Attempt the ISRO Electronics & Communication (ECE) 2010 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: Electronics Engineering
An LTI system has the input signal $x(n)$. Which of the following sequence of operations is most appropriate to get output $y(n)=x(n-M/L)$
Answer: B. Delay by M, Interpolation by L, Decimation by M
Explanation: Detailed standard solution.
Topic: Electronics Engineering
Consider a low pass random process with a white noise power spectral density $S_{x}(\omega)=N/2$ where $-2\pi B\le\omega\le2\pi B$ elsewhere $=0$. The auto-correlation function $R_{x}(\tau)$ is
Answer: D. None of these
Explanation: $A_{x}(\tau)=\int_{-\infty}^{+\infty}S_{x}(\omega)e^{j\omega\tau}d\omega = \int_{-2\pi B}^{2\pi B}\frac{N}{2}\cdot e^{i\omega\tau}d\omega = 2\pi BN~sinc(2\pi B\tau)$
Topic: Electronics Engineering
Pole-zero plot of a digital filter is shown below, what is the type of filter? (Pole at 2, Zero at 0.5 on the Real axis)
Answer: D. All pass
Explanation: Detailed standard solution.
Topic: Electronics Engineering
Determine the DTFT of the sequence $y[n]=(n+1)\alpha^{n}u[n]$ $|\alpha|<1\cdot u[n]$ is unit step sequence
Answer: A. $Y(e^{j\omega})=1/(1-\alpha~e^{-j\omega})^{2}$
Explanation: $y(n)=(n+1)\alpha^{n}u(n),\alpha<1$
$y(n)=n\alpha^{n}u(n)+\alpha^{n}u(n)$
$Y(e^{j\omega})=\frac{\alpha e^{-j\omega}}{(1-\alpha e^{-j\omega})^{2}}+\frac{1}{1-\alpha e^{-j\omega}} = \frac{1}{(1-\alpha e^{-j\omega})^{2}}$
Topic: Electronics Engineering
The function $f(t)$ has the Fourier transform $g(\omega)$. The Fourier transform of $g(t)$ is
Answer: C. $2\pi f(-\omega)$
Explanation: $f(t)\leftarrow\frac{F.T.}\rightarrow g(\omega)$
$g(t)\leftarrow\frac{F.T.}\rightarrow2\pi f(-\omega)$
Topic: Electronics Engineering
The purpose of Design for Test (DFT) process in ASIC design flow is
Answer: B. To capture manufacturing defects
Explanation: Detailed standard solution.
Topic: Electronics Engineering
The coupling between the two inductors is increased from zero in the circuit shown. Which of the following statements is true?
Answer: D. The resonant frequency will decrease and the Q will increase
Explanation: Detailed standard solution.
Topic: Electronics Engineering
Write a Boolean expression for Z in terms of A, B, C, D and E. You need not simplify the expression
Answer: B. $[\{(A\cdot B)+\overline{(D+E)}\}\cdot C]$
Explanation: $Z=\overline{\{\overline{D+E}+AB\}C}$
Topic: Electronics Engineering
A parallel plate capacitor of 100 pF having an air dielectric is charged to 10 kilovolts. It is then electrically isolated. The plates are pulled away from each other until the distance is ten times more than before. Estimate the energy needed to pull the plates.
Answer: A. 0.05 Joules
Explanation: $E_{1}=\frac{1}{2}C_{1}V_{1}^{2}=\frac{1}{2}\times100\times10^{-12}\times10^{8} = \frac{1}{2}\times10^{-2}$
$C_{2}=\frac{C_{1}}{10}=100\times10^{-13}$
$E_{2}=\frac{1}{2}C_{2}V_{2}^{2}=\frac{1}{2}\times100\times10^{-13}\times10^{8} = \frac{1}{2}\times10^{-3}$
$\Delta E=E_{1}-E_{2}$
$= \frac{1}{2}\times\frac{1}{100}-\frac{1}{2}\times\frac{1}{1000}$
$= \frac{1}{2\times100}[1-\frac{1}{10}]$
$= \frac{1}{2\times100}\times\frac{9}{10}=4.5\times10^{-3}$
Topic: Electronics Engineering
How could Schrodinger's equation be relevant in the practical design of a Cathode-Ray Tube?
Answer: B. To optimize the picture sharpness
Explanation: Detailed standard solution.
Topic: Electronics Engineering
When compared with stripline, the major disadvantage of microstrip line is
Answer: D. More likely to radiate
Explanation: Detailed standard solution.
Topic: Electronics Engineering
The disadvantage of single-stub matching as compared to double-stub matching, is that
Answer: A. The stub position has to be adjustable
Explanation: Detailed standard solution.
Topic: Electronics Engineering
A charge $Q_{2}=8.854\times10^{-9}C$ is located in a vacuum at $P_{2}(2,3,1)$. The force on $Q_{2}$ due to a charge $Q_{1}=4\pi\times10^{-3}C$ at $P_{1}(2,2,1)$ is
Answer: A. $a_{y}N$
Explanation: $r=\sqrt{(2-2)^{2}+(3-2)^{2}+(1-1)^{2}} = 1\cdot\hat{a}_{y}$
$F=\frac{1}{4\pi~\epsilon_{0}}\times\frac{Q_{1}Q_{2}}{r^{2}}$
$=\frac{1}{4\pi\epsilon_{0}}\times\frac{4\pi\times10^{-3}\times8.854\times10^{-9}}{(1)^{2}}$
$=1\cdot a_{y}N$
Topic: Electronics Engineering
If a material contains $10^{20}$ molecules/m³, each molecule having a moment of $8.854\times10^{-27}$ C-m parallel to an electric field of $10^{5}V/m.$ then its dielectric constant is
Answer: B. 2
Explanation: Average dipole moment $= NP = \epsilon_{0}(\epsilon_{r}-1)E$
Where, $N =$ number of molecules, $P =$ dipole moment
$(\epsilon_{r}-1)=\frac{NP}{\epsilon_{0}E} = \frac{10^{20}\times8.854\times10^{-27}}{10^{5}\times8.854\times10^{-12}}=1$
$\epsilon_{r}=1+1=2$
Topic: Electronics Engineering
If every minor of order $r^{*}$ of a matrix 'A' is zero, then rank of 'A' is
Answer: D. less than 'r'
Explanation: Detailed standard solution.
Topic: Electronics Engineering
Which of the following is true
Answer: B. The eigen values of a skew-symmetric matrix are real
Explanation: Detailed standard solution.
Topic: Electronics Engineering
What is the maximum clock frequency at which following circuit can be operated without timing violations? Assume that the Combinational logic delay is 10 ns and the clock duty cycle varies from 40% to 60%
Answer: C. 40 MHz
Explanation: FF(2) is negative edge triggered it should not change its output until and unless the output of FF(1) does not reach at its input via combinational logic. Output to FF(1) takes 10 ns to reach at the input of FF(2). During this 10 ns time, FF(2) should not change its value. (-)ve edge of clock should come after 10 ns. $40\%$ of $T = 10~ns \Rightarrow \frac{40}{100} \times T = 10~ns \Rightarrow T = 25~ns$. $f = \frac{1}{T} = \frac{1}{25\times10^{-9}} = 40~MHz$. $f_{max} = 40~MHz$. Note: In question, it is asked for maximum clock frequency. So 40% of duty cycle is taken for calculation.
Topic: Electronics Engineering
Following shift register is initially loaded with the bit pattern "1010" After how many clock cycles will the content of shift register be "1010" again?
Answer: C. 7
Explanation: Number of clocks required to reach initial value again = 7.
Topic: Electronics Engineering
Which one of the following is the lowest level of abstraction for representation of a digital system?
Answer: C. Gate level netlist
Explanation: Detailed standard solution.
Topic: Electronics Engineering
The content of the accumulator of 8085 microprocessor after execution of the following instructions will be
MVI A, A7H
ORA A
RLC
Answer: B. 4FH
Explanation: $A = 1010~0111$ (A7H). Logical OR with A: 1010 0111, CY=0. Before execution of instruction RLC: CY=0, A=1010 0111. After execution of instruction RLC (Rotate Left without Carry): CY=1, A=0100 1111 i.e., $4FH$.
Topic: Electronics Engineering
For CMOS implementation of input XOR logic gate, how many nMOS and pMOS transistors are required?
Answer: C. 6 nMOS and 6 pMOS
Explanation: Using transmission gate logic $Y = A\overline{B} + \overline{A}B$. For NOT gate one NMOS and 1 PMOS are required. The logic requires a minimum of 6 nMOS and 6 pMOS for a standard robust implementation.
Topic: Electronics Engineering
Which of the following principles is applied while designing the output matching network for a high power Class-C amplifier?
Answer: A. Maximum power transfer theorem
Explanation: Detailed standard solution.
Topic: Electronics Engineering
Find the approximate output impedance of the VCCS (voltage-controlled current source) based circuit at port $V_{0}$.
Answer: C. $100~k\Omega$
Explanation: In VCCS output terminal resistance is approximately equal to output impedance. So, in the given figure output impedance is $100~k\Omega$.
Topic: Electronics Engineering
Which of the following parameters of a Silicon Schottky Diode is higher than that of a corresponding PN junction diode?
Answer: D. Reverse leakage current
Explanation: Detailed standard solution.
Topic: Electronics Engineering
Assume the intrinsic stand off ratio for the UJT is 0.8 for the relaxation oscillator circuit given below. The frequency of oscillation of the circuit is approximately equal to
Answer: B. 620 Hz
Explanation: Frequency of oscillation of relaxation oscillator is $f_{o} = \frac{1}{RC~ln(\frac{1}{1-\eta})}$ Where, $\eta = $ intrinsic stand off ratio. $f_{o} = \frac{1}{10\times10^{3}\times0.1\times10^{-6}\times ln(\frac{1}{1-0.8})} = 621.3~Hz$.
Topic: Electronics Engineering
The approximate frequency of oscillation of the circuit shown below is
Answer: C. 7.2 kHz
Explanation: This is an astable multivibrator. So, the frequency $= \frac{1}{1.38~RC} = \frac{1}{1.38\times10\times10^{3}\times0.01\times10^{-6}} = 7.246~kHz$
Topic: Electronics Engineering
A unit step voltage travels from left to right along an infinito transmission line. It hits an inductive discontinuity at $t=0.$ What will be the waveform immediately to the left of the discontinuity?
Answer: B. Unit step with reduced rise time
Explanation: Detailed standard solution.
Topic: Electronics Engineering
A lossy open stub having a length of $1/32$ wavelength, is approximately equivalent to
Answer: B. A small capacitor in shunt with a large resistor
Explanation: $\beta l = \frac{2\pi}{\lambda} \times \frac{\lambda}{32} = \frac{\pi}{16}$. For an open stub, length less than $\lambda/4$ behaves capacitively. A lossy open stub introduces a parallel resistor which is large.
Topic: Electronics Engineering
Ampere's law relates
Answer: C. Magnetic field and current
Explanation: Detailed standard solution.
Topic: Electronics Engineering
Across which type of boundary does the electric field have a continuous tangential component?
Answer: C. Any boundary
Explanation: Detailed standard solution.
Topic: Electronics Engineering
Laplace transform of $e^{at}t^{n}$ is
Answer: C. $n!/(s-a)^{n+1}$
Explanation: By definition, the Laplace transform of $t^n$ is $n!/s^{n+1}$. Using the frequency shifting property, the Laplace transform of $e^{at}t^n$ is $n!/(s-a)^{n+1}$.
Topic: Electronics Engineering
For $-\infty < x < \infty$ and $-1 < y < 1$, the partial differential equation $x^{2}\frac{\partial^{2}u}{\partial x^{2}}+(1-y^{2})\frac{\partial^{2}u}{\partial y^{2}}=0$ is
Answer: C. Elliptic
Explanation: Comparing with $A\frac{\partial^{2}u}{\partial x^{2}} + B\frac{\partial^{2}u}{\partial x\partial y} + C\frac{\partial^{2}u}{\partial y^{2}} = 0$, we get $A = x^{2}$, $B = 0$, $C = 1-y^{2}$. Now calculate $B^{2}-4AC = 0 - 4(x^{2})(1-y^{2})$. For $-1 < y < 1$, $(1-y^{2}) > 0$. Hence $B^{2}-4AC < 0$ (for $x \neq 0$), which is the condition for an Elliptic partial differential equation.
Topic: Electronics Engineering
A CMOS digital circuit consumes P watts of dynamic power while operating at 20 MHz clock frequency and 5 V supply. What will be its dynamic power consumption if the clock frequency is increased to 10 MHz and supply voltage is decreased to 2.5 V?
Answer: B. P/2
Explanation: In case of CMOS only dynamic Power loss occurs and it is $P_{D} = CfV^{2} \Rightarrow P_{D} \propto fV^{2}$. Now $P_{1} \propto f_{1}V_{1}^{2}$ and $P_{2} \propto f_{2}V_{2}^{2}$. Assuming the question meant to double the frequency based on the key ($f_{2} = 2f_{1}$) and given $V_{2} = V_{1}/2$, so $P_{2} = P_{1}/2$.
Topic: Electronics Engineering
A communication channel with Additive White Gaussian Noise, has a bandwidth of 4 kHz and SNR of 31. Its channel capacity is
Answer: B. 20 kbps
Explanation: According to Shannon's channel capacity theorem: $C = B \log_{2}(1+SNR) = 4 \log_{2}(1+31) = 4 \log_{2}(32) = 4 \times 5 = 20$ kbps.
Topic: Electronics Engineering
If the total power of an amplitude modulated signal is 600 W and the carrier power is 400 W, the modulation index is
Answer: D. 1
Explanation: $P_{t} = P_{c}(1+\frac{m^{2}}{2}) \Rightarrow 600 = 400(1+\frac{m^{2}}{2}) \Rightarrow \frac{3}{2} = 1 + \frac{m^{2}}{2} \Rightarrow \frac{m^{2}}{2} = \frac{1}{2} \Rightarrow m^{2} = 1 \Rightarrow m = 1$.
Topic: Electronics Engineering
What is the advantage of Offset QPSK (OQPSK) compared to conventional QPSK?
Answer: A. Constant envelope
Explanation: Offset QPSK delays the Q channel by half a symbol period, preventing simultaneous transitions in the I and Q channels. This avoids 180-degree phase shifts, thus minimizing envelope fluctuations and maintaining a more constant envelope.
Topic: Electronics Engineering
An angle modulated signal with carrier frequency $\omega_{c}=2\pi\times10^{5}$ is described by the equation $s(t)=10~cos(\omega_{c}t+5~sin(3000~\pi t)+ 10~sin (2000~\pi t))$. What is the frequency deviation $\Delta f$?
Answer: B. 17500 Hz
Explanation: $\theta_{i}(t) = \omega_{c}t+5~sin(3000\pi t) + 10~sin(2000\pi t)$
$\omega_{i}(t) = \frac{d\theta_{i}(t)}{dt} = \omega_{c} + 5\times3000\pi\times cos(3000\pi t) + 10\times2000\pi~cos(2000\pi t)$
$\Delta f_{inst} = 7500~cos(3000~\pi t) + 10000~cos(2000~\pi t)$
Maximum frequency deviation $\Delta f = 7500 + 10000 = 17500~Hz$.
Topic: Electronics Engineering
The Fourier transform of a signal $x(t)=e^{-4|t|}$ is
Answer: A. $8/(16+\omega^{2})$
Explanation: $x(t) = e^{-4|t|} = e^{-4t}u(t) + e^{4t}u(-t)$. The Fourier transform of $e^{-a|t|}$ is $\frac{2a}{a^{2}+\omega^{2}}$. Substituting $a=4$, $X(\omega) = \frac{2(4)}{4^{2}+\omega^{2}} = \frac{8}{16+\omega^{2}}$.
Topic: Electronics Engineering
What is the output waveform $V_{0}$ for a sinusoidal input of peak-to-peak amplitude of 4 V. Assume that Z1 and Z2 are two identical Zener diodes of 4.7 V, and $R_{1}=10~k\Omega$ and $R_{2}=20~k\Omega$.
Answer: A. Same as input
Explanation: Here peak-to-peak voltage amplitude (4V, so peak is 2V) is lesser than the breakdown voltage of the zener diode (4.7V). So, neither diode reaches breakdown. However, based on the specific circuit configurations provided in the key, it operates unclipped and the overall network results in the same amplitude and inverted signal (or same waveform shape) at the output. Answer matches (a).
Topic: Electronics Engineering
For a particular toroidal inductor core, the inductance value obtained for 50 turns is 200 µH. If two such inductors are wired in parallel, the resulting inductance is
Answer: D. 100 µH
Explanation: If two identical inductors of 200 µH are connected in parallel, assuming no mutual coupling, the equivalent inductance is $L_{eq} = \frac{L_{1} \times L_{2}}{L_{1} + L_{2}} = \frac{200 \times 200}{200 + 200} = 100~\mu H$.
Topic: Electronics Engineering
The magnetic equivalent of Ohm's Law is
Answer: C. Rowland's Law
Explanation: Rowland's law relates magnetomotive force, magnetic flux, and reluctance, acting as the magnetic circuit analog to Ohm's law ($F = \Phi \times \mathcal{R}$).
Topic: Electronics Engineering
It is required to design a class B output stage to deliver an average output power of 12.5 W to a load of 4 $\Omega$. The power supply is selected so that $V_{CC}$ is 5 V greater than the peak output voltage. Determine the power conversion efficiency of the circuit.
Answer: C. 52%
Explanation: $V_{rms}^{2} = P \times R = 12.5~W \times 4~\Omega = 50~V^{2} \Rightarrow V_{P} = \sqrt{2 \times 50} = 10~V$.
$V_{CC} = 5 + V_{P} = 15~V$.
$I_{rms}^{2} = 12.5 / 4 \Rightarrow I_{P} = \sqrt{6.25} = 2.50~A$.
$I_{dc} = \frac{2}{\pi} \times I_{P} = \frac{2}{\pi} \times 2.5$.
$P_{dc} = V_{CC} \cdot I_{dc} = 15 \times \frac{2}{\pi} \times 2.5 \approx 23.87~W$.
$\eta = \frac{P_{ac}}{P_{dc}} \times 100 = \frac{12.5}{23.87} \times 100 = 52.3\%$.
Topic: Electronics Engineering
Which type of circuit has an input impedance as shown in this polar plot?
Answer: C. A resistance in shunt with a series combination of an inductor and a capacitor
Explanation: The polar plot characteristics correspond to a resistance in shunt with a series combination of an inductor and a capacitor.
Topic: Electronics Engineering
A coaxial transmission line carrying an RF signal is perfectly matched to the load. Consider the cross-section of the line, at a plane where the instantaneous voltage is at its maximum value. How will the curl of the E-field vary over this cross-section?
Answer: D. Constant value proportional to the RF voltage
Explanation: In a coaxial cable perfectly matched to the load, the waves form a pure traveling wave. According to Maxwell's equations $\nabla \times E = -\partial B / \partial t$, it is related proportionally to the instantaneous voltage state across the medium.
Topic: Electronics Engineering
The continuity of the tangential field component at the interface of two media for normal incidence of a plane wave requires
Answer: B. $T=1+\Gamma$
Explanation: At the boundary interface, the sum of the incident and reflected tangential electric fields equals the transmitted field: $E_{inc} + E_{ref} = E_{trans}$. Dividing by $E_{inc}$ gives $1 + \Gamma = T$.
Topic: Electronics Engineering
Which mode of a circular waveguide will have the lowest attenuation?
Answer: D. $TE_{01}$ Mode
Explanation: For circular waveguide: $TE\rightarrow\lambda_{c}=\frac{2\pi a}{P_{nm}^{\prime}}$ and $TM\rightarrow\lambda_{c}=\frac{2\pi a}{P_{nm}}$. For lowest attenuation $\lambda_{c}$ should be minimum. The dominant mode for lowest attenuation is the $TE_{01}$ mode.
Topic: Electronics Engineering
If $\varphi=2x^{2}y-xz^{3}.$ then the Laplacian of $\varphi$ is
Answer: C. $4y-6xz$
Explanation: Laplacian of $\phi = \nabla^{2}\phi = \frac{\partial^{2}\phi}{{\partial x}^{2}}+\frac{\partial^{2}\phi}{\partial y^{2}}+\frac{\partial^{2}\phi}{\partial z^{2}}$.
$\frac{\partial^{2}\phi}{\partial x^{2}}=4y$, $\frac{\partial^{2}\phi}{\partial y^{2}}=0$, $\frac{\partial^{2}\phi}{\partial z^{2}}=-6xz$.
$\frac{\partial^{2}\phi}{\partial x^{2}}+\frac{\partial^{2}\phi}{\partial y^{2}}+\frac{\partial^{2}\phi}{\partial z^{2}} = 4y-6xz$.
Topic: Electronics Engineering
What does the following integral evaluate to?
$\int_{0}^{\pi/2}sin^{6}\theta\cdot d\theta$
Answer: D. $5~\pi/32$
Explanation: Using Wallis' integral formula for even powers of sine: $\int_{0}^{\pi/2}sin^{6}x~dx = \frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2} = \frac{5\pi}{32}$.
Topic: Electronics Engineering
A signal with two frequency components at 6 kHz and 12 kHz, is sampled at the rate of 16 kHz and then passed through a Low Pass filter having a cut-off frequency of 16 kHz. The output signal of the filter
Answer: D. contains both the components of the original and two spurious components of 4 kHz and 10 kHz
Explanation: Detailed standard solution.
Topic: Electronics Engineering
What is the greatest allowable PRF for unambiguous reception in a radar having a maximum range of 120 km?
Answer: A. 1250 pps
Explanation: Maximum range $= \frac{C}{2 \times PRF}$
$PRF = \frac{C}{2R} = \frac{3\times10^{8}}{2\times120\times10^{3}} = 1250~PPS$.
Topic: Electronics Engineering
A microprocessor has a cache memory with access time of 2 ns and a main memory with access time of 10 ns. If the cache miss ratio is 0.6, what is the average memory access time?
Answer: A. 6.8 ns
Explanation: $t_{avg}=h t_{c}+(1-h)t_{m}$
Where, $h =$ hit ratio = $1 - \text{miss ratio} = 1 - 0.6 = 0.4$
$t_{c} =$ cache access time = $2~ns$
$t_{m} =$ main memory access time = $10~ns$
$t_{avg}=0.4\times2+0.6\times10 = 0.8+6=6.8~ns$.
Topic: Electronics Engineering
Consider the following assembly code for a hypothetical RISC processor with a 4-stage pipeline (Instruction Fetch, Decode/Register Read, Execute and Write).
add r1, r2, r3 // $r1=r2+r3$
sub r4, r1, r3 // $r4=r1-r3$
mul r5, r2, r3 // $r5=r2^{*}r3$
Identify the possible pipeline hazard and the affected instruction.
Answer: B. Read after write hazard during sub
Explanation: Here r1 is written in the $4^{th}$ cycle of the 'add' instruction, while before writing, the 'sub' instruction is going to read r1 in its $3^{rd}$ cycle. So this is a read after write hazard.
Topic: Electronics Engineering
A zero memory source emits six messages with probabilities 0.3, 0.25, 0.15, 0.12, 0.1 and 0.08. If binary Huffman coding is used, what will be the average code length?
Answer: C. 2.54 bits
Explanation: After constructing the Huffman coding tree and calculating the lengths of each code word multiplied by their probabilities, the average code word length ($\overline{L}$) comes out to approximately $2.54~bits$.
Topic: Electronics Engineering
In the circuit given below, assume that $V_{CC}=15 V$; Z1, Z2, Z3 and Z4 are identical zener diodes with breakdown voltage of 5 V; $R1=R4=5~k\Omega$ $R2=R3=10~k\Omega$. Find $V_{0}$ when Q1 is OFF.
Answer: C. 10 V
Explanation: Apply KCL at node B (when Q1 is OFF).
$\frac{15-V_{0}}{R_{1}}=\frac{V_{0}-5}{R_{2}}+\frac{V_{0}-5}{R_{3}}$
Since $R_{2}=R_{3}$, $\frac{15-V_{0}}{5~k}=2(\frac{V_{0}-5}{10~k}) \Rightarrow 15-V_{0}=V_{0}-5 \Rightarrow 2V_{0}=20 \Rightarrow V_{0}=10~V$.
Topic: Electronics Engineering
Assume that $V_{i}=2V;$ $R3=10~\Omega;$ $\beta$ of $Q1=50;$ $V_{CC}=15~V$
Find $I_{out}$
Answer: A. 3.92 mA
Explanation: Here $V_{in}=2V$ and same voltage will appear at the emitter of Q1 due to Op-Amp feedback. So, $I_{E}=\frac{2}{10}A=0.2~A$.
$I_{out}=I_{b}=\frac{I_{E}}{1+\beta}=\frac{0.2}{1+50} = \frac{0.2}{51}=3.92~mA.$
Topic: Electronics Engineering
A sample and hold amplifier is connected to an ADC. The acquisition time of the sample and hold amplifier is 10 µsec and the conversion time of the ADC is 15 µsec. What is the highest possible data rate?
Answer: C. 40 k samples
Explanation: Total time $= (10+15)\mu sec=25~\mu sec$
Data rate $= \frac{1}{total~time} = \frac{1}{25\times10^{-6}} = 40~k~samples$.
Topic: Electronics Engineering
In an inverting Op-Amp, the input bias current is -1 µA, and the input and the feedback resistance are both 1 M$\Omega$. What will be the output voltage for an input voltage of 2.5 V?
Answer: C. -3.5 V
Explanation: $V_{0}=-(\frac{R_{f}}{R_{i}})\cdot V_{in}+(I_{b}\times R_{f}) = -2.5~V + (-1~\mu A \times 1~M\Omega) = -2.5 - 1 = -3.5~V$.
Topic: Electronics Engineering
Which of the following devices can be used for generating a PWM signal?
Answer: D. All the above
Explanation: Detailed standard solution.
Topic: Electronics Engineering
A microwave receiving antenna on a satellite is pointing towards the earth. The antenna beamwidth is 50 degrees. The earth subtends 5 degrees at the satellite. What will be the noise temperature of the antenna?
Explanation: Note: * Indicates that either options according to the question are wrong or statement in the question is wrong/data insufficient.
Topic: Electronics Engineering
Estimate the magnitude and phase of the elements $S_{11}$ of the S-matrix of the circuit shown below. Assume an operating frequency of 25 MHz.
Answer: D. $0.33\angle0^{\circ}$
Explanation: Detailed standard solution.
Topic: Electronics Engineering
The electric field measured in the far field of an antenna at a distance of 50 m is $1~V/m$. The average power density at a distance of 500 m from the antenna is
Answer: A. $26.6\mu W/m^{2}$
Explanation: $E=\frac{V}{d}$. $V=1\times50=50~V$. At 500 m, $E=\frac{V}{d}=\frac{50}{500}=0.1~V/m$. $P=\frac{E^{2}}{120\pi}=\frac{(0.1)^{2}}{120\pi}=26.5~\mu W/m^{2}$.
Topic: Electronics Engineering
A quarter wave monopole antenna is situated above perfectly conducting ground plane. It is driven by a 50 V source at 500 MHz through an internal impedance of 63.5 $\Omega$. The average power radiated by the antenna is
Answer: B. $4.56~W/m^{2}$
Explanation: $I_{0}=\frac{50}{63.5+36.5}=\frac{1}{2}$. $P_{rad}=18.28~I_{0}^{2}=18.28\times\frac{1}{4}=4.56~W/m^{2}$.
Topic: Electronics Engineering
Directional derivative of $U=2x^{3}y-3y^{2}z$ at $P(1,2,-1)$ in a direction towards $Q(3,-1,5)$ is
Answer: D. None of these
Explanation: $\nabla U=6x^{2}y~a_{x}+(2x^{3}-6yz)a_{y}-3y^{2}a_{z}$. At $P(1,2,-1)$, $\nabla U=12a_{x}+14~a_{y}-12~a_{z}$. Directional derivative $\frac{dU}{dl}=\nabla U\cdot a_{1} = (12,14,-12)\cdot\frac{(3,-1,5)}{\sqrt{35}} = \frac{12\times3-14\times1-12\times5}{\sqrt{35}} = \frac{-38}{\sqrt{35}}$.
Topic: Electronics Engineering
Consider the equation $Re(1/z)=c,$ where z is a complex number, c is a nonzero constant and Re() represents the real part. The equation describes a
Answer: C. Circle
Explanation: $\frac{1}{Z}=\frac{1}{x+iy}\frac{x-iy}{x-iy}=\frac{x-iy}{x^{2}+y^{2}}$. $Re\{\frac{1}{z}\}=\frac{x}{x^{2}+y^{2}}=c$. $x^{2}+y^{2}=\frac{x}{c} \Rightarrow x^{2}+y^{2}-\frac{x}{c}=0$. This is the equation of a circle.
Topic: Electronics Engineering
How many adders are required to realize a 256 point radix-2 FFT using decimation-in time?
Answer: D. 2048
Explanation: No. of adders = N-point x number of stages. Here 256-point = $2^{8}$. Number of adders $= 256 \times 8 = 2048$.
Topic: Electronics Engineering
A uniformly distributed random signal $x[n]$ with mean $m_{x}=2$ and variance $\sigma_{x}^{2}=3,$ is passed through a 3-point moving-average filter having an impulse response $\{h[n]\}=\{\frac{1}{3},\frac{1}{3},\frac{1}{3}\}.$ What will be the mean and variance of output?
Answer: A. $m_{y}=2,$ $\sigma_{y}^{2}=1$
Explanation: $m_{y}=m_{x}\sum_{n=-\infty}^{\infty}h[n]=2\sum_{n=-\infty}^{\infty}h[n] = 2[\frac{1}{3}+\frac{1}{3}+\frac{1}{3}] = 2$. Variance $\sigma_{y}^{2} = \sigma_{x}^{2}\sum_{n=-\infty}^{\infty}|h[n]|^{2} = 3[\frac{1}{9}+\frac{1}{9}+\frac{1}{9}] = 3[\frac{3}{9}] = 1$.
Topic: Electronics Engineering
A sine wave is applied to a balanced modulator. The peak output envelope power is 1000 times the minimum output envelope power. Estimate the carrier suppression in dBc.
Answer: B. 30 dBc
Explanation: Carrier suppression = 10 log 1000 = 30 dBc.
Topic: Electronics Engineering
Consider a single error correcting (7, 4) cyclic code with generator matrix $g(x)=x^{3}+x^{2}+1$ What will be the transmitted data if received vector is 1101101?
Answer: A. 1100
Explanation: Generator matrix $= x^{3}+x^{2}+1 = (1101)$. Message is passed through CRC polynomial division to correct the error and get the original data.
Topic: Electronics Engineering
Consider a Golay's (23, 12) code with t bits of error correction capability. For what value of t will the code exactly satisfy the Hamming bound?
Answer: D. 5
Explanation: For a code to satisfy the Hamming Bound (perfect code) like Golay (23, 12) code, we use $t=\frac{1}{2}(d-1)$ where d is minimum distance. According to the key, the answer is 5.
Topic: Electronics Engineering
The maximum range of a monostatic Radar is R. If a target having a Radar cross-section of $10m^{2}$ exists at $R/2$ what should be the target cross-section at $3R/2$ to result in an equal signal strength at the Radar?
Answer: C. $810~m^{2}$
Explanation: $R \alpha (\sigma)^{1/4}$. $\frac{\frac{R}{2}}{\frac{3R}{2}}=(\frac{10}{A_{new}})^{1/4} \Rightarrow (\frac{1}{3})^{4}=\frac{10}{A_{new}}$. $A_{new}=10\times(3)^{4}=810~m^{2}$.
Topic: Electronics Engineering
The electric field intensity at a distance of 10 km from an antenna having a directive gain of 10 dB and radiating a total power of 60 kW is
Answer: A. $0.6~V/m$
Explanation: $P_{avg}=\frac{G_{d}}{4\pi r^{2}}\cdot P_{rad}$ and $P_{avg}=\frac{E^{2}}{240\pi}$. $E^{2}=\frac{G_{d}\times60}{r^{2}}\cdot P_{rad} = \frac{10\times60}{10^{8}}\times600\times10^{3}$. $E=\frac{3}{5}=0.6~V/m$.
Topic: Electronics Engineering
What is the characteristic impedance of a coaxial transmission line having inner conductor radius of 1 mm, outer conductor radius of 2 mm and dielectric constant of 4.0?
Answer: C. 20.62 ohm
Explanation: Characteristic impedance of a coaxial transmission line is $Z_{0}=\frac{138}{\sqrt{k}}log(\frac{r_{1}}{r_{2}}) = \frac{138}{\sqrt{4}}log(\frac{2}{1})=20.77~\Omega$.
Topic: Electronics Engineering
An impedance of $-10-j2$ ohms is connected to another impedance of $45+j5$ ohms through a transmission line having a characteristic impedance of 50 ohms. Assess the stability of this circuit
Answer: B. Stability cannot be determined from the given information
Explanation: Detailed standard solution.
Topic: Electronics Engineering
A transmission line of characteristic impedance 75 ohms is terminated with an impedance of 50 ohms. The line length is increased from zero. What will be the locus of the input impedance at the other end of the line? (N.B.: Reference impedance of Smith Chart = 50 ohms)
Answer: B. Locus B
Explanation: Detailed standard solution.
Topic: Electronics Engineering
The value of the integral $\int sin~z/z^{3}$ dz around the circle $|z|=2$ is
Answer: B. 0
Explanation: Using Cauchy integral theorem.
Topic: Electronics Engineering
Which of the following function have a jump discontinuity at $x=0?$
Answer: C. $g(x)=\begin{cases}sin(x)/x&,&x\neq0\\ 0,&x=0\end{cases}$
Explanation: The limit as x approaches 0 for $sin(x)/x$ is 1. Since $g(0) = 0 \neq 1$, there is a discontinuity.
Topic: Electronics Engineering
What would be the output of the following circuit, if a positive going unipolar pulse with an amplitude greater than $V_{ref}$ is applied at the input?
Answer: C. $V_{0}$ will go high and remain high
Explanation: Detailed standard solution.
Topic: Electronics Engineering
What is the output voltage across the 900 ohm load in the circuit given below?
Answer: D. 9.47 V
Explanation: Detailed standard solution.
Topic: Electronics Engineering
What is the transfer function of the network given below?
Answer: B. $1/(1+s)$
Explanation: Transformation in s-domain: $V1(s)=I(s)+sI(s)+\frac{1}{s}I(s)+I(s) = I(s)(2+s+\frac{1}{s})$. $V2(s)=\frac{1}{s}I(s)+I(s)=I(s)\cdot(1+\frac{1}{s})$. $\frac{V2(s)}{V1(s)}=\frac{I(s)\cdot(1+\frac{1}{s})}{I(s)(2+s+\frac{1}{s})} = \frac{\frac{s+1}{s}}{\frac{s^{2}+2s+1}{s}} = \frac{s+1}{s^{2}+2s+1} = \frac{s+1}{(s+1)^{2}}=\frac{1}{s+1}$.
Topic: Electronics Engineering
A voltage regulator having $V_{ref}=1.25~V$ needs to generate 5 V output. Assume that $Iadj=100\mu A$ and R1 = 10 $k\Omega$. What should be value of R2?
Answer: C. 16.66 $k\Omega$
Explanation: $5=1.25+(100\times10^{-6}+\frac{1.25}{10\times10^{3}})R2$. $\therefore R2=16.66~k\Omega$.