2008|Electronics and Comm (GATE Exam)-Previous Question Paper Solution

Description: GATE Exam Previous Year Question Paper Solution Electronics and Communication (ECE) - 2008
Number of Questions: 85
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Tags: Matrices and Determinants Electronics and Communication Engineering - EC Analog Circuits Communications Differential Calculus Electronic Devices Vector Calculus
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The two numbers represented in signed 2s complement form are P + 11101101 and Q = 11100110. If Q is subtracted from P, the value obtained in signed 2s complement is

  1. 1000001111

  2. 00000111

  3. 11111001

  4. 111111001


Correct Option: B
Explanation:

In the following circuit, the comparators output is logic “1” if V1 > V2 and is logic “0” otherwise. The D / A conversion is done as per the relation VDAC = $\sum_{n=0}^3 2^{n-1}b_n$ Volts, where b3 (MSB), b1, b2 and b0 (LSB) are the counter outputs. The counter starts from the clear state.

The stable reading of the LED displays is

  1. 06

  2. 07

  3. 12

  4. 13


Correct Option: D
Explanation:

In the following circuit, the comparators output is logic “1” if V1 > V2 and is logic “0” otherwise. The D / A conversion is done as per the relation VDAC = $\sum_{n=0}^3 2^{n-1}b_n$ Volts, where b3 (MSB), b1, b2 and b0 (LSB) are the counter outputs. The counter starts from the clear state.

The magnitude of the error between VDAC and Vin at steady state (in volts) is

  1. 0.2

  2. 0.3

  3. 0.5

  4. 1.0


Correct Option: B
Explanation:

The $V_{ADC} - V_in$ at steady state is = 6.5 - 6.2 = 0.3V

The logic function implemented by the following circuit at the terminal OUT is

  1. P NOR Q

  2. P NAND Q

  3. P OR Q

  4. P AND Q


Correct Option: D
Explanation:

An 8085 executes the following instructions: 2710 LXI H, 30A0 H 2713 DAD H 2414 PCHL All address and constants are in Hex. Let PC be the contents of the program counter and HL be the contents of the HL register pair just after executing PCHL. Which of the following statements is correct?

  1. PC = 2715 H, HL = 30A0H

  2. PC = 30A0H, HL = 2715 H

  3. PC = 6140 H, HL = 6140 H

  4. PC = 6140 H, HL = 2715 H


Correct Option: C
Explanation:

For the circuit shown in the figure, D has a transition from 0 to 1 after CLK changes from 1 to 0. Assume gate delays to be negligible. Which of the following statements is true?

  1. Q goes to 1 at the CLK transition and stays at 1.

  2. Q goes to 0 at the CLK transition and stays 0.

  3. Q goes to 1 at the CLK tradition and goes to 0 when D goes to 1.

  4. Q goes to 0 at the CLK transition and goes to 1 when D goes to 1.


Correct Option: A
Explanation:

For each of the positive edge-triggered J - K flip flop used in the following figure, the propagation delay is $\Delta$t.

Which of the following wave forms correctly represents the output at Q1?


Correct Option: B
Explanation:

For the circuit shown in the following, I0 - I3 are inputs to the 4 : 1 multiplexers, R (MSB) and S are control bits. The output Z can be represented by

  1. PQ + P$\bar Q$S + $\overline{QRS}$

  2. P$\bar Q$ + PQ$\bar R$+ $\overline{PQS}$

  3. P$\overline{QR}$+ $\bar P$QR + PARS + $\overline{QRS}$

  4. PQ$\bar R$+ PQR$\bar S$+ P$\overline{QR}$S + $\overline{QRS}$


Correct Option: A
Explanation:

Which of the following Boolean Expressions correctly represents the relation between P, Q, R and M1?

  1. M1 = (P OR Q) XOR R

  2. M1 = (P AND Q) X OR R

  3. M1 = (P NOR Q) X OR R

  4. M1 = (P XOR Q) XOR R


Correct Option: D
Explanation:

A rectangular waveguide of internal dimensions (a = 4 cm and b = 3 cm) is to be operated in TE11 mode. The minimum operating frequency is

  1. 6.25 GHz

  2. 6.0 GHz

  3. 5.0 GHz

  4. 3.75 GHz


Correct Option: A
Explanation:

For a Hertz dipole antenna, the half power beam width (HPBW) in the E-plane is

  1. 360o

  2. 180o

  3. 90o

  4. 45o


Correct Option: C
Explanation:

The beam-width of Hertizian dipole is 180o and its half power beam-width is 90o.

One end of a loss-less transmission line having the characteristic impedance of 75 $\Omega$ and length of 1 cm is short-circuited. At 3 GHz, the input impedance at the other end of transmission line is

  1. 0

  2. Resistive

  3. Capacitive

  4. Inductive


Correct Option: D
Explanation:

Consider the following assertions:

S1 : For Zener effect to occur, a very abrupt junction is required. S2 : For quantum tunneling to occur, a very narrow energy barrier is required.

Which of the above assertions is/are correct?

  1. Only S2 is true.

  2. S1 and S2 are both true, but S2 is not a reason for S1.

  3. Only S1 is true.

  4. Both S1 and S2 are false.


Correct Option: A
Explanation:

 Options (1) is correct.

Which of the following options is true?

  1. A silicon wafer heavily doped with boron is a p+ substrate.

  2. A silicon wafer lightly doped with boron is a p+ substrate.

  3. A silicon wafer heavily doped with arsenic is a p+ substrate.

  4. A silicon wafer lightly doped with arsenic is a p+ substrate.


Correct Option: A
Explanation:

Trivalent impurities are used for making p- type semiconductors. So, Silicon wafer heavily doped with boron is a p+ substrate.

For static electric and magnetic fields in an inhomogeneous source-free medium, which of the following represents the correct form of Maxwell`s equations?

  1. $\nabla$. E = 0, $\nabla$ X B = 0

  2. $\nabla$. E = 0,$\nabla$. B = 0

  3. $\nabla$ X B = 0,$\nabla$ x B = 0

  4. $\nabla$ x E = 0,$\nabla$. B = 0


Correct Option: D
Explanation:

A uniform plane wave in the free space is normally incident on an infinitely thick dielectric slab (dielectric constant $\epsilon$= 9). The magnitude of the reflection coefficient is

  1. 0

  2. 0.3

  3. 0.5

  4. 0.8


Correct Option: C
Explanation:

The cross section of a JFET is shown in the following figure. Let Vc be − 2V and let VP be the initial pinch -off voltage. If the width W is doubled (with other geometrical parameters and doping levels remaining the same), then the ratio between the mutual trans conductances of the initial and the modified JFET is

  1. 4

  2. $\dfrac{1}{2} \left( \dfrac{1-\sqrt 2 \ V_p}{1-\sqrt {1/2} V_p} \right) $

  3. $ \left( \dfrac{1-\sqrt 2 \ V_p}{1-\sqrt {1/2} V_p} \right) $

  4. $ \left[ \dfrac{1- (2 - \sqrt V_p)} {1- (\sqrt {1/2} V_p) } \right] $


Correct Option: C
Explanation:

The drain current of MOSFET in saturation is given by ID = K (VGS-VT) where K is constant. The magnitude of the transconductance gm is

  1. $\dfrac{K(V_{GS}- V_T)^2}{V_{DS}}$

  2. 2K (VGS - VT)

  3. $\dfrac{I_d}{V_{GS} - V_{DS}}$

  4. $\dfrac{K (V_{GS} - V_T)^2 }{V_{GS}}$


Correct Option: B
Explanation:

$g_m = \dfrac{\partial I_P}{\partial V_{GS} } = \dfrac{\partial}{\partial V_{GS} } K (V_{GS} - V_T)^2 = 2K (V_{GS} - V_T)$

The measured trans conductance gm of an NMOS transistor operating in the linear region is plotted against the gate voltage VG at a constant drain voltage VD. Which of the following figures represents the expected dependence of gm on VG?


Correct Option: C
Explanation:

Consider the amplitude modulated (AM) signal Ac cos$\omega_0$t + 2cos$\omega_m$t cos$\omega_0$t. For demodulating the signal using envelope detector, the minimum value of Ac should be

  1. 2

  2. 1

  3. 0.5

  4. 0


Correct Option: A
Explanation:

The probability density function (pdf) of random variable is as shown below:


Correct Option: A
Explanation:

CDF is the integration of PDF. Plot in option (1) is the integration of plot given in question.

A memory less source emits n symbols each with a probability p. The entropy of the source as a function of n

  1. increases as log n

  2. decreases as log $\dfrac{1}{n}$

  3. increases as n

  4. increases as n log n


Correct Option: A
Explanation:

Noise with double-sided power spectral density on K over all frequencies is passed through a RC low pass filter with 3 dB cut-off frequency of fc. The noise power at the filter output is

  1. K

  2. Kfc

  3. k$\pi$fc

  4. $\infty$


Correct Option: C
Explanation:

Consider a binary symmetric channel (BSC) with probability of error being p. To transmit a bit, say 1, we transmit a sequence of three 1s. The receiver will represent 1 if at least two bits are 1. The probability that the transmitted bit will be received in error is

  1. p3 + 3p2 (1 - p)

  2. p3

  3. (1 - p3)

  4. p3 + p2 (1 - p)


Correct Option: D
Explanation:

Four messages band limited to W, W, 2W and 3 W respectively are to be multiplexed using Time Division Multiplexing (TDM). The minimum bandwidth required for transmission of this TDM signal is

  1. 2 W

  2. 3 W

  3. 6 W

  4. W


Correct Option: D
Explanation:

Consider the frequency modulated signal 10 cos [2$\pi$ x 105 t + 5 sin (2$\pi$ x 1500 t) + 7.5 sin (2$\pi$ x 1000 t)] with carrier frequency of 105 Hz. The modulation index is

  1. 12.5

  2. 10

  3. 7.5

  4. 5


Correct Option: B
Explanation:

The signal cos$\omega$c t + 0.5 cos$\omega_m$t sin$\omega_c$t + is

  1. FM only

  2. AM only

  3. AM and FM both

  4. None of these


Correct Option: C
Explanation:

null

A speed signal, band limited to 4 kHz and peak voltage varying between + 5 V and - 5 V, is sampled at the Nyquist rate. Each sample is quantized and represented by 8 bits.

If the bits 0 and 1 are transmitted using bipolar pulses, the minimum bandwidth required for distortion free transmission is

  1. 64 kHz

  2. 32 kHz

  3. 8 kHz

  4. 4 kHz


Correct Option: B
Explanation:

A speed signal, band limited to 4 kHz and peak voltage varying between + 5 V and - 5 V are sampled at the Nyquist rate. Each sample is quantized and represented by 8 bits. What is the number of quantization levels required to reduce the quantization noise by a factor of 4?

  1. 1024

  2. 512

  3. 256

  4. 64


Correct Option: B
Explanation:

A speed signal, band limited to 4 kHz and peak voltage varying between + 5 V and - 5 V, is sampled at the Nyquist rate. Each sample is quantised and represented by 8 bits.

Assuming the signal to be uniformly distributed between its peak to peak value, the signal to noise ratio at the quantiser output is

  1. 16 dB

  2. 32 dB

  3. 48 dB

  4. 4 dB


Correct Option: C
Explanation:

A discrete time linear shift - invariant system has an impulse response h [n] with h [0] = 1, h [1] = - 1, h [2] = 2 and zero otherwise. The system is given an input sequence x [n] with x [0] = x [2] = 1 and zero otherwise. The number of non zero samples in the output sequence y [n] and the value of y [2] are respectively

  1. 5, 2

  2. 6, 2

  3. 6, 1

  4. 5, 3


Correct Option: D
Explanation:

The impulse response h (t) of a linear time invariant continuous time system is described by h (t) = exp ($\alpha$t) u (t) + exp ($\beta$t) u (- t) where u (- t) denotes the unit step function, and $\alpha$ and $\beta$ are real constants. The system is stable if

  1. $\alpha$ is positive and $\beta$ is positive

  2. $\alpha$ is negative and $\beta$ is negative

  3. $\alpha$ is positive and $\beta$ is negative

  4. $\alpha$ is negative and $\beta$ is positive


Correct Option: D
Explanation:

In the following network, the switch is closed at t = 0- and the sampling starts from t = 0. The sampling frequency is 10 Hz.

The samples x (n), n = (0, 1, 2....) are given by

  1. 5 (1 - $e^{-0.05n}$)

  2. 5$e^{-0.05n}$

  3. 5 (1 - $e^{-\delta n}$)

  4. 5$e^{-\delta n}$


Correct Option: B
Explanation:

In the following network, the switch is closed at t = 0- and the sampling starts from t = 0. The sampling frequency is 10 Hz.

The expression and the region of convergence of the z −transform of the sampled signal are

  1. $\dfrac{5z}{z-e^5}, \left| z \right| \lt e^{-5}$

  2. $\dfrac{5z}{z-e^{0.05}}, \left| z \right| \lt e^{-0.05}$

  3. $\dfrac{5z}{z-e^{-0.05}}, \left| z \right| \lt e^{-0.05}$

  4. $\dfrac{5z}{z-e^5}, \left| z \right| \lt e^{-5}$


Correct Option: C
Explanation:

The impulse response h (t) of linear time - invariant continuous time system is given by h (t) = exp (- 2t) u (t) , where u (t) denotes the unit step function.

The output of this system, to the sinusoidal input x (t) = 2 cos 2t for all time t, is

  1. 0

  2. $2^{-0.25} cos(2t - 0.125\pi)$

  3. $2^{-0.5} cos(2t - 0.125\pi)$

  4. $2^{-0.5} cos(2t - 0.25\pi)$


Correct Option: D
Explanation:

The input and output of a continuous time system are respectively denoted by x (t) and y (t). Which of the following descriptions corresponds to a casual system?

  1. y (t) = x (t - 2) + x (t + 4)

  2. y (t) = (t - 4) x (t + 1)

  3. y (t) = (t + 4) x (t - 1)

  4. y (t) = (t + 5) x (t + 5)


Correct Option: C
Explanation:

The impulse response h (t) of linear time - invariant continuous time system is given by h (t) = exp (- 2t) u (t) , where u (t) denotes the unit step function.

The frequency response H $(\omega)$of this system in terms of angular frequency$(\omega)$, is given by H $(\omega)$ =

  1. $\dfrac{1}{1+j2\omega}$

  2. $\dfrac{sin\omega}{\omega}$

  3. $\dfrac{1}{2+j\omega}$

  4. $\dfrac{j\omega}{2+j\omega}$


Correct Option: C
Explanation:

A linear, time - invariant, causal continuous time system has a rational transfer function with simple poles at s = - 2 and s = - 4 and one simple zero at s = - 1. A unit step u (t) is applied at the input of the system. At steady state, the output has constant value of 1. The impulse response of this system is

  1. [exp (- 2t) + exp (- 4t)] u (t)

  2. [- 4 exp (- 2t) - 12 exp (- 4t) - exp (- t)] u (t)

  3. [- 4 exp (- 2t) + 12 exp (- 4t) u (t)

  4. [- 0.5 exp (- 2t) + 1.5 exp (- 4t)] u (t)


Correct Option: C
Explanation:

{x (n)} is a real - valued periodic sequence with a period N. x (n) and X (k) form N - point Discrete Fourier Transform (DFT) pairs. The DFT Y (k) of the sequence y (n) = $\dfrac{1}{n} \displaystyle \sum_{r = 0} ^{N-1} x (r) \times (n+r)$is

  1. |X (k)|2

  2. $\dfrac{1}{N} \displaystyle \sum_{r = 0} ^{N-1} x (r) \times (k+r)$

  3. $\dfrac{1}{N} \displaystyle \sum_{r = 0} ^{N-1} x (r) \times (k+r)$

  4. 0


Correct Option: A
Explanation:

Let (x) t be the input and (y) t be the output of a continuous time system. Match the system properties P1, P2 and P3 with system relations R1, R2, R3, R ||| |---|---| | Properties| Relations| | P1 : Linear but NOT time - invariant| R1 : y (t) = t2 x (t)| | P2 : Time - invariant but NOT linear| R2 : y (t) = |t| x (t)| | P3 : Linear and time - invariant| R3 : y (t) = |x (t)|| | | R4 : y (t) = x (t - 5)|

  1. (P1, R1), (P2, R3), (P3, R4)

  2. (P1, R2), (P2, R3), (P3, R4)

  3. (P1, R3), (P2, R1), (P3, R2)

  4. (P1, R1), (P2, R2), (P3, R3)


Correct Option: B
Explanation:

The signal x (t) is described by x (t) =$ \begin{cases} 1 & for \ -1 \le t \le + 1 \\ 0 & otherwise \end{cases}

$ Two of the angular frequencies at which its Fourier transform becomes zero are

  1. $\pi$, 2$\pi$

  2. 0.5$\pi$, 1.5$\pi$

  3. 0, $\pi$

  4. 2$\pi$, 2.5$\pi$


Correct Option: A
Explanation:

In the following graph, the number of trees (P) and the number of cut-set (Q) are

  1. P = 2 Q = 2

  2. P = 2 Q = 6

  3. P = 4 Q = 6

  4. P = 4 Q = 10


Correct Option: C
Explanation:

The following series if RLC circuit with zero conditions is excited by a unit impulse. For t > 0, the output voltage vc (t) is

  1. $\dfrac{2}{\sqrt 3} \left( e^{\dfrac{-t}{2}t} - e^{\dfrac{\sqrt 3}{2}t} \right) $ V

  2. $\dfrac{2}{\sqrt 3} te^{\dfrac{-1}{2}t}$ V

  3. $\dfrac{2}{\sqrt 3} e^{\dfrac{-t}{2}t} cos \left( \dfrac{\sqrt 3}{2}t \right) $ V

  4. $\dfrac{2}{\sqrt 3} e^{\dfrac{-t}{2}t} sin \left( \dfrac{\sqrt 3}{2}t \right) $ V


Correct Option: D
Explanation:

The following series if RLC circuit with zero conditions is excited by a unit impulse. For t > 0, the voltage across the resistor is

  1. $\dfrac{1}{\sqrt 3} \left( e^{\dfrac{\sqrt 3}{2}} - e^{-\dfrac{1}{2}t} \right) $ V

  2. $e^{\dfrac{1}{2}t} \left[ cos(\dfrac{\sqrt 3}{2}t) - \dfrac{1}{\sqrt 3}sin(\dfrac{\sqrt 3 t}{2}) \right] $ V

  3. $\dfrac{2}{\sqrt 3} e^{\dfrac{-1}{2}} sin \left( \dfrac{\sqrt 3}{2}t \right) $ V

  4. $\dfrac{2}{\sqrt 3} e^{\dfrac{-1}{2}} cos \left( \dfrac{\sqrt 3}{2}t \right) $ V


Correct Option: B
Explanation:

A two-port network shown below is excited by external DC source. The voltage and the current are measured with voltmeters V1, V2 and ammeters A1, A2 (all assumed to be ideal), as indicated

Under following conditions, the readings obtained are: (i) S1 - open, S2 - closed A1 = 0, V1 = 4.5 V, V2 = 1.5 V, A2 = 1 A (ii) S1 - open, S2 - closed A1 = 4 A, V1 = 6 V, V2 = 6 V, A2 = 0

The z-parameter matrix for this network is

  1. $\left[ \begin{array} \ 1.5 & 1.5 \\ 4.5 & 1.5 \end{array} \right] $

  2. $\left[ \begin{array} \ 1.5 & 4.5 \\ 1.5 & 4.5 \end{array} \right] $

  3. $\left[ \begin{array} \ 1.5 & 4.5 \\ 1.5 & 1.5 \end{array} \right] $

  4. $\left[ \begin{array} \ 4.5 & 1.5 \\ 1.5 & 4.5 \end{array} \right] $


Correct Option: C
Explanation:

A two-port network shown below is excited by external DC source. The voltage and the current are measured with voltmeters V1, V2 and ammeters A1, A2 (all assumed to be ideal), as indicated

Under following conditions, the readings obtained are: (i) S1 - open, S2 - closed A1 = 0, V1 = 4.5 V, V2 = 1.5 V, A2 = 1 A (ii) S1 - open, S2 - closed A1 = 4 A, V1 = 6 V, V2 = 6 V, A2 = 0

The h-parameter matrix for this network is

  1. $\left[ \begin{array} \ -3 & 3 \\ -1 & 0.67 \end{array} \right] $

  2. $\left[ \begin{array} \ -3 & -1 \\ 3 & 0.67 \end{array} \right] $

  3. $\left[ \begin{array} \ 3 & 3 \\ 1 & 0.67 \end{array} \right] $

  4. $\left[ \begin{array} \ 3 & 1 \\ -3 & -0.67 \end{array} \right] $


Correct Option: A
Explanation:

The driving point impedance of the following network is given by Z (s) = $\dfrac{0.2s}{s^2 + 0.1s + 2}$

The component values are

  1. L = 5 H, R = 0.5 $\Omega$, C = 0.1 F

  2. L = 0.1 H, R = 0.5 $\Omega$, C = 5 F

  3. L = 5 H, R = 2 $\Omega$, C = 0.1 F

  4. L = 0.1 H, R = 2 $\Omega$, C = 5 F


Correct Option: D
Explanation:

In the following circuit, switch S is closed at t = 0. The rate of change of current $\dfrac{di}{dt} (0^+)$ is given by

  1. 0

  2. $\dfrac{R_s I_s}{L}$

  3. $\dfrac{(R+R_s) I_s}{L}$

  4. $\infty$


Correct Option: B
Explanation:

The Thevenin equivalent impedance Zth between the nodes P and Q in the following circuit is

  1. 1

  2. 1 + s + $\dfrac{1}{s}$

  3. 2 + s + $\dfrac{1}{s}$

  4. $\dfrac{s^2 + s + 1}{s^2 + 2s +1}$


Correct Option: A
Explanation:

The circuit shown in the figure is used to charge the capacitor C alternately from two current sources as indicated. The switches S1 and S2 are mechanically coupled and connected as follows: For 2nT $\le$ t $\le$ (2n + 1) T, (n = 0, 1, 2, ...) S1 to P1 and S2 to P2 For (2n + 1) T $\le$ t $\le$ (2n + 2) T, (n = 0, 1, 2, ...) S1 to Q1 and S2 to Q2

Assume that the capacitor has zero initial charge. Given that u (t) is a unit step function, the voltage vc (t) across the capacitor is given by

  1. $\displaystyle \sum_{n=1}^\infty (-1)^n \ tu \ (t - nT)$

  2. u (t) + 2$\displaystyle \sum_{n=1}^\infty (-1)^n \ u \ (t - nT)$

  3. t u (t) + 2$\displaystyle \sum_{n=1}^\infty (-1)^n \ u \ (t - nT)$(t - nT)

  4. $\displaystyle \sum_{n=1}^\infty [0.5 - e^{-(t-2nT)} + 0.5 e^{-(t-2nT)} ]$


Correct Option: C
Explanation:

How many solutions does the equation sin (z) = 10 have?

  1. No real or complex solution

  2. Exactly two distinct complex solutions

  3. A unique solution

  4. An infinite number of complex solutions


Correct Option: A
Explanation:

Sin (z) can have value between - 1 to + 1. Thus, no solution.

For real values of x, the minimum value of the function f (x) = exp (x) + exp (- x) is

  1. 2

  2. 1

  3. 0.5

  4. 0


Correct Option: A
Explanation:

Which of the following functions would have only odd powers of x in its Taylor series expansion about the point x = 0?

  1. sin x3

  2. sin x2

  3. cos x3

  4. cos x2


Correct Option: A
Explanation:

$sin x = x + \frac{x^3}{3!}+\frac{x^5}{5!}+ ... \\ cos x = 1 + \frac{x^2}{2!}+\frac{x^4}{4!}+ ... \\ \text{Thus only $(x^3)$ will have odd power of x.}$

The pole-zero given below corresponds to a

  1. Law pass filter

  2. High pass filter

  3. Band filter

  4. Notch filter


Correct Option: C
Explanation:

Which of the following is a solution to the differential equation$\frac{dx(t)}{dt} + 3x (t)$= 0?

  1. x (t) = 3et

  2. x (t) = 2e-3t

  3. x (t) = -$\frac{3}{2} t^2$

  4. x (t) = 3t2


Correct Option: B
Explanation:

$\text{We have} \hspace{2cm} \frac{dx(t)}{dt} + 3x(t) = 0 \\ or \hspace{3.3cm} (D + 3)x(t) = 0 \\ Since m = -3, \hspace{1cm} x(t) = C e^{-3t}$

All the four entries of the 2 x 2 matrix = $\begin{bmatrix} \ p11 & p12 \ \ p21 & p22 \ \end{bmatrix}$ matrix are non-zero, and one of its eigen value is zero. Which of the following statements is true?

  1. p11p12 - p12p21 = 1

  2. p11p22 - p12p21 = -1

  3. p11p22 - p12p21 = 0

  4. p11p22 + p12p21 = 0


Correct Option: C
Explanation:

The product of Eigen value is equal to the determinant of the matrix. Since one of the Eigen value is zero, the product of Eigen value is zero, thus determinant of the matrix is zero. Thus p11 p22 - p12 p21 = 0

Which of the following is NOT associated with a p - n junction?

  1. Junction Capacitance

  2. Charge Storage Capacitance

  3. Depletion Capacitance

  4. Channel Length Modulations


Correct Option: D
Explanation:

Channel length modulation is not associated with p - n junction. It is being associated with MOSFET in which effective channel length decreases, producing the phenomenon called channel length modulation.

Step responses of a set of three second-order underdamped systems all have the same percentage overshoot. Which of the following diagrams represents the poles of the three systems?


Correct Option: C
Explanation:

Transfer function for the given pole zero plot is: $$ \dfrac{(s+Z_1)(s+Z_2)}{(s+P_1)(s+P_2)} $$ From the plot $R_e \ (P_1 \ and \ P_2) > (Z_1 \ and Z_2)$

So, these are two lead compensator. Hence both high pass filters and the system is high pass filter.

A silicon wafer has 100 nm of oxide on it and is furnace at a temperature above 10000 C for further oxidation in dry oxygen. The oxidation rate

  1. is independent of current oxide thickness and temperature

  2. is independent of current oxide thickness but depends on temperature

  3. slows down as the oxide grows

  4. is zero as the existing oxide prevents further oxidation


Correct Option: D
Explanation:

Oxidation rate is zero because the existing oxide prevents the further oxidation.

In the following limiter circuit, an input voltage Vi = 10 sin 100 $\pi$t is applied. Assume that the diode drop is 0.7 V when it is forward biased. The zener breakdown voltage is 6.8 V. The maximum and minimum values of the output voltage respectively are

  1. 6.1 V, - 0.7 V

  2. 0.7 V, - 7.5 V

  3. 7.5 V, - 0.7 V

  4. 7.5 V, - 7.5 V


Correct Option: C
Explanation:

The residue of the function f (z) =$\frac{1}{(z + 2)^2(z - 2)^2}$ at z = 2 is

    • $\frac{1}{32}$
    • $\frac{1}{16}$
  1. $\frac{1}{16}$

  2. $\frac{1}{32}$


Correct Option: A
Explanation:

Px (x) = M exp (- 2 |x|) - Nexp (- 3 |x|) is the probability density function for the real random variable X, over the entire x axis, M and N are both positive real numbers. The equation relating M and N is

  1. M - $\frac{2}{3}$N = 1

  2. 2M + $\frac{1}{3}$N = 1

  3. M + N = 1

  4. M + N = 3


Correct Option: A

The recursion relative to solve x = e-x using Newton-Raphson method is

  1. xn+1 = $e^{-X_n}$

  2. xn+1 = xn -$e^{-X_n}$

  3. xn+1 = (1 + xn) $\frac{e^{-X_n}}{1 + e^{-X_n}}$

  4. xn+1 = $\frac{x_n^2 - e^{-X_n} (1 - X_n) - 1}{X_n - e ^{-X_n}}$


Correct Option: C
Explanation:

Group I lists a set of four transfer functions. Group II gives a list of possible step response y (t). Match the step responses with the corresponding transfer functions.

  1. P - 3, Q - 1, R - 4, S - 2

  2. P - 3, Q - 2, R - 4, S - 1

  3. P - 2, Q - 1, R - 4, S - 2

  4. P - 3, Q - 4, R - 1, S - 2


Correct Option: D
Explanation:

The value of the integral of the function g(x, y) = 4x3 + 10y4 along a straight line segment from the point (0, 0) to the point (1, 2) in the x-y plane is

  1. 33

  2. 35

  3. 40

  4. 56


Correct Option: A
Explanation:

Consider the matrix P = $\begin{bmatrix} \ 0 & 1 \ \ -2 & -3 \ \end{bmatrix}$. The value of ep is

  1. $\begin{bmatrix} \ 2e^{-2}-3e^{-1} & e^{-1}-e^{-2} \ \ 2e^{-2}-2e^{-1} & 5e^{-2}-e^-{1} \ \end{bmatrix}$

  2. $\begin{bmatrix} \ e^{-1}+e^{-1} & 2e^{-2}-e^{-1} \ \ 2e^{-1}-4e^{2} & 3e^{-1}+2e^{-2} \ \end{bmatrix}$

  3. $\begin{bmatrix} \ 5e^{-2}-e^{-1} & 3e^{-1}-e^{-2} \ \ 2e^{-2}-6e^{-1} & 4e^{-2}-6^{-1} \ \end{bmatrix}$

  4. $\begin{bmatrix} \ 2e^{-1}-e^{-2} & e^{-1}-e^{-2} \ \ -2e^{-1}+2e^{-2} & -e^{-1}+2e^{-2} \ \end{bmatrix}$


Correct Option: D
Explanation:

Consider points P and Q in the x - y plane, with P = (1, 0) and Q (0, 1). The line integral 2$\int\limits_P^o (x dx + ydy)$along the semicircle with the line segment PQ as its diameter

  1. is - 1

  2. is 0

  3. is 1

  4. depends on the direction (clockwise or anti - clockwise) of the semicircle


Correct Option: B
Explanation:

$I = 2 \int\limits_P^Q (xdx + ydy) \\ = 2 \int\limits_P^Q xdx + 2 \int\limits_P^Q ydy \\ = 2 \int\limits_1^0 xdx + 2 \int\limits_1^0 ydy = 0$

In the Taylor series expansion of exp (x) + sin (x) about the point x = $\pi$, the coefficient of (x - $\pi$)2 is

  1. exp ($\pi$)

  2. 0.5 exp ($\pi$)

  3. exp ($\pi$) + 1

  4. exp ($\pi$) - 1


Correct Option: B
Explanation:

The number of open right half plane of G (s) = $ \dfrac{10}{s^5+2s^4+3s^3+6s^2+5s+3} $

  1. 0

  2. 1

  3. 2

  4. 3


Correct Option: C
Explanation:

A signal flow graph of a system is given below:

The set of equalities that corresponds to this signal flow graph is


Correct Option: C
Explanation:

Group I gives two possible choices for the impedance Z in the diagram. The circuit elements in Z satisfy the conditions R2C2 > R1C1. The transfer functions $\dfrac{V_0}{V_i}$represents a kind of controller.

Match the impedances in Group I with the type of controllers in Group II.

  1. Q - 1, R - 2

  2. Q - 1, R - 3

  3. Q - 2, R - 3

  4. Q - 3, R - 2


Correct Option: B
Explanation:

A certain system has transfer function G (s) = $\dfrac{s+8}{s^2+\alpha s-4}$ where$\alpha$ is a parameter. Consider the standard negative unity feedback configuration as shown below:

Which of the following statements is true?

  1. The closed loop systems is never stable for any value of $\alpha$.

  2. For some positive value of $\alpha$, the closed loop system is stable, but not for all positive values.

  3. For all positive values of $\alpha$, the closed loop system is stable.

  4. The closed loop system is stable for all values of $\alpha$, both positive and negative.


Correct Option: C
Explanation:

The magnitude of frequency responses of an underdamped second order system is 5 at 0 rad/sec and peaks to $\dfrac{10}{\sqrt3}$ at 5 $\sqrt2$ rad/sec. The transfer function of the system is

  1. $ \dfrac{500}{(s^2+10s+100)} $

  2. $ \dfrac{375}{(s^2+5s+75)} $

  3. $ \dfrac{720}{(s^2+12s+144)} $

  4. $ \dfrac{1125}{(s^2+25s+225)} $


Correct Option: A
Explanation:

An astable multivibrator circuit using IC 555 timer is shown below. Assume that the circuit is oscillating steadily.

The voltage Vc across the capacitor varies between

  1. 3 V to 5 V

  2. 3 V to 6 V

  3. 3.6 V to 6 V

  4. 3.6 V to 5 V


Correct Option: A

Silicon is doped with boron to a concentration of 4 x 1017 atoms cm3. Assume the intrinsic carrier concentration of silicon to be 1.5 x 1010 cm3 and the value of kT/q to be 25 mV at 300 K. Compared to undopped silicon, the fermi level of doped silicon

  1. goes down by 0.31 eV

  2. goes up by 0.13 eV

  3. goes down by 0.427 eV

  4. goes up by 0.427 eV


Correct Option: C
Explanation:

Two identical NMOS transistors M1 and M2 are connected as shown below. V bias is chosen so that both transistors are in saturation. The equivalent gm of the pair is defined to be $\frac{\partial I_{out}}{\partial V_i}$at constant Vout The equivalent gm of the pair is

  1. the sum of individual gm's of the transistors

  2. the product of individual gm's of the transistors

  3. nearly equal to the gm of M1

  4. nearly equal to $\frac{g_m}{g_o}$of M2


Correct Option: C
Explanation:

The current in both transistor are equal. Thus $g_m$ is decide by $M_1.$

For the circuit shown in the following figure, transistor M1 and M2 are identical NMOS transistors. Assume the M2 is in saturation and the output is unloaded.

  1. Ix = Ibias + Is

  2. Ix = Ibias

  3. Ix = Ibias - $\bigg(V_{DD} - \frac{V_{out}}{R_E}\bigg)$

  4. Ix = Ibias - Is


Correct Option: B
Explanation:

$\text{By current mirror} \\ I_x = \frac{\Big(\frac{W}{L}\Big)2}{\Big(\frac{W}{L}\Big)_1}I{bias} \\ \text{Since MOSFETs are identical,} \\ \text{Thus} \hspace{1cm} \Big(\frac{W}{L}\Big)2 = \Big(\frac{W}{L}\Big)_2 \\ \text{Hence} \hspace{1cm} I_x = I{bias}$

Consider the Schmidt trigger circuit shown below: A triangular wave which goes from - 12 V to 12 V is applied to the inverting input of OPMAP. Assume that the output of the OPAMP swings from + 15 V to - 15 V. The voltage at the non-inverting input switches between

  1. − 12 V to + 12 V

    • 7.5 V to 7.5 V
    • 5 V to + 5 V
  2. 0 V and 5 V


Correct Option: C
Explanation:

The OPAMP circuit shown above represents a

  1. high pass filter

  2. low pass filter

  3. band pass filter

  4. band reject filter


Correct Option: B
Explanation:

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

  1. radius as well as operating wavelength are halved

  2. radius as well as operating wavelength are doubled

  3. radius is halved and operating wavelength is doubled

  4. radius is doubled and operating wavelength is halved


Correct Option: C
Explanation:

Consider the following circuit using an ideal OPAMP. The I - V characteristic of the diode is described by the relation I = I0$e^{\bigg(\frac{V}{V_1}1\bigg)}$where VT = 25 mV, I0 = 1$\mu$A and V is the voltage across the diode (taken as positive for forward bias). For an input voltage Vi = - 1 V, the output voltage V0 is

  1. 0 V

  2. 0.1 V

  3. 0.7 V

  4. 1.1 V


Correct Option: B
Explanation:

In the following transistor circuit, VBE = 0.7 V, r3 = 25 mV / IE and $\beta$ and all the capacitances are very large.

The value of DC current IE is

  1. 1 mA

  2. 2 mA

  3. 5 mA

  4. 10 mA


Correct Option: A
Explanation:

In the following transistor circuit, VBE = 0.7 V, r3 = 25 mV / IE, and $\beta$ and all the capacitances are very large

The mid-band voltage gain of the amplifier is approximately

    • 180
    • 120
    • 90
    • 60

Correct Option: D
Explanation:

At 20 GHz, the gain of a parabolic dish antenna of 1 metre and 70% efficiency is

  1. 15 dB

  2. 25 dB

  3. 35 dB

  4. 45 dB


Correct Option: D
Explanation:

$\lambda = \frac{c}{f} = \frac{3 \times 10^8}{20 \times 10^9} = \frac{3}{200} \\ \text{Gain} \hspace{1cm} G_P = \eta \pi ^2(\frac{D}{\lambda})^2 = 0.7 \times \pi^2 \bigg(\frac{1}{\frac{3}{100}}\bigg)^2 = 30705.4 \\ = 44.87 dB$

The system of linear equations 4x + 2y = 7 and 2x + y = 6 has

  1. a unique solution

  2. no solution

  3. an infinite number of solutions

  4. exactly two distinct solutions


Correct Option: B
Explanation:

$\text{The given system is} \\ \begin{bmatrix} \ 4 & 2 \ \ 2 & 1 \ \end{bmatrix} \begin{bmatrix} \ x \ \ y \ \end{bmatrix} = \begin{bmatrix} \ 7 \ \ 6 \ \end{bmatrix} \\ \text{We have} \hspace{1cm} A = \begin{bmatrix} \ 4 & 2 \ \ 2 & 1 \ \end{bmatrix} \\ \text{and} \hspace{1cm} |A| = \begin{vmatrix} \ 4 & 2 \ \ 2 & 1 \ \end{vmatrix} = 0 \hspace{1cm} \text{Rank of matrix $\rho(A) < 2$} \\ \text{Now} \hspace{1cm} C = \begin{vmatrix} \ 4 & 2 & | & 7\ \ 2 & 1 & | & 6\ \end{vmatrix} \hspace{1cm} \text{Rank of matrix $\rho(C) = 2$} \\ \text{Since $\rho{A} \neq \rho (C)$ there is no solution.}$

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