Two resistors are connected in a circuit loop of area , as shown in the figure below. The circuit loop is placed on the plane. When a time-varying magnetic flux, with fluxdensity (in Tesla), is applied along the positive z-axis, the magnitude of current I (in Amperes, rounded off to two decimal places) in the loop is __________ .

Given: Loop area, Magnetic field, Total resistance in the loop = We use Faraday's Law of electromagnetic induction: where is the magnetic flux. So, So, the induced EMF is: (The negative sign indicates direction, but we are interested in magnitude of current) Now, using Ohm's Law:

Consider a narrow band signal, propagating in a lossless dielectric medium , with phase velocity and group velocity . Which of the following statement is true? ( is the velocity of light in vacuum.)

Phase velocity, Group velocity, Here, Hence,

A two-port network has scattering parameters given by . If the port-2 of the two port is short circuited, the parameter for the resultant one port network is

When output is short, Substituting in first equation,

A. $\frac{S_{11}-S_{11}S_{22}+S_{12}S_{21}}{1+S_{22}}$

B. $\frac{S_{11}+S_{11}S_{22}-S_{12}S_{21}}{1+S_{22}}$

C. $\frac{S_{11}+S_{11}S_{22}+S_{12}S_{21}}{1-S_{22}}$

D. $\frac{S_{11}-S_{11}S_{22}+S_{12}S_{21}}{1-S_{22}}$

If the scattering matrix [S] of a two port network is then the network is

For reciprocal networks, For symmetrical networks, For antimetrical networks, For lossless reciprocal networks, and

D. neither lossless nor reciprocal

A load of 50 is connected in shunt in a 2-wire transmission line of as shown in the figure. The 2-port scattering parameter matrix (s-matrix) of the shunt element is

The line is terminated with at the center and so, matched on both the sides.

A. $\begin{bmatrix} -\frac{1}{2} &\frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \end{bmatrix}$

B. $\begin{bmatrix} 0 & 1\\ 1&0 \end{bmatrix}$

C. $\begin{bmatrix} -\frac{1}{3} &\frac{2}{3} \\ \frac{2}{3} & -\frac{1}{3} \end{bmatrix}$

D. $\begin{bmatrix} \frac{1}{4} &-\frac{3}{4} \\ -\frac{3}{4} & \frac{1}{4} \end{bmatrix}$

GATE ECE · Electromagnetics

Antennas And Miscellaneous Topics – GATE ECE ElectromagneticsPractice Questions & PYQs

Generate GATE-level questions on Antennas and S-Parameters. Focus on: 1. Antenna parameters: Radiation pattern, Gain, Directivity, and Return loss. 2. Antenna types: Dipole, Monopole, and Antenna arrays. 3. S-parameters: Scattering matrix, Reciprocity, and Losslessness. 4. Radar and Optical Fibers: Radar cross-section, Numerical aperture, and Fiber modes.

10 questions · 10 PYQs · 0 AI practice · GATE ECE 2027

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Q1GATE 2025NAT

Two resistors are connected in a circuit loop of area $5 \mathrm{~m}^{2}$ , as shown in the figure below. The circuit loop is placed on the $x-y$ plane. When a time-varying magnetic flux, with fluxdensity $\mathrm{B}(\mathrm{t})=0.5 \mathrm{t}$ (in Tesla), is applied along the positive z-axis, the magnitude of current I (in Amperes, rounded off to two decimal places) in the loop is __________ .

🚀 Solve in Practice Mode

📖 Explanation

Given: Loop area, $A = 5\ \text{m}^2$ Magnetic field, $B(t) = 0.5t\ \text{T}$ Total resistance in the loop = $2\ \Omega + 2\ \Omega = 4\ \Omega$ We use Faraday's Law of electromagnetic induction: $\mathcal{E} = -\frac{d\Phi}{dt}$ where $\Phi = B(t) \cdot A$ is the magnetic flux. So, $\Phi(t) = B(t) \cdot A = 0.5t \cdot 5 = 2.5t$ $\frac{d\Phi}{dt} = \frac{d}{dt}(2.5t) = 2.5\ \text{V}$ So, the induced EMF is: $\mathcal{E} = -2.5\ \text{V}$ (The negative sign indicates direction, but we are interested in magnitude of current) Now, using Ohm's Law: $I = \frac{|\mathcal{E}|}{R} = \frac{2.5}{4} = 0.625\ \text{A}$

Q2GATE 2023MCQ

Consider a narrow band signal, propagating in a lossless dielectric medium $\left(\epsilon_{r}=4, \mu_{r}=1\right)$ , with phase velocity $v_{p}$ and group velocity $v_{g}$ . Which of the following statement is true? ( $c$ is the velocity of light in vacuum.)

🚀 Solve in Practice Mode

📖 Explanation

Phase velocity, $V_{p}=\frac{\omega}{\beta}=\frac{\omega}{\omega \sqrt{\mu \epsilon}}=\frac{1}{\sqrt{\mu_{0} \epsilon_{0}}}, \frac{1}{\sqrt{\mu_{r} \epsilon_{r}}}=\frac{C}{\sqrt{\mu_{r} \epsilon_{r}}}$ $\therefore \quad V_{p} \lt C$ Group velocity, $V_{g}=\frac{d \omega}{d \beta}=\frac{V_{p}}{1-\frac{\omega}{V_{p}} \frac{d V_{p}}{d \omega}}$ Here, $\quad V_{p} \neq f(\omega)$ $\therefore \quad V_{g}=V_{p} \lt C$ Hence, $\quad V_{p} \lt C$ $\quad \quad V_{g} \lt C$

Q3GATE 2017NAT

An optical fiber is kept along the $\hat{z}$ direction. The refractive indices for the electric fields along $\hat{x}$ and $\hat{y}$ directions in the fiber are $n_{x}$ =1.5000 and $n_{y}$ =1.5001, respectively ( $n_{x} \neq n_{y}$ due to the imperfection in the fiber cross-section). The free space wavelength of a light wave propagating in the fiber is 1.5 $\mu$ m. If the light wave is circularly polarized at the input of the fiber, the minimum propagation distance after which it becomes linearly polarized, in centimeter, is ___________.

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📖 Explanation

Initially the wave is circularly polarized. So, the initial phase difference between field components in $\hat{a}_{x}$ direction and $\hat{a}_{y}$ direction is $\frac{\pi}{2}$ To become linearly polarized, the wave must travel a minimum distance, such that, the phase difference at that point between the field components in $\hat{a}_{x}$ direction and $\hat{a}_{y}$ direction is $\pi$ (i.e., the travel of this minimum distance should provide an additional phase difference of $\pi / 2$ between $\hat{a}_{x}$ and $\hat{a}_{y}$ field components). So,

\begin{aligned} z_{min}k_z&\sim z_{min}k_y=\frac{\pi}{2} \\ z_{min}\left[ \frac{\omega }{V_{px}}\sim \frac{\omega }{V_{py}} \right]&= \frac{\pi}{2} \\ 2 \pi z_{min}\left[ \frac{f}{c}\sqrt{\varepsilon _{rx}}\sim \frac{f}{c}\sqrt{\varepsilon _{ry}}\right]&=\frac{\pi}{2} \\ \frac{4 z_{min}}{\lambda _0}[n_x\sim n_y]&=1 \\ z_{min}&=\frac{\lambda _0}{4(n_x\sim n_y)} \\ &=\frac{1.5}{4(1.5\sim 1.5001)}\mu m \\ &= \frac{1.5}{4(0.0001)}\mu m=\frac{1.5}{4}cm\\ z_{min}&=0.375cm \end{aligned}

Q4GATE 2016NAT

A radar operating at 5 GHz uses a common antenna for transmission and reception. The antenna has a gain of 150 and is aligned for maximum directional radiation and reception to a target 1 km away having radar cross-section of 3 $m^{2}$ . If it transmits 100 kW, then the received power (in $\mu$ W) is__________

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📖 Explanation

\begin{aligned} f&=5 \mathrm{GHz}, G=150, R=1 \mathrm{km} \\ P_{t}&=100 \mathrm{kW}, \sigma=3 \mathrm{m}^{2} \\ P_{r}&= \frac{(G \lambda)^{2} \cdot \sigma}{(4 \pi)^{3} R^{4}} \times P_{t} \\ &=\frac{\left(150 \times \frac{3 \times 10^{8}}{5 \times 10^{9}}\right)^{2} \cdot 3}{(4 \pi)^{3} \cdot 10^{12}} \times 10^{5} \\ &=\frac{81 \times 3 \times 10^{5}}{(4 \pi)^{3} \times 10^{12}} \\ &= 1.224 \times 10^{-8}=0.012 \mu \mathrm{W} \end{aligned}

Q5GATE 2016NAT

Light from free space is incident at an angle $\theta _{i}$ to the normal of the facet of a step-index large core optical fibre. The core and cladding refractive indices are $n _{1}=1.5 \; and \; n_{2}=1.4$ , respectively. The maximum value of $\theta _{i}$ (in degrees) for which the incident light will be guided in the core of the fibre is ________

🚀 Solve in Practice Mode

📖 Explanation

\begin{aligned} & \sin \alpha_{\max } &=\sqrt{n_{1}^{2}-n_{2}^{2}}=\sqrt{1.5^{2}-1.4^{2}} \\ \Rightarrow & \alpha_{\max } &=\sin ^{-1}(0.5385)=32.58^{\circ} \end{aligned}

Q6GATE 2014MCQ

A two-port network has scattering parameters given by

[S]=\begin{bmatrix} S_{11} &S_{12} \\ S_{21} & S_{22} \end{bmatrix}

. If the port-2 of the two port is short circuited, the $S_{11}$ parameter for the resultant one port network is

🚀 Solve in Practice Mode

📖 Explanation

\begin{array}{l} b_{1}=s_{11} \cdot a_{1}+s_{12} \cdot a_{2} \qquad\ldots(i)\\ b_{2}=s_{21} \cdot a_{1}+s_{22} a_{2}\qquad\ldots(ii) \end{array}

When output is short, $a_{2}=-b_{2}$

\begin{aligned} -a_{2} &=s_{21} \cdot a_{1}+s_{22} a_{2} \\ \Rightarrow \quad a_{2} &=\frac{-s_{21} \cdot a_{1}}{1+s_{22}} \end{aligned}

Substituting in first equation,

\begin{aligned} b_{1} &=s_{11} \cdot a_{1}+s_{12}\left(\frac{-s_{21} \cdot a_{1}}{1+s_{22}}\right) \\ &=\frac{s_{11} \cdot a_{1}+s_{11} \cdot s_{22} a_{1}-s_{12} \cdot s_{21} a_{1}}{1+s_{22}} \\ \frac{b_{1}}{a_{1}} &=s_{11}=\frac{s_{11}+s_{11} \cdot s_{22}-s_{12} \cdot s_{21}}{1+s_{22}} \end{aligned}

Q7GATE 2010MCQ

If the scattering matrix [S] of a two port network is

|S|=\begin{bmatrix} 0.2\angle 0^{\circ}&0.9\angle 90^{\circ} \\ 0.9\angle 90^{\circ}& 0.1\angle 90^{\circ} \end{bmatrix}

then the network is

🚀 Solve in Practice Mode

📖 Explanation

For reciprocal networks, $S_{12}=S_{21}$ For symmetrical networks, $S_{11}=S_{22}$ For antimetrical networks, $S_{11}=-S_{22}$ For lossless reciprocal networks, $\left|S_{11}\right|=\left|S_{22}\right|$ and $\left|S_{11}\right|^{2}+\left|S_{12}\right|^{2}=1$

Q8GATE 2008MCQ

In the design of a single mode step index optical fibre close to upper cut-off, the single-mode operation is not preserved if

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Q9GATE 2007MCQ

A load of 50 $\Omega$ is connected in shunt in a 2-wire transmission line of $Z_{0} = 50\Omega$ as shown in the figure. The 2-port scattering parameter matrix (s-matrix) of the shunt element is

🚀 Solve in Practice Mode

📖 Explanation

The line is terminated with $50 \Omega$ at the center and so, matched on both the sides.

Q10GATE 2004MCQ

In a microwave test bench, why is the microwave signal amplitude modulated at 1 kHz

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Antennas And Miscellaneous Topics – GATE ECE ElectromagneticsPractice Questions & PYQs

10 questions · 10 PYQs · 0 AI practice · GATE ECE 2027