GATE ECE · Electromagnetics

Waveguides – GATE ECE ElectromagneticsPractice Questions & PYQs

Generate GATE-level questions on Waveguides. Focus on: 1. Rectangular waveguides: TM and TE modes, Boundary conditions. 2. Cut-off frequencies, Dispersion relations, and Phase/Group velocity in guides. 3. Power transmission and Attenuation in waveguides.

26 questions · 20 PYQs · 0 AI practice · GATE ECE 2027

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Q1GATE 2022MSQ

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 _______.

🚀 Solve in Practice Mode

📖 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.

Q2GATE 2021NAT

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.

🚀 Solve in Practice Mode

📖 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$

Q3GATE 2021MCQ

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

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📖 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 $\epsilon_{r 2}=1 ; \mu_{r 2}=4$ $\Gamma=\frac{\eta_{2}-\eta_{1}}{\eta_{2}+\eta_{1}}=\frac{120 \pi(2)-120 \pi}{120 \pi(2)+120 \pi}=\frac{1}{3}$

Explanation Figure 1

Q4GATE 2021NAT

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 _____

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📖 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, $V_{p}=\frac{C}{\sqrt{1-\left(\frac{f_{C 10}}{f}\right)^{2}}}$ $\frac{V_{p}}{C}=\frac{1}{\sqrt{1-\left(\frac{f_{C 10}}{f}\right)^{2}}}=\frac{1}{\sqrt{1-\left(\frac{1.875}{3.4}\right)^{2}}}=1.198$

Q5GATE 2019NAT

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 ___________

🚀 Solve in Practice Mode

📖 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, $\frac{w}{h}=\sqrt{3}=1.732$

Q6GATE 2019MCQ

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

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📖 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}

Q7GATE 2018NAT

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 _______.

🚀 Solve in Practice Mode

📖 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}

Q8GATE 2017MCQ

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.

🚀 Solve in Practice Mode

📖 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 $8.19 \mathrm{GHz} \leq f \leq 12.45 \mathrm{GHz}$

Q9GATE 2016MCQ

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

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📖 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}

Q10GATE 2015MCQ

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

🚀 Solve in Practice Mode

📖 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 $T E_{21}$

Q11GATE 2015NAT

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 __________.

🚀 Solve in Practice Mode

📖 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}

Q12GATE 2014NAT

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 ____.

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📖 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}

Q13GATE 2014NAT

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______.

🚀 Solve in Practice Mode

📖 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}

Q14GATE 2012MCQ

The magnetic field along the propagation direction inside a rectangular waveguide with the crosssection shown in the figure is $H_{z}=3 cos(2.094 \times 10^{2}x)cos(2.618 \times 10^{2}y)$ $cos(6.283 \times 10^{10}t-\beta z)$ The phase velocity $v_{p}$ of the wave inside the waveguide satisfies

🚀 Solve in Practice Mode

📖 Explanation

From the given data

\begin{aligned} \omega_{c} &=\left(\sqrt{\left(\frac{m \pi}{a}\right)^{2}+\left(\frac{n\pi}{b}\right)^{2}}\right) c \\ f_{c} &=(\sqrt{(2.09)^{2}+(2.618)^{2}}) \times \frac{3 \times 10^{10}}{2 \times 3.14} \\ &=16 \mathrm{GHz} \\ f &=\frac{6.283 \times 10^{10}}{2 \times 3.14}=10 \mathrm{GHz} \\ \text { Hence, } \quad f& \lt f_{c^{\prime}} \gamma=\alpha+j 0\\ \beta=0, \quad \quad v_{p}&=\frac{\omega}{\beta}=\infty \end{aligned}

Q15GATE 2011MCQ

The modes in a rectangular waveguide are denoted by $\frac{TE_{mn}}{TM_{mn}}$ where m and n are the eigen numbers along the larger and smaller dimensions of the waveguide respectively. Which one of the following statements is TRUE?

🚀 Solve in Practice Mode

📖 Explanation

In case of rectangular wave guide $T E_{\text {rm }}$ exists for all values of m and n except m=0 and n=0 . For $T M_{m n}$ Io exist both values of mand n must be non zero

Q16GATE 2009MCQ

Which of the following statements is true regarding the fundamental mode of the metallic waveguides shown ?

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📖 Explanation

P is coaxial line and support TEM wave $\therefore$ P has no cutoff frequency Q and R are wave-guides and cutoff frequency each depends upon their dimensions.

Q17GATE 2008MCQ

A rectangular waveguide of internal dimensions (a=4cm and b=3cm) is to be operated in $TE_{11}$ mode. The minimum operating frequency is

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📖 Explanation

Minimum operating frequency,

\begin{aligned} f_{0}&=\frac{c}{2 \pi} \sqrt{\left(\frac{m \pi}{a}\right)^{2}+\left(\frac{n \pi}{b}\right)^{2}}\\ \text{For }\;T E_{11}&\;\text{ mode}\\ f_{0} &=\frac{3 \times 10^{8}}{2} \sqrt{\left(\frac{1}{0.04}\right)^{2}+\left(\frac{1}{0.03}\right)^{2}} \\ &=\frac{3 \times 10^{8}}{2} \times \frac{0.05}{0.04 \times 0.03}=6.25 \mathrm{GHz} \end{aligned}

Q18GATE 2007MCQ

The $\vec{E}$ field in a rectangular waveguide of inner dimension $a \times b$ is given by $\vec{E}=\frac{\omega \mu }{h^{2}}(\frac{\pi}{a})H_{0}sin(\frac{2\pi x}{a})^{2}sin(\omega t-\beta z)\hat{y}$ where $H_{0}$ is a constant, and a and b are the dimensions along the x-axis and the y-axis respectively. The mode of propagation in the waveguide is

🚀 Solve in Practice Mode

📖 Explanation

The wave is E(x,z,t) with no propagation in y-direction. This is possible only in TE waves when $n=0$ $\left(\cos \frac{n \pi}{b} y=1\right)$

Q19GATE 2007MCQ

An air-filled rectangular waveguide has inner dimensions of 3cm x 2cm. The wave impedance of the TE20 mode of propagation in the waveguide at a frequency of 30 GHz is (free space impedance $\eta_{0}$ = 377 $\Omega$ )

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} \text { For } TE_{20}, \quad \lambda_{C}&=a=3 \times 10^{-2} \mathrm{m} \\ f_{C}&= \frac{C}{\lambda_{C}}=\frac{3 \times 10^{8}}{3 \times 10^{-2}}=10^{10} \mathrm{Hz} \\ \eta_{T E}&=\frac{\eta}{\sqrt{1-\left(\frac{f_{c}}{f}\right)^{2}}} =\frac{377}{\sqrt{1-\left(\frac{10^{10}}{3 \times 10^{10}}\right)^{2}}}\\ \eta_{T E}&=400 \Omega \end{aligned}

Q20GATE 2006MCQ

A rectangular wave guide having $TE_{10}$ mode as dominant mode is having a cut off frequency 18-GHz for the mode $TE_{30}$ . The inner broad-wall dimension of the rectangular wave guide is

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} \mathrm{TE}_{30^{\prime}} \quad \lambda_{c}&=\frac{2 a}{m}=\frac{2 a}{3} \\ \lambda_{c} &=\frac{c}{t}=\frac{3 \times 10^{8}}{18 \times 10^{9}}=\frac{1}{60} \\ \frac{1}{60} &=\frac{2 a}{3} \\ \Rightarrow \quad a &=\frac{1}{40}=2.5 \mathrm{cm}=\frac{5}{2} \mathrm{cm} \end{aligned}

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