Attempt the ISRO Electronics & Communication (ECE) 2017-2 previous year paper in real CBT exam mode or practice mode with verified answers and explanations.
All ISRO Electronics & Communication (ECE) papers
80 questions. Answers are shown only where the official answer key is available.
Topic: Digital Signal Processing
What will be the minimum numbers of tap require to realize a FIR filter having $f_{pass}=10$ kHz and $f_{stop}=15$ kHz, 0.1 dB pass band ripple and 60 dB attenuation in stop band. Sampling frequency is 200 kHz.
Answer: A. 110
Explanation: Based on the official answer key, the correct option is (a). No detailed mathematical derivation is provided in the reference.
Topic: Signals and Systems
The system with the transfer function $\frac{Y(s)}{X(s)}=\frac{s}{s+p}$ has an output $y(t)=\cos(2t-\frac{\pi}{3})$ for the input signal $x(t)=p\cos(2t-\frac{\pi}{2})$. Then, the system parameter p is
Answer: C. $2/\sqrt{3}$
Explanation: $\frac{Y(s)}{X(s)}=\frac{s}{s+p}=H(s)$
$\angle H(s)=90^\circ-\tan^{-1}(\frac{\omega}{p})$
$\angle y(t)=\angle x(t)+\angle H(s)$
$-\frac{\pi}{3}=-\frac{\pi}{2}+90^\circ-\tan^{-1}(\frac{\omega}{p})$
where $\omega=2$. So, $-\frac{\pi}{3}=-\frac{\pi}{2}+\frac{\pi}{2}-\tan^{-1}(\frac{2}{p})$
$\tan^{-1}(\frac{2}{p})=\frac{\pi}{3}$
$\frac{2}{p}=\tan\frac{\pi}{3}$
$p=\frac{2}{\sqrt{3}}$
Topic: Control Systems
The approximate Bode magnitude plot of a minimum-phase system is shown in the figure below. The transfer function of the system is
(Figure indicates an initial magnitude of 20 dB up to $\omega=0.1$, then a slope up to $\omega=10$ reaching 140 dB, then a slope up to $\omega=100$ reaching 160 dB, and constant thereafter).
Answer: C. $10^8\frac{(s+0.1)^3}{(s+10)^2(s+100)}$
Explanation: Slope after $\omega=0.1$ is +60 dB/dec. Thus, there are 3 zeros at 0.1.
Slope after $\omega=10$ changes from +60 dB/dec to +20 dB/dec (since it goes from 140 dB to 160 dB over one decade). This implies two poles at $\omega=10$.
Slope after $\omega=100$ is 0 dB/dec, implying 1 pole at $\omega=100$.
Gain is 20 dB for $\omega<0.1$, so $20\log_{10}K=20 \Rightarrow K=10$.
$T.F.=\frac{10(\frac{s}{0.1}+1)^3}{(\frac{s}{10}+1)^2(\frac{s}{100}+1)}=\frac{10 \times 10^3 \times (s+0.1)^3}{\frac{(s+10)^2}{100} \times \frac{(s+100)}{100}}=\frac{10^8(s+0.1)^3}{(s+10)^2(s+100)}$.
Topic: Network Theory
What is the power loss in the 10 Ω resistor in the Network shown in figure?
Answer: B. 15.13 W
Explanation: Using mesh analysis:
$i_3=\frac{11(230-25)+3(-69-40)-8(15+80)}{(11 \times 205)-(3 \times 109)-(8 \times 95)}=\frac{1425}{1168}=1.22$ A
Power absorbed by 10 Ω resistor will be $P_{10\Omega}=i_3^2(10\Omega)=(1.22)^2 \times 10\text{ W}=14.88$ W. The nearest answer is option (b) 15.13 W.
Topic: Network Theory
The voltmeter in the circuit shown in the figure is ideal. The transformer has two identical windings with perfect coupling. The reading on the voltmeter will be
Answer: D. Zero
Explanation: As the voltmeter is ideal and coupling is perfect, the voltage across secondary terminals will be the same as on the primary side (in terms of magnitude and phase, considering dot convention and identical windings). Hence, the potential difference across the voltmeter will be zero.
Topic: Digital Electronics
If input to T flip flop is 200 Hz signal, then what will be the output signal frequency if four T flip flops are connected in cascade
Answer: D. None of the above
Explanation: Input frequency = 200 Hz. Output frequency when 4 T flip/flop stages are cascaded = $\frac{F_t}{2^4}=\frac{200}{16}=12.5$ Hz. Note: The official solution text calculates it as $\frac{200}{2 \times 4}=25$ Hz, which also leads to (d) None of the above.
Topic: Digital Electronics
Simplify the below function represented as sum of minterms
$F(A, B, C, D, E)=\Sigma(0, 1, 2, 3, 8, 9, 16, 17, 20, 21, 24, 25, 28, 29, 30, 31)$
Answer: A. $A\overline{D}+(\overline{C+D})+\overline{A}\overline{B}\overline{C}+(\overline{\overline{A}+\overline{B}+\overline{C}})$
Explanation: Using K-map simplification, the terms are derived as $\overline{A}\overline{B}\overline{C}+\overline{A}\overline{C}\overline{D}+A\overline{C}\overline{D}+A\overline{D}+ABC=A\overline{D}+C\overline{D}+\overline{A}\overline{B}C+ABC$, which equates equivalently to option (a).
Topic: Analog Electronics
All transistor in the N output current mirror in figure given below are matched with a finite gain $\beta$ and early voltage $V_A=\infty$. The expression for load current is
Answer: A. $\frac{I_{ref}}{(1+\frac{(1+N)}{\beta(\beta+1)})}$
Explanation: $I_{ref}=I_C+\frac{(N+1)I_B}{(\beta+1)}$.
$I_{ref}=I_B(\beta+\frac{(N+1)}{\beta+1})$.
$I_B=\frac{I_{ref}}{(\beta+\frac{(N+1)}{\beta+1})}$.
$I_1=\beta I_B=\frac{\beta I_{ref}}{(\beta+\frac{(N+1)}{\beta+1})}=\frac{I_{ref}}{(1+\frac{(N+1)}{\beta(\beta+1)})$.
Topic: Control Systems
From the figure, obtain state equation
Answer: B. $[\dot{X}]=\begin{bmatrix} -2 & 4 \\ 0 & -3 \end{bmatrix}[X]+\begin{bmatrix} 0 \\ 1 \end{bmatrix}u$
Explanation: From the signal flow graph: $\dot{x}_1=-2x_1+4x_2$ and $\dot{x}_2=-3x_2+u$. Hence, $[\dot{X}]=\begin{bmatrix} -2 & 4 \\ 0 & -3 \end{bmatrix}[X]+\begin{bmatrix} 0 \\ 1 \end{bmatrix}u$.
Topic: Communication Systems
In an ADC, the minimum Effective Number of Bits (ENOB) requires to represent each quantization level to achieve Signal to Noise and Distortion Ratio (SINAD) of 70 dB is
Answer: D. 12 bits
Explanation: SNR required = 70 dB.
$SNR \le 1.8+6m$ dB.
$70 \le 1.8+6m$ dB.
$m=11.36$ bits $\approx$ 12 bits.
Topic: Electromagnetics
Two isotropic antennas are separated by a distance of two wavelengths. If both the antennas are fed with currents of equal phase and magnitude, the number of lobes in the radiation pattern in the horizontal plane are
Answer: D. 8
Explanation: Phase difference, $\Psi=\alpha+\beta d\cos\theta$.
$\beta=\frac{2\pi}{\lambda}$, $d=2\lambda$.
$\Psi=\frac{2\pi}{\lambda} \times 2\lambda\cos\theta=4\pi\cos\theta$.
$E_T=2E_0\cos(\frac{\Psi}{2})=2E_0\cos(2\pi\cos\theta)$.
This term is maximum for $\theta=$ 0°, 60°, 120°, 180°, 240°, 270°, 300°. Thus, there is a total of 8 lobes.
Topic: Electromagnetics
The half-power beam width (HPBW) of an antenna in the two orthogonal planes are 120° and 40° respectively. The directivity of the antenna is approximately equals to
Answer: D. 8.5 dB
Explanation: HPBW in two orthogonal planes = 120° and 40°.
Directivity = $10\log_{10}(\frac{41253}{120 \times 40})$ (Standard approximation uses 41253 sq. degrees).
Directivity $\approx 8.6$ dB. Nearest option is 8.5 dB.
Topic: Communication Systems
Two resistors $R_1$ and $R_2$ (in ohms) at temperatures $T_1$ K and $T_2$ K respectively, are connected in series. Their equivalent noise temperature is
Answer: D. $(R_1T_1+R_2T_2)/(R_1+R_2)$
Explanation: Noise spectrum across $R_1=2kT_1R_1$.
Noise spectrum across $R_2=2kT_2R_2$.
Total noise spectrum = $2kT_1R_1+2kT_2R_2$.
Equivalent noise temperature = $\frac{\text{Total noise spectrum}}{2k \times \text{Total resistance}}=\frac{2kR_1T_1+2kR_2T_2}{2k(R_1+R_2)}=\frac{R_1T_1+R_2T_2}{R_1+R_2}$.
Topic: Signals and Systems
Laplace transform of $e^{-\alpha t}f(t)$ is
Answer: C. $F(s+\alpha)$
Explanation: $x(t)=e^{-\alpha t}f(t)$.
$X(s)=\int_{-\infty}^{\infty} e^{-\alpha t}f(t)e^{-st} dt=\int_{-\infty}^{\infty} f(t)e^{-(s+\alpha)t} dt=F(s+\alpha)$.
Topic: Electromagnetics
A monochromatic plane wave of wavelength 500 μm is propagating in the direction as shown in the figure below. $\vec{E}_i, \vec{E}_r$ and $\vec{E}_t$ incident, reflected and transmitted electric field vectors associated with the wave. The expression for $\vec{E}_t$ and $\vec{E}_r$ are
Answer: B. $\frac{E_0}{\sqrt{2}}(\hat{a}_x-\hat{a}_y)e^{-j\frac{2\pi \times 10^4(x+y)}{5\sqrt{2}}}$ V/m and $-0.10\frac{E_0}{\sqrt{2}}(\hat{a}_x+\hat{a}_y)e^{-j\frac{2\pi \times 10^4(x-y)}{5\sqrt{2}}}$ V/m
Explanation: Based on the Fresnel equations for reflection and transmission coefficients and given permittivities, $E_t=\frac{E_0}{\sqrt{2}}(\hat{a}_x-\hat{a}_y)e^{-j\frac{2\pi \times 10^4(x+y)}{5\sqrt{2}}}$ and $E_r=0.1\frac{E_0}{\sqrt{2}}(-\hat{a}_x-\hat{a}_y)e^{-j\frac{2\pi \times 10^4(x-y)}{5\sqrt{2}}}$. Note: The official key maps to (a), but mathematical derivation yields (b).
Topic: Control Systems
Consider a unity feedback system having an open loop transfer function
$G(j\omega)=\frac{k}{j\omega(j0.2\omega+1)(j0.05\omega+1)}$
Find open loop gain (k) with gain margin of 20 dB
Answer: B. 2.5
Explanation: $G(j\omega)=\frac{k}{j\omega(j0.2\omega+1)(j0.05\omega+1)}$
Gain margin $= 20 \text{ dB} = 10$
$\angle G(j\omega) = -90^{\circ} - \tan^{-1}(0.2\omega) - \tan^{-1}(0.05\omega)$
Equating it equal to $-180^{\circ}$ for phase cross over frequency.
$-180^{\circ} = -90^{\circ} - \tan^{-1}\left(\frac{0.2\omega_p+0.05\omega_p}{1-\frac{\omega_p^2}{100}}\right)$
$1-\frac{\omega_p^2}{100}=0 \Rightarrow \omega_p=10 \text{ rad/sec}$
$|G(j\omega)|_{\omega=\omega_p} = \frac{1}{\text{Gain Margin}}$
$\frac{1}{10} = \frac{K}{10\sqrt{2^2+1^2}\sqrt{(0.5)^2+1}}$
$K = \sqrt{5}\sqrt{1.25} = 2.5$
Topic: Control Systems
The open loop transfer function of a unity feedback system is
$G(s)=\frac{K}{s(s²+s+2)(s+3)}$
The range of K for which the system is stable is
Answer: A. $\frac{21}{44}>K>0$
Explanation: $G(s)=\frac{K}{s(s²+s+2)(s+3)}$
Characteristic equation:
$(s³+s²+2s)(s+3)+K=0$
$s^4+4s^3+5s^2+6s+K=0$
By forming the Routh array, for the system to be stable:
$\frac{(21-4K)4}{14}>0 \text{ and } K>0$
$\frac{21}{4}>K>0$
(Note: The option contains a typo as $\frac{21}{44}$, but mathematically it corresponds to $\frac{21}{4}$ as derived from the Routh array)
Topic: Analog Electronics
For the CE (Common emitter) circuit shown, what will be the value of $I_E$ and $V_{CE}$?
$V_{CC}=10 \text{ V}$, $R_C=2 \text{ k}\Omega$, $V_{BE}=0.7 \text{ V}$
Answer: D. 3.02 mA, 4 V
Explanation: $I_B \times 220 \text{ k}\Omega = 4 - 0.7$
$I_B = 0.015 \text{ mA}$
Let, $V_{CE} = 4 \text{ V}$
$I_C = \frac{10-4}{2 \text{ k}\Omega} = 3 \text{ mA}$
$I_E = 3 \text{ mA} + 0.015 \text{ mA} = 3.015 \text{ mA} \approx 3.02 \text{ mA}$
Topic: Signals and Systems
A signal $x(t)=\text{sinc}(\alpha t)$, where $\alpha$ is a real constant, is the input to a Linear Time Invariant system whose impulse response is $h(t)=\text{sinc}(\beta t)$, where $\beta$ is a real constant. If $\min(\alpha, \beta)$ denotes the minimum of $\alpha$ and $\beta$ and similarly, $\max(\alpha, \beta)$ denotes the maximum of $\alpha$ and $\beta$, and K is a constant, which one of the following statements is true about the output of the system (here, $\text{sinc}(x)=\frac{\sin(\pi x)}{\pi x}$).
Answer: A. It will be of the form $K \text{ sinc}(\gamma t)$ where $\gamma=\min(\alpha,\beta)$
Explanation: $x(t) = \text{sinc}(\alpha t) \longleftrightarrow X(\omega)$ is a rectangular pulse of width $\alpha$.
$h(t) = \text{sinc}(\beta t) \longleftrightarrow H(\omega)$ is a rectangular pulse of width $\beta$.
$Y(\omega) = X(\omega)H(\omega)$. The multiplication of two rectangular pulses of different widths in the frequency domain results in a rectangular pulse with a width equal to the minimum of the two widths. Therefore, the time-domain output is a sinc function with the parameter $\min(\alpha, \beta)$.
Topic: Communication Systems
A message signal given by $m(t)=(\frac{1}{2})\cos(\omega_1 t)-(\frac{1}{2})\sin(\omega_2 t)$ is amplitude modulated with a carrier of frequency $\omega_c$ to generate $s(t)=[1+m(t)]\cos(\omega_c t)$. What is the power efficiency achieved by this modulation scheme?
Answer: C. 20%
Explanation: Power of message signal $P_m = \frac{(1/2)^2}{2} + \frac{(-1/2)^2}{2} = \frac{1}{8} + \frac{1}{8} = \frac{1}{4}$.
Carrier power $P_c = \frac{1^2}{2} = \frac{1}{2}$.
Sideband power $P_{sb} = \frac{P_m}{2} = \frac{1/4}{2} = \frac{1}{8}$.
Total power $P_t = P_c + P_{sb} = \frac{1}{2} + \frac{1}{8} = \frac{5}{8}$.
Efficiency $\eta = \frac{P_{sb}}{P_t} = \frac{1/8}{5/8} = \frac{1}{5} = 20\%$.
Topic: Communication Systems
If $E_b$, the energy per bit of a binary digital signal, is $10^{-5}$ Ws (Watt-Second) and the one-sided power spectral density of the white noise, $N_0 = 10^{-6} \text{ W/Hz}$, then the output SNR of the matched filter is
Answer: D. 13 dB
Explanation: Energy per bit $E_b = 10^{-5} \text{ Ws}$.
Noise $N_0 = 10^{-6} \text{ W/Hz}$.
$SNR = 10 \log_{10}\left(\frac{2E_b}{N_0}\right) = 10 \log_{10}\left(\frac{2 \times 10^{-5}}{10^{-6}}\right) = 10 \log_{10}(20) = 13 \text{ dB}$.
Topic: Communication Systems
At a given probability of error, binary coherent FSK is inferior to binary coherent PSK by
Answer: B. 3 dB
Explanation: $P_e \text{ for FSK} = \frac{1}{2}\text{erfc}\sqrt{\frac{E_b}{2N_0}}$
$P_e \text{ for PSK} = \frac{1}{2}\text{erfc}\sqrt{\frac{E_b}{N_0}}$
Hence, for a given $P_e$, PSK is better than FSK by a factor of 2, which is $10 \log_{10}(2) = 3 \text{ dB}$.
Topic: Digital Signal Processing
If $z_0$ is a zero of a (real-valued) linear-phase FIR filter then following is/are also zero/zeros of a (real-valued) linear-phase FIR filter,
Answer: C. $1/Z_0$, $Z_0^*$ and $1/z_0^*$
Explanation: For a linear phase FIR filter, $h(n)$ should be symmetric. Using the symmetry condition $h(n)=h(N-1-n)$ in z-domain: $H(z)=z^{-(N-1)}H(1/z)$. If $H(z_0)=0$, then $H(1/z_0)=0$. Also, roots of a polynomial with real coefficients exist in complex conjugate pairs. Thus, $H(Z_0^*)=0$ and $H(1/Z_0^*)=0$.
Topic: Network Theory
In the network shown below, switch is opened at $t=0$ after long time. The current $i_L(t)$ for $t>0$ is given as
Answer: D. $e^{-2t}(2\sin t-4\cos t)\text{ A}$
Explanation: For $t<0$, $i_L = -4 \text{ A}$, $V_c = 8 \text{ V}$. For $t>0$, using the equivalent circuit in Laplace domain and superposition theorem:
$I_L(s) = \frac{-4(s+2)}{s²+4s+5} + \frac{2}{s²+4s+5} = \frac{-4(s+2)}{(s+2)^2+1} + \frac{2}{(s+2)^2+1}$
Taking inverse Laplace transform:
$I_L(t) = -4e^{-2t}\cos t + 2e^{-2t}\sin t = e^{-2t}(2\sin t - 4\cos t) \text{ A}$.
Topic: Network Theory
The Q factor of a RLC circuit is 5 at its resonance frequency of 1 kHz. Find the bandwidth of the circuit.
Answer: B. 200 Hz
Explanation: $Q \text{ factor} = 5$
$f_R = 1 \text{ kHz}$
$\text{Bandwidth} = \frac{f_R}{Q} = \frac{1000}{5} = 200 \text{ Hz}$.
Topic: Network Theory
For the circuit shown in figure, find the frequency at which this circuit will be at resonance
Answer: D. 0.5 rad/sec
Explanation: $Z_{in} = j\omega + \frac{-j/2\omega}{1 - \frac{j}{2\omega}} = j\omega + \frac{-j}{2\omega-j} = j\omega + \frac{(-j)(2\omega+j)}{4\omega²+1}$
$Z_{in} = \frac{1+j(4\omega³-\omega)}{4\omega²+1}$
For resonance $Z_{in}$ should be real, so its imaginary part is zero.
$4\omega³-\omega=0 \Rightarrow \omega=0.5 \text{ rad/sec}$.
Topic: Digital Electronics
The following CMOS transistor based circuit with A, B, C as input and X, Y as output represents which circuit?
Answer: C. Positive edge trigger S-R Flip-Flop
Explanation: When $C=1$ and $A=1$, $B=1 \Rightarrow X=0$ and $Y=0$. Thus it represents an SR flip-flop with positive edge trigger.
Topic: Digital Electronics
Minimum number of complementary CMOS transistors pair will be required to implement function, $F=ABC+(\overline{A+B+C})$ are
Answer: D. 9
Explanation: 9 CMOS pairs are required to construct the given logical function $F = ABC + (\overline{A+B+C})$ efficiently with full complementary CMOS logic.
Topic: Digital Electronics
The CMOS circuit shown below implements the function
Answer: C. $\overline{(A+B)C+D}$
Explanation: In the pull-down network, transistors A and B are in parallel, which are in series with C, and this entire combination is in parallel with D. Since CMOS logic inherently implements the inverted output, the function is $Y = \overline{(A+B)C+D}$.
Topic: Analog Electronics
Transistors $Q_1$ and $Q_2$ are identical and $\beta>>1$ in the circuit shown in the figure below. The output voltage is $(V_t=0.026 \text{ V})$.
Answer: B. $\log_{10}(\frac{v_2}{v_1}\frac{R_1}{R_2})$
Explanation: $\frac{V_1}{R_1}=I_{C1}$ and $\frac{V_2}{R_2}=I_{C2}$
$I_{C1} = I_S e^{V_{BE1}/V_T} \Rightarrow V_{BE1} = V_T \ln(\frac{I_{C1}}{I_S})$
$I_{C2} = I_S e^{V_{BE2}/V_T} \Rightarrow V_{BE2} = V_T \ln(\frac{I_{C2}}{I_S})$
$V_0 = \frac{333}{20}(-V_{BE1}+V_{BE2}) = \frac{333}{20}V_T \ln\left(\frac{V_2}{R_2}\times\frac{R_1}{V_1}\right)$
$= \frac{333}{20} \times 0.026 \times 2.303 \log_{10}\left(\frac{v_2}{v_1}\frac{R_1}{R_2}\right) \approx \log_{10}\left(\frac{v_2}{v_1}\frac{R_1}{R_2}\right)$.
Topic: Microprocessors
Consider following 8085 microprocessor program
MVI A, DATA1
ORA A
JM DISPLAY
OUT PORT1
CMA
DISPLAY:
ADI 01H
OUT PORT1
HLT
If $DATA1=A7H$, the output of PORT1 is
Answer: D. 59H
Explanation: DATA 1 = A7 H. A7 H in binary is 10100111.
ORA A -> A7 H (or) A7 H = A7 H. Since the MSB is 1, the sign flag is set, so JM DISPLAY will branch to DISPLAY.
However, if we follow the official solution logic given, it appears the jump is NOT taken or the logic is evaluated sequentially as:
A7 H = 10100111.
CMA gives 1's complement: 01011000 = 58 H.
ADI 01 H adds 1: 01011000 + 00000001 = 01011001 = 59 H.
Output is 59H.
Topic: Signals and Systems
The DTFT of a sequence $x[n]$ is given by $X(e^{j\omega})$. Since $X(e^{j\omega})$ is period function of $\omega$, it can be expressed classical Fourier series as, $X(e^{j\omega})=\sum_{n=-\infty}^{\infty}C_{n}e^{jn\omega_{0}\omega}$ where $\omega_{0}$ is a fundamental frequency. Which of the following statement is correct?
Answer: C. $\omega_{0}=1, C_{n}=x[-n]$
Explanation: $X(e^{j\omega})=\sum_{n=-\infty}^{\infty}x[n]e^{-j\omega n}$
Given $X(e^{j\omega})=\sum_{n=-\infty}^{\infty}C_{n}e^{jn\omega_{0}\omega}$
Replacing n by (-n) in the first equation:
$X(e^{j\omega})=\sum_{n=-\infty}^{\infty}x[-n]e^{j\omega n}$
Comparing it with the given series, we get $C_{n}=x[-n]$ and $\omega_{0}=1$.
Topic: Engineering Mathematics
Evaluate $\int_{0}^{1}\frac{\ln(x+1)}{x^{2}+1}\cdot dx$
Answer: B. $\frac{\pi}{8}\ln 2$
Explanation: Let $x=\tan y$, then $dx=\sec^{2}y dy$.
$\int_{0}^{1}\frac{\ln(x+1)}{x^{2}+1}dx = \int_{0}^{\pi/4}\frac{\ln(1+\tan y)}{\sec^{2}y}\sec^{2}y dy = \int_{0}^{\pi/4}\ln(1+\tan y)dy = I$.
Using the property $\int_{0}^{a}f(x)dx = \int_{0}^{a}f(a-x)dx$:
$I = \int_{0}^{\pi/4}\ln(1+\tan(\pi/4-y))dy = \int_{0}^{\pi/4}\ln(1+\frac{1-\tan y}{1+\tan y})dy = \int_{0}^{\pi/4}\ln(\frac{2}{1+\tan y})dy$.
Adding both expressions for I:
$2I = \int_{0}^{\pi/4}\ln(2)dy \Rightarrow I = \frac{\pi}{8}\ln(2)$.
Topic: Signals and Systems
Fourier transform of a real and odd function is
Answer: C. Imaginary and odd
Explanation: Based on the properties of the Fourier transform, if a signal in the time domain is real and odd, its Fourier transform is purely imaginary and odd.
Topic: Signals and Systems
Let $F(w)$ be the Fourier Transform of a function $f(t)$. The $F(0)$ is
Answer: A. $\int_{-\infty}^{\infty}f(t)dt$
Explanation: $F(\omega)=\int_{-\infty}^{\infty}f(t)e^{-j\omega t}dt$. Putting $\omega=0$, we get $F(0)=\int_{-\infty}^{\infty}f(t)e^{0}dt = \int_{-\infty}^{\infty}f(t)dt$.
Topic: Electromagnetics
A transmission line with a characteristic impedance of 100 $\Omega$ is used to match a 50 $\Omega$ section to a 200 $\Omega$ section. If the matching is to be done both at 500 MHz and 1.2 GHz, the length of the transmission line can be approximately,
Answer: C. 1.35 m
Explanation: $Z_{1}=50~\Omega$, $Z_{2}=200~\Omega$. $Z_{0}=\sqrt{Z_{1}Z_{2}} = \sqrt{50 \times 200} = 100~\Omega$. Hence, quarter wave matching is suitable, and length will be odd multiples of $\lambda/4$.
At $f=500\text{ MHz}$, $\lambda = \frac{3\times 10^{8}}{5\times 10^{8}} = 0.6\text{ m}$. $\lambda/4 = 0.15\text{ m}$.
At $f=1.2\text{ GHz}$, $\lambda = \frac{3\times 10^{8}}{12\times 10^{8}} = 0.25\text{ m}$. $\lambda/4 = 0.0625\text{ m}$.
Checking multiples: $0.15 \times 9 = 1.35\text{ m}$ and $0.0625 \times 21 = 1.35\text{ m}$. Both match at 1.35 m.
Topic: Control Systems
System has some poles lying on imaginary axis is
Answer: D. Marginally stable
Explanation: If a system has non-repeated poles lying on the imaginary axis (and no poles in the right half of the s-plane), its impulse response contains oscillatory terms of constant amplitude, making the system marginally stable.
Topic: Control Systems
The open-loop DC gain of a unity negative feedback system with closed loop transfer function $(s+4)/(s^{2}+7s+13)$ is
Answer: C. $4/9$
Explanation: Closed loop T.F. $= \frac{G(s)}{1+G(s)} = \frac{s+4}{s^{2}+7s+13}$.
Open loop T.F. $G(s) = \frac{\text{Numerator}}{\text{Denominator} - \text{Numerator}} = \frac{s+4}{s^{2}+7s+13 - (s+4)} = \frac{s+4}{s^{2}+6s+9}$.
Open loop DC gain is obtained by putting $s=0$: $G(0) = \frac{4}{9}$.
Topic: Control Systems
The unit impulse response of a system is $h(t)=e^{-t}, t\ge 0$. For this system, the steady-state value of the output for unit step input is equal to
Answer: C. 1
Explanation: Input $x(t)=u(t) \Rightarrow X(s)=\frac{1}{s}$.
$h(t)=e^{-t} \Rightarrow H(s)=\frac{1}{s+1}$.
Output $Y(s) = X(s)H(s) = \frac{1}{s(s+1)} = \frac{1}{s} - \frac{1}{s+1}$.
$y(t) = 1 - e^{-t}$.
For steady state, $y(\infty) = \lim_{t\to\infty} (1 - e^{-t}) = 1$.
Topic: Control Systems
A system has fourteen poles and two zeros. Its high frequency asymptote in its magnitude plot having a slope of
Answer: B. -240 dB/decade
Explanation: Number of poles = 14.
Number of zeros = 2.
High frequency asymptote's slope $= 20 \times (\text{Number of zeros} - \text{Number of poles}) \text{ dB/decade} = 20(2 - 14) = -240 \text{ dB/decade}$.
Topic: Communication Systems
Consider a system shown in the figure. Let $X(f)$ and $Y(f)$ denote the Fourier transforms of $x(t)$ and $y(t)$ respectively. The ideal HPF has the cut-off frequency 10 kHz.
(The figure shows $x(t)$ passing through a balanced modulator with 10 kHz carrier, then a HPF with 10 kHz cut-off, then another balanced modulator with 13 kHz carrier to produce $y(t)$). The positive frequencies where $Y(f)$ has spectral peaks are
Answer: B. 2 kHz and 24 kHz
Explanation: Peaks of $X(f)$ are at 1 kHz and -1 kHz initially.
Output of the first balanced modulator (DSB-SC with $f_{c1}=10\text{ kHz}$) has peaks at $f_{c1} \pm 1\text{ kHz} = 11\text{ kHz}$ and $9\text{ kHz}$.
The ideal HPF with cut-off 10 kHz will pass only the 11 kHz component.
The second balanced modulator (DSB-SC with $f_{c2}=13\text{ kHz}$) will shift this 11 kHz component to $f_{c2} \pm 11\text{ kHz}$.
Positive frequencies = $13\text{ kHz} + 11\text{ kHz} = 24\text{ kHz}$ and $13\text{ kHz} - 11\text{ kHz} = 2\text{ kHz}$.
Topic: Signals and Systems
Let $x(t)=rect(t-\frac{1}{2})$ for $-\frac{1}{2}\le t\le\frac{1}{2}$ and zero otherwise, then Fourier Transform of $x(t)+x(-t)$ will be given by
Answer: C. $2sinc(\frac{\omega}{2})cos(\frac{\omega}{2})$
Explanation: Note: Based on the official key, the question is marked as having a discrepancy (*). The mathematical derivation yields $Y(\omega) = 2sinc(\frac{\omega}{2\pi})cos(\frac{\omega}{2})$. If the sinc function is defined as $sinc(x) = \frac{\sin(x)}{x}$, the transform involves $2sinc(\frac{\omega}{2})cos(\frac{\omega}{2})$. None of the options perfectly matched standard conventions in the official key, but derived logic points towards form (c).
Topic: Network Theory
In the network shown below, switch is moved from position a to b at $t=0$. The current $i_{L}(t)$ for $t>0$ is given as
Answer: C. $(3-9t)e^{-5t}A$
Explanation: For $t<0$, the inductor is shorted by position a. $i_{L}(0^-) = 3\text{ A}$.
For $t>0$, taking the Laplace transform of the circuit:
$(\frac{4}{s} - I_L)6 + 6 = \frac{12}{s} + I_L(2s + 14 + \frac{50}{s})$
$\frac{12}{s} + 6 = I_L(2s + 20 + \frac{50}{s})$
$I_L = \frac{6s+12}{2s^2+20s+50} = \frac{3s+6}{s²+10s+25} = \frac{3s+6}{(s+5)^2} = \frac{3(s+5) - 9}{(s+5)^2} = \frac{3}{s+5} - \frac{9}{(s+5)^2}$.
Taking the inverse Laplace transform gives $i_L(t) = 3e^{-5t} - 9te^{-5t} = (3-9t)e^{-5t}\text{ A}$.
Topic: Network Theory
The voltage gain $v_{out}/v_{in}$ of a circuit shown below is zero. If $\omega=333.33~rad/s,$ the value of C is
(The circuit is a bridged-T network with mutual inductance $M=0.09\text{ H}$, $L_1=0.12\text{ H}$, $L_2=0.27\text{ H}$, $k=0.5$).
Answer: D. $33.33~\mu F$
Explanation: When $v_{out}/v_{in} = 0$, the transmission through the parallel paths cancels out. The condition for the null in this bridged-T network with mutual coupling simplifies to $3M = \frac{1}{\omega² C}$.
$M = k\sqrt{L_1 L_2} = 0.5\sqrt{0.12 \times 0.27} = 0.5 \times 0.18 = 0.09\text{ H}$.
$C = \frac{1}{3M\omega²} = \frac{1}{3(0.09)(1000/3)^2} = \frac{1}{0.27 \times 1000000/9} = \frac{9}{270000} = \frac{1}{30000}\text{ F} = 33.33~\mu F$.
Topic: Analog Electronics
Output voltage $V_{0}$ of the circuit shown in figure below. (The input voltages are $V_{1}=2.5~V$ and $V_{2}=1~V$)
Answer: C. -4.5 V
Explanation: Using superposition or nodal analysis at the op-amp terminals:
Voltage at non-inverting terminal $V_+ = V_2 \times \frac{1.5\text{k}}{3.3\text{k} + 1.5\text{k}} = 1 \times \frac{1.5}{4.8} = \frac{1.5}{4.8}\text{ V}$.
Output voltage $V_0 = -V_1 \left(\frac{2.2\text{k}}{1\text{k}}\right) + V_+ \left(1 + \frac{2.2\text{k}}{1\text{k}}\right)$.
$V_0 = -2.5(2.2) + \left(\frac{1.5}{4.8}\right)(1 + 2.2) = -5.5 + \frac{1.5}{4.8}(3.2) = -5.5 + 1 = -4.5\text{ V}$.
Topic: Digital Electronics
Without any additional circuitry an 8:1 MUX can be used to obtain
Answer: D. All functions of 4 variables
Explanation: An 8:1 MUX has 3 select lines which can be used to connect 3 of the 4 variables. The 8 data inputs can be connected to the 4th variable, its complement, 0, or 1. Hence, using an 8:1 MUX, we can obtain all functions of 4 variables.
Topic: Digital Electronics
The circuit shown in figure converts
(The figure shows a logic circuit where the MSB passes directly to the output, and each subsequent output bit is the XOR of the current input bit and the previous output bit).
Answer: D. Gray to Binary code
Explanation: The circuit diagram shows a cascade of XOR gates where the MSB is retained as is, and each subsequent bit is formed by XORing the current input bit with the previous generated output bit. This is the standard logic implementation for converting a Gray code to Binary code.
Topic: Digital Electronics
Which of the following statement is true for Programmable Logic array (PLA)?
Answer: C. Fused programmable AND array and Fused programmable OR array
Explanation: For a Programmable Logic Array (PLA), both the AND array and the OR array are fused programmable, allowing maximum flexibility in implementing Boolean functions.
Topic: Digital Electronics
When transistors are used in digital circuits they usually operate in the
Answer: C. Saturation and cut-off regions
Explanation: Transistors used in digital circuits operate primarily as switches. Therefore, they operate in the saturation region (representing logic level 'ON' or 0/1 depending on the logic family) and the cut-off region (representing logic level 'OFF').
Topic: Electronic Devices and Circuits
Two initially identical samples A & B of pure germanium are doped with donors to concentrations of $1\times10^{20}$ and $3\times10^{20}$ respectively. If the hole concentration in A is $9\times10^{12}$ then the hole concentration in B at the same temperature will be
Answer: A. $3\times10^{12}m^{-3}$
Explanation: Using the mass-action law, $n\cdot p = n_i^2$. For sample A: $N_{DA} \times P_A = n_i^2 \Rightarrow 1\times10^{20} \times 9\times10^{12} = n_i^2 = 9\times10^{32}$. For sample B at the same temperature, $n_i^2$ remains constant. So, $N_{DB} \times P_B = n_i^2 \Rightarrow 3\times10^{20} \times P_B = 9\times10^{32} \Rightarrow P_B = \frac{9\times10^{32}}{3\times10^{20}} = 3\times10^{12}m^{-3}$.
Topic: Electronic Devices and Circuits
The built in potential (diffusion potential) in a p-n junction
Answer: D. all of the above
Explanation: The built-in potential is given by $V_0 = V_T \ln\left(\frac{N_A N_D}{n_i^2}\right)$. It is equal to the difference in the Fermi level of the two sides. As doping concentrations ($N_A$ or $N_D$) increase, $V_0$ increases. According to the provided solution key, it is also stated that as temperature increases, $V_0$ increases. Thus, all statements are considered correct.
Topic: Digital Signal Processing
Let $Y(k)$ be the 5-point DFT of the sequence $y(n)=\{1~2~3~4~5\}$. What is the 5-point DFT of the sequence $Y(k)$?
Answer: C. [5 25 20 15 10]
Explanation: Using the duality property of DFT, $\text{DFT}(\text{DFT}(y(n))) = N \cdot y(-n \bmod N)$. Here, $N=5$. The time-reversed sequence modulo 5 is $y(-n \bmod 5) = \{1, 5, 4, 3, 2\}$. Therefore, the DFT of $Y(k)$ is $5 \times \{1, 5, 4, 3, 2\} = \{5, 25, 20, 15, 10\}$.
Topic: Engineering Mathematics
Let A be the series $\sum_{n=1}^{\infty}\frac{(-1)^{n}}{\log(n+2)}$ and B be the series $\sum_{n=2}^{\infty}\left(\frac{3n-4}{3n+2}\right)^{\frac{n+1}{3}}$ for real numbers. Then which of the following is true
Answer: D. Series A is conditionally convergent and series B is divergent
Explanation: Based on the official answer key, the correct option is (d). No detailed mathematical derivation is provided in the reference.
Topic: Engineering Mathematics
A test has 5 multiple-choice questions. Each question has 4 answer options (A, B, C, D). What is the probability that a student will choose "B" for at least three questions if he/she leaves no questions blank?
Answer: C. 53/512
Explanation: Probability that a student will answer B is $p = 1/4$. Probability that a student will not answer B is $q = 3/4$. The probability of choosing B for at least three questions is given by the binomial distribution: $P(X \ge 3) = ^5C_3 p³ q² + ^5C_4 p^4 q^1 + ^5C_5 p^5 q^0 = ^5C_3(1/4)^3(3/4)^2 + ^5C_4(1/4)^4(3/4) + ^5C_5(1/4)^5 = \frac{10 \times 9}{1024} + \frac{5 \times 3}{1024} + \frac{1}{1024} = \frac{90+15+1}{1024} = \frac{106}{1024} = \frac{53}{512}$.
Topic: Electromagnetics
Which of the following has the highest skin depth?
Answer: A. Al
Explanation: Skin depth $\delta = \sqrt{\frac{2}{\omega \mu \sigma}}$. Skin depth is inversely proportional to the square root of conductivity ($\sigma$). Since Aluminum (Al) has the lowest conductivity among the given options (Ag, Cu, Au, Al), it will have the highest skin depth.
Topic: Electromagnetics
The electric field vector of a wave is given as $\vec{E}=E_{0}e^{j(\omega t+3x-4y)}\cdot\frac{8\vec{a}_{x}+6\vec{a}_{y}+5\vec{a}_{z}}{\sqrt{125}}V/m$. Its frequency is 10 GHz. The phase velocity in Y-direction will be
Answer: B. $1.5\times10^{10}m/s$
Explanation: From the given phase term $(\omega t + 3x - 4y)$, the wave vector component in the y-direction is $K_y = 4$. The frequency $f = 10^{10}$ Hz, so $\omega = 2\pi \times 10^{10}$ rad/s. The phase velocity in the y-direction is $v_{py} = \frac{\omega}{K_y} = \frac{2\pi \times 10^{10}}{4} \approx 1.57 \times 10^{10} \text{ m/s}$. The closest derived value in the options based on the official solution key is approximately $1.5 \times 10^{10} \text{ m/s}$ (option b).
Topic: Electromagnetics
The electric field of a plane wave propagating in a lossless non-magnetic medium is given by the following equation
$\vec{E}(z,t)=3\cos(2\pi\times10^{9}t+\beta z)\hat{a}_{x}+2\cos(2\pi\times10^{9}t+\beta z+\frac{\pi}{2})\hat{a}_{y}$
The type of wave polarization is
Answer: A. Right hand elliptical
Explanation: The amplitudes are $E_x = 3$ and $E_y = 2$. Because the amplitudes are unequal, the polarization is elliptical. The phase difference between the y-component and x-component is $\pi/2$. At $z=0, t=0$, $\vec{E} = 3\hat{a}_x$. At $\omega t = \pi/2$, $\vec{E} = -2\hat{a}_y$. The vector rotates from the +x axis to the -y axis. Considering the wave propagates in the -z direction (due to the $+\beta z$ term), pointing the thumb in the -z direction shows the fingers curl from +x to -y (clockwise). Hence, it is right-hand elliptical.
Topic: Electromagnetics
A ring of radius R carries a linear charge density $\lambda$. It is rotating with angular speed $\omega$. The magnetic field at its center is
Answer: B. $\frac{\mu_{0}\lambda\omega}{2}$
Explanation: Total charge on the ring is $q = 2\pi R \cdot \lambda$. The current equivalent to the rotating charge is $I = \frac{q}{T} = \frac{2\pi R \lambda}{2\pi/\omega} = \lambda R \omega$. The magnetic field at the center of a circular current loop is $B = \frac{\mu_0 I}{2R} = \frac{\mu_0 (\lambda R \omega)}{2R} = \frac{\mu_0 \lambda \omega}{2}$.
Topic: Communication Systems
The signal $m(t)$ as shown is applied both to a phase modulator (with $k_{p}$ as the phase constant) and a frequency modulator with ($k_{f}$ as the frequency constant) having the same carrier frequency. The ratio $k_{p}/k_{f}$ (in rad/Hz) for the same maximum phase deviation is
(Assuming $m(t)$ is a square wave alternating between +2 and -2 with a period of 4 seconds as per the source).
Answer: B. $4\pi$
Explanation: For phase modulation, the phase deviation is $k_p m(t)$. For frequency modulation, the phase deviation is $2\pi k_f \int m(t) dt$. For the same maximum phase deviation: $k_p \cdot |m(t)|_{\max} = 2\pi k_f \cdot |\int m(t) dt|_{\max}$. The maximum value of $m(t)$ is 2. The maximum value of its integral (which forms a triangular wave) is $2 \times (T/2) = 2 \times 2 = 4$. So, $k_p(2) = 2\pi k_f(4) \Rightarrow k_p = 4\pi k_f \Rightarrow \frac{k_p}{k_f} = 4\pi$.
Topic: Communication Systems
A message $m(t)$ bandlimited to the frequency $f_{m}$ has a power of $P_{m}$. The power of the output signal in the figure is
(The figure shows $m(t)\cos\omega_0t$ multiplied by $\cos(\omega_0t + \theta)$, followed by an ideal low pass filter with a cutoff of $f_m$ and passband gain of 1).
Answer: D. $\frac{P_m \cos² \theta}{4}$
Explanation: The output of the multiplier is $A(t) = m(t)\cos(\omega_0 t) \cdot \cos(\omega_0 t + \theta) = \frac{m(t)}{2}[\cos(\theta) + \cos(2\omega_0 t + \theta)]$. After passing through the ideal low pass filter with a cut-off frequency $f_m$, the high-frequency term is removed. The output signal is $y(t) = \frac{m(t)}{2}\cos(\theta)$. The power of $y(t)$ is $\frac{\text{Power of } m(t) \times \cos²(\theta)}{4} = \frac{P_m \cos²(\theta)}{4}$.
Topic: Signals and Systems
The input $x(t)$ and output $y(t)$ of a system are related as $y(t)=\int_{-\infty}^{t}x(t)\cos(4t)dt$. The system is
Answer: D. not time-invariant and not stable
Explanation: $y(t-t_0) = \int_{-\infty}^{t-t_0}x(t)\cos(4t)dt$. However, $y^{\prime}(t) = \int_{-\infty}^{t}x(t-t_0)\cos(4t)dt$. Since $y(t-t_0) \ne y^{\prime}(t)$, the system is not time-invariant. Furthermore, if $x(t)$ is bounded, $\cos(4t)$ is also bounded, but integrating a bounded non-zero signal from $-\infty$ to $t$ can result in an unbounded signal, making it not stable.
Topic: Network Theory
For the circuit of the figure the inductor current $i_{L}$ just before $t=0$ is
Answer: C. 4 A
Explanation: According to the official key, the answer is (c) 4 A. Assuming the current divider rule applies before the switch action, $i_L$ splits between the parallel resistive branches. If the resistors are $2~\Omega$ and $3~\Omega$ respectively, $i_L = 10 \times \frac{2}{2+3} = 4 \text{ A}$.
Topic: Network Theory
The network shown in below figure consist of two coupled coils and a capacitor. At $t=0$, the switch is closed connecting a voltage generator, $V_{in}=V~\sin(\frac{t}{\sqrt{MC}})$. What will be the value of $\frac{dv_{a}}{dt}(0^{+})$?
Answer: A. $(\frac{V}{L})\sqrt{\frac{M}{C}}$
Explanation: By taking the Laplace transform of the circuit and solving for $V_a(s)$, and then taking the inverse Laplace transform, we get $V_a(t) = V\sqrt{\frac{M}{L}} \sin\left(\frac{t}{\sqrt{LC}}\right)$. Differentiating with respect to $t$, $\frac{dV_a(t)}{dt} = \frac{V}{L}\sqrt{\frac{M}{C}} \cos\left(\frac{t}{\sqrt{LC}}\right)$. Substituting $t=0$, we obtain $\frac{dV_a(0^+)}{dt} = \frac{V}{L}\sqrt{\frac{M}{C}}$.
Topic: Network Theory
The Thevenin's equivalent circuit of the network shown in figure is across a-b is
Answer: B. $2~\Omega$
Explanation: For Thevenin's equivalent resistance, short the independent voltage source and connect a 1 V test source across terminals a-b. This yields $I_2 = 1/6 \text{ A}$, meaning the dependent source $2I_2 = 1/3 \text{ V}$. The current from the 1V source is $I_1 = I_2 + 2/3 = 1/6 + 1/3 = 1/2 \text{ A}$. The equivalent resistance $R_{th} = V_{test}/I_1 = 1 / (1/2) = 2~\Omega$.
Topic: Network Theory
The average power consumed by the following circuit is
(Given $R=3.33~\Omega$, $X_{L}=2.5~\Omega$, $V_{rms}=20\angle 53.13^{\circ}V$)
Answer: C. 120 W
Explanation: The average power is dissipated only by the resistive component. $P_{avg} = \frac{V_{rms}^2}{R} = \frac{(20)^2}{3.33} = \frac{400}{3.33} = 120 \text{ W}$.
Topic: Control Systems
The given figure shows the pole zero pattern of a filter in the s-plane. The filter in question is a
Answer: B. Band pass filter
Explanation: The pole-zero plot shows poles clustered around a center frequency $\omega_0$ and $-\omega_0$ on the imaginary axis, with a zero at the origin. This configuration characterizes a band pass filter.
Topic: Analog Electronics
Class C amplifier operates
Answer: D. Less than half of the cycle of input signal
Explanation: A Class C amplifier is biased such that the output current flows for less than one-half of the input signal cycle (conduction angle < 180 degrees).
Topic: Analog Electronics
A particular amplifier circuit used for frequency doubling is
Answer: A. push-push
Explanation: A push-push amplifier configuration combines the outputs of two devices in phase, causing the fundamental frequencies to cancel while the second harmonics add up, making it ideal for frequency doubling.
Topic: Analog Electronics
The configuration of given figure is a
(Figure shows an Op-amp with a series-parallel RC network in the positive feedback loop)
Answer: D. Wein bridge oscillator
Explanation: The circuit contains an op-amp with a series RC and a parallel RC network in its positive feedback path, which is the classic structure of a Wien bridge oscillator.
Topic: Electronic Devices and Circuits
For current flowing through semiconductor, which of the following statement is true
Answer: C. Conduction Current + Diffusion Current
Explanation: The total current flowing through a semiconductor is the sum of the drift current (caused by the electric field, often referred to here as conduction current) and the diffusion current (caused by the concentration gradient of charge carriers).
Topic: Digital Electronics
The Gray code for $(A5)_{16}$ is equivalent to
Answer: C. 11011111
Explanation: $(A5)_{16}$ translates to the binary sequence $(10100101)_2$. To convert binary to Gray code, keep the MSB same and XOR consecutive bits: $1$, $(1\oplus 0)=1$, $(0\oplus 1)=1$, $(1\oplus 0)=1$, $(0\oplus 0)=0$, $(0\oplus 1)=1$, $(1\oplus 0)=1$, $(0\oplus 1)=1$. The exact Gray code is 11110111. Option (c) in the official paper was marked correct despite a typo in the provided options text.
Topic: Communication Systems
In Direct Broadcast System (DBS),
Answer: A. MPEG-2 is used for video compression
Explanation: Direct Broadcast Systems (DBS), such as direct-to-home satellite television, typically use the MPEG-2 digital video compression standard to compress the video signal for transmission.
Topic: Digital Signal Processing
If analog sampling frequency of a band limited signal is doubled then corresponding digital sampling frequency will be
Answer: C. $\pi/2$
Explanation: Based on the official answer key, the correct option is (c). No detailed mathematical derivation is provided in the reference.
Topic: Communication Systems
In communication system, if for a given rate of information transmission requires channel bandwidth, $B_{1}$ and signal-to-noise ratio $SNR_{1}$. If the channel bandwidth is doubled for same rate of information then new signal-to-noise ratio will be
Answer: C. $\sqrt{SNR_{1}}$
Explanation: Using Shannon's Channel Capacity theorem: $C = B \log_2(1 + SNR) \approx B \log_2(SNR)$ for large SNR. If the bandwidth is doubled for the same capacity, $C = 2B \log_2(SNR')$. Equating them gives $B \log_2(SNR) = 2B \log_2(SNR') \Rightarrow \log_2(SNR') = \frac{1}{2}\log_2(SNR) \Rightarrow SNR' = \sqrt{SNR}$.
Topic: Communication Systems
Output SNR of a 10 bit PCM was found to be 30 dB, desired SNR is 42 dB. To achieve desired SNR by increasing the number of quantization levels, then new levels will be
Explanation: The SNR for PCM is roughly given by $SNR_{PCM} = K + 6n \text{ dB}$. To increase the SNR from 30 dB to 42 dB (an increase of 12 dB), the number of bits (n) must be increased by 2. The original was 10 bits, so the new number of bits is 12. The number of quantization levels would be $2^{12} = 4096$. None of the given options match this answer.
Topic: Electromagnetics
Indicate which one of the following modes do NOT exist in a rectangular resonant cavity
Answer: A. TE110
Explanation: In a rectangular resonant cavity, the $TE_{110}$ mode does not exist because a transverse electric mode requires at least one non-zero index in the direction of wave propagation within a completely closed cavity structure.
Topic: Electromagnetics
A long solenoid of radius R, having N turns per unit length carries a time dependent current $I(t)=I_{0}\sin(\omega t)$. The magnitude of induced electric field at a distance $R/2$ radially from the axis of the solenoid is
Answer: B. $\frac{R}{4}\mu_{0}NI_{0}\omega \cos(\omega t)$
Explanation: According to Faraday's law, $\oint E \cdot dl = -\frac{d\phi}{dt}$. The magnetic flux $\phi$ enclosed by a loop of radius $P$ (where $P < R$) is $B \cdot A = (\mu_0 N I)(\pi P²)$. So, $E(2\pi P) = \mu_0 N \pi P² \frac{dI}{dt} \Rightarrow E = \frac{\mu_0 N P}{2} \frac{dI}{dt}$. Since $I(t) = I_0 \sin(\omega t)$, we have $\frac{dI}{dt} = I_0 \omega \cos(\omega t)$. At distance $P = R/2$, $E = \frac{\mu_0 N (R/2)}{2} I_0 \omega \cos(\omega t) = \frac{R}{4}\mu_0 N I_0 \omega \cos(\omega t)$.
Topic: Electromagnetics
Penetration depth of magnetic field inside a superconductor is
Answer: B. London depth of penetration
Explanation: The characteristic length that describes the exponential decay of a magnetic field as it penetrates into a superconductor is known as the London penetration depth.
Topic: Electromagnetics
A parallel plate air-filled capacitor has plate area of $10^{-4} m^{2}$ and plate separation of $10^{-3} m$. It is connected to a 2 V, 1.8 GHz source. The magnitude of the displacement current is $(\epsilon_{0}=1/36\pi\times10^{-9}F/m)$
Answer: B. 20 mA
Explanation: Displacement current $I_d = C \frac{dV}{dt}$. Capacitance $C = \frac{\epsilon_0 A}{d}$. Since $V = V_0 \sin(\omega t)$, the magnitude of the displacement current is $|I_d| = \omega C V_0$. So, $|I_d| = 2\pi(1.8 \times 10^9) \left[\frac{(1/36\pi \times 10^{-9})(10^{-4})}{10^{-3}}\right](2) = (3.6\pi \times 10^9) \left(\frac{10^{-13}}{36\pi \times 10^{-3}}\right) (2) = 20 \times 10^{-3} \text{ A} = 20 \text{ mA}$.
Topic: Electromagnetics
Two rectangular waveguide have dimensions of $1\text{ cm}\times0.5\text{ cm}$ and $1\text{ cm}\times0.25\text{ cm}$ respectively. Their respective cut-off frequencies will be
Answer: C. 15 GHz and 15 GHz
Explanation: The dominant mode for a rectangular waveguide is typically $TE_{10}$, and its cut-off frequency is determined only by the broader dimension 'a'. $f_c = \frac{c}{2a}$. For the first waveguide, $a_1 = 1 \text{ cm}$, so $f_{c1} = \frac{3 \times 10^8}{2 \times 10^{-2}} = 15 \text{ GHz}$. For the second waveguide, $a_2 = 1 \text{ cm}$, so $f_{c2} = \frac{3 \times 10^8}{2 \times 10^{-2}} = 15 \text{ GHz}$.