Attempt the ISRO Electronics & Communication (ECE) 2016 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: Analog Circuits
The approximate output Frequency of 555 Oscillator for $R_{A}=R_{B}=2.2~\text{k}\Omega$ and $C=2000~\text{pF}$
Answer: B. 109.3 kHz
Explanation: The time period of the IC-555 working in astable mode can be given as
$T_{total}=0.693\times(R_{A}+2R_{B})C$
$=0.693\times 3 \times 2.2\times 10^{3}\times 2000\times 10^{-12}$
$=9.147\times 10^{-6} \text{ s}$
Now, $f=\frac{1}{T}$
$f=109.3~\text{kHz}$
Hence option (b).
Topic: Digital Electronics
A counter is designed using J-K Flip-Flops as shown in figure. Define its count sequence
Explanation: From the timing diagram and state transitions of the specific circuit provided, the possible counter states $(Q_{1}Q_{2}Q_{3})$ are 000 and 100 only. Since the options do not accurately reflect this sequence, this question is typically marked as having an error or insufficient data matching the choices.
Topic: Digital Electronics
A 1 MHz clock signal is applied to a J-K Flip-Flop with $J=K=1$. What is the frequency of the Flip-Flop output signal?
Answer: B. 500 kHz
Explanation: When $J=K=1$, the J-K flip-flop acts as a toggle flip-flop. It toggles its state on every active clock edge, which divides the input clock frequency by 2.
$f_{out}=\frac{f_{in}}{2}=\frac{10^{6}}{2}\text{ Hz}=500~\text{kHz}$
Topic: Digital Electronics
How many inputs and outputs does a full adder have?
Answer: A. 3, 2
Explanation: A full adder is a combinational logic circuit that performs the addition of three bits. It has 3 inputs (typically denoted as $A$, $B$, and $C_{in}$) and 2 outputs (Sum $S$ and Carry Out $C_{out}$).
Topic: Digital Electronics
Which shift register counter requires the most decoding circuitry?
Answer: A. Johnson Counter
Explanation: For ripple and ring counters, the output state directly reflects the clock number (or count) in a relatively straightforward manner (binary or one-hot). Hence, decoder circuit design is easier. For a Johnson counter, the output states follow a twisted-ring sequence that does not reflect the binary count value directly. Therefore, its decoder circuit design requires more logic gates and becomes more complex.
Topic: Digital Electronics
A 10-bit DAC has a step size of 10 mV. What is its Full scale output voltage and the percentage resolution?
Answer: C. 10.23 V, 0.1%
Explanation: Full Scale Voltage $V_{FS} = (2^{n}-1) \times \text{Step size} = (2^{10}-1) \times 10~\text{mV} = (1024-1) \times 0.01~\text{V} = 10.23~\text{V}$.
Percentage Resolution $= \frac{\text{Step Size}}{V_{FS}} \times 100 = \frac{10~\text{mV}}{10.23~\text{V}} \times 100 = 0.0977\% \simeq 0.1\%$.
Topic: Digital Electronics
If the input signals (A and B) and output signal are as below then the circuit element is (Output is HIGH only when A is LOW and B is LOW)
Answer: C. NOR Gate
Explanation: In the given timing diagram, Output $= 1$ when and only when both the inputs are zero ($A=0, B=0$). This truth table corresponds to the NOR logic.
Topic: Digital Electronics
For a 10-bit digital ramp ADC using 500 kHz clock, the maximum conversion time is
Answer: C. $2046~\mu\text{s}$
Explanation: For a digital ramp (counter-type) ADC, the maximum conversion time occurs for the full-scale analog input.
Maximum conversion time $= (2^{n}-1)T_{clk} = \frac{(2^{n}-1)}{f_{clk}} = \frac{(2^{10}-1)}{500\times 10^{3}}~\text{s} = \frac{1023}{500\times 10^{3}}~\text{s} = 2046~\mu\text{s}$.
Topic: Digital Signal Processing
In Digital Filters, how many interpolated data points are inserted between samples when performing 4X over sampling?
Answer: B. 3
Explanation: When performing an $L$-times oversampling (interpolation), $L-1$ zeros (or interpolated data points) are inserted between consecutive original samples. For 4X oversampling, $L=4$, therefore $4-1 = 3$ interpolated points are inserted.
Topic: Communication Systems
A PAM source generates four symbols 3 V, 1 V, -1 V and -3 V with probability of $p(3)=0.2$, $p(1)=0.3$, $p(-1)=0.3$, $p(-3)=0.2$ respectively. The variance for this source will be
Answer: A. 4.2 V
Explanation: Mean $E[X]=\overline{X}=0$ since the PDF of 'X' is symmetric about the origin.
$\text{Variance} = \sigma_{x}^{2} = E[(X-\overline{X})^{2}] = E[X^{2}] = \sum x_{i}^{2}p(x_{i})$
$= (-3)^{2}(0.2) + (-1)^{2}(0.3) + (1)^{2}(0.3) + (3)^{2}(0.2)$
$= 9(0.2) + 1(0.3) + 1(0.3) + 9(0.2) = 1.8 + 0.3 + 0.3 + 1.8 = 4.2~\text{V}^{2}$ (The unit is conventionally $\text{V}^{2}$, but listed as V in the options).
Topic: Digital Electronics
The output equivalent circuit of following circuit is (two NAND gates used as NOT gates feeding a third NAND gate)
Answer: C. OR
Explanation: The first two NAND gates act as inverters, producing $\overline{A}$ and $\overline{B}$. These are fed to a third NAND gate. The final output is $Y = \overline{\overline{A} \cdot \overline{B}}$. By De Morgan's Law, $Y = A + B$, which represents an OR gate.
Topic: Communication Systems
A binary source in which 0's occurs 3 times as often as 1's. Then its entropy in bits/symbol will be
Answer: C. 0.81 bits/symbol
Explanation: Let $p(1) = x$, then $p(0) = 3x$. Since probabilities sum to 1, $x + 3x = 1 \Rightarrow 4x = 1 \Rightarrow x = 0.25$.
$p(1) = 0.25 = \frac{1}{4}$ and $p(0) = 0.75 = \frac{3}{4}$.
Entropy $H(X) = p(0)\log_{2}\left(\frac{1}{p(0)}\right) + p(1)\log_{2}\left(\frac{1}{p(1)}\right)$
$= \frac{3}{4}\log_{2}\left(\frac{4}{3}\right) + \frac{1}{4}\log_{2}(4) = \frac{3}{4}(2 - 1.585) + \frac{1}{4}(2) = 0.311 + 0.5 = 0.811~\text{bits/symbol}$.
Topic: Communication Systems
Maximum value of signal to noise ratio of an 8-bit ADC with an input range of 10 V will be
Answer: D. 49.8 dB
Explanation: For an $n$-bit ADC, the maximum Signal-to-Quantization Noise Ratio (SQNR) is approximated by the formula:
$SQNR \approx 6n + 1.8~\text{dB}$
For $n=8$ bits:
$SQNR = 6(8) + 1.8 = 48 + 1.8 = 49.8~\text{dB}$.
Topic: Electromagnetics
The available power required at a receiving antenna is $10^{-6}$ Watts. Transmitting and receiving antennas have gain of 40 dB each. The carrier frequency is 4 GHz and the distance between them is 30 miles. The required transmitter power is
Answer: A. 0.64 W
Explanation: Using the Friis transmission equation: $P_{r}=\frac{P_{t}G_{t}G_{r}\lambda^{2}}{(4\pi R)^{2}} \Rightarrow P_{t}=\frac{P_{r}(4\pi R)^{2}}{\lambda^{2}G_{t}G_{r}}$
Given: $P_{r}=10^{-6}~\text{W}$, $f=4~\text{GHz} \Rightarrow \lambda=\frac{c}{f}=\frac{3\times10^{8}}{4\times10^{9}}=0.075~\text{m}$.
$R=30~\text{miles} = 30\times 1.609~\text{km} = 48.27\times 10^{3}~\text{m}$.
$G_{t}=G_{r}=40~\text{dB} = 10^{4}$.
$P_{t}=\frac{10^{-6} \times (4\pi \times 48.27\times 10^{3})^{2}}{(0.075)^{2} \times 10^{4} \times 10^{4}} \approx 0.64~\text{W}$.
Topic: Communication Systems
An FM signal at 10.7 MHz IF needs to be digitized for demodulation in a digital domain. If the bandwidth of this signal is 200 kHz, the maximum usable sampling frequency is
Answer: C. 400 kHz
Explanation: Center frequency $f_{c}=10.7~\text{MHz}$, Bandwidth $BW=200~\text{kHz} = 0.2~\text{MHz}$.
Upper frequency $f_{H} = f_{c} + \frac{BW}{2} = 10.7 + 0.1 = 10.8~\text{MHz}$.
The maximum usable sampling frequency $f_{s}$ for bandpass sampling is given by $f_{s}=\frac{2f_{H}}{k}$, where $k = \lfloor \frac{f_{H}}{BW} \rfloor$.
$k = \lfloor \frac{10.8}{0.2} \rfloor = \lfloor 54 \rfloor = 54$.
$f_{s}=\frac{2 \times 10.8~\text{MHz}}{54} = \frac{21.6~\text{MHz}}{54} = 0.4~\text{MHz} = 400~\text{kHz}$.
Topic: Communication Systems
A satellite system employs QPSK modulation with 40% excess bandwidth per carrier including guard band. The voice channels use 64 kbps PCM coding. The number of channels supported by 36 MHz bandwidth of the transponder in bandwidth limited case will be
Answer: B. 800
Explanation: $(BW)_{QPSK}=R_{s}(1+\alpha)=1.4 R_{s}$. Since 40% excess bandwidth is used, $\alpha=0.4$.
$R_{s}=\frac{R_{b}}{\log_{2}M}=\frac{R_{b}}{\log_{2}4} = \frac{R_{b}}{2} = 32~\text{kbps}$.
So, BW for each QPSK channel $= 1.4 \times 32~\text{kHz} = 44.8~\text{kHz}$.
Total number of channels possible $= \lfloor \frac{36000}{44.8} \rfloor = 803$. Nearest matching option is 800.
Topic: Communication Systems
Hamming codes are used for
Answer: B. Signal error correction
Explanation: Hamming codes are used for single bit error corrections (listed as 'Signal error correction' in the given options as a possible typo for 'Single'). They are not used for burst-error corrections.
Topic: Electronic Devices and Circuits
Which statement is correct for Schottky diode?
Answer: D. The storage time $t_{s}$ is almost zero
Explanation: A Schottky diode is a majority carrier device. Since current is purely due to majority carriers, there is no minority carrier charge storage, making the reverse recovery time (or storage time $t_{s}$) almost zero.
Topic: Electronic Devices and Circuits
Photodiodes operate at
Answer: C. Reverse bias
Explanation: Photodiodes are typically operated in the reverse bias region because the reverse saturation current is directly proportional to the incident light intensity, allowing for effective light detection.
Topic: Electronic Devices and Circuits
The dynamic resistance of diode varies as
Answer: B. $1/I$
Explanation: The dynamic resistance of the diode is given as $r_{d}=\frac{V_{T}}{I_{DC}}$, where $V_{T}=\frac{kT}{q}$ is the thermal voltage and $I_{DC}$ is the DC current flowing through the diode. Therefore, $r_{d} \propto \frac{1}{I_{DC}}$.
Topic: Electronic Devices and Circuits
In a p-n diode, with the increase reverse bias, the reverse current
Answer: C. remain constant
Explanation: In an ideal p-n diode, with the increase in reverse bias voltage (below the breakdown region), the reverse saturation current depends primarily on temperature and remains practically constant.
Topic: Communication Systems
The Nyquist filter has impulse response given by
Answer: A. $h(t)=\frac{\sin(\frac{\pi t}{T})}{(\frac{\pi t}{T})}$
Explanation: The frequency response of an ideal Nyquist filter is a rectangular pulse: $H(f) = T\text{rect}(fT)$. Using the Fourier Transform pair $\text{sinc}(t) \leftrightarrow \text{rect}(f)$, we get $\text{sinc}(\frac{t}{T}) \leftrightarrow T\text{rect}(fT)$. So, $h(t) = \text{sinc}(\frac{t}{T}) = \frac{\sin(\frac{\pi t}{T})}{(\frac{\pi t}{T})}$.
Topic: Communication Systems
Shown signal space diagram is for
(Diagram shows two constellation points on the orthogonal axes $\phi_{1}$ and $\phi_{2}$, specifically at $(A\sqrt{T/2}, 0)$ and $(0, A\sqrt{T/2})$ with a distance of $A\sqrt{T}$ between them.)
Answer: B. Binary FSK
Explanation: The diagram displays two orthogonal signals located on the $\phi_{1}$ and $\phi_{2}$ axes. Orthogonal signaling is characteristic of a Binary FSK (BFSK) constellation.
Topic: Analog Circuits
Negative feedback in amplifiers
Answer: B. improves the signal to noise ratio at the output
Explanation: Negative feedback reduces the system noise (introduced within the amplifier stages) in the closed-loop system by a factor of $(1+A\beta)$, thus generally improving the signal-to-noise ratio at the output of the amplifier.
Topic: Analog Circuits
Output response of a diode clipper circuit shown in figure will be
(The circuit is a series diode clipper where $V_i$ connects to the positive terminal of a battery $V_B$, then a resistor $R$, and finally a diode $D_1$ pointing downwards to ground. Output $V_o$ is taken across the diode.)
Answer: A. Output is zero for $V_{i} \le V_{B}$ and $V_{i}-V_{B}$ for $V_{i} > V_{B}$
Explanation: Case 1 ($D_{1}$ ON): Considering the diode to be ideal, $V_{D}=0$, so $V_{out}=0$. Case 2 ($D_{1}$ OFF): For $D_{1}$ to be off, $I=0$, and applying KVL yields $-V_{in} + V_{B} + V_{out} = 0 \Rightarrow V_{out} = V_{in} - V_{B}$. This occurs when $V_{in} - V_{B} > 0 \Rightarrow V_{in} > V_{B}$. Thus, the output is zero for $V_{in} \le V_{B}$ and follows $(V_{in} - V_{B})$ for $V_{in} > V_{B}$.
Topic: Electronic Devices and Circuits
For an n-channel MOSFET, if condention parameter $(k_{n})$ is $0.249~\text{mA/V}^{2}$, gate to source voltage $V_{GS}$ is $2V_{TN}$ where $V_{TN}=0.75~\text{V}$. The current will be
Answer: C. 0.140 mA
Explanation: Given, $V_{GS}=2V_{TN}$ and $k_{n}=0.249~\text{mA/V}^{2}$. Assuming the MOSFET is in the saturation region:
$I_{D} = k_{n}(V_{GS}-V_{TN})^{2} = 0.249(2V_{TN}-V_{TN})^{2} = 0.249(0.75)^{2} = 0.249 \times 0.5625 = 0.1400625~\text{mA} \simeq 0.140~\text{mA}$.
Topic: Electronic Devices and Circuits
If $\alpha=0.98$, $I_{CBO}=6~\mu\text{A}$, and $I_{B}=100~\mu\text{A}$ for a transistor, then the value of $I_{C}$ will be
Answer: D. 5.2 mA
Explanation: First, calculate $\beta$: $\beta = \frac{\alpha}{1-\alpha} = \frac{0.98}{1-0.98} = 49$.
The collector current is given by $I_{C} = \beta I_{B} + (1+\beta)I_{CBO}$.
$I_{C} = 49 \times 100~\mu\text{A} + (50) \times 6~\mu\text{A} = 4900~\mu\text{A} + 300~\mu\text{A} = 5200~\mu\text{A} = 5.2~\text{mA}$.
Topic: Network Theory
A unit step voltage $2u(t-\theta)$ is applied in a series RC circuit with $R=2~\Omega$, $C=1\text{F}$. Assuming zero initial conditions, find $i(t)$
Answer: C. $i(t)=u(t-\theta)e^{-0.5(t-\theta)}\text{ A}$
Explanation: Given $v_{s}(t)=2u(t-\theta)$, its Laplace transform is $V_{s}(s)=\frac{2e^{-\theta s}}{s}$.
$I(s)=\frac{V_{s}(s)}{R+\frac{1}{Cs}} = \frac{\frac{2e^{-\theta s}}{s}}{2+\frac{1}{s}} = \frac{2e^{-\theta s}}{2s+1} = \frac{e^{-\theta s}}{s+0.5}$.
Taking the inverse Laplace transform, $i(t)=u(t-\theta)e^{-0.5(t-\theta)}\text{ A}$.
Topic: Communication Systems
The autocorrelation of a wide sense stationary random process is given by $e^{-2|\tau|}$. The peak value of the spectral density is
Answer: B. 1
Explanation: The Autocorrelation function $R_{X}(\tau) = e^{-2|\tau|}$.
The Power Spectral Density (PSD) is the Fourier Transform of the autocorrelation function:
$S_{X}(\omega) = \mathcal{F}\{e^{-2|\tau|}\} = \frac{2(2)}{2^2+\omega²} = \frac{4}{4+\omega^{2}}$.
The peak value of $S_{X}(\omega)$ occurs at $\omega=0$. Thus, $S_{X}(0) = \frac{4}{4} = 1$.
Topic: Material Science
Which of the following gives piezo electric effect?
Answer: B. PVDF
Explanation: Polyvinylidene fluoride (PVDF) is a highly non-reactive thermoplastic fluoropolymer that exhibits strong piezoelectricity, often several times greater than that of quartz.
Topic: Material Science
The mean free path for electron drift increases with
Answer: A. Purity
Explanation: Based on the official answer key, the correct option is (a). High purity in a material reduces the number of scattering centers (impurities/defects), thereby increasing the mean free path for electron drift.
Topic: Material Science
Pure Metals generally have
Answer: B. High Conductivity and High temperature coefficient
Explanation: Pure metals have high electrical conductivity due to a large number of free electrons. They also have a high positive temperature coefficient of resistance because increased temperature significantly increases lattice scattering.
Topic: Optical Instrumentation
A typical optical fiber has
Answer: A. high refractive index core and low refractive index cladding
Explanation: For total internal reflection to occur within the optical fiber, the refractive index of the core ($n_1$) must be greater than the refractive index of the cladding ($n_2$).
Topic: Material Science
The ceramic materials are
Answer: D. All of the above
Explanation: Ceramics are generally inorganic and non-metallic solids. They are characterized by very high hardness and strength, brittle nature, extremely high melting points, and good electrical and thermal insulation.
Topic: Material Science
The best definition of a superconductor is
Answer: A. It is a material showing perfect conductivity and Meissner effect below a critical temperature
Explanation: A superconductor is defined not just by its zero electrical resistance (perfect conductivity) below a critical temperature, but also by the expulsion of magnetic fields from its interior, which is known as the Meissner effect.
Topic: Signals and Systems
A periodic function of half wave symmetry is necessarily
Answer: C. neither odd nor even
Explanation: It is possible that a periodic function with half-wave symmetry (where $x(t) = -x(t \pm T/2)$) is neither purely even nor purely odd. It simply means that its Fourier series will only contain odd harmonics, regardless of whether it exhibits even or odd symmetry overall.
Topic: Signals and Systems
The Fourier transform of a Gaussian time pulse is
Answer: C. Gaussian
Explanation: The Fourier transform of a Gaussian pulse in the time domain is always a Gaussian function in the frequency domain.
Topic: Communication Systems
The covariance function $C_x(\tau)$ of a stationary stochastic process $x(t)$ is said to be positive definite. This means that
Answer: A. $C_x(\tau)\ge0$ for all $\tau$
Explanation: The covariance function $C_x(\tau)$ of a stationary process $x(t)$ is said to be positive definite when $C_x(\tau)\ge0$ for all '$\tau$'. (Note: Mathematically, positive semi-definiteness for a covariance function generally implies its Fourier transform is non-negative, but according to the given ISRO solution context, option (a) is marked as the answer).
Topic: Electromagnetics
If a vertical dipole antenna is used in conjunction with a loop antenna for direction finding, the field pattern obtained will be
Answer: A. [Image Option 1: Cardioid]
Explanation: The combined radiation pattern of a vertical dipole (which produces a circular omnidirectional pattern) and a loop antenna (which produces a figure-eight pattern) results in a cardioid (heart-shaped) radiation pattern, which is highly useful in direction finding.
Topic: Communication Systems
The coding system typically used in digital telemetry is
Answer: C. PCM (Pulse Code Modulation)
Explanation: In a digital telemetry system, the sensor picks up the physical quantity and converts it into an electrical analog form. To transmit this analog data using a digital system, it must first be converted into digital form. For this purpose, Pulse Code Modulation (PCM) is used. PAM, PPM, and PDM are pulse-analog modulation schemes, not purely digital techniques.
Topic: Electronic Devices and Circuits
The temperature coefficient of resistance for a thermistor is
Answer: C. high and negative
Explanation: Thermistors (specifically NTC type, which are the most common) typically exhibit a high negative temperature coefficient of resistance, meaning their resistance drops significantly as temperature increases.
Topic: Electronic Devices and Circuits
Compared to field effect photo transistor, bipolar photo transistors are
Answer: A. more sensitive and faster
Explanation: Based on the official answer key, the correct option is (a). Bipolar photo transistors typically offer higher sensitivity due to their inherent internal current gain ($h_{FE}$) compared to field-effect counterparts in standard applications.
Topic: Electromagnetics
A uniform plane wave in air impinge at $45^\circ$ angle on a lossless dielectric material with dielectric constante $\epsilon_r$. The transmitted wave propagates in a $30^\circ$ direction with respect to the normal. The value of $\epsilon_r$ is
Answer: C. 2
Explanation: According to Snell's law: $n_1\sin\theta_1 = n_2\sin\theta_2$.
For air, $n_1 = \sqrt{\epsilon_{r1}} = 1$.
$\sin 45^\circ = \sqrt{\epsilon_r}\sin 30^\circ$
$\frac{1}{\sqrt{2}} = \sqrt{\epsilon_r} \left(\frac{1}{2}\right)$
$\sqrt{\epsilon_r} = \frac{2}{\sqrt{2}} = \sqrt{2}$.
Squaring both sides gives $\epsilon_r = 2$.
Topic: Electronic Devices and Circuits
If a donor type impurity is added to the semiconductor, then at a given temperature, the Fermi Level
Answer: C. Moves towards the conduction band
Explanation: The Fermi level position is given by $E_C - E_F = kT \ln\left(\frac{N_C}{N_D}\right)$. As the donor concentration ($N_D$) increases, the term $kT \ln\left(\frac{N_C}{N_D}\right)$ decreases, which means the difference between $E_C$ and $E_F$ becomes smaller. Thus, the Fermi level moves towards the conduction band.
Topic: Electromagnetics
Which of the following statement is correct?
Answer: C. Ampere's circuital law states that the line integral of $\overline{H}$ about any closed path is exactly equal to the direct current enclosed by that path
Explanation: Option (c) correctly states Ampere's Circuital Law for static magnetic fields: $\oint \overline{H} \cdot d\overline{l} = I_{enclosed}$. Option (a) is incorrect because electrons and holes move in opposite directions under an electric field. Option (b) is incorrect because Electric flux density $D = \epsilon_0 E + P$.
Topic: Electromagnetics
Magnetic field intensity (H), within a magnetic material where, $M=150~\text{A/m}$, $\mu=1.5\times10^{-5}\text{H/m}$, $\mu_{r}=30$;
Explanation: Given, $M=150~\text{A/m}$ and $\mu_r = 30$.
We know that magnetization $M = \chi_M H$, where $\chi_M$ is the magnetic susceptibility.
$\chi_M = \mu_r - 1 = 30 - 1 = 29$.
Magnetic field intensity $H = \frac{M}{\chi_M} = \frac{150}{29} \approx 5.172~\text{A/m}$.
None of the options match the correct calculated value.
Topic: Electromagnetics
A steel pipe is constructed of a material for which $\mu_{r}=200$ and $\sigma=5\times10^{6}\text{mho/m}$. The outer and inner radii are 8 and 6 mm respectively and the length is 80 m. If the total current carried by the pipe is $2\cos(10^{4}\pi t)\text{ A}$, then the skin depth will be
Answer: A. $0.225\times10^{-3}\text{m}$
Explanation: From the current expression, $\omega = 10^4\pi$, so $f = \frac{10^4\pi}{2\pi} = 5 \times 10^3~\text{Hz}$.
Skin depth $\delta = \frac{1}{\sqrt{\pi f \mu \sigma}} = \frac{1}{\sqrt{\pi \times (5 \times 10^3) \times (200 \times 4\pi \times 10^{-7}) \times (5 \times 10^6)}}$.
$\delta = \frac{1}{\sqrt{20\pi² \times 10^3}} \approx 0.225 \times 10^{-3}~\text{m}$.
Topic: Electronic Devices and Circuits
A small amount of phosphorus is added to silicon so that there are $2.5\times10^{17}$ conduction electrons per cubic meter and $9.0\times10^{14}$ holes per cubic meter. Let $\mu_{e}=0.13~\text{m}^{2}/\text{V.s}$ and $\mu_{h}=0.05~\text{m}^{2}/\text{V.s}$. The resistivity in $\Omega\text{-m}$ will be
Answer: B. 192.042
Explanation: Phosphorus makes the material n-type.
$n = 2.5 \times 10^{17}~\text{m}^{-3}$ and $p = 9.0 \times 10^{14}~\text{m}^{-3}$.
Conductivity $\sigma = nq\mu_e + pq\mu_h \approx nq\mu_e$ (since $n \gg p$).
Resistivity $\rho = \frac{1}{\sigma} = \frac{1}{2.5 \times 10^{17} \times 1.602 \times 10^{-19} \times 0.13}$.
$\rho \approx 192.067~\Omega\text{-m}$. Option (b) is the closest.
Topic: Electronic Devices and Circuits
Silicon dioxide is used in ICs
Answer: C. To control the location of diffusion and to protect and insulate the Si Surface
Explanation: In IC fabrication, the $\text{SiO}_2$ layer serves as a high-quality insulating layer, provides surface passivation to protect the silicon surface, and acts as a photomask to selectively block dopants and control the location of diffusion.
Topic: Engineering Mathematics
Evaluate $\int_{0}^{1}\int_{0}^{\sqrt{1+x^{2}}}\frac{dy\cdot dx}{(1+x^{2}+y^{2})}$
Answer: B. $\frac{\pi}{4}[\log(1+\sqrt{2})]$
Explanation: Let $I = \int_{0}^{1} \left[ \int_{0}^{\sqrt{1+x^{2}}} \frac{dy}{(1+x^{2})+y^{2}} \right] dx$.
Using $\int \frac{1}{a²+u²}du = \frac{1}{a}\tan^{-1}(\frac{u}{a})$, where $a = \sqrt{1+x²}$:
$I = \int_{0}^{1} \left[ \frac{1}{\sqrt{1+x²}} \tan^{-1}\left(\frac{y}{\sqrt{1+x²}}\right) \right]_0^{\sqrt{1+x²}} dx = \int_0^1 \frac{1}{\sqrt{1+x²}} [\tan^{-1}(1) - \tan^{-1}(0)] dx$.
$I = \frac{\pi}{4} \int_0^1 \frac{1}{\sqrt{1+x²}} dx = \frac{\pi}{4} [\ln(x+\sqrt{1+x²})]_0^1 = \frac{\pi}{4} \ln(1+\sqrt{2})$.
Topic: Analog Circuits
Both transistor $T_{1}$ and $T_{2}$ in figure have a threshold voltage of 1 V. The device parameters $K_{1}$ and $K_{2}$ of $T_{1}$ and $T_{2}$ are 36 $\mu\text{A/V}^{2}$ and 9 $\mu\text{A/V}^{2}$ respectively. The output voltage $V_{0}$ is
Answer: C. 3 V
Explanation: Both transistors have their gates connected to their drains, ensuring they operate in the saturation region. Therefore, $I_{D1} = I_{D2}$.
$K_1 (V_{GS1} - V_T)^2 = K_2 (V_{GS2} - V_T)^2$.
From the circuit, $V_{GS1} = 5 - V_0$ and $V_{GS2} = V_0$.
$36(5 - V_0 - 1)^2 = 9(V_0 - 1)^2$.
$4(4 - V_0)^2 = (V_0 - 1)^2$.
Taking the square root: $2(4 - V_0) = V_0 - 1 \Rightarrow 8 - 2V_0 = V_0 - 1 \Rightarrow 3V_0 = 9 \Rightarrow V_0 = 3~\text{V}$.
Topic: Control Systems
A ramp input applied to a unity feedback system results in 5% steady state error. The type number and zero frequency gain of the system are
Answer: A. 1 and 20
Explanation: A finite, non-zero steady-state error for a ramp input implies the system is Type 1.
The steady-state error for a Type 1 system with a ramp input is $e_{ss} = \frac{1}{K_v}$.
Given $e_{ss} = 5\% = 0.05$, we have $0.05 = \frac{1}{K_v} \Rightarrow K_v = \frac{1}{0.05} = 20$.
The velocity error constant $K_v$ corresponds to the gain of the system.
Topic: Control Systems
The magnitude of frequency response of an under-sampled second order system is 5 at 0 rad/sec and peaks to $10/\sqrt{3}$ at $5\sqrt{2}~\text{rad/sec}$. The transfer function of the system is
Answer: A. $\frac{500}{(s^{2}+10s+100)}$
Explanation: Checking option (a): $T(s) = \frac{500}{s²+10s+100}$.
At $\omega=0$, $T(0) = \frac{500}{100} = 5$.
At $\omega = 5\sqrt{2}~\text{rad/sec}$, magnitude $|T(j\omega)| = \frac{500}{\sqrt{(100 - (5\sqrt{2})^2)^2 + (10 \times 5\sqrt{2})^2}}$.
$|T(j5\sqrt{2})| = \frac{500}{\sqrt{(100-50)^2 + (50\sqrt{2})^2}} = \frac{500}{\sqrt{2500 + 5000}} = \frac{500}{\sqrt{7500}} = \frac{500}{50\sqrt{3}} = \frac{10}{\sqrt{3}}$.
This perfectly matches the given conditions.
Topic: Control Systems
The system with the open loop transfer function $G(s)H(s)=\frac{1}{s(s^{2}+s+1)}$ has a gain margin of
Answer: B. zero
Explanation: To find the phase crossover frequency $\omega_{pc}$, equate the imaginary part of $G(j\omega)H(j\omega)$ to zero.
$G(j\omega)H(j\omega) = \frac{1}{j\omega(-\omega²+j\omega+1)} = \frac{1}{-\omega² + j\omega(1-\omega²)}$.
For the imaginary part to be zero, $1 - \omega² = 0 \Rightarrow \omega_{pc} = 1~\text{rad/sec}$.
Magnitude at $\omega = 1$ is $|G(j1)H(j1)| = \left|\frac{1}{-1^2 + j(1)(1-1^2)}\right| = \left|\frac{1}{-1}\right| = 1$.
Gain Margin = $-20\log_{10}(|G(j\omega_{pc})H(j\omega_{pc})|) = -20\log_{10}(1) = 0~\text{dB}$.
Topic: Control Systems
Given $f(t)=L^{-1}\left[\frac{3s+1}{s^{3}+4s^{2}+(k-3)s}\right]$. If $\lim_{t\rightarrow\infty}f(t)=1$ then the value of k is
Answer: D. 4
Explanation: (Note: The term $(k-3)$ in the denominator is presumed to be $(k-3)s$ to yield a valid non-infinite steady state based on FVT).
Using the Final Value Theorem: $\lim_{t \to \infty} f(t) = \lim_{s \to 0} sF(s)$.
$sF(s) = s \left[ \frac{3s+1}{s(s²+4s+(k-3))} \right] = \frac{3s+1}{s²+4s+(k-3)}$.
Taking the limit as $s \to 0$, we get $\frac{1}{k-3}$.
Given $\lim_{t \to \infty} f(t) = 1$, we have $\frac{1}{k-3} = 1 \Rightarrow k-3 = 1 \Rightarrow k=4$.
Topic: Engineering Mathematics
A person on a trip has a choice between private car and public transport. The probability of using a private car is 0.45. While using public transport, further choice available are bus and metro. Out of which the probability of commuting by a bus is 0.55. In such a situation, the probability (rounded upto two decimals) of using a car, bus and metro respectively would be
Answer: A. 0.45, 0.30 and 0.25
Explanation: Probability of using a private car = 0.45.
Probability of using public transport = $1 - 0.45 = 0.55$.
Probability of using a bus (given public transport) = $0.55 \times 0.55 = 0.3025 \approx 0.30$.
Probability of using a metro (given public transport) = $0.55 \times (1 - 0.55) = 0.55 \times 0.45 = 0.2475 \approx 0.25$.
Topic: Engineering Mathematics
A real root of equation $x^{3}-5x-7=0$ by the method of false position correct to three decimal places is
Answer: A. 2.747
Explanation: Let $f(x) = x³ - 5x - 7 = 0$.
$f(2) = 8 - 10 - 7 = -9$.
$f(3) = 27 - 15 - 7 = 5$.
The root lies between 2 and 3.
Using the false position method, $x_1 = \frac{a f(b) - b f(a)}{f(b) - f(a)} = \frac{2(5) - 3(-9)}{5 - (-9)} = \frac{10 + 27}{14} = 2.643$.
$f(2.643) = -1.752$. Since $f(2.643) < 0$, the root lies between 2.643 and 3.
Iterating further yields $x_2 \approx 2.736$, and eventually converges to $2.747$.
Topic: Engineering Mathematics
Find the z- transform of $\cosh(n\theta)$
Answer: A. $\frac{z(z-\cosh\theta)}{(z^{2}-2z\cosh\theta+1)}$
Explanation: Using the exponential definition, $\cosh(n\theta) = \frac{e^{n\theta} + e^{-n\theta}}{2}$.
$Z[\cosh(n\theta)] = \frac{1}{2} \left( Z[e^{n\theta}] + Z[e^{-n\theta}] \right) = \frac{1}{2} \left( \frac{z}{z-e^{\theta}} + \frac{z}{z-e^{-\theta}} \right)$.
$= \frac{1}{2} \left[ \frac{z(z-e^{-\theta}) + z(z-e^{\theta})}{z² - z(e^{\theta}+e^{-\theta}) + 1} \right] = \frac{1}{2} \left[ \frac{2z^2 - z(e^{\theta}+e^{-\theta})}{z² - 2z\cosh\theta + 1} \right] = \frac{z² - z\cosh\theta}{z² - 2z\cosh\theta + 1} = \frac{z(z-\cosh\theta)}{z² - 2z\cosh\theta + 1}$.
Topic: Engineering Mathematics
Newton-Raphson iteration formula for finding $\sqrt[3]{C},$ where $C>0$ is
Answer: C. $x_{n+1}=\frac{2x_{n}^{3}+C}{3x_{n}^{2}}$
Explanation: Let $x = \sqrt[3]{C}$, which implies $x³ = C$, or $f(x) = x³ - C = 0$.
The derivative is $f'(x) = 3x^2$.
The Newton-Raphson formula is $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$.
$x_{n+1} = x_n - \frac{x_n^3 - C}{3x_n^2} = \frac{3x_n^3 - x_n^3 + C}{3x_n^2} = \frac{2x_n^3 + C}{3x_n^2}$.
Topic: Engineering Mathematics
The value of $\lim_{x\rightarrow8}\left(\frac{x^{1/3}-2}{x-8}\right)$ is
Answer: A. $\frac{1}{12}$
Explanation: This is a limit of the form $\frac{0}{0}$ as $x \to 8$.
Using L'Hôpital's rule, differentiate the numerator and the denominator:
$\lim_{x\to 8} \frac{\frac{d}{dx}(x^{1/3}-2)}{\frac{d}{dx}(x-8)} = \lim_{x\to 8} \frac{\frac{1}{3}x^{-2/3}}{1} = \frac{1}{3}(8)^{-2/3} = \frac{1}{3}(2^3)^{-2/3} = \frac{1}{3}(2^{-2}) = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12}$.
Topic: Control Systems
Find the values of $K_{P}$, $K_{I}$, and $K_{D}$ for PID controller whose characteristics equation has real roots at 10, $\xi=0.8$ and $\omega_{n}=2~\text{rad/s}$
Answer: A. $K_{D}=1.8, K_{P}=8, K_{I}=10$
Explanation: For PID controller, $G_c(s) = K_p + K_D s + \frac{K_I}{s}$.
The characteristic equation is $1 + G_c(s)G(s) = 0$.
$1 + \left(K_P + K_D s + \frac{K_I}{s}\right) \left(\frac{4}{s²+6s+1}\right) = 0$.
$s³ + 6s^2 + s + 4K_P s + 4K_D s² + 4K_I = 0$.
$s³ + (6+4K_D)s² + (1+4K_P)s + 4K_I = 0$ ... (i)
Given roots are $s = -10$ and complex roots corresponding to $\xi=0.8, \omega_n=2$.
The desired characteristic equation is $(s² + 2(0.8)(2)s + 2^2)(s+10) = 0$.
$(s² + 3.2s + 4)(s+10) = s³ + 13.2s^2 + 36s + 40 = 0$ ... (ii)
Comparing (i) and (ii):
$6 + 4K_D = 13.2 \Rightarrow K_D = 1.8$.
$1 + 4K_P = 36 \Rightarrow K_P \approx 8$.
$4K_I = 40 \Rightarrow K_I = 10$.
Topic: Control Systems
The transfer function $(\frac{1+0.5s}{1+s})$ represents a
Answer: A. lag network
Explanation: Comparing the given transfer function with the standard transfer function of a compensator $T(s) = \frac{\alpha(1+\tau s)}{1+\alpha\tau s}$, we have $\tau = 0.5$ and $\alpha\tau = 1$, which gives $\alpha = 2$. Since $\alpha > 1$, the given compensator is a lag network.
Topic: Electronic Devices and Circuits
The wavelength of emitted radiation by electron while transiting from one energy state of -15 eV to other energy state of -5 eV is
Answer: B. 1240 Å
Explanation: The energy gap $E_G = -5 - (-15) = 10~\text{eV}$.
Wavelength $\lambda = \frac{1.24}{E_G}~\mu\text{m} = \frac{1.24}{10} \times 10^{-6}~\text{m} = 0.124~\mu\text{m} = 1240~\text{Å}$.
Topic: Network Theory
According to maximum power transfer theorem, maximum power transfer occurs when
Answer: D. Load R is equal to the R of network looking back at it from load terminals with all sources being replaced by their respective internal resistance
Explanation: The Maximum Power Transfer Theorem states that maximum power is delivered to a load when the load resistance equals the Thevenin equivalent resistance of the network, which is the resistance looking back from the load terminals with all independent sources replaced by their internal resistances.
Topic: Electronic Devices and Circuits
How does the dynamic resistance of diode vary with temperature?
Answer: A. Directly proportional
Explanation: Dynamic resistance $R_d = \frac{dV}{dI} = \frac{\eta V_T}{I}$.
Thermal voltage $V_T = \frac{T}{11600}~\text{V}$.
Thus, $R_d \propto T$ (Directly proportional).
Topic: Electromagnetics
The electric field component of a time harmonic plane EM wave traveling in a non-magnetic lossless dielectric medium has amplitude of $2~\text{V/m}$. If the relative permittivity of the medium is 4, the magnitude of the time-average power density vector (in $\text{W/m}^2$) is
Answer: A. $\frac{1}{30\pi}$
Explanation: $E = 2~\text{V/m}, \epsilon_r = 4$.
Intrinsic impedance $\eta = \frac{120\pi}{\sqrt{\epsilon_r}} = \frac{120\pi}{2} = 60\pi~\Omega$.
Average power density $P_{avg} = \frac{1}{2} \frac{E²}{\eta} = \frac{1}{2} \frac{2^2}{60\pi} = \frac{1}{30\pi}~\text{W/m}^2$.
Topic: Electromagnetics
The force on a point charge +q kept at a distance d from the surface of an infinite grounded metal plate in a medium of permittivity $\epsilon$ is
Answer: C. $\frac{q^{2}}{16\pi\epsilon d^{2}}$ towards the plate
Explanation: By the method of images, the grounded plate is replaced by an image charge $-q$ at a distance $d$ behind the plate. The distance between the original charge and the image charge is $2d$. The force is $F = \frac{q \times (-q)}{4\pi\epsilon(2d)^2} = -\frac{q²}{16\pi\epsilon d²}$. The negative sign indicates an attractive force towards the plate.
Topic: Electromagnetics
A material has conductivity of $10^{-2}\text{ mho/m}$ and a relative permittivity of 4. The frequency at which the conduction current in the medium is equal to the displacement current is
Answer: A. 45 MHz
Explanation: Conduction current density equals displacement current density when $\sigma = \omega \epsilon$.
$f = \frac{\sigma}{2\pi \epsilon} = \frac{10^{-2}}{2\pi \epsilon_0 \epsilon_r} = \frac{10^{-2}}{2\pi \times (10^{-9}/36\pi) \times 4} = \frac{10^{-2} \times 18}{10^{-9} \times 4} = 4.5 \times 10^7~\text{Hz} = 45~\text{MHz}$.
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: Each pole contributes a slope of -20 dB/decade and each zero contributes +20 dB/decade.
Total slope = $20 \times (\text{Number of zeros} - \text{Number of poles}) = 20 \times (2 - 14) = -240~\text{dB/decade}$.
Topic: Communication Systems
A 1 mW video signal having a bandwidth of 100 MHz is transmitted to a receiver through a cable that has 40 dB loss. If the effective one-sided noise spectral density at the receiver is $10^{-20}\text{ Watt/Hz}$, then the signal-to-noise ratio at the receiver is
Answer: A. 50 dB
Explanation: Input power $P_{in} = 1~\text{mW} = -30~\text{dBW}$.
Loss in cable = 40 dB, so received signal power $P_{out} = -30 - 40 = -70~\text{dBW}$.
Noise power $P_N = N_0 \times BW = 10^{-20} \times 100 \times 10^6 = 10^{-12}~\text{W} = -120~\text{dBW}$.
$SNR = P_{out} - P_N = -70 - (-120) = 50~\text{dB}$.
Topic: Microprocessors
When the 8051 is reset and the line in HIGH, the program counter points to the first program instruction in the
Answer: A. internal code memory
Explanation: When the 8051 microcontroller is reset and the $\overline{EA}$ (External Access) line is HIGH, the Program Counter fetches the first instruction from the internal code memory starting at address 0000H.
Topic: Microprocessors
How is the status of the carry, auxiliary carry and parity flag affected if write instruction?
MOV A, #9C
ADD A, #64H
Answer: B. $CY=1, AC=1, P=0$
Explanation: 9CH = 1001 1100
64H = 0110 0100
Addition: 1001 1100 + 0110 0100 = 1 0000 0000 (Result is 00H with a carry).
Since there is a carry out of D7, CY=1.
There is a carry from D3 to D4, so AC=1.
For 8051, P=1 when there is an odd number of 1's in the accumulator. Since 00H has zero 1's, P=0.
Topic: Digital Electronics
In given Combinational logic, X is given by
(The diagram shows a 4:1 MUX with inputs $I_0=0, I_1=1, I_2=1, I_3=0$ selected by A, B outputting $F_o$, connected to a second 4:1 MUX with same inputs selected by $F_o, C$.)
Answer: A. $X=A\overline{B}\overline{C}+\overline{A}B\overline{C}+\overline{A}\overline{B}C+ABC$
Explanation: For the first 4:1 MUX, $F_o = \overline{A}B(1) + A\overline{B}(1) = A \oplus B$.
For the second 4:1 MUX, $X = \overline{F_o}C(1) + F_o\overline{C}(1) = F_o \oplus C = A \oplus B \oplus C$.
Expanding the XOR of 3 variables gives $X = A\overline{B}\overline{C} + \overline{A}B\overline{C} + \overline{A}\overline{B}C + ABC$.
Topic: Analog Circuits
Each Transistor in Darlington pair (as shown in figure) has $h_{fe}=100$. Overall $h_{FE}$ of composite transistor neglecting leakage current is,
Answer: C. 10100
Explanation: Let $\beta_1 = 100$ and $\beta_2 = 100$.
According to the provided official solution, the overall $\beta = \frac{I_{C2}}{I_{B1}}$. Since $I_{C2} = \beta_2 I_{B2} = \beta_2(\beta_1 + 1)I_{B1}$, the overall $\beta = 100(100+1) = 10100$.
Topic: Network Theory
In a given network (shown in Figure), a steady state is reached with switch k open. At t=0 which is closed. Determine the values of $I_{1}$, $I_{2}$ and $I_{3}$ at $t=0^{+}$
Answer: A. 1 A, 1/3 A, 0 A
Explanation: For $t<0$, steady state is reached. The inductor acts as a short circuit and capacitors act as open circuits. Total resistance $= 2\Omega + 4\Omega = 6\Omega$. $I_L(0^-) = 6\text{V} / 6\Omega = 1\text{A}$.
By voltage division logic given in the reference, the voltage across the 0.5 F capacitor is $V_{C1}(0^-) = 4/3\text{V}$.
At $t=0^+$, the inductor current cannot change instantaneously, so $I_1 = 1\text{A}$. Also, $I_3 = 0\text{A}$ (as KVL violates otherwise, according to the official solution logic). The current $I_2 = \frac{V_{C1}(0^+)}{4\Omega} = \frac{4/3}{4} = 1/3\text{A}$.
Topic: Digital Electronics
Simplify Boolean function represented in sum of product of min-terms,
$F(x,y,z)=\Sigma(0,2,4,5,6)$
Answer: A. $z'+xy'$
Explanation: Using a 3-variable K-map:
Minterms 0, 2, 4, 6 form a quad, which simplifies to $\overline{z}$.
Minterms 4 and 5 form a pair, which simplifies to $x\overline{y}$.
Therefore, the simplified function is $F = \overline{z} + x\overline{y}$ (or $z'+xy'$).
Topic: Control Systems
A system described by the transfer function $H(s)=\frac{1}{s^{3}+\alpha s^{2}+ks+3}$ is stable. The constraints on $\alpha$ and k are
Answer: B. $\alpha>0$, $\alpha k>3$
Explanation: The characteristic equation is $s³ + \alpha s² + ks + 3 = 0$.
Constructing the Routh array:
$s³$: 1, k
$s²$: $\alpha$, 3
$s^1$: $\frac{\alpha k - 3}{\alpha}$, 0
$s^0$: 3
For the system to be stable, all elements in the first column must be strictly positive. Thus, $\alpha > 0$ and $\frac{\alpha k - 3}{\alpha} > 0 \Rightarrow \alpha k > 3$.
Topic: Control Systems
Consider a unity feedback system whose open loop transfer function is
$G(s)=\frac{k}{s(s^{2}+2s+2)}$
The Nyquist plot for this system is
Answer: B. [Plot starting from negative imaginary axis and ending at origin]
Explanation: $G(j\omega) = \frac{K}{j\omega(-\omega²+2j\omega+2)}$.
At $\omega=0$: $|GH(0)| = \infty, \angle GH(0) = -90^\circ$.
At $\omega=\infty$: $|GH(\infty)| = 0, \angle GH(\infty) = -270^\circ$.
The Nyquist plot starts at $-\infty$ on the imaginary axis ($-90^\circ$) and spirals into the origin at an angle of $-270^\circ$ (approaching along the positive imaginary axis). Option (b) matches this profile.
Topic: Communication Systems
Eye Diagram gives an idea of
Answer: D. All of the above
Explanation: An eye diagram can be used to measure timing synchronization, clock jitter effects, intersymbol interference, and additive noise (which relates to SNR). The modulation scheme can also be inferred from the eye diagram patterns.
Topic: Control Systems
The signal flow diagram for a certain feedback control system is shown in figure. Now consider the following set of equations for the nodes
1. $x_{2}=a_{1}x_{1}+a_{9}x_{3}$
2. $x_{3}=a_{2}x_{2}+a_{8}x_{4}$
3. $x_{4}=a_{3}x_{3}+a_{5}x_{2}$
4. $x_{5}=a_{4}x_{4}+a_{6}x_{2}$
Which of the following are correct?
Answer: D. 1, 2 and 4
Explanation: Based on the signal flow graph:
Node $x_2 = a_1 x_1 + a_9 x_3$ (Correct)
Node $x_3 = a_2 x_2 + a_8 x_4$ (Correct)
Node $x_4 = a_3 x_3 + a_5 x_2 + a_7 x_4$ (The given equation 3 misses the $a_7 x_4$ term, hence incorrect)
Node $x_5 = a_4 x_4 + a_6 x_2$ (Correct)
Thus, statements 1, 2, and 4 are correct.