ISRO Electronics & Communication (ECE) 2009 Question Paper & Solutions
Attempt the ISRO Electronics & Communication (ECE) 2009 previous year paper in real CBT exam mode or practice mode with verified answers and explanations.
All ISRO Electronics & Communication (ECE) papers
80 questions.
Question 1
Topic: Electronics Engineering
A 5 bit DAC has a current output. For a digital input of 10100, an output current of 10 mA is produced. What will be the output current for a digital input of 11101?
- A. 14.5 mA
- B. 10 mA
- C. 100 mA
- D. Not possible to calculate
Answer: A. 14.5 mA
Explanation: output = Resolution × D
(Here D = decimal equivalent of given binary value).
For 10100 (which is 20 in decimal), output is 10 mA. So, Resolution = $10~mA / 20 = 0.5~mA$.
For 11101 (which is 29 in decimal), output = $0.5~mA \times 29 = 14.5~mA$.
Question 2
Topic: Electronics Engineering
The ethernet protocol uses
- A. CSMA/CA
- B. SCPC
- C. CSMA/CD
- D. Slotted ALOHA
Answer: C. CSMA/CD
Explanation: Detailed standard solution.
Question 3
Topic: Electronics Engineering
A micro controller differs from a microprocessor in terms of
- A. I/O interfaces and instruction decoding
- B. Memory configuration and I/O interfaces
- C. Data bus width and clock speed
- D. Memory configuration and instruction decoding
Answer: B. Memory configuration and I/O interfaces
Explanation: Detailed standard solution.
Question 4
Topic: Electronics Engineering
An 8-bit microcontroller has an external RAM with the memory map from 8000H to 9FFFH. The number of bytes RAM can store is
- A. 8193
- B. 8192
- C. 8191
- D. 8000
Answer: B. 8192
Explanation: Address range given is: 8000H to 9FFFH
Number of bytes $= 9FFFH - 8000H + 1 = (2000)_{H}$ or $(8192)_{d}$
Question 5
Topic: Electronics Engineering
An audio frequency of 15 KHz is frequency modulated with a deviation of 75 KHz. The resulting bandwidth is
- A. 180 KHz
- B. 150 KHz
- C. 210 KHz
- D. 240 KHz
Answer: A. 180 KHz
Explanation: $\beta=\frac{\Delta f}{f_{m}}=\frac{75}{15}=5$
$B.W.=2(\beta+1)f_{m} = 2(5+1)15~kHz = 180~kHz$
Question 6
Topic: Electronics Engineering
A 1 kW carrier is Amplitude Modulated to a depth of 60%. The total power in the modulated carrier is
- A. 1 kW
- B. 1.06 kW
- C. 1.6 kW
- D. 1.18 kW
Answer: D. 1.18 kW
Explanation: $P_{t}=P_{c}(1+\frac{\mu^{2}}{2}) = 1(1+0.18)KW = 1.18~KW.$
Question 7
Topic: Electronics Engineering
The $y_{21}$ parameter of the network shown in the given figure will be
- A. $1/6$
- B. $-1/2$
- C. $-1/3$
- D. $1/3$
Answer: B. $-1/2$
Explanation: Convert star to delta. The equivalent delta network has a $6~\Omega$ resistor in parallel with the top $3~\Omega$ resistor. This reduces to a $\pi$-network. So $Y_{12} = -\frac{1}{2}$ mho.
Question 8
Topic: Electronics Engineering
An inductance of 1 H is realized using air core with 100 turns. What will be the inductance if the number of turns are doubled?
- A. 1 H
- B. 0.5 H
- C. 2 H
- D. None of these
Answer: D. None of these
Explanation: $L=\frac{\mu_{0}kN^{2}A}{l}$ if the number of turns is doubled then inductance becomes four times. Hence, new inductance $= 4~H$.
Question 9
Topic: Electronics Engineering
An inductor of inductance 0.1 H, carrying current of 6 Amps will store energy of
- A. 6 Joules
- B. 36 Joules
- C. 1.8 Joules
- D. 3.6 Joules
Answer: C. 1.8 Joules
Explanation: $E=\frac{1}{2}LI^{2}=\frac{1}{2}\times0.1\times36 = 1.8$ Joules
Question 10
Topic: Electronics Engineering
A coil has self resistance of $5~\Omega$ and inductance of $1~\mu H$. The value of Q at 1 MHz frequency is
- A. 200
- B. 100
- C. 40
- D. None of these
Answer: D. None of these
Explanation: $Q=\frac{\omega L}{R}=\frac{2\pi fL}{R} = \frac{2\pi\times1\times10^{6}\times1\times10^{-6}}{5}=1.2566$
Question 11
Topic: Electronics Engineering
The closed loop transfer function of a unity feedback control system is $G(s)=\frac{1}{(s+2)^{2}}$. The closed loop transfer function will have poles at
- A. -2, $\pm j$
- B. -2, -1
- C. -2, -2
- D. -2, 2
Answer: C. -2, -2
Explanation: $G(s)=\frac{1}{(s+2)^{2}}$
There are repeated poles at $s=-2$.
Question 12
Topic: Electronics Engineering
The dominant mode of propagation in microstrip line in low frequency approximation is
- A. $TM_{11}$
- B. Quasi-TEM
- C. TEM
- D. $TE_{01}$
Answer: B. Quasi-TEM
Explanation: Detailed standard solution.
Question 13
Topic: Electronics Engineering
Consider a transmission line of characteristic impedance 50 ohms and the line is terminated at one end by $+j50$ ohms, the VSWR produced in the transmission line will be
- A. +1
- B. zero
- C. -1
- D. infinity
Answer: D. infinity
Explanation: Reflection coefficient $(\rho) = \frac{Z_{L}-Z_{0}}{Z_{L}+Z_{0}} = \frac{j50-50}{j50+50}$
$|\rho|=1$ therefore $VSWR=\infty$
Question 14
Topic: Electronics Engineering
The divergence of magnetic flux density is
Answer: C. 0
Explanation: $\nabla\cdot B=0$
Question 15
Topic: Electronics Engineering
A straight line and circle of radius are given. A chord is drawn at random to this circle parallel to the given line. Expected length of the chord is
- A. 2a
- B. $\pi a$
- C. $\pi a/2$
- D. None of these
Answer: C. $\pi a/2$
Explanation: A chord must be always less than the diameter, so expected length of the chord is less than 2a. $\frac{\pi a}{2}=1.57a<2a$. So, correct option is $\frac{\pi a}{2}$.
Question 16
Topic: Electronics Engineering
There is a function $f(x)$, such that $f(0)=1$ and $f^{\prime}(0)=-1$ and $f(x)$ is positive for all values of x, then
- A. $f^{\prime\prime}(x)<0$ for all x
- B. $-1<f^{\prime\prime}(x)<0$ for all x
- C. $-2<P^{\prime\prime}(x)<-1$ for all x
- D. None of these
Answer: D. None of these
Explanation: $f(x)=e^{-x}\Rightarrow f(0)=e^{-0}=1$
$f^{\prime}(x)=-e^{-x}\Rightarrow f^{\prime}(0)=-e^{-0}=-1$
$f(x)=e^{-x}is+ve~for\forall x$
$f^{\prime\prime}(x)=e^{-x}is+ve~for~\forall x$
Question 17
Topic: Electronics Engineering
Interrupt latency is the time elapsed between:
- A. Occurrence of an interrupt and its detection by the CPU
- B. Assertion of an interrupt and the start of the associated ISR
- C. Assertion of an interrupt and the completion of the associated ISR
- D. Start and completion of associated ISR
Answer: B. Assertion of an interrupt and the start of the associated ISR
Explanation: Detailed standard solution.
Question 18
Topic: Electronics Engineering
Inverse Fourier transform of signal $X(j\omega)=2\pi\delta(\omega)+\pi\delta(\omega-4\pi)+\pi\delta(\omega+4\pi)$
- A. $1+cos~4\pi t$
- B. $\pi(1-cos~4\pi t)$
- C. $2\pi(1-cos~4\pi t)$
- D. $2\pi(1+cos~4\pi t)$
Answer: A. $1+cos~4\pi t$
Explanation: $X(j\omega)=2\pi\delta(\omega)+\pi\delta(\omega-4\pi)+\pi\delta(\omega+4\pi)$
$1 \leftarrow\frac{F.T.}\rightarrow 2\pi\delta(\omega)$
$cos~\omega_{0}t\leftarrow\frac{F.T.}\rightarrow\pi[\delta(\omega-\omega_{0})+\delta(\omega+\omega_{0})]$
Question 19
Topic: Electronics Engineering
Determine the discrete-time Fourier Transform for the signal $x[n]=[-2,-1,0,1,2]$ ↑ $n=0$
- A. $2~j(2~sin~2~\omega+sin~\omega)$
- B. $-2j(2~sin~2~\omega+sin~\omega)$
- C. 2(2cos 2 - cos )
- D. -2(2cos 2 - cos )
Answer: B. $-2j(2~sin~2~\omega+sin~\omega)$
Explanation: $x(e^{j\omega})=\sum_{n=-2}^{2}x[n]e^{-j\omega n}$
$=-2e^{j2\omega}-e^{j\omega}+0+e^{-j\omega}+2e^{-j2\omega}$
$=-2(cos~2~\omega+j~sin~2~\omega)-(cos~\omega+j~sin~\omega) +cos~\omega-j~sin~\omega+2(cos~2~\omega-j~sin~2~\omega)$
$=-2~cos~2\omega-2j~sin~2\omega-cos~\omega-j~sin~\omega+cos~\omega -j~sin~\omega+2~cos~2\omega-2j~sin~2\omega$
$=-4j~sin~2\omega-2j~sin~\omega = -2j(2~sin~2\omega+sin~\omega)$
Question 20
Topic: Electronics Engineering
Auto-correlation of a sinusoid $s(t)=A~sin(\omega t)$ is
- A. $(A^{2}/2)$ sin(\omega\tau)
- B. $(A/2)$ sin(\omega t)
- C. $(A^{2}/2)cos(cor)$
- D. $(A/2)$ cos(\omega\tau)
Answer: C. $(A^{2}/2)cos(cor)$
Explanation: $s(t)=A~sin(\omega t)$
$R_{x}(\tau)=E[x(t+\tau)x(t)] = E[A^{2}sin(\omega t+\omega\tau)sin(\omega t)]$
$= E[\frac{A^{2}}{2}cos(\omega\tau)-\frac{A^{2}}{2}cos(2\omega t+\omega\tau)] = \frac{A^{2}}{2}cos(\omega\tau)$
Question 21
Topic: Electronics Engineering
A source produces 4 symbols with probabilities $1/2$, $1/4$ $1/8$ $1/8$ For this source a practical coding scheme has an average code word length of 2 bits/symbol. The efficiency of the code is
- A. 1
- B. $7/8$
- C. $1/2$
- D. $1/4$
Answer: B. $7/8$
Explanation: Efficiency $(\eta)=\frac{H(s)}{\overline{L}}$
Where, $H(s)=$ entropy and $\overline{L}=$ Average code word length
$H(s)=\sum_{i=0}^{N-1}P_{i}log_{2}^{(\frac{1}{P_{i}})}$
$=\frac{1}{2}log_{2}^{2}+\frac{1}{4}log_{2}^{4}+\frac{1}{8}log_{2}^{8}+\frac{1}{8}log_{2}^{8} = \frac{1}{2}+\frac{1}{2}+\frac{3}{8}+\frac{3}{8}=\frac{7}{4}.$
So, efficiency $(\eta)=\frac{\frac{7}{4}}{2}=\frac{7}{8}.$
Question 22
Topic: Electronics Engineering
When signal frequency is 2000 kHz and IF is 455 kHz, the image frequency could be
- A. 2910 kHz
- B. 1645 kHz
- C. 2455 kHz
- D. 2090 kHz
Answer: A. 2910 kHz
Explanation: Image frequency $=f_{s}\pm2f_{IF} = (2000+910)kHz = 2910~kHz$
Question 23
Topic: Electronics Engineering
The Qof RLC tuned circuit with source voltage V in the figure is 100 at resonant frequency of 100 kHz. What is the voltage $V_{L}$ across inductor at resonance?
- A. 50 V
- B. $V_{L}-V_{C}$
- C. 100 V
- D. $V_{R}-V_{C}$
Answer: C. 100 V
Explanation: $V_{L}=QV=100~V$
Question 24
Topic: Electronics Engineering
A5 V reference is drawn from the circuit shown in the figure. Zener diode of 400 mW and 5 V with firing current of 5mA is used. The value $R_{s}$ is
- A. 50 ohms
- B. 500 ohms
- C. 75 ohms
- D. 470 ohms
Answer: A. 50 ohms
Explanation: $R_{s}=\frac{10-5}{100}k\Omega = \frac{5}{100}\times1000~\Omega=50~\Omega.$
Question 25
Topic: Electronics Engineering
Find the equivalent resistance R for the network shown in the figure.
- A. 200 ohms
- B. 101 ohms
- C. 220 ohms
- D. 100 ohms
Answer: D. 100 ohms
Explanation: We have to calculate the equivalent resistance from the left side.
$R_{eq}=\frac{200\times200}{400}=100~\Omega$
Question 26
Topic: Electronics Engineering
The voltage across last resistor is V. All resistors are unity. Then $V_{s}$ is given by
- A. 13 V
- B. 8 V
- C. 4 V
- D. None of these
Answer: B. 8 V
Explanation: We start from left, two same resistors are in series, so voltage drop is same across both the resistors and again sum of these resistors are parallel to 1 Ω resistor so equal voltage and proceed in the same way. So, $V_{s}=8~V$
Question 27
Topic: Electronics Engineering
Natural frequency of a unity feedback control system of transfer function $G(s)=\frac{10}{s(s+1)}i\epsilon$
- A. $0.5~rad/sec$
- B. $3.16~rad/sec$
- C. $4.6~rad/sec$
- D. None of these
Answer: B. $3.16~rad/sec$
Explanation: $1+G(s)=0$
$1+\frac{10}{s(s+1)}=0 \Rightarrow s^{2}+s+10=0$
$\omega_{n}^{2}=10 \Rightarrow \omega_n = \sqrt{10}=3.16~rad/sec$
Question 28
Topic: Electronics Engineering
Which of the following conditions will not guarantee distortionless line?
- A. $R=0$ $G=0$
- B. Very low frequency range $(R>>\omega L$ $G>>\omega C)$
- C. $R/L=G/C$
- D. Very high frequency range $(R<<\omega L$ $G<<\omega C)$
Answer: B. Very low frequency range $(R>>\omega L$ $G>>\omega C)$
Explanation: Detailed standard solution.
Question 29
Topic: Electronics Engineering
For a line characteristic impedance $Z_{0}$ terminated in a load $Z_{R}$ such that $Z_{R}>Z_{0},$ the Voltage Standing Wave Ratio (VSWR) is given by
- A. $Z_{R}/Z_{0}$
- B. $Z_{0}$
- C. $Z_{R}$
- D. $Z_{0}/Z_{R}$
Answer: D. $Z_{0}/Z_{R}$
Explanation: Reflection coefficient $(\Gamma)=\frac{Z_{0}-Z_{R}}{Z_{0}+Z_{R}}$
$VSWR=\frac{1+|\Gamma|}{1-|\Gamma|}=\frac{1+\frac{Z_{0}-Z_{R}}{Z_{0}+Z_{R}}}{1-(\frac{Z_{0}-Z_{R}}{Z_{0}+Z_{R}})} = \frac{Z_{0}+Z_{R}+Z_{0}-Z_{R}}{Z_{0}+Z_{R}-Z_{0}+Z_{R}}=\frac{Z_{0}}{Z_{R}}.$
Question 30
Topic: Electronics Engineering
The beamwidth of a reflector antenna of diameter $=70$ cm at 30 GHz is
- A. 3.7 degree
- B. 0.5 degree
- C. 1.1 degree
- D. 1 degree
Answer: D. 1 degree
Explanation: Given, $d=70~cm$ $f=30~GHz$
$\lambda=\frac{3\times10^{8}}{30\times10^{9}}\times100~cm=1~cm.$
Beamwidth (0) $=\frac{K\lambda}{d}=\frac{70\times1}{70}=1$ ($\cdot\cdot K=70$)
Question 31
Topic: Electronics Engineering
$T_{r}=r/(1+r^{2}+r^{4})$ $\sum_{1}^{\infty}T_{r}=$
- A. $1/4$
- B. 1
- C. None of the these
- D. $1/2$
Answer: D. $1/2$
Explanation: $T_{r}=\frac{r}{1+r^{2}+r^{4}}$
$\sum_{r=1}^{\infty}T_{r}=\frac{1}{3}+\frac{2}{21}+\frac{3}{91}+\frac{4}{273}+\cdot\cdot\cdot+\infty$
It will converges for the higher values, so the approximate summation upto 4th term is $\sum_{r=1}^{\infty}T_{r}=0.47\mathcal{G}=0.5$
Question 32
Topic: Electronics Engineering
A man with n keys wants open a lock. He tries his keys at random. The expected number of attempts for his success is (keys are replaced after every attempt)
- A. $n/2$
- B. n
- C. $\sqrt{n}$
- D. None of the above
Answer: B. n
Explanation: $E(x)=\Sigma xP(x) = 1\times\frac{1}{n}+2(\frac{n-1}{n})\times\frac{1}{n}+3(\frac{n-1}{n})^{2}\times\frac{1}{n}\cdot\cdot\cdot = n$
Question 33
Topic: Electronics Engineering
If a counter having 10 FFs initially at 0, what count will it hold after 2060 pulses?
- A. 000 000 1100
- B. 000 001 1100
- C. 000 001 1000
- D. 000 000 1110
Answer: A. 000 000 1100
Explanation: Detailed standard solution.
Question 34
Topic: Electronics Engineering
A memory system of size 16 K bytes is required to be designed using memory chips which have 12 address lines and 4 data lines each. Then the number of such chips required to design the memory system is
Answer: C. 8
Explanation: Number of chips required $= \frac{16\times2^{10}\times8}{2^{12}\times4} = 8$
Question 35
Topic: Electronics Engineering
Consider the following program for 8085
XRA A
LXI B, 0007H
Loop: DCXB
JNZ LOOP
The loop will be executed
- A. 8 times
- B. once
- C. 7 times
- D. infinite times
Answer: B. once
Explanation: After execution of XRA A instruction, the zero flag is set and DCX B instruction will not alter the zero flag. So when processor will execute JNZ instruction, this statement is false, and processor will not go inside the loop further. Finally, processor will execute the loop only once.
Question 36
Topic: Electronics Engineering
A certain JK FF has $t_{pd}=12$ ns. The largest MOD counter that can be constructed from such FFs and still operate to 10 MHz is
Answer: B. 256
Explanation: $f_{max}=\frac{1}{nt_{pd}}$ Where, $n=$ number of FFs, $t_{pd}=$ propagation delay of FF, $f_{max}=$ maximum operating frequency.
$10\times10^{6}=\frac{1}{(n\times12)ns}$
$n=\frac{1}{10\times12\times10^{6}\times10^{-9}}=8.33\simeq8.$
Mod-counter $=2^{n}=256$
Question 37
Topic: Electronics Engineering
Convolution of $x(t+5)$ with impulse function $\delta(t-7)$ is equal to
- A. $x(t-2)$
- B. $x(t+12)$
- C. $x(t-12)$
- D. $x(t+2)$
Answer: A. $x(t-2)$
Explanation: $x(t)*\delta(t-t_{0})=x(t_{0})$
$x(t+5)*\delta(t-7)=x(t-2)$
Question 38
Topic: Electronics Engineering
The purpose of a Cassegrain feed in a parabolic reflector antenna is to
- A. Achieve higher antenna gain
- B. Reduce the antenna size
- C. Reduce the beam width
- D. Ease of locating the feed at convenient point
Answer: D. Ease of locating the feed at convenient point
Explanation: Detailed standard solution.
Question 39
Topic: Electronics Engineering
A transfer function $G(s)$ has pole-zero plot as shown in the figure. Given that the steady state function gain is 2, the transfer function $G(s)$ will be given by
- A. $\frac{5(s+1)}{s^{2}+4s+4}$
- B. $\frac{2(s+1)}{s^{2}+4s+5}$
- C. $\frac{10(s+1)}{s^{2}+4s+5}$
- D. $\frac{10(s+1)}{(s+2)^{2}}$
Answer: C. $\frac{10(s+1)}{s^{2}+4s+5}$
Explanation: Zeros at $S=-1$. Poles at $S=-2\pm j$. $Gain(k)=2(dc~gain)$. $G(s)H(s)=\frac{10(s+1)}{s^{2}+4s+5}$
Question 40
Topic: Electronics Engineering
The impedance of a series RLC circuit is
- A. $\sqrt{R^{2}+X_{L}^{2}+X_{C}^{2}}$
- B. $\sqrt{R^{2}+X_{L}^{2}-X_{C}^{2}}$
- C. $\sqrt{R^{2}+(X_{L}+X_{C})^{2}}$
- D. $\sqrt{R^{2}+(X_{L}-X_{C})^{2}}$
Answer: D. $\sqrt{R^{2}+(X_{L}-X_{C})^{2}}$
Explanation: Detailed standard solution.
Question 41
Topic: Electronics Engineering
The advantage of $\pi/4$ QPSK compared to conventional QPSK is?
- A. $\pi/4$ QPSK requires less bandwidth compared to conventional QPSK
- B. $\pi/4$ QPSK produces lower envelope variation compared to conventional QPSK
- C. $\pi/4$ QPSK receiver design is simple compared to conventional QPSK
- D. There is no advantage
Answer: A. $\pi/4$ QPSK requires less bandwidth compared to conventional QPSK
Explanation: Detailed standard solution.
Question 42
Topic: Electronics Engineering
A moving coil iron ammeter may be compensated for frequency errors by
- A. Series inductance
- B. Shunt resistance
- C. Series resistor
- D. Shunt capacitor
Answer: D. Shunt capacitor
Explanation: Detailed standard solution.
Question 43
Topic: Electronics Engineering
Find the value of $V_{0}$ in the figure
- A. 1/2 V
- B. $-1/2$ V
- C. 0 V
- D. $-3/2$ V
Answer: A. 1/2 V
Explanation: $2=2I_{1}+1$
$I_{1}=\frac{1}{2}$
$V_{o}=+\frac{1}{2}V$
Question 44
Topic: Electronics Engineering
What is the approximate skin depth in copper at 100 MHz?
- A. 0.1 micron
- B. 10 microns
- C. 10 mm
- D. 100 mm
Answer: B. 10 microns
Explanation: Skin depth $(\delta)=\frac{1}{\sqrt{\pi f \mu\sigma}}$
Where, Frequency $(f)=100MHz$, Conductivity $(\sigma)=5.8\times10^{7}S$, $\mu=4\pi\times10^{-7}$.
$\delta=\frac{1}{\sqrt{\pi\times100\times10^{6}\times4\pi\times10^{-7}\times5.8\times10^{7}}} = 6.6\times10^{-6}$ as per option approximate answer is (b).
Question 45
Topic: Electronics Engineering
While designing a low noise amplifier, what is the importance of the noise resistance of a transistor?
- A. It gives the effective impedance offered by the input with respect to noise current
- B. It defines the criterion for a conjugate matching of input impedance
- C. It tells us the resistance that would generate the same amount of noise at room temperature
- D. It tells us how rapidly the noise figure increases as we move away from the optimum source impedance
Answer: D. It tells us how rapidly the noise figure increases as we move away from the optimum source impedance
Explanation: Detailed standard solution.
Question 46
Topic: Electronics Engineering
Unattenuated radiation field at the surface of the earth of a quarter-wave monopole will exist if the earth surface is
- A. Lossy dielectric
- B. Perfect insulator
- C. Perfectly conducting
- D. None of these
Answer: C. Perfectly conducting
Explanation: Detailed standard solution.
Question 47
Topic: Electronics Engineering
Given that $A=\begin{bmatrix}1&0&0\\ 0&2&0\\ 0&0&3\end{bmatrix}$ Evaluate $A^{3}-6A^{2}+11A-10I$
- A. Null matrix
- B. Identity matrix
- C. -4I
- D. None of these
Answer: C. -4I
Explanation: $A=\begin{bmatrix}1&0&0\\ 0&2&0\\ 0&0&3\end{bmatrix}$, $A^{2}=\begin{bmatrix}1&0&0\\ 0&4&0\\ 0&0&9\end{bmatrix}$, $A^{3}=\begin{bmatrix}1&0&0\\ 0&8&0\\ 0&0&27\end{bmatrix}$
$A^{3}-6A^{2}+11A-10I=\begin{bmatrix}-4&0&0\\ 0&-4&0\\ 0&0&-4\end{bmatrix} = -4\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix}=-4I$
Question 48
Topic: Electronics Engineering
$\lim_{x\rightarrow0}x~\log~\sin(x)$ is
- A. 0
- B. $\frac{1}{2}$
- C. 1
- D. 2
Answer: A. 0
Explanation: Detailed standard solution.
Question 49
Topic: Electronics Engineering
A 12 bit ADC is operating with $1~\mu s$ clock period and the total conversion time is seen to be $14~\mu s$. The ADC must be of
- A. Flash type
- B. Counting type
- C. Integrating type
- D. Successive approximation type
Answer: D. Successive approximation type
Explanation: Detailed standard solution.
Question 50
Topic: Electronics Engineering
Which of the following types of devices is not field programmable?
- A. FPGA
- B. ASIC
- C. CPLD
- D. PLD
Answer: D. PLD
Explanation: Note: The provided official key states (d), although ASIC (Application Specific Integrated Circuit) is generally the device that is not field programmable.
Question 51
Topic: Electronics Engineering
Which is the correct order of different process steps for a typical FPGA design?
- A. Functional simulation, Synthesis, Place & Route, Timing Verification
- B. Functional simulation, Timing Verification, Synthesis, Place & Route
- C. Timing Verification, Synthesis, Functional simulation, Place & Route
- D. Synthesis, Functional simulation, Timing Verification, Place & Route
Answer: B. Functional simulation, Timing Verification, Synthesis, Place & Route
Explanation: Detailed standard solution.
Question 52
Topic: Electronics Engineering
The theoretical dividing line between Reduced Instruction Set Computing (RISC) microprocessor and Complex Instructions Set Computing (CISC) microprocessor is
- A. Instruction execution rate to be one instruction per clock cycle
- B. Number of address and data lines
- C. Number of pins in the chip
- D. None of the above
Answer: A. Instruction execution rate to be one instruction per clock cycle
Explanation: Detailed standard solution.
Question 53
Topic: Electronics Engineering
In a pulsed radar, the peak radar transmitter power is increased by a factor of 81 keeping all other radar parameters unchanged. The maximum range capability of the radar increases by a factor of
Answer: D. 3
Explanation: Range $(R_{max})\propto(P_{T})^{1/4}$. The increase in maximum range is $(81)^{1/4} = 3$.
Question 54
Topic: Electronics Engineering
A radar transmitter has a PRF of 1500 PPS and a pulse width of 1 micro second. Its duty cycle is
- A. 0.0075
- B. 0.015
- C. 0.0015
- D. 0.00075
Answer: C. 0.0015
Explanation: Duty cycle $= PRF \times \text{Pulse Width} = 1500 \times 1 \times 10^{-6} = 0.0015$
Question 55
Topic: Electronics Engineering
Which of the following technology results in least power dissipation?
- A. CMOS
- B. ECL
- C. TTL
- D. NMOS
Answer: A. CMOS
Explanation: Detailed standard solution.
Question 56
Topic: Electronics Engineering
A power MOSFET is a
- A. Current controlled device
- B. Frequency controlled device
- C. Voltage controlled device
- D. None of the above
Answer: C. Voltage controlled device
Explanation: Detailed standard solution.
Question 57
Topic: Electronics Engineering
The advantage of differential PSK (DPSK) over coherent PSK is
- A. DPSK requires less bandwidth compared coherent PSK
- B. DPSK receiver design is simple compared to coherent PSK
- C. DPSK Bit Error rate is lower than coherent PSK
- D. For same bandwidth, DPSK bit rate is higher compared to coherent PSK
Answer: A. DPSK requires less bandwidth compared coherent PSK
Explanation: Note: The provided key marks (a) as the answer. In standard literature, the primary advantage of DPSK is a simpler receiver design (b) as it avoids carrier recovery.
Question 58
Topic: Electronics Engineering
An ideal current meter should have
- A. Zero resistance
- B. Finite resistance
- C. Infinite resistance
- D. Very large resistance
Answer: A. Zero resistance
Explanation: Detailed standard solution.
Question 59
Topic: Electronics Engineering
For the network shown in the figure $V_{i}(t)$ is the input and $i(t)$ is the output. The transfer function $\frac{I(s)}{V_{i}(s)}$ of the network is
- A. $\frac{C}{RCs^{2}+LCs+1}$
- B. $\frac{C}{LCs^{2}+RCs+1}$
- C. $\frac{Cs}{RCs^{2}+LCs+1}$
- D. $\frac{Cs}{LCs^{2}+RCs+1}$
Answer: D. $\frac{Cs}{LCs^{2}+RCs+1}$
Explanation: Apply KVL to the circuit:
$V_{i}(t)=Ri(t)+L\frac{di(t)}{dt}+\frac{1}{C}\int i(t)dt$
Taking Laplace transform:
$V_{i}(s)=RI(s)+LSI(s)+\frac{1}{CS}I(s)$
$\frac{I(s)}{V_{i}(s)}=\frac{1}{R+Ls+\frac{1}{Cs}} = \frac{Cs}{LCs^{2}+RCs+1}$
Question 60
Topic: Electronics Engineering
The length of a lossy shorted stub is $1/32~\lambda$. What will be the nature of its input impedance?
- A. A low inductive reactance in series with a low resistance
- B. A low capacitive reactance in shunt with a high resistance
- C. A high capacitive reactance in series with a low resistance
- D. A high inductive reactance in series with a low resistance
Answer: A. A low inductive reactance in series with a low resistance
Explanation: Detailed standard solution.
Question 61
Topic: Electronics Engineering
Refer to the circuit in the figure. Which of the following statements is NOT true about this circuit?
- A. It is a low pass filter
- B. It has a finite attenuation at infinite frequency
- C. Its phase shift varies monotonically with the frequency
- D. It has one pole and one zero
Answer: C. Its phase shift varies monotonically with the frequency
Explanation: Detailed standard solution.
Question 62
Topic: Electronics Engineering
Which of the following is the correct continuity equation:
- A. $\nabla\cdot J=\frac{\partial p}{\partial t}$
- B. $\nabla\cdot J=-\frac{\partial\rho}{\partial t}$
- C. $\nabla\cdot J=0$
- D. $\nabla\cdot J=\frac{\partial E}{\partial t}$
Answer: B. $\nabla\cdot J=-\frac{\partial\rho}{\partial t}$
Explanation: Continuity equation is $\nabla\cdot J=\frac{-\partial\rho}{\partial t}$
Question 63
Topic: Electronics Engineering
$\alpha, \beta$ are the roots of the equation $x^{2}+x+1=0$ Then $\alpha^{3n}+\beta^{3n}$ is
Answer: B. 2
Explanation: $\alpha,\beta=\frac{-1\pm j\sqrt{3}}{2}$
So $\alpha$ and $\beta$ are cube roots of unity. $\alpha^{3}=\beta^{3}=1$
So, $(\alpha^{3})^{n}+(\beta^{3})^{n} = 1+1=2$
Question 64
Topic: Electronics Engineering
A point moving in the complex plane satisfies the following relation $z^{2}+z^{*^{2}}=8$ where $z^{*}$ stands for the complex conjugate of z. The difference of the distances of the moving point from $(2\sqrt{2},0)$ and $(-2\sqrt{2},0)$ is
Answer: C. 4
Explanation: Detailed standard solution.
Question 65
Topic: Electronics Engineering
The greatest negative number which can be stored in a computer that has 8-bit word length and uses 2's complement arithmetic is
- A. -256
- B. -255
- C. -128
- D. -127
Answer: C. -128
Explanation: Detailed standard solution.
Question 66
Topic: Electronics Engineering
Which of the following ADCs uses over sampling in its operation
- A. Sigma-delta ADC
- B. Counter ramp converter
- C. Successive Approximation Register ADC
- D. Flash Convertor
Answer: A. Sigma-delta ADC
Explanation: Detailed standard solution.
Question 67
Topic: Electronics Engineering
A mealy state machine's output depends on
- A. State and outputs
- B. Inputs
- C. State
- D. State and inputs
Answer: D. State and inputs
Explanation: Mealy type model: In this model, the output is a function of the present state and the present input.
Moore type model: In this model, the output is a function of the present state only.
Question 68
Topic: Electronics Engineering
The characteristic equation of the T-FF is given by
- A. $Q^{+}=\overline{T}Q$
- B. $Q^{+}=T\overline{Q}$
- C. $Q^{+}=TQ$
- D. $Q^{+}=T\overline{Q}+Q\overline{T}$
Answer: D. $Q^{+}=T\overline{Q}+Q\overline{T}$
Explanation: Detailed standard solution.
Question 69
Topic: Electronics Engineering
Percentage modulation of an AM wave having a power content of 8 KW at carrier frequency and 2 KW in each of its side band is
- A. 60%
- B. 70%
- C. 100%
- D. 80%
Answer: C. 100%
Explanation: $P_{c}=\frac{1}{2}A_{c}^{2} \Rightarrow 8\times1000=\frac{1}{2}A_{c}^{2} \Rightarrow A_{c}^{2}=16000$
$P_{USB}=P_{LSB}=\frac{1}{8}\mu^{2}A_{c}^{2} \Rightarrow 2\times1000=\frac{1}{8}\times\mu^{2}\times16000$
$\mu=1 \Rightarrow \mu \% = 100\%$
Question 70
Topic: Electronics Engineering
Approximate equivalent noise temperature (deg. K) of an amplifier with a noise factor of 1.04 is
- A. 301.6
- B. 11.6
- C. 278.4
- D. 5.8
Answer: B. 11.6
Explanation: $T_{e}=T(F-1)$
Where, $T_{e}=$ equivalent noise temperature, $T=$ room temperature (290°K), $F=$ noise factor
$T_{e}=290(1.04-1)=11.6$
Question 71
Topic: Electronics Engineering
A differential amplifier is used in the input stage of Op-Amps to achieve very high
- A. open-loop gain
- B. bandwidth
- C. slew rate
- D. CMRR
Answer: D. CMRR
Explanation: Detailed standard solution.
Question 72
Topic: Electronics Engineering
Which two-port parameters are best suited for analyzing a series-shunt feedback circuit?
- A. Z-Parameters
- B. h-Parameters
- C. Y-Parameters
- D. S-Parameters
Answer: B. h-Parameters
Explanation: Detailed standard solution.
Question 73
Topic: Electronics Engineering
For the circuit shown in figure, the switch is closed at $t=0$ After some time when the current in the inductor was 6 A, the rate of change of current through the inductor was 4 A/sec. The value of the inductor is
- A. 0.5 H
- B. 1.5 H
- C. 1.0 H
- D. Indeterminate
Answer: A. 0.5 H
Explanation: $V=Ri+L\frac{di}{dt}$
$20=3\times6+L\times4 \Rightarrow L=\frac{2}{4}=0.5~H$
Question 74
Topic: Electronics Engineering
In the circuit shown in the below figure, the value of output $V_{0}$ is
- A. +6 V
- B. -9 V
- C. -6 V
- D. +9 V
Answer: B. -9 V
Explanation: $V_{0}=-R_{f}(\frac{V_{a}}{R_{a}}+\frac{V_{b}}{R_{b}}+\frac{V_{c}}{R_{c}}) = -120(\frac{1}{40}+\frac{2}{80}+\frac{3}{120}) = -9V$
Question 75
Topic: Electronics Engineering
The characteristics equation of a closed loop control system is given as $s^{2}+4s+16=0.$ The resonant frequency in radian/sec of the system is
- A. 2
- B. $2\sqrt{3}$
- C. 4
- D. $2\sqrt{2}$
Answer: D. $2\sqrt{2}$
Explanation: $s^{2}+4s+16=0$
$\omega_{n}^{2} = 16 \Rightarrow \omega_{n} = 4$
$2\xi\omega_{n}=4 \Rightarrow 2\xi\times4=4 \Rightarrow \xi=0.5$
$\omega_{r}=\omega_{n}\sqrt{1-2\xi^{2}} = 4\sqrt{1-2(0.5)^{2}} = 4\sqrt{0.5} = 2\sqrt{2}$
Question 76
Topic: Electronics Engineering
Which point in the Smith Chart is labeled with the WRONG impedance?
N.B. - Reference $Z_{0}$ is 50 $\Omega$
- A. $10+j10$
- B. $50+j50$
- C. $200-j200$
- D. $25-j50$
Answer: D. $25-j50$
Explanation: On the same resistance circle two different resistance value can't lie. According to the data $50+j50$ point is right but $25-j50$ is wrong point on the circle.
Question 77
Topic: Electronics Engineering
If a line is terminated in an open circuit, the VSWR is
Answer: A. $\infty$
Explanation: Detailed standard solution.
Question 78
Topic: Electronics Engineering
Unit of relative permeability is
- A. Henry/meter
- B. Henry
- C. Henry-meter
- D. It is dimensionless
Answer: D. It is dimensionless
Explanation: Detailed standard solution.
Question 79
Topic: Electronics Engineering
$f(x)=[\tan^{2}x]$
([] stands for greatest integer function)
- A. $f(x)$ continuous at $x=0$
- B. limit $f(x)$ does not exist as x tend to 0
- C. $f^{\prime}(0)=1$
- D. $f(x)$ not derivable at $x=0$
Answer: D. $f(x)$ not derivable at $x=0$
Explanation: Detailed standard solution.
Question 80
Topic: Electronics Engineering
Number of real values of $(a+ib)^{1/n} + (a-ib)^{1/n}$ is
- A. 0
- B. 1
- C. n
- D. None of the above
Answer: B. 1
Explanation: Detailed standard solution.