# Waveguides

 Question 1
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 1 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 2
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 2 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 3
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 1.1 B 0.55 C 0.65 D 0.75
GATE EC 2018   Electromagnetics
Question 3 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 4
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 4 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 5
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 5 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 6
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 6 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}
 Question 7
An air-filled rectangular waveguide of internal dimensions $a \; cm \times b \; cm \; (a \gt b)$ has a cutoff frequency of 6 GHz for the dominant $TE_{10}$ mode. For the same waveguide, if the cutoff frequencyof the $TM_{11}$ mode is 15 GHz, the cutoff frequency of the $TE_{01}$ mode in GHz is __________.
 A 13.75 B 10.35 C 18.25 D 20.5
GATE EC 2015-SET-2   Electromagnetics
Question 7 Explanation:
Given, $f_{c} T E_{10}$ mode is $6 \mathrm{GHz}$
$\Rightarrow \quad \frac{c}{2 a}=\frac{3 \times 10^{10}}{2 \times a}=6 \times 10^{9}$
$\Rightarrow a=2.5 \mathrm{cm}$
Given $f_{c}$ for $T M_{11}$ mode is $15 \mathrm{GHz}$
\begin{array}{l} \Rightarrow \frac{c}{2} \sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}}=15 \times 10^{9} \\ \begin{array}{l} \frac{3 \times 10^{10}}{2} \sqrt{\frac{1}{6.25}+\frac{1}{b^{2}}}=15 \times 10^{9} \\ \Rightarrow \quad b=\frac{2.5}{\sqrt{5.25}} \end{array} \\ \begin{aligned} \Rightarrow \quad & \quad 2 \end{aligned} \end{array}
Now $f_{c}$ for $T E_{01}$ mode is,
\begin{aligned} f_{c} &=\frac{c}{2 b}=\frac{3 \times 10^{10}}{2.5} \sqrt{5.25} \\ &=13.75 \mathrm{GHz} \end{aligned}
 Question 8
The longitudinal component of the magnetic field inside an air-filled rectangular waveguide made of a perfect electric conductor is given by the following expression
$H_{z}(x,y,z,t)=0.1 cos(25\pi x) cos(30.3 \pi y)$ $cos(12 \pi *10^{9}t-\beta z) (A/m)$
The cross-sectional dimensions of the waveguide are given as a = 0.08 m and b = 0.033 m. The mode of propagation inside the waveguide is
 A $TM_{12}$ B $TM_{21}$ C $TE_{21}$ D $TE_{12}$
GATE EC 2015-SET-1   Electromagnetics
Question 8 Explanation:
\begin{aligned} H_{z}(x, y, z, t) &=0.1 \cos (25 \pi x) \cos (30.3 \pi y) \\ & \cos \left(12 \pi \times 10^{9} t-\beta z\right)^{-1 / m}\\ a &=0.08 m, b=0.033 m \end{aligned}
Axial component is $\mathrm{H} \Rightarrow$ the propagating mode is
$T E_{\text {mn}}$, m, n can be found by
$\begin{array}{l} \quad \frac{m \pi}{a} x=25 \pi x \\ \Rightarrow \quad \frac{m}{0.08}=25 \\ \Rightarrow \quad m=2 \\ \Rightarrow \quad \frac{n\pi}{b} y=30.3 \pi y \\ \Rightarrow \quad \frac{n}{0.033}=30.3 \\ \Rightarrow \quad n=1 \end{array}$
$\therefore$ The mode of propagation is $T E_{21}$
 Question 9
Consider an air filled rectangular waveguide with a cross-section of 5cm x 3cm. For this waveguide, the cut-off frequency (in MHz) of $TE_{21}$ mode is ____.
 A 7810 B 8750 C 5750 D 6750
GATE EC 2014-SET-3   Electromagnetics
Question 9 Explanation:
Given $a=5 \mathrm{cm}=5,10^{-2} \mathrm{m}$
$\begin{array}{c} b=3 \mathrm{cm}=3 \times 10^{-2} \mathrm{m} \\ f_{c}=\frac{C}{2 \pi} \sqrt{\left(\frac{n m}{a}\right)^{2}+\left(\frac{r m}{b}\right)^{2}} \\ f_{C T E_{21}}=\frac{C}{2} \sqrt{\left(\frac{2}{5 \mathrm{cm}}\right)^{2}+\left(\frac{1}{3 \mathrm{cm}}\right)^{2}} \\ =\frac{3 \times 10^{8}}{2} \sqrt{\left(\frac{2 \times 100}{5}\right)^{2}+\left(\frac{100}{3}\right)^{2}} \\ f_{c}=7810 \mathrm{MHz} \end{array}$
 Question 10
For a rectangular waveguide of internal dimensions $a \times b(a \gt b)$, the cur-off frequency for the $TE_{11}$ mode is the arithmetic mean of the cut-off frequencies for $TE_{10}$ mode and $TE_{20}$ mode. If a = $\sqrt{5}$ cm, the value of b (in cm) is______.
 A 1 B 2 C 3 D 4
GATE EC 2014-SET-2   Electromagnetics
Question 10 Explanation:
Cut-off frequency of a rectangular waveguide is
\begin{aligned} f_{c}&=\frac{c}{2 \pi} \sqrt{\left(\frac{m \pi}{a}\right)^{2}+\left(\frac{n\pi}{b}\right)^{2}} \\ f_{c} \text{ for }\quad T E_{10}&=\frac{c}{2 \pi} \sqrt{\left(\frac{1 \pi}{a}\right)^{2}}=\frac{c}{2 a} \\ f_{c} \text{ for }\quad T E_{20}&=\frac{c}{2 \pi} \sqrt{\left(\frac{2 \pi}{a}\right)^{2}}=\frac{c}{2 a} \\ f_{c} \text { for } \quad T E_{11} &=\frac{c}{2 \pi} \sqrt{\left(\frac{\pi}{a}\right)^{2}}+\left(\frac{\pi}{b}\right)^{2} \\ &=\frac{c}{2} \sqrt{\left(\frac{1}{a}\right)^{2}+\left(\frac{1}{b}\right)^{2}} \\ \text { Given, } f_{C T_{11}} &=\frac{f_{C_{I E 10}}+f_{C I E_{20}}}{2} \end{aligned}
\begin{aligned} \frac{c}{2} \sqrt{\left(\frac{1}{a}\right)^{2}+\left(\frac{1}{b}\right)^{2}}&=\frac{\frac{c}{2 a}+\frac{c}{a}}{2} \\ \sqrt{\left(\frac{1}{a}\right)^{2}+\left(\frac{1}{b}\right)^{2}}&=\frac{1}{2 a}+\frac{1}{a}=\frac{3}{2 a} \\ \left(\frac{1}{a}\right)^{2}+\left(\frac{1}{b}\right)^{2}&=\frac{9}{4 a^{2}} \\ \Rightarrow \quad\left(\frac{1}{b}\right)^{2} &=\frac{5}{4 a^{2}} \\ b &=\frac{2 a}{\sqrt{5}}=\frac{2 \times \sqrt{5}}{\sqrt{5}} \\ &=2 \mathrm{cm} \end{aligned}
There are 10 questions to complete.