f_{c(10)}
f_{c(10)}\lt 1.25 f_{c(10)} \lt \left[f_{c(01)}=2 f_{c(10)}\right]
So, for the given frequency, the waveguide will work in \mathrm{TE}_{10} mode.
\begin{aligned} \text { So, } \lambda_{g} &=\frac{\lambda_{0}}{\sqrt{1-\left(\frac{f_{c(10)}}{f}\right)^{2}}} \\ &=\frac{c / f}{\sqrt{1-\left(\frac{1}{1.25}\right)^{2}}}=\frac{c / f}{0.6} \\ \frac{c}{(1.25) f_{c(10)}(0.6)} &=\lambda_{g}=4 \mathrm{cm} \\ \frac{c}{f_{c(10)}} &=3 \times 10^{-2}=2 a \\ a &=1.5 \mathrm{cm} \\ b &=\frac{a}{2}=0.75 \mathrm{cm} \end{aligned}
Question 8
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 dominatnt TE_{10} mode but not higher than 95% of the next higher cutoff frequency. The range of the allowable operating frequency f is.
A
8.19 GHz \leq f \leq 13.1 GHz
B
8.19 GHz \leq f \leq 12.45 GHz
C
6.55 GHz \leq f \leq 13.1 GHz
D
1.64 GHz \leq f \leq 10.24 GHz
GATE EC 2017-SET-2   Electromagnetics
Question 8 Explanation: 
a=2.29 \;\mathrm{cm}, b=1.02 \mathrm{cm}
Waveguide is operating in (TE _{10}) dominant mode
f_{c}\left(T E_{10}\right)=\frac{c}{2 a}=\frac{3 \times 10^{10}}{2 \times 2.29}=6.55 \mathrm{GHz}
\Rightarrow 25 \% above cut-off frequency
=1.25 \times 6.55 \mathrm{GHz}=8.1875 \mathrm{GHz}
Next higher order mode is \mathrm{TE}_{20}\left(\because a_{2} \gt 2b\right)
\begin{aligned} f_{c}\left(T E_{20}\right) &=\frac{c}{a}=\frac{3 \times 10^{10}}{2.29}=13.1 \mathrm{GH}_{2} \\ \Rightarrow 95 \% \text { of } f_{c}\left(\mathrm{TE}_{20}\right) &=0.95 \times 13.1 \\ &=12.445 \mathrm{GHz} \end{aligned}
Range of allowable operating frequency
8.19 \mathrm{GHz} \leq f \leq 12.45 \mathrm{GHz}
Question 9
Consider an air-filled rectangular waveguide with dimensions a = 2.286 cm and b = 1.016 cm. The increasing order of the cut-off frequencies for different modes is
A
TE_{01} \lt TE_{10} \lt TE_{11} \lt TE_{20}
B
TE_{20} \lt TE_{11} \lt TE_{10} \lt TE_{01}
C
TE_{10} \lt TE_{20} \lt TE_{01} \lt TE_{11}
D
TE_{10} \lt TE_{11} \lt TE_{20} \lt TE_{01}
GATE EC 2016-SET-3   Electromagnetics
Question 9 Explanation: 
\begin{aligned} a &=2.286 \mathrm{cm} \\ b &=1.016 \mathrm{cm} \\ f_{c_{T E_{10}}} &=\frac{c}{2 a} \\ f_{c_{T E_{01}}} &=\frac{c}{2 b} \\ f_{c_{T E_{20}}} &=\frac{c}{a} \\ f_{c_{T E_{11}}} &=\frac{c}{2} \sqrt{\left(\frac{1}{a}\right)^{2}+\left(\frac{1}{b}\right)^{2}} \\ \text{As }a \gt 2 b \Rightarrow f_{c_{T E_{10}}} & \lt f_{c_{T E_{20}}} \lt f_{c_{T E_{01}}} \lt f_{c_{T E_{11}}} \end{aligned}
Question 10
Consider an air-filled rectangular waveguide with dimensions a = 2.286 cm and b = 1.016 cm. At 10 GHz operating frequency, the value of the propagation constant (per meter) of the corresponding propagating mode is __________
A
55
B
214
C
158
D
321
GATE EC 2016-SET-3   Electromagnetics
Question 10 Explanation: 
Operation frequency =10 \mathrm{GHz}
Cut off frequency for \mathrm{TE}_{10}, \mathrm{f}_{c}=\frac{\mathrm{c}}{2 \mathrm{a}}
\begin{aligned} f_{c} &=\frac{3 \times 10^{8}}{2 \times 2.286 \times 10^{-2}} \\ &=6.57 \mathrm{GHz} \end{aligned}
Next mode T E_{20} has f_{c}=13 \mathrm{GHz}
10 \mathrm{GHz} operates only in dominant mode.
\begin{aligned} r &=\sqrt{\left(\frac{m \pi}{a}\right)^{2}-\omega^{2} \mu \epsilon} \\ &=\sqrt{\left(\frac{\pi}{a}\right)^{2}-\frac{(2 \pi f)^{2}}{\left(3 \times 10^{8}\right)^{2}}} \\ &=\pi \sqrt{\left(\frac{1}{2.286 \times 10^{-2}}\right)^{2}-\left(\frac{2 \times 10 \times 10^{9}}{3 \times 10^{8}}\right)^{2}} \\ &=j 158\left(\mathrm{m}^{-1}\right) \end{aligned}
There are 10 questions to complete.