2007|Electronics and Comm (GATE Exam)-Previous Question Paper Solution

Description: GATE Exam Previous Year Question Paper Solution Electronics and Communication (ECE) - 2007
Number of Questions: 84
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Tags: Communications Electronics and Communication Engineering - EC Electronic Devices Analog Circuits Differential Calculus Matrices and Determinants Vector Calculus
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The Boolean function Y = AB + CD is to be realized using only 2-input NAND gates. The minimum number of gates required is

  1. 2

  2. 3

  3. 4

  4. 5


Correct Option: B
Explanation:

Y = AB + CD = $\overline{AB}.\overline{CD}$

This is SOP form and we require only 3 NAND gates

An 8085 assembly language program is given below. Line 1: MVI A, B5H 2: MVI B, 0EH 3: XRI 69H 4: ADD B 5: ANI 9BH 6: CPI 9FH 7: STA 3010H 8: HLT

The contents of the accumulator just after execution of the ADD instruction in line 4 will be

  1. C3H

  2. EAH

  3. DCH

  4. 69H


Correct Option: B
Explanation:

An 8085 assembly language program is given below. Line 1: MVI A, B5H 2: MVI B, 0EH 3: XRI 69H 4: ADD B 5: ANI 9BH 6: CPI 9FH 7: STA 3010H 8: HLT

After execution of line 7 of the program, the status of the CY and Z flags will be

  1. CY = 0, Z = 0

  2. CY = 0, Z = 1

  3. CY = 1, Z = 0

  4. CY = 1, Z = 1


Correct Option: C
Explanation:

The CY = 1 and Z = 0

The circuit diagram of a standard TTL NOT gate is shown in the figure. When Vi = 2.5 V, the modes of operation of the transistors will be

  1. Q1 : reverse active; Q2 : normal active; Q3 : saturation; Q4 : cut-off

  2. Q1 : reverse active; Q2 : saturation; Q3 : saturation; Q4 : cut-off

  3. Q1 : normal active; Q2 : cut-off; Q3 : cut-off; Q4 : saturation

  4. Q1 : saturation; Q2 : saturation;Q3 : saturation; Q4 : normal active


Correct Option: B
Explanation:

Two 5-bit binary numbers, i.e. X = 01110 and Y = 11001 are represented in two’s complement format. The sum of X and Y represented in two’s complement format using 6 bits is

  1. 100111

  2. 001000

  3. 000111

  4. 101001


Correct Option: C
Explanation:

The boolean expression Y = $\bar A \bar B \bar C D + \bar ABC\bar D + A\bar B \bar C D + AB\bar C \bar D$ can be minimized to

  1. Y = $\bar A \bar B \bar C D + \bar A B \bar C + A \bar C D$

  2. Y = $\bar A \bar B \bar C D + BC \bar D + A \bar B \bar C D$

  3. Y = $\bar A BC \bar D + \bar B \bar C D + A\bar B \bar C D$

  4. Y = $\bar A BC \bar D + \bar B \bar C D + AB \bar C \bar D$


Correct Option: D
Explanation:

In the digital -to- Analog converter circuit shown in the figure below, VR = 10 V and R = 10 k$\Omega$.

The Voltage V0 is

    • 0.781 V
    • 1.562 V
    • 3.125 V
    • 6.250 V

Correct Option: C
Explanation:

The net current in inverting terminal of OP - amp is $$L = \dfrac{1}{4}+ \dfrac{1}{16} = \dfrac{5I}{16}$$

So that $V_0 = -R \times \dfrac{5I}{16} = -3.125$

An 8255 chip is interfaced to an 8085 microprocessor system as an I/O mapped I/O as shown in the figure. The address lines A0 and A1 of the 8085 are used by the 8255 chip to decode internally its three ports and the Control register. The address lines A3 to A7 as well as the IO$\sqrt M$signal are used for address decoding. The range of addresses for which the 8255 chip would get selected is

  1. F8H - FBH

  2. F8H - FCH

  3. F8H - FFH

  4. F0H - F7H


Correct Option: C
Explanation:

The following binary values were applied to the X and Y inputs of the NAND latch shown in the figure in the sequence indicated below: X = 0, Y = 1; X = 0, Y = 0; X = 1, Y = 1. The corresponding stable P, Q outputs will be

  1. P = 1, Q = 0; P = 1, Q = 0; P = 1, Q = 0 or P = 0, Q = 1

  2. P = 1, Q = 0; P = 0, Q = 1; or P = 0, Q = 1; P = 0, Q = 1

  3. P = 1, Q = 0; P = 1, Q = 1; P = 1, Q = 0 or P = 0, Q = 1

  4. P = 1, Q = 0; P = 1, Q = 1; P = 1, Q = 1


Correct Option: C
Explanation:

For the circuit shown below, the counter state (Q1, Q0) follows the sequence

  1. 00, 01, 10, 11, 00 …

  2. 00, 01, 10, 00, 01 …

  3. 00, 01, 11, 00, 01 …

  4. 00, 10, 11, 00, 10 …


Correct Option: A
Explanation:

In the following circuit, X is given by

  1. X = $A\bar B \bar C+ \bar A B \bar C + \bar A \bar B C + ABC$

  2. X = $\bar A BC + A\bar B C + AB\bar C + \bar A \bar B \bar C$

  3. X = $AB + BC+ AC$

  4. X = $\bar A \bar B + \bar B \bar C + \bar B \bar C + \bar A \bar C$


Correct Option: A
Explanation:

The $\vec E$ field in a rectangular waveguide of inner dimensions a$\times$b is given by $\vec E$= $\dfrac{\omega \mu}{h^2} \left( \dfrac{\pi}{a} \right)$H0 sin $\left( \dfrac{2\pi x}{a} \right)^2$sin ($\omega t - \beta z$$\widehat y$
Where H0 is a constant, and a and b are the dimensions along the x-axis and the y-axis respectively. The mode of propagation in the waveguide is

  1. TE20

  2. TM11

  3. TM20

  4. TE10


Correct Option: A
Explanation:

A load of 50$\Omega$is connected in shunt in a 2-wire transmission line of Z0 = 50 $\Omega$as shown in the figure. The 2-port scattering parameter matrix (S-matrix) of the shunt element is

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

  2. $ \left[ \begin{array} \ 0 & 0 \\ 1 & 1 \end{array} \right] $

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

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


Correct Option: C
Explanation:

In a p+n junction diode under reverse bias, the magnitude of electric field is maximum at

  1. the edge of the depletion region on the p-side

  2. the edge of the depletion region on the n-side

  3. the p+n junction

  4. the centre of the depletion region on the n-side


Correct Option: C
Explanation:

The electric field has the maximum value at the junction of $p^+ n$.

Group I lists four types of p-n junction diodes. Match each device in Group I with one of the option in Group II to indicate the bias condition of that device in its normal mode of operation.||| |---|---| | Group I| Group II| | (P) Zener Diode| (1) Forward bias| | (Q) Solar cell| (2) Reverse bias| | (R) LASER diode| | | (S) Avalanche Photodiode| |

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

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

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

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


Correct Option: B
Explanation:

The parallel branches of a 2-wire transmission line are terminated in 100 $\Omega$ and 200 $\Omega$ resistors as shown in the figure. The characteristic impedance of the line is Z0 = 50 $\Omega$ and each section has a length of $\dfrac{\lambda}{4}$. The voltage reflection coefficient at the input is

    • j $\dfrac{7}{5}$
  1. $\dfrac{-5}{7}$

  2. j$\dfrac{5}{7}$

  3. $\dfrac{5}{7}$


Correct Option: D
Explanation:

A plane wave of wavelength $\lambda$ is travelling in a direction making an angle 30° with positive x-axis and 90° with positive y-axis. The $\vec E$ field of the plane wave can be represented as (E0 is constant)

  1. $\vec E$ = $\widehat Y$ E0 ej $\left( \omega t + \dfrac{\pi}{\lambda}x \times \dfrac{\sqrt 3 \pi}{\gamma} z \right)$

  2. $\vec E$ = $\widehat Y$ E0 ej $\left( \omega t - \dfrac{\pi}{\lambda} \times \dfrac{\sqrt 3 \pi}{\gamma} z \right)$

  3. $\vec E$ = $\widehat Y$E0 ej $\left( \omega t + \dfrac{\sqrt 3 \pi}{\gamma} \times \dfrac{\pi}{\lambda} z \right)$

  4. $\vec E$ = $\widehat Y$ E0 ej $\left( \omega t - \dfrac{\pi}{\lambda} x + \dfrac{\sqrt 3 \pi}{\gamma} z \right)$


Correct Option: A
Explanation:

A $\dfrac{\lambda}{2}$ dipole is kept horizontally at a height of $\dfrac{\lambda_0}{2}$ above a perfectly conducting infinite ground plane. The radiation pattern in the plane of the dipole ($\vec E$ plane) looks approximately as


Correct Option: B
Explanation:

A right circularly polarised (RCP) plane wave is incident at an angle of 60° to the normal, on an air-dielectric interface. If the reflected wave is linearly polarised, the relative dielectric constant $\epsilon_{r2}$ is

  1. $\sqrt 2$

  2. $\sqrt 3$

  3. 2

  4. 3


Correct Option: D
Explanation:

The DC current gain ($\beta$)of a BJT is 50. Assuming that the emitter injection efficiency is 0.995, the base transport factor is

  1. 0.980

  2. 0.985

  3. 0.990

  4. 0.995


Correct Option: B
Explanation:

The electron and hole concentrations in an intrinsic semiconductor are ni per cm3 at 300 K. Now, if acceptor impurities are introduced with a concentration of NA per cm3 where (where NA >> ni), the electron concentration per cm3 at 300 K will be

  1. ni

  2. ni + NA

  3. NA - ni

  4. $\dfrac{n^2i}{N_A}$


Correct Option: D
Explanation:

The figure shows the high-frequency capacitance-voltage (C-V) characteristics of a Metal/SiO2 /silicon (MOS) capacitor having an area of 1 $\times$10-4 cm2. Assume that the permittivities ($\epsilon_0 \epsilon_r$) of silicon and SiO2 are 1 $\times$ 10-12 F / cm and 3.5 $\times$10-13 F / cm respectively.

The gate oxide thickness in the MOS capacitor is

  1. 50 nm

  2. 143 nm

  3. 350 nm

  4. 1 $\mu$m


Correct Option: A
Explanation:

The figure shows the high-frequency capacitance-voltage (C-V) characteristics of a Metal/SiO2 /silicon (MOS) capacitor having an area of 1 $\times$10-4 cm2. Assume that the permittivities ($\epsilon_0 \epsilon_r$) of silicon and SiO2 are 1 $\times$ 10-12 F / cm and 3.5 $\times$10-13 F / cm respectively.

Consider the following statements about the C-V characteristics plot: S1: The MOS capacitor has an n-type substrate. S2: If positive charges are introduced in the oxide, the C-V plot will shift to the left. Which of the following is true?

  1. Both S1 and S2 are true.

  2. S1 is true and S2 is false.

  3. S1 is false and S2 is true.

  4. Both S1 and S2 are false.


Correct Option: C
Explanation:

The figure shows the high-frequency capacitance-voltage (C-V) characteristics of a Metal/SiO2 /silicon (MOS) capacitor having an area of 1 $\times$10-4 cm2. Assume that the permittivities ($\epsilon_0 \epsilon_r$) of silicon and SiO2 are 1 $\times$ 10-12 F / cm and 3.5 $\times$10-13 F / cm respectively.

The maximum depletion layer width in silicon is

  1. 0.143 $\mu$m

  2. 0.857 $\mu$m

  3. 1 $\mu$m

  4. 1.143 $\mu$m


Correct Option: B
Explanation:

A p +n junction has a built-in potential of 0.8 V. The depletion layer width at a reverse bias of 1.2V is 2 $\mu$m. For a reverse bias of 7.2 V, the depletion layer width will be

  1. 4 $\mu$m

  2. 4.9 $\mu$m

  3. 8 $\mu$m

  4. 12 $\mu$m


Correct Option: A
Explanation:

The raised cosine pulse p(t) is used for zero ISI in digital communications. The expression for p(t) with unity roll - off factor is given by p(t) =$\dfrac{sin \ 4\pi Wt}{4\pi Wt(1 - 16 w^2t^2)}$ The value of p(t) at t = $\dfrac{1}{4W}$is

    • 0.5
  1. 0

  2. 0.5

  3. $\infty$


Correct Option: C
Explanation:

In a GSM system, 8 channels can co-exist in 200 KHz bandwidth using TDMA. A GSM based cellular operator is allocated 5 MHz bandwidth. Assuming a frequency reuse factor of $\dfrac{1}{5}$ i.e a five-cell repeat pattern, the maximum number of simultaneous channels that can exist in one cell is

  1. 200

  2. 40

  3. 25

  4. 5


Correct Option: B
Explanation:

If R($\tau$) is the autocorrelation function of a real, wide-sense stationary random process, then which of the following is NOT true?

  1. R($\tau$) = R(-$\tau$)

  2. |R($\tau$)| $\le$ R(0)

  3. R($\tau$) = –R(–$\tau$)

  4. The mean square value of the process is R(0)


Correct Option: C
Explanation:

Autocorrelation is an even function.

In delta modulation, the slope overload distortion can be reduced by

  1. decreasing the step size

  2. decreasing the granular noise

  3. decreasing the sampling rate

  4. increasing the step size


Correct Option: D
Explanation:

Slope overload distortion can be reduced by increasing the step size $\dfrac{\Delta}{T_\pi} \ge slope \ of \ x(t)$

If S(f) is the power spectral density of a real, wide-sense, stationary random process, then which of the following relations is true?

  1. S(0) $\ge$ S(f)

  2. S(f) $\ge$ 0

  3. S(–f) = –S(f)

  4. $\int_{-\infty}^\infty S(f) \ df = 0$


Correct Option: B
Explanation:

Power spectral density is non-negative. Thus, it is always zero or greater than zero.

Two 4-ray signal constellations are shown. It is given that $\phi_1$ and $\phi_2$ constitute an orthonormal basis for the two constellations. Assume that the four symbols in both the constellations are equiprobable. Let $\dfrac{N_0}{2}$ denote the power spectral density of white Gaussian noise.

If these constellations are used for digital communications over an AWGN channel, then which of the following statements is true?

  1. Probability of symbol error for constellation 1 is lower.

  2. Probability of symbol error for constellation 1 is higher.

  3. Probability of symbol error is equal for both the constellations.

  4. The value of 0 N will determine which of the two constellations has a lower probability of symbol error.


Correct Option: A
Explanation:

Noise power is same for both which is $\dfrac{N_0}{2}$

Thus probability of error will be lower for the constellation 1 as it has higher signal energy.

Two 4-ray signal constellations are shown. It is given that $\phi_1$ and $\phi_2$ constitute an orthonormal basis for the two constellations. Assume that the four symbols in both the constellations are equiprobable. Let $\dfrac{N_0}{2}$ denote the power spectral density of white Gaussian noise.

The ratio of the average energy of constellation 1 to the average energy of constellation 2 is

  1. 4a2

  2. 3.5

  3. 2

  4. 8


Correct Option: B
Explanation:

By using solution we get                        Energy of constellation 1 = 14 a^2                      Energy of constellation 2 = 4 a^2                                                  Ratio = 14 a^2/4 a^2 = 3.5                      

In the following scheme, if the spectrum M(f) of m(t) is as shown, then the spectrum Y(f) of y(t) will be


Correct Option: B
Explanation:

If E denotes expectation, then the variance of a random variable X is given by

  1. E [x2] – E2 [x]

  2. E [x2] + E2 [x]

  3. E [x2]

  4. E2 [x]


Correct Option: A
Explanation:

The variance of a random variable $x$ is given by $E[X^2] - D^Q[X]$

During transmission over a certain binary communication channel, bit errors occur independently with probability p. The probability of AT MOST one bit in error in a block of n bits is given by

  1. pn

  2. 1 – pn

  3. np(1 – p)n-1 + (1 – p)n

  4. 1 – (1 – p)n


Correct Option: C
Explanation:

A Hilbert transformer is a

  1. non-linear system

  2. non-causal system

  3. time-varying system

  4. low-pass system


Correct Option: A
Explanation:

A Hilbert transformer is a non-linear system.

In a Direct Sequence CDMA System, the chip rate is 1.2288 $\times$ 106 chips per second. If the processing gain is desired to be at least 100, then the data rate

  1. must be less than or equal to 12.288 $\times$ 103 bits per sec

  2. must be greater than 12.288 $\times$ 103 bits per sec

  3. must be exactly equal to 12.288 $\times$ 103 bits per sec

  4. can take any value less than 122.88 $\times$ 103 bits per sec


Correct Option: A
Explanation:

An input to a 6-level quantizer has the probability density function f(x) as shown in the figure. Decision boundaries of the quantizer are chosen so as t maximize the entropy of the quantizer output. It is given that 3 consecutive decision boundaries are -1, 0 and 1.

Assuming that the reconstruction levels of the quantizer are the mid-points of the decision boundaries, the ratio of signal power to quantization noise power is

  1. $\dfrac{152}{9}$

  2. $\dfrac{64}{3}$

  3. $\dfrac{76}{3}$

  4. 28


Correct Option: A
Explanation:

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An input to a 6-level quantizer has the probability density function f(x) as shown in the figure. Decision boundaries of the quantizer are chosen so as to maximize the entropy of the quantizer output. It is given that 3 consecutive decision boundaries are -1, 0 and 1.

The values of a and b are

  1. a = $\dfrac{1}{6}$and b = $\dfrac{1}{12}$

  2. a = $\dfrac{1}{5}$and b = $\dfrac{3}{40}$

  3. a = $\dfrac{1}{4}$and b = $\dfrac{1}{16}$

  4. a = $\dfrac{1}{3}$and b = $\dfrac{1}{24}$


Correct Option: A
Explanation:

The z-transform X [z] of a sequence x[n] is given by x[z] =$\dfrac{0.5}{1-2z^{-1}}$. It is given that the region of convergence of X[z] includes the unit circle. The value of x[0] is

    • 0.5
  1. 0

  2. 0.25

  3. 0.5


Correct Option: B
Explanation:

The 3-dB bandwidth of the low-pass signal e-t u(t), where u(t) is the unit step function, is given by

  1. $\dfrac{1}{2\pi} HZ$

  2. $\dfrac{1}{2\pi} \sqrt{ \sqrt{2} - 1HZ }$

  3. $\infty$

  4. 1 HZ


Correct Option: A
Explanation:

A 5-point sequence x (n) is given as X [–3] = 1, x [–2] = 1, x [–1] = 0, x [0] = 5, x [1] = 1. If x($e^{j\omega}$) denotes the discrete – time fourier transform of x [n], what is the value of $\displaystyle \int_{-\pi}^\pi x(e^{j\omega})$$d\omega$?

  1. 5

  2. 10$\pi$

  3. 16$\pi$

  4. 5 + j10$\pi$


Correct Option: B
Explanation:

If the Laplace transform of a signal y (t) is y(s) = $\dfrac{1}{s(s-1)}$, its final value is

    • 1
  1. 0

  2. 1

  3. unbounded


Correct Option: D
Explanation:

The frequency response of a linear, time-invariant system is given by H(f) = $\dfrac{5}{1 + j10\pi f}$ The step response of the system is

  1. 5 (1- e-5t) u (t)

  2. 5 $\left( 1 - e^{-\dfrac{t}{5}} \right) u(t)$

  3. $\dfrac{1}{5}$(1- e-5t) u (t)

  4. $\dfrac{1}{5}$$\left( 1 - e^{\dfrac{t}{5}} \right) u(t)$


Correct Option: B
Explanation:

The RC circuit shown in the figure is

  1. a low-pass filter

  2. a high-pass filter

  3. a band-pass filter

  4. a band-reject filter


Correct Option: C
Explanation:

For the circuit shown in the figure, the Thevenin voltage and resistance looking into X - Y are

  1. $\dfrac{4}{3}$V, 2$\Omega$

  2. 4V, $\dfrac{2}{3}$$\Omega$

  3. $\dfrac{4}{3}$V, $\dfrac{2}{3}$$\Omega$

  4. 4V, 2$\Omega$


Correct Option: D
Explanation:

In the AC network shown in the figure, the phasor voltage VAB (in Volts) is

  1. 0

  2. 5 $\angle$300

  3. 12.5 $\angle$300

  4. 17 $\angle$300


Correct Option: D
Explanation:

An independent voltage source in series with an impedance ZS = RS + jXS delivers a maximum average power to a load impedance ZL when

  1. ZL = RS + jXS

  2. ZL = RS

  3. ZL = jXS

  4. ZL = RS - jXS


Correct Option: D
Explanation:

According to maximum $Z_L = Z_B^* = (R_B - jX_B)$

In the circuit shown, VC is 0 volts at t = 0 sec. For t > 0, the capacitor current ic (t) where t is in seconds, is given by

  1. 0.50 exp (-25t) mA

  2. 0.25 exp (-25t) mA

  3. 0.50 exp (-12.5t) mA

  4. 0.25 exp (-6.25t) mA


Correct Option: A
Explanation:

Two series resonant filters are as shown in the figure. Let the 3-dB bandwidth of Filter 1 be B1 and that of Filter 2 be B2. The value of $\dfrac{B_1}{B_2}$ is

  1. 4

  2. 1

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

  4. $\dfrac{1}{4}$


Correct Option: D
Explanation:

For |x| << 1, coth (x) can be approximated as

  1. x

  2. x2

  3. $\frac{1}{x}$

  4. $\frac{1}{x^2}$


Correct Option: C
Explanation:

$cothx = \frac{cosh x}{sinh x} \\ \text{as} |x| < < 1, coshx \approx 1 \hspace{0.2cm} and \hspace{0.2cm} sinh x \approx x \\ \text{Thus} \hspace{0.2cm} coth x \approx \frac{1}{x}$

$lim_{\theta \rightarrow 0} \frac{sin(\theta / 2)}{\theta}$ is

  1. 0.5

  2. 1

  3. 2

  4. not defined


Correct Option: A
Explanation:

$lim_{\theta \rightarrow 0} \frac{sin(\theta / 2)}{\theta} = lim_{\theta \rightarrow 0} \frac{sin(\theta / 2)}{2(\theta / 2)} = \frac{1}{2} lim_{\theta \rightarrow 0} \frac{sin(\theta / 2)}{(\theta / 2)} = \frac{1}{2} = 0.5$

The following plot shows a function y which varies linearly with x. The value of the integral I = $\int\limits_1^2 y \hspace{0.2cm}dx$is

  1. 1.0

  2. 2.5

  3. 4.0

  4. 5.0


Correct Option: B
Explanation:

$\text{The given plot is straight line with equation is} \\ \frac{x}{-1}+ \frac{y}{1} = 1 \\ or \hspace{1cm} y = x + 1 \\ Now \hspace{1cm} I = \int\limits_1^2 ydx = \int\limits_1^2 (x + 1) dx \\ = \bigg[\frac{(x+1^2)}{2}\bigg]^2 = \frac{9}{2} - \frac{4}{2} = 2.5$

Which one of the following functions is strictly bounded?

  1. $\frac{1}{x^2}$

  2. ex

  3. x2

  4. $e^{-x^2}$


Correct Option: D
Explanation:

$\text{We have} \hspace{1cm} lim_{x \rightarrow 0} \frac{1}{x^2} = \infty \\ \hspace{2.7cm} lim_{x \rightarrow \infty} x^2 = \infty \\ \hspace{2.7cm} lim_{x \rightarrow \infty} e^{-x} = \infty \\ \hspace{2.7cm} lim_{x \rightarrow \infty} e^{-x^2} = 0 \\ \hspace{2.7cm} lim_{x \rightarrow 0} e^{-x^2} = 1 \hspace{1cm} \text{Thus $e^{-x^2}$ is strictly bounded.}$

For the function e-x, the linear approximation around x = 2 is

  1. (3 -x) e-2

  2. 1 - x

  3. [ 3 + 2$\sqrt{2}$r - ( 1+ $\sqrt{2}$x ] e-2

  4. e-2


Correct Option: A
Explanation:

$\text{We have} f(x) = e^{-x} = e^{-(x-2)-2} = e^{-(x-2)} e^{-2} \\ \hspace{2.5cm} = \bigg[1 - (x - 2) + \frac{(X - 2)^2}{2!}... \bigg] e^{-2} \\ \hspace{2.5cm} = \bigg[1 - (x - 2) \bigg] e^{-2}\hspace{2cm} \text{Neglecting higher powers}$

The correct full wave rectifier circuit is


Correct Option: C
Explanation:

In a transconductance amplifier, it is desirable to have

  1. a large input resistance and a large output resistance

  2. a large input resistance and a small output resistance

  3. a small input resistance and a large output resistance

  4. a small input resistance and a small output resistance


Correct Option: A
Explanation:

In the transconductance amplifier, it is desirable to have large input resistance and large output resistance. ideally, Id = Gm.Vs Id = Gm.Vi(Rd/Rd+RL) if Rd>>RL, Id = Gm Vi Vi = Vs if Ri >> Rs

It is given that X1, X2, …..XM are M non-zero orthogonal vectors. The dimension of the vector space spanned by the 2M vectors X1, X2, …XM , - XM is

  1. 2M

  2. M + 1

  3. M

  4. dependent on the choice of X1, X2, …..XM


Correct Option: C

Consider the function f(x) = x2 -x -2. The maximum value of f (x) in the closed interval [-4, 4] is

  1. 18

  2. 10

    • 2.25
  3. indeterminate


Correct Option: A
Explanation:

An examination consists of two papers, Paper 1 and Paper 2. The probability of failing in Paper 1 is 0.3 and that in Paper 2 is 0.2. Given that a student has failed in Paper 2, the probability of failing in Paper 1 is 0.6. The probability of a student failing in both the papers is

  1. 0.5

  2. 0.18

  3. 0.12

  4. 0.06


Correct Option: C
Explanation:

$\text{Probability of failing in paper 1 is} \hspace{1cm} P(A) = 0.3 \\ \text{Possibility of failing in paper 2 is} \hspace{1cm} P(B) = 0.2 \\ Probability of failing in paper 1, when \\ \text{Student has failed in paper 2 is} \hspace{1cm} P(\frac{A}{B}) = 0.6 \\ \text{We know that} \\ P(\frac{A}{B}) = \frac{(P \cap B)}{P(B)} \\ or \hspace{1cm} P(A \cap B) = P(B) P (\frac{A}{B}) = 0.6 \times 0.2 = 0.12$

The equation x3 -x2 + 4x - 4 = 0 is to be solved using the Newton-Raphson method. If x = 2 is taken as the initial approximation of the solution, the next approximation using this method will be

  1. $\frac{2}{3}$

  2. $\frac{4}{3}$

  3. 1

  4. $\frac{3}{2}$


Correct Option: B
Explanation:

$\text{We have} \\ f(x) = x^3 - x^2 + 4x -4 \\ f(x) = 3x^2 - 2x +4 \\ \text{Taking $x_0 - 2$ in Newton - Raphson method} \\ x_1 = x_0 - \frac{f(x_0)}{f(x_0)} = 2 - \frac{2^3 - 2^2 + 4(2) - 4}{3(2)^2 - 2(2) + 4} = \frac{4}{3}$

Three functions f1 (t), f2 (t) and f3 (t) which are zero outside the interval [0, T] are shown in the figure. Which of the following statements is correct?

  1. f1 (t) and f2 (t) are orthogonal.

  2. f1 (t) and f3 (t) are orthogonal.

  3. f2 (t) and f3 (t) are orthogonal.

  4. None of these


Correct Option: C
Explanation:

For two orthogonal signals f(x) and g(x)

$$\int_{- \infty}^{+ \infty} f(x) g(x) dx =0$$

i.e common area between f(x) and g(x) is zero.

If C is a closed curve enclosing a surface S, the magnetic field intensity $\overrightarrow{H}$, the current density $\overrightarrow{j}$and the electric flux density $\overrightarrow{D}$ are related by

  1. $\int \int_s \overrightarrow{H} . d \overrightarrow{s} = ∯_c\bigg(\overrightarrow{j} + \frac{\partial \overrightarrow{D}}{\partial t}\bigg). d \overrightarrow{t}$

  2. $\int_s \overrightarrow{H}.d \overrightarrow{l} = ∯_s\bigg(\overrightarrow{j} + \frac{\partial \overrightarrow{D}}{\partial t}\bigg) . d \overrightarrow{S}$

  3. $\int \int_s \overrightarrow{H} . d \overrightarrow{s} = $

  4. $\oint_C \overrightarrow{H} . d \overrightarrow{l} \oint_c = \iint_s \bigg(\overrightarrow{j} + \frac{\partial \overrightarrow{D}}{\partial t}\bigg) . d \overrightarrow{S}$


Correct Option: D
Explanation:

$\nabla \times H = J + \frac{\partial D}{\partial t} \hspace{4cm} \text{Maxwell Equations} \\ \iint_s \nabla \times H . ds = \iint-s (J+ \frac{\partial D}{\partial t}).ds \hspace{0.7cm} \text{Integral form} \\ \oint H . dl = \iint_s(J+ \frac{\partial D}{\partial t}).ds \hspace{2.4cm} \text{Integral form}$

The solution of the differential equation K2 $\frac{d^2y}{dx^2}$ = y – y2 under the boundary conditions (i) y = y1 at x = 0 and (ii) y = y2 at x = $\infty$, where k, y1 and y2 are constants, is

  1. y = (y1 – y2) exp (– x/k2) + y2

  2. y = (y2 – y1) exp (– x/k) + y1

  3. y = (y1 – y2) sinh (– x/k) + y1

  4. y = (y1 – y2) exp (– x/k) + y2


Correct Option: D
Explanation:

If a semi-circular contour D of radius 2 is as shown in the figure given below, the value of the integral $\int_D N \frac{1}{(S^2 -1)}ds$ is

  1. j$\pi$

  2. – j$\pi$

  3. $\pi$

  4. $\pi$


Correct Option: A
Explanation:

For the BJT circuit shown, assume that the $\beta$ of the transistor is very large and VBE = 0.7 V. The mode of operation of the BJT is

  1. cut-off

  2. saturation

  3. normal active

  4. reverse active


Correct Option: B
Explanation:

For the op-amp circuit shown in the figure, V0 is

  1. – 2 V

  2. – 1 V

  3. – 0.5 V

  4. 0.5 V


Correct Option: C
Explanation:

For the Zener diode shown in the figure, the Zener voltage at the knee is 7 V. The knee current is negligible and the Zener dynamic resistance is 10 $\Omega$. If the input voltage (Vi) ranges from 10 V to 16 V, what will be the range of out put voltage (V0)?

  1. 7.00 V to 7.29 V

  2. 7.14 V to 7.29 V

  3. 7.14 V to 7.43 V

  4. 7.29 V to 7.43 V


Correct Option: C
Explanation:

In the Op-Amp circuit shown, assume that the diode current follows the equation I = Is exp (V/VT). For Vi = 2V, V0 = V01, and for Vi = 4 V, V0 = V02. The relationship between V01 and V02 is

  1. V02 = $\sqrt{2}$V01

  2. V02 = e2 V01

  3. V02 = V01 In 2

  4. V01 - V02 = VT In 2


Correct Option: D
Explanation:

Group I lists four different semiconductor devices. Match each device in Group I with its characteristic property in Group II

Group I Group II
(P) BJT (1) Population inversion
(Q) MOS capacitor (2) Pinch-off voltage
(R) LASER diode (3) Early effect
(S) JFET (4) Flat-band voltage
  1. P-3, Q-1, R-4, S-2

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

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

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


Correct Option: C
Explanation:

In BJT as the B-C reverse bias voltage increases, the B-C space region width increases which $x_B$ (i.e neutral base width) > A change in neutral base width will change the collector current. A reduction in base width will causes the gradient in minority carrier concentration to increase, which in turn causes an increased in the diffusion current. This effect is known as base modulation as early effect.

In JFET the gate to source voltage that must be applied to achieve pinch off voltage is described as pinch off voltage and is also called as turn voltage or threshold voltage.

In LASER population inversion occurs on the condition when concentration of electrons in one energy state is greater than that in lower energy state, i.e a non equilibrium condition.

In MOS capacitor, flat band voltage is the gate voltage that must be applied to create flat ban condition in which there is no space region in semiconductor under oxide.

Therefore BJT : Early effect

MOS capacitor : Flat - band voltage

LASER diode : Population inversion

JFET : Pinch-off voltage

If the closed-loop transfer function of a control system is given as T(s) = $$\dfrac{s-5}{(s+2)(s+3)}$$, then it is

  1. an unstable system

  2. an uncontrollable system

  3. a minimum phase system

  4. a non-minimum phase system


Correct Option: D
Explanation:

In a minimum phase system, all the poles as well as zeros are on the left half of the s-plane. In given system as there is right half zero (s = 5), the system is a non-minimum phase system.

In the CMOS inverter circuit shown, if the transconductance parameters of the NMOS and PMOS transistor are kn = kp = $\mu$nC0x $\frac{W_n}{L_n}$=$\mu$p Cox $\frac{W_p}{L_p}$= 40 $\mu$A/ V2 and their threshold voltage are VTHn = |VTHp| = 1V, the current I is

  1. 0 A

  2. 25 $\mu$A

  3. 45 $\mu$A

  4. 90 $\mu$A


Correct Option: D
Explanation:

The open loop transfer function of a plant is given as G(s) = $\dfrac{1}{s^2-1}$. If the plant is operated in a unity feedback configuration, the lead compensator that an stabilize this control system is

  1. $\dfrac{10(s-1)}{s+2}$

  2. $\dfrac{10(s-1)}{s+2}$

  3. $\dfrac{10(s+2)}{s+10}$

  4. $\dfrac{2(s+2)}{s+10}$


Correct Option: A
Explanation:

The asymptotic Bode plot of a transfer function is as shown in the figure. The transfer function G(s) corresponding to this Bode plot is

  1. $\dfrac{1}{(s+1)(s+20)}$

  2. $\dfrac{1}{s(s+1)(s+20)}$

  3. $\dfrac{100}{s(s+1)(s+20)}$

  4. $\dfrac{100}{s(s+1)(1+0.05s)}$


Correct Option: D
Explanation:

A control system with a PD controller is shown in the figure. If the velocity error constant KV = 1000 and the damping ratio $\xi$= 0.5, the values of KP and KD are

  1. Kp = 100, KD = 0.09

  2. Kp = 100, KD = 0.9

  3. Kp = 10, KD = 0.09

  4. Kp = 10, KD = 0.9


Correct Option: B
Explanation:

A transmission line of characteristic impedance 50 Ω is terminated by a 50 Ωload. When excited by a sinusoidal voltage source at 10 GHz, the phase difference between two points spaced 2 mm apart on the line is found to be $\frac{2 \pi}{\lambda}$ radians. The phase velocity of the wave along the line is

  1. 3 x 108 m/s

  2. 0.8 x 108 m/s

  3. 1.2 x 108 m/s

  4. 1.6 x 108 m/s


Correct Option: D
Explanation:

An air-filled rectangular waveguide has inner dimensions of 3 cm X 2 cm. The wave impedance of the TE20 mode of propagation in the waveguide at a frequency of 30 GHz is (free space impedance $\eta_0$= 377$\Omega$)

  1. 308$\Omega$

  2. 355$\Omega$

  3. 400$\Omega$

  4. 461$\Omega$


Correct Option: C
Explanation:

The transfer function of a plant is T (s) = $ \dfrac{5}{(s+5)(s^2+s+1)} $. The second - order approximation of T(s) using dominate pole concept is

  1. $ \dfrac{1}{(s+5)(s+1)} $

  2. $ \dfrac{5}{(s+5)(s+1)} $

  3. $ \dfrac{5}{(s^2+s+1)} $

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


Correct Option: D
Explanation:

Consider the Op-Amp circuit shown in the figure.

The transfer function V0 (s) / Vi (s) is

  1. $\frac{1- sRC}{1+ sRC}$

  2. $\frac{1+ sRC}{1- sRC}$

  3. $\frac{1}{1- sRC}$

  4. $\frac{1}{1+ sRC}$


Correct Option: A
Explanation:

The state space representation of a separately excited DC servo motor dynamics is given as $ \left[ \begin{array} \dfrac{d \omega }{ dt } \\ \dfrac{di_s}{dt} \end{array} \right] = \left[ \begin{array} \ -1 & 1 \\ -1 & -10 \end{array} \right] \left[ \begin{array} \ \omega \\ i_s \end{array} \right] $ + $ \left[ \begin{array} \ 0 \\ 10 \end{array} \right] u $

  1. $\dfrac{10}{s^2+11s+11}$

  2. $\dfrac{1}{s^2+11s+11}$

  3. $\dfrac{10+10}{s^2+11s+11}$

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


Correct Option: A
Explanation:

Consider the Op-Amp circuit shown in the figure.

If Vi = V1 sin ($\omega t$) and V0 = V2 sin ($\omega t$+ $\phi$), the minimum and maximum values of $\phi$(in radians) are respectively

  1. $\frac{- \pi}{2} and \frac{\pi}{2}$

  2. 0 and $\frac{\pi}{2}$

  3. -$\pi$ and 0

  4. $\frac{-\pi}{2}$ and 0


Correct Option: C
Explanation:

Consider a linear system whose state space representation is x and (t) = Ax (t). If the initial state vector of the system is x (0) =$ \left[ \begin{array} \ 1 \\ -2 \end{array} \right] $, the system response is x (t) = $ \left( \begin{array} \ e^{-2t} \\ -2 e^{-2t} \end{array} \right) $. If the initial state vector of the system changes, the system response becomes x(t) = $ \left[ \begin{array} \ e^{-t} \\ -e^{-6} \end{array} \right] $.

The system matrix A is

  1. $ \left[ \begin{array} \ 0 & 1 \\ -1 & 1 \end{array} \right] $

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

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

  4. $ \left[ \begin{array} \ 0 & 1 \\ -2 & -3 \end{array} \right] $


Correct Option: D
Explanation:

As shown in previous solution the system matrix is $ \left[ \begin{array} \ 0 & 1 \\ -2 & -3 \end{array} \right] $

Consider a linear system with state space representation is x and (t) = Ax (t). If the initial state vector of the system is x (0) = $ \left[ \begin{array} \ 1 \\ -2 \end{array} \right] $, the system response is x (t) = $ \left( \begin{array} \ e^{-2t} \\ -2 e^{-2t} \end{array} \right) $ . If the initial state vector of the system changes, the system response becomes x(t) = $ \left[ \begin{array} e^{-t} \\ -e^{-6} \end{array} \right] $ .

The eigen value and eigen vector pairs ($\lambda_i, V_i$) for the system are

  1. $ \left( \begin{array} \ -1, & \left[ \begin{array} \ 1 \\ -1 \end{array} \right] \end{array} \right) $and $ \left( \begin{array} \ -2, & \left[ \begin{array} \ 1 \\ -2 \end{array} \right] \end{array} \right) $

  2. $ \left( \begin{array} \ -1, & \left[ \begin{array} \ 1 \\ -1 \end{array} \right] \end{array} \right) $and $ \left( \begin{array} \ 2, & \left[ \begin{array} \ 1 \\ -2 \end{array} \right] \end{array} \right) $

  3. $ \left( \begin{array} \ 1, & \left[ \begin{array} \ 1 \\ -1 \end{array} \right] \end{array} \right) $and $ \left( \begin{array} \ -2, & \left[ \begin{array} \ 1 \\ -2 \end{array} \right] \end{array} \right) $

  4. $ \left( \begin{array} \ -2, & \left[ \begin{array} \ 1 \\ -1 \end{array} \right] \end{array} \right) $and $ \left( \begin{array} \ -1, & \left[ \begin{array} \ 1 \\ -2 \end{array} \right] \end{array} \right) $


Correct Option: A
Explanation:

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