2005|Electronics and Comm (GATE Exam)-Previous Question Paper Solution

Description: GATE Exam Previous Year Question Paper Solution Electronics and Communication (ECE) - 2005
Number of Questions: 90
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Tags: Differential Calculus Signals and System Electronics and Communication Engineering - EC Transform Theory Network Graphs Electronic Devices
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The Boolean function f implemented in figure using two input multiplexers is

  1. $A\bar BC + AB\bar C$

  2. $ABC + A\bar B\bar C$

  3. $\bar ABC + \bar A\bar B\bar C$

  4. $\bar A\bar BC + \bar AB\bar C$


Correct Option: A
Explanation:

 

The transistors used in a portion of the TTL gate shown in figure have $\beta$ = 100 ,the base-emitter voltage is 0.7 V for a transistor in active region and 0.75 V for a transistor in saturation. If the sink current I = 1 mA and the output is at logic 0, then the current IR will be equal to

  1. 0.65 mA

  2. 0.70 mA

  3. 0.75 mA

  4. 1.00 mA


Correct Option: C
Explanation:

The present output Qn of an edge triggered JK flip-flop is logic 0. If J = 1, then Qn + 1

  1. cannot be determined

  2. will be logic 0

  3. will be logic 1

  4. will race around


Correct Option: C
Explanation:

Consider an 8085 microprocessor system. The following program starts at location 0100H. LXI SP, 00FF LXI H, 0701 MVI A, 20H SUB M

The content of accumulator when the program counter reaches 0109H is

  1. 20H

  2. 02H

  3. 00H

  4. FFH


Correct Option: C
Explanation:

Consider an 8085 microprocessor system. If in addition following code exists from 0109H onwards. ORI 40H ADD M What will be the result in the accumulator after the last instruction is executed?

  1. 40H

  2. 20H

  3. 60H

  4. 42H


Correct Option: C
Explanation:

Decimal 43 in Hexadecimal and BCD number system is respectively

  1. B2, 0100 0011

  2. 2B, 0100 0011

  3. 2B, 0011 0100

  4. B2, 0100 0100


Correct Option: B
Explanation:

What memory address range is NOT represented by chip #1 and chip #2 in figure? A0 to A15 in this figure are the address lines and CS means Chip Select.

  1. 0100 - 02FF

  2. 1500 - 16FF

  3. F900 - FAFF

  4. F800 - F9FF


Correct Option: D
Explanation:

null

The Boolean expression for the truth table shown is

A B C f 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 0 1 1 0 1 1 1 1 0 Find the value of f.

  1. $B(A + C)(\bar A + \bar C)$

  2. $B(A + \bar C)(\bar A + C)$

  3. $\bar B(A + \bar C)(\bar A + C)$

  4. $\bar B(A + \bar C)(\bar A + C)$


Correct Option: A
Explanation:

We have $$ f = \bar A BC + AB\bar C \\ = B(\bar A C + A\bar C) \\ = B(A+C)(\bar A + \bar C)$$

Figure shows a ripple counter using positive edge triggered flip-flops. If the present state of counter is Q2Q1Q0 = 011, then its next state (Q2Q1Q0) will be

  1. 010

  2. 100

  3. 111

  4. 101


Correct Option: B
Explanation:

Since $T_2T_1T_0$ is at 111, at every clock $Q_2Q_1Q_0$ will be changed. Ir present state is 011, the next state will be 100.

Which one of the following represents the electric field lines for the TE02 mode in the cross-section of a hollow rectangular metallic waveguide?


Correct Option: D
Explanation:

Correct Answer: In $T E_{02}$ mode, the waves travel horizontally.

The magnetic field intensity vector of a plane wave is given by $\bar H (x,y,z,t)$ = 10 sin (50000t + 0.004x + 30)$\widehat a_y$ where $\widehat a_y$denotes the unit vector in y direction. The wave is propagating with a phase velocity

  1. 5 x 104 m/s

    • 3 x 108 m/s
    • 1.25 x 107 m/s
  2. 3 x 108 m/s


Correct Option: C
Explanation:

For an n-channel MOSFET and its transfer curve shown in figure, the threshold voltage is

  1. 1 V and the device is in active region

    • 1 V and the device is in saturation region
  2. 1 V and the device is in saturation region

    • 1 V and the device is in active region

Correct Option: C
Explanation:

Refractive index of glass is 1.5. Find the wavelength of a beam of light with a frequency of 1014 Hz in glass. Assume velocity of light is 3 × 108 m/s in vacuum.

  1. 3 $\mu$m

  2. 3 mm

  3. 2 $\mu$m

  4. 1 m


Correct Option: C
Explanation:

Characteristic impedance of a transmission line is 50$\Omega$. Input impedance of the open circuited line is Zoc = 100 + j150$\Omega$. When the transmission line is short-circuited, the value of the input impedance will be

  1. 50 $\Omega$

  2. 100 + j150 $\Omega$

  3. 7.69 + j11.54 $\Omega$

  4. 7.69 - j11.54 $\Omega$


Correct Option: D
Explanation:

Voltage standing wave pattern in a lossless transmission line with characteristic impedance 50 $\Omega$ and a resistive load is shown in figure.

The reflection coefficient is given by

    • 0.6
    • 1
  1. 0.6

  2. 0


Correct Option: A
Explanation:

The reflection coefficient is $\Gamma = \dfrac{Z_L - Z_0}{Z_L + Z_0} = \dfrac{12.5 - 50}{12.5+50} = -0.6$

Voltage standing wave pattern in a lossless transmission line with characteristic impedance 50 $\Omega$ and a resistive load is shown in figure.

The value of the load resistance is

  1. 50 $\Omega$

  2. 200 $\Omega$

  3. 12.5 $\Omega$

  4. 0 $\Omega$


Correct Option: C
Explanation:

A silicon sample A is doped with 1018 atoms/cm3 of Boron. Another sample B of identical dimensions is doped with 1018 atoms/cm3 of Phosphorus. The ratio of electron to hole mobility is 3. The ratio of conductivity of the sample A to B is

  1. 3

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

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

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


Correct Option: B
Explanation:

A MOS capacitor is made using p-type substrate in the accumulation mode. The dominant charge in the channel is due to the presence of

  1. holes

  2. electrons

  3. positively charged ions

  4. negatively charged ions


Correct Option: A
Explanation:

In accumulation mode for NMOS having p - substrate, when positive voltage is applied at the gate, this will induce negative charge near p - type surface beneath the gate. When VGS is made sufficiently large, an inversion of electrons is formed and this in effect forms an n - channel.

A Silicon PN junction at a temperature of 20°C has a reverse saturation current of 10 pico-Amperes (pA). The reverse saturation current at 40°C for the same bias is approximately

  1. 30 pA

  2. 40 pA

  3. 50 pA

  4. 60 pA


Correct Option: B
Explanation:

The reverse saturation current doubles for every $10^\circ C$ rise in temperature as follows $I_0 (T) = I_{01} \times 2^{(T-T_1)/10}$

Thus at $40^\circ C, \ I_0 = 40\ pA$

A signal as shown in figure is applied to a matched filter. Which of the following represents the output of this matched filter?


Correct Option: C
Explanation:

 

Match the following:

Group 1 Group 2
$P - { 1 + km (t) } A sin(\omega_0 t)$ W – phase modulation
$Q - km(t) A sin(\omega_0 t)$ X – Frequency modulation
$R - A sin{\omega_0 t + km(t) }$ Y – Amplitude modulation
$S - A sin{\omega_0 t + k \int_{-\infty}^t m(t) dt }$ Z – DSB–SC modulation
  1. P – Z, Q – Y, R – X, S – W

  2. P – W, Q – X, R – Y, S – Z

  3. P – X, Q – W, R – Z, S – Y

  4. P – Y, Q – Z, R – W, S – X


Correct Option: D
Explanation:

Noise with uniform power spectral density of NoW/Hz is passed through a filter $H(\omega) = 2 \ exp(-j\omega t_d)$ followed by an ideal low-pass filter of bandwidth B Hz. The output noise power in Watts is

  1. 2NoB

  2. 4NoB

  3. 8NoB

  4. 16NoB


Correct Option: B
Explanation:

Which of the following analog modulation schemes requires the minimum transmitted power and minimum channel bandwidth?

  1. VSB

  2. DSB-SC

  3. SSB

  4. AM


Correct Option: C
Explanation:

An output of a communication channel is a random variable $\nu$ with the probability density function as shown in figure. The mean square value of $\nu$ is

  1. 4

  2. 6

  3. 8

  4. 9


Correct Option: C
Explanation:

A carrier is phase modulated (PM) with frequency deviation of 10 kHz by a single tone frequency of 1 kHz. If the single tone frequency is increased to 2 kHz, assuming that phase deviation remains unchanged, the bandwidth of the PM signal is

  1. 21 kHz

  2. 22 kHz

  3. 42 kHz

  4. 44 kHz


Correct Option: D
Explanation:

A symmetric three-level midtread quantizer is to be designed assuming equiprobable occurrence of all quantization levels.

The quantization noise power for the quantization region between -a and +a in the figure is

  1. $\dfrac{4}{81}$

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

  3. $\dfrac{5}{81}$

  4. $\dfrac{2}{81}$


Correct Option: A
Explanation:

A device with input x(t) and output y(t) is characterized by: y(t) = x2(t). An FM signal with frequency deviation of 90 kHz and modulating signal bandwidth of 5 kHz is applied to this device. The bandwidth of the output signal is

  1. 370 kHz

  2. 190 kHz

  3. 380 kHz

  4. 95 kHz


Correct Option: A
Explanation:

A symmetric three-level midtread quantizer is to be designed assuming equiprobable occurrence of all quantization levels.

If the input probability density function is divided into three regions as shown in figure, the value of a in the figure is

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

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

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

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


Correct Option: B
Explanation:

As the area under pdf curve must be unity, all three regions are equivaprobable. Thus area under each region must be $\dfrac{1}{3}$ $2a \times \dfrac{1}{4} = \dfrac{1}{3} \rightarrow a = \dfrac{2}{3}$

 

A sequence x(n) has non-zero values as shown in the figure.

The Fourier transform of y(2n) will be

  1. $e^{-j2\omega} [cos4\omega + 2cos 2 \omega +2]$

  2. $[cos2\omega + 2cos \omega +2]$

  3. $e^{-j\omega} [cos2\omega + 2cos \omega +2]$

  4. $e^{j2\omega} [cos2\omega + 2cos \omega +2]$


Correct Option: C
Explanation:

A sequence x(n) has non-zero values as shown in figure.

The sequence $y(n) = \begin{cases} x\left( \dfrac{n}{2} -1 \right) & \text{for n even} \\ 0 & \text{for n odd} \end{cases} $ will be


Correct Option: A
Explanation:

Which of the following can be impulse response of a causal system?


Correct Option: B
Explanation:

For casual system       h(t) = 0          for t<= 0          

A signal $x(n) = sin(\omega_on + \phi)$ is the input to a linear time invariant system having a frequency response $H(e^{j\omega})$. If the output of the system is $Axn - n_0$, then the most general form of $\angle H (e^{j\omega})$ will be

  1. $-n_0 \omega_0 + \beta$ for any arbitrary real $\beta$

  2. $-n_0 \omega_0 + 2\pi k$ for any arbitrary integer $k$

  3. $n_0 \omega_0 + 2\pi k$ for any arbitrary integer $k$

  4. $-n_0 \omega_0 + \phi$


Correct Option: B
Explanation:

The function x(t) is shown in figure. Even and odd parts of a unit-step function u(t) are respectively

  1. $\dfrac{1}{2}. \dfrac{1}{2} \times (t)$

  2. $-\dfrac{1}{2}. \dfrac{1}{2} \times (t)$

  3. $\dfrac{1}{2}. -\dfrac{1}{2} \times (t)$

  4. $-\dfrac{1}{2}. -\dfrac{1}{2} \times (t)$


Correct Option: A
Explanation:

The output y(t) of a linear time invariant system is related to its input x(t) by the following equation: $y(t) = 0.5 \times (t-t_d + T) + x(t-t_d) + 0.5 \times (t-t_d-T)$. The filter transfer function $H(\omega)$ of such a system is given by

  1. $(1+cos\omega T) e^{-j\omega t_d}$

  2. $(1+ 0.5cos\omega T) e^{-j\omega t_d}$

  3. $(1+cos\omega T) e^{j\omega t_d}$

  4. $(1+ 0.5cos\omega T) e^{-j\omega t_d}$


Correct Option: A
Explanation:

For a signal x(t), the Fourier transform is X(f). Then the inverse Fourier transform of X(3f + 2) is given by

  1. $ \dfrac{1}{2} \times (\dfrac{1}{2}) e^{j3\pi t} $

  2. $ \dfrac{1}{3} \times (\dfrac{1}{3}) e^{-j4\pi t} $

  3. $ 3 \times (3t) e^{-j4\pi t}$

  4. x(3t + 2)


Correct Option: B
Explanation:

The power in the signal s(t) = 8 cos $\left( 20\pi t - \dfrac{\pi}{2} \right)$ + 4 sin $(15 \pi t)$ is

  1. 40

  2. 41

  3. 42

  4. 82


Correct Option: A
Explanation:

Choose the function $f(t); -\infty \lt 1 \lt +\infty$for which a Fourier series cannot be defined.

  1. 3 sin (25t)

  2. 4 cos (20t + 3) + 2sin (10t)

  3. exp (-|t|) sin(25t)

  4. 1


Correct Option: C
Explanation:

Let $x(n) = \left( \dfrac{1}{2} \right) ^ n u(n), \ y(n) = x^2(n)$ and $Y (e^{j\omega})$ be the Fourier transform of y(n)  Then $Y(e^{j0})$ is

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

  2. 2

  3. 4

  4. $\dfrac{4}{3}$


Correct Option: D
Explanation:

Match the following and choose the correct combination:

 
Group 1 Group 2
E. continuous and aperiodic signal 1. Fourier representation is continuous and periodic
F. continuous and periodic signal 2. Fourier representation is discrete and periodic
G. discrete and aperiodic signal 3. Fourier representation is continuous and periodic
H. discrete and periodic signal 4. Fourier representation is discrete and periodic
  1. E – 3, F – 2, G – 4, H – 1

  2. E – 1, F – 3, G – 2, H – 4

  3. E – 1, F – 2, G – 3, H – 4

  4. E – 1, F – 3, G – 4, H – 2


Correct Option: D
Explanation:

For continuous and aperiodic signal, Fourier representation is continuous and periodic.

For continuous and periodic signal, Fourier representation is discrete and periodic.

For discrete and aperiodic signal, Fourier representation is discrete and periodic.

For discrete and periodic signal, Fourier representation is discrete and periodic.

The ABCD parameters of an ideal n : 1 transformer shown in figure are $ \left[ \begin{array} \ n & 0 \\ 0 & x \end{array} \right]

$. The value of X will be

  1. n

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

  3. n2

  4. $\dfrac{1}{n^2}$


Correct Option: B
Explanation:

Impedance Z as shown in the figure is

  1. $j29\Omega$

  2. $j9\Omega$

  3. $j19\Omega$

  4. $j39\Omega$


Correct Option: B
Explanation:

The h parameters of the circuit shown in figure are

  1. $\left[ \begin{array} \ 0.1 & 0.1 \\ -0.1 & 0.3 \end{array} \right]$

  2. $\left[ \begin{array} \ 10 & -1 \\ 1 & 0.05 \end{array} \right]$

  3. $\left[ \begin{array} \ 30 & 20 \\ 20 & 20 \end{array} \right]$

  4. $\left[ \begin{array} \ 10 & 1 \\ -1 & 0.05 \end{array} \right]$


Correct Option: D
Explanation:

In a series RLC circuit, R = 2k$\Omega$, L = 1H, and C = $\dfrac{1}{400} \mu F$. The resonant frequency is

  1. $2 \times 10^4 Hz$

  2. $\dfrac{1}{\pi} \times 10^4 Hz$

  3. $10^4 Hz$

  4. $2\pi \times 10^4 Hz$


Correct Option: B
Explanation:

For the circuit in the figure, the instantaneous current i1 (t) is

  1. $\dfrac{10\sqrt 3}{2} \angle 90^\circ Amps$

  2. $\dfrac{10\sqrt 3}{2} \angle -90^\circ Amps$

  3. $5 \angle 60^\circ Amps$

  4. $5 \angle -60^\circ Amps$


Correct Option: A
Explanation:

The first and the last critical frequency of an RC-driving point impedance function must respectively be

  1. a zero and a pole

  2. a zero and a zero

  3. a pole and a pole

  4. a pole and a zero


Correct Option: C
Explanation:

The condition on R, L and C such that the step response y(t) in figure has no oscillations, is

  1. $R \ge \dfrac{1}{2}\sqrt{\dfrac{L}{C}}$

  2. $R \ge \sqrt{\dfrac{L}{C}}$

  3. $R \ge 2\sqrt{\dfrac{L}{C}}$

  4. $R \ge \dfrac{1}{\sqrt{LC}}$


Correct Option: C
Explanation:

A square pulse of 3 volts amplitude is applied to C - R circuit shown in figure. The capacitor is initially uncharged. The output voltage v0 at time t = 2 sec is

  1. 3 V

  2. -3V

  3. 4 V

    • 4V

Correct Option: B
Explanation:

The maximum power that can be transferred to the load resistor RL of 100$\Omega$ from the voltage source of 5 V is __________

  1. 1 W

  2. 10 W

  3. 0.25 W

  4. 0.5 W


Correct Option: C
Explanation:

If R1 = R2 = R4 and R3 = 1. 1R in the bridge circuit shown in figure, then the reading in the ideal voltmeter connected between a and b is

  1. 0.238 V

  2. 0.138 V

  3. -0.238 V

  4. 1 V


Correct Option: C
Explanation:

For the circuit shown in figure, Thevenin's voltage and Thevenin's equivalent resistance at terminals a - b is

  1. $5V \ and \ 2\Omega$

  2. $7.5V \ and \ 2.5\Omega$

  3. $4V \ and \ 2\Omega$

  4. $3V \ and \ 2.5\Omega$


Correct Option: B
Explanation:

The following differential equation has $3\frac{d^2y}{dt^2}+4\bigg(\frac{dy}{dt}\bigg)^3+y^2 + 2 = x$

  1. degree = 2, order = 2

  2. degree = 1, order = 2

  3. degree = 4, order = 3

  4. degree = 2, order = 3


Correct Option: B
Explanation:

Order is the highest derivative term present in the equation and degree is the power of highest derivative term.

Order = 2, degree = 1

A fair dice is rolled twice. The probability that an odd number will follow an even number is

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

  2. $\frac{1}{6}$

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

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


Correct Option: D
Explanation:

Probability of coming odd number is $\frac{1}{2}$ and the probability of coming even number is $\frac{1}{2}$. Both the events are independent to each other, thus probability of coming odd number after as even number is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

A solution of the following differential equation is given by $\frac{d^2y}{dx^2}-5\frac{dy}{dx}+6y = 0$

  1. $y = e^{2x} + e^{-3x}$

  2. $y = e^{2x} + e^{3x}$

  3. $y = e^{-2x} + e^{3x}$

  4. $y = e^{-2x} + e^{-3x}$


Correct Option: B
Explanation:

The primary reason for the widespread use of silicon in semiconductor device technology is

  1. abundance of silicon on the surface of Earth

  2. larger bandgap of silicon in comparison to germanium

  3. favourable properties of silicon-dioxide (SiO2)

  4. lower melting point


Correct Option: A
Explanation:

Silicon is abundant on the surface of Earth in the from of SiO2.

The effect of current shunt feedback in an amplifier is to

  1. increase the input resistance and decrease the output resistance

  2. increase both input and output resistances

  3. decrease both input and output resistances

  4. decrease the input resistance and increase the output resistance


Correct Option: D
Explanation:

The given solution is right.                    Here  Ri  = input resistance without feedback                              Rif  = input resistance with feedback                              R of  = output resistance with feedback                  

A linear system is equivalently represented by two sets of state equations; $\bar X = AX + BU$ and W = CW + DU. The eigen values of the representations are also computed as $[\lambda]$ and $[\mu]$. Which of the following statements is true?

  1. $[\lambda] = [\mu] \ and \ X =W $

  2. $[\lambda] = [\mu] \ and \ X \ne W $

  3. $[\lambda] \ne [\mu] \ and \ X = W $

  4. $[\lambda] \ne [\mu] \ and \ X \ne W $


Correct Option: C
Explanation:

Let $A = \begin{bmatrix} \ 2 & -0.1 \ \ 0 & 3 \ \end{bmatrix} \hspace{0.2cm} and \hspace{0.2cm} A^{-1}\begin{bmatrix} \ \frac{1}{2} & a \ \ 0 & b \ \end{bmatrix}$, then (a + b) =

  1. $\dfrac{7}{20}$

  2. $\dfrac{3}{20}$

  3. $\dfrac{19}{60}$

  4. $\dfrac{11}{20}$


Correct Option: A
Explanation:

Which one of the following polar diagrams corresponds to a lag network?


Correct Option: D
Explanation:

The derivative of the symmetric function drawn in figure will look like


Correct Option: C
Explanation:

For x > 0 the slope of given curve is negative. Only (C) satisfy this condition.

The value of the integral $I = \frac{1}{\sqrt{2x}} \int\limits_0^\infty exp \Big(- \frac{x^2}{8}\Big)$ dx is

  1. 1

  2. $\pi$

  3. 2

  4. 2$\pi$


Correct Option: A
Explanation:

Many circles are drawn in a Smith chart used for transmission line calculations. The circles shown in figure represent

  1. unit circles

  2. constant resistance circles

  3. constant reactance circles

  4. constant reflection coefficient circles


Correct Option: C
Explanation:

The given figure represent constant reactance circle.

Given an orthogonal matrix $\begin{bmatrix} \ 1 & 1 & 1 & 1 \ \ 1 & 1 & -1 & -1 \ \ 1 & -1 & 0 & 0 \ \ 0 & 0 & 1 & -1 \end{bmatrix}$$\begin{bmatrix} \ AA^T \ \end{bmatrix}^{-1}$ is

  1. $\begin{bmatrix} \ \frac{1}{4} & 0 & 0 & 0 \ \ 0 & \frac{1}{4} & 0 & 0 \ \ 0 & 0 & \frac{1}{2} & 0 \ \ 0 & 0 & 0 & \frac{1}{2} \ \end{bmatrix}$

  2. $\begin{bmatrix} \ \frac{1}{2} & 0 & 0 & 0 \ \ 0 & \frac{1}{2} & 0 & 0 \ \ 0 & 0 & \frac{1}{2} & 0 \ \ 0 & 0 & 0 & \frac{1}{2} \ \end{bmatrix}$

  3. $\begin{bmatrix} \ 1 & 0 & 0 & 0 \ \ 0 & 1 & 0 & 0 \ \ 0 & 0 & 1 & 0 \ \ 0 & 0 & 0 & 1 \ \end{bmatrix}$

  4. $\begin{bmatrix} \ \frac{1}{4} & 0 & 0 & 0 \ \ 0 & \frac{1}{4} & 0 & 0 \ \ 0 & 0 & \frac{1}{4} & 0 \ \ 0 & 0 & 0 & \frac{1}{4} \ \end{bmatrix}$


Correct Option: C
Explanation:

$\text{From orthogonal matrix} \\ \begin{bmatrix} \ AA^T \ \end{bmatrix} = I \\ \text{Since the inverse of I is I, thus} \\ \begin{bmatrix} \ AA^T \ \end{bmatrix} = I^{-1} = I $

In what range should Re(s) remain so that the Laplace transform of the function $e^{(x + 2)t+5}$ exists?

  1. Re(s) > a + 2

  2. Re(s) > a + 7

  3. Re(s) < 2

  4. Re(s) > a + 5


Correct Option: A
Explanation:

$\text{We have} \hspace{0.2cm} f(t) = e^{(a+2)t+5} = e^5 . e^{(a+2)t} \\ \text{Taking laplace transform we get} \\ F(s) = e^5 \begin{bmatrix} \ \frac{1}{s - (a+2)} \end{bmatrix} \hspace{2cm} \text{Thus} \hspace{0.2cm}Re(s) > (a+2)$

For an npn transistor connected as shown in figure, VBE = 0.7 Volts. Given that reverse saturation current of the junction at room temperature 300°K is 10-13 A, the emitter current is

  1. 30 mA

  2. 39 mA

  3. 49 mA

  4. 20 mA


Correct Option: C
Explanation:

$I_E = I_s \bigg(\frac{V_m}{e m V_r} - 1\bigg) = 10^{-13} \bigg(\frac{0.7}{e^{1 \times 26 \times 10^{-1}}}\bigg) = 49 mA$

The circuit using a BJT with $\beta$ = 50 and VBE = 0.7 V is shown in figure. The base current IB and collector voltage VC are respectively

  1. $43 \mu A\hspace{0.2cm} and \hspace{0.2cm}11.4 \hspace{0.1cm} volts$

  2. $40 \mu A\hspace{0.2cm} and \hspace{0.2cm}16 \hspace{0.1cm} volts$

  3. $45 \mu A\hspace{0.2cm} and \hspace{0.2cm}11 \hspace{0.1cm} volts$

  4. $45 \mu A\hspace{0.2cm} and \hspace{0.2cm}10 \hspace{0.1cm} volts$


Correct Option: B
Explanation:

Match the following and choose the correct combination

Group 1 Group 2
E. Newton -Raphson method 1. Solving nonlinear equations
F. Runge-Kutta method 2. Solving linear simultaneous equations
G. Simpson's Rule 3. Solving ordinary differential equations
H. Gauss elimination 4. Numerical integration
5. Interpolation
6. Calculation of Eigen values
  1. E - 6, F - 1, G - 5, H - 3

  2. E - 1, F - 6, G - 4, H - 3

  3. E - 1, F - 3, G - 4, H - 2

  4. E - 5, F - 3, G - 4, H - 1


Correct Option: C
Explanation:

Newton - Raphson $\rightarrow \hspace{0.2cm}$ Method solving nonlinear eq.

Runge - kutta Method $\rightarrow \hspace{0.2cm}$ Solving ordinary differential eq.

Simpson's Rule $\rightarrow \hspace{0.2cm}$ Numerical Integration

Gauss eliminatio $\rightarrow \hspace{0.2cm}$ Solving linear simultaneous eq.

In the derivation of expression for peak percent overshoot, $M_p = exp \left( \dfrac{-\pi\xi}{\sqrt{1-\xi^2}} \right) \times 100 \% $, which of the following conditions is not required?

  1. System is linear and time invariant.

  2. The system transfer function has a pair of complex conjugate poles and no zeroes.

  3. There is no transportation delay in the system.

  4. The system has zero initial conditions.


Correct Option: C
Explanation:

The polar diagram of a conditionally stable system for open loop gain K = 1 is shown in figure. The open loop transfer function of the system is known to be stable. The closed loop system is stable for

  1. $k<5\ and \ \dfrac{1}{2} < k < \dfrac{1}{8}$

  2. $k<\dfrac{1}{8}\ and \ \dfrac{1}{2} < k < 5$

  3. $k<\dfrac{1}{8}\ and \ 5 < k$

  4. $k>\dfrac{1}{8}\ and \ k < 5$


Correct Option: B
Explanation:

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A ramp input applied to an unity feedback system results in 5% steady state error. The type number and zero frequency gain of the system are respectively

  1. 1 and 20

  2. 0 and 20

  3. 0 and $\dfrac{1}{20}$

  4. 1 and $\dfrac{1}{20}$


Correct Option: A
Explanation:

Both transistors T1 and T2 in figure have a threshold voltage of 1 Volt. The device parameters K1 and K2 of T1 and T2 are, respectively, 36 A/V2 and 9 A/V2. The output voltage V0 is

  1. 1 V

  2. 2 V

  3. 3 V

  4. 4 V


Correct Option: D

The Zener diode in the regulator circuit shown in figure has a Zener voltage of 5.8 Volts and a Zener knee current of 0.5 mA. The maximum load current drawn from this circuit ensuring proper functioning over the input voltage range between 20 and 30 Volts, is

  1. 23.7 mA

  2. 14.2 mA

  3. 13.7 mA

  4. 24.2 mA


Correct Option: A
Explanation:

$\text{The maximum load current will be at maximum input voltage i.e} \\ V_{max} = 30 V i.e \\ or \hspace{1cm} \frac{30 - 5.8}{1K} = I_L = 0.5 m \\ or \hspace{1cm} I_L = 24.2 - 0.5 = 23.7 mA$

Given the ideal operational amplifier circuit shown in figure indicate the correct transfer characteristics assuming ideal diodes with zero cut-in voltage.


Correct Option: B
Explanation:

$\text{Only one diode will be in ON conditions} \\ \text{When lower diode is in ON condition, then} \\ V_m = \frac{2k}{2.5k} V_{sat} = \frac{2}{2.5}10 = 8 V \\ \text{when upper diode is in ON condition} \\ V_m = \frac{2k}{2.5k} V_{sat} = \frac{2}{4}(-10) = -5 V$

An unity feedback system is given as $G(s) = \dfrac{K(1-s)}{s(s+3)}$ Indicate the correct root locus diagram.


Correct Option: A
Explanation:

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In an ideal differential amplifier shown in figure, a large value of RE

  1. increases both the differential and common-mode gains

  2. increases the common-mode gain only

  3. decreases the differential mode gain only

  4. decreases the common-mode gain only


Correct Option: D
Explanation:

$\text{Common mode gain} \\ A_{CM} = -\frac{R_C}{2R_E} \\ \text{And differential mode gain} \\ A_{DM} = -gmR_C \\ \text{Thus only common mode gain depends on $R_E$ and for large value of $$R_E it decreases.}$

The voltage eo indicated in figure has been measured by an ideal voltmeter. Which of the following can be calculated?

  1. Bias current of the inverting input only

  2. Bias current of the inverting and non-inverting inputs only

  3. Input offset current only

  4. Both the bias currents and the input offset current


Correct Option: C
Explanation:

A Silicon PN junction diode under reverse bias has depletion region of width 10 m. The relative permittivity of Silicon, $\epsilon_r = 11.7$ and the permittivity of free space $\epsilon_0 = 8.85 \times 10^{-12} F/m$. The depletion capacitance of the diode per square metre is

  1. $100 \mu F$

  2. $10 \mu F$

  3. $1 \mu F$

  4. $20 \mu F$


Correct Option: B
Explanation:

$C = \frac{\epsilon_0 \epsilon_r A}{d} \\ or \hspace{0.2cm} \frac{C}{A} = \frac{\epsilon_0 \epsilon_r}{d} = \frac{8.85 \times 10^{-12} \times 11.7}{10 \times 10^{-6}} = 10.35 \mu F$

The Op-amp circuit shown in the figure below is a filter. The type of filter and its cut-off frequency are respectively

  1. high pass, 10000 rad/sec

  2. low pass, 1000 rad/sec

  3. high pass, 1000 rad/sec

  4. low pass, 10000 rad/sec


Correct Option: C
Explanation:

At low frequency capacitor is open circuit and voltage acr s non-inverting terminal is zero. At high frequency capacitor act as short circuit and all input voltage appear at non-inverting terminal. Thus, this is high pass circuit.

$$\omega = \frac{1}{RC} = \frac{1}{1 \times 10^3 \times 1 \times 10^{-6} = }=1000 rad/sec$$

A double integrator plant, $G(s) = \dfrac{k}{s^2} \ H(s) = 1$ is to be compensated to achieve the damping ratio $\xi$ = 0.5, and an undamped natural, $\omega_n = 5\ rad/s$. Which one of the following compensator Gc(S) will be suitable?

  1. $\dfrac{s+3}{s+9.9}$

  2. $\dfrac{s+9.9}{s+3}$

  3. $\dfrac{s-6}{s+8.33}$

  4. $\dfrac{s+6}{s}$


Correct Option: A
Explanation:

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Two identical and parallel dipole antennas are kept apart by a distance of $\frac{\lambda}{4}$ in the H-plane. They are fed with equal current but the right most antenna has a phase shift of + 90o. The radiation pattern is given as


Correct Option: A
Explanation:

Given rd = 20 k$\Omega$, IDSS = 10 mA, Vp = -8 V.

ID and IDS under DC conditions are respectively

  1. 5.625 mA and 8.75 V

  2. 4.500 mA and 11.00 V

  3. 7.500 mA and 5.00 V

  4. 6.250 mA and 7.50 V


Correct Option: A
Explanation:

Given rd = 20 k$\Omega$, IDSS = 10 mA, Vp = -8 V.

Zi and Z0 of the circuit are respectively

  1. 2 M$\Omega$ and 2 k$\Omega$

  2. $2 M \Omega \hspace{0.2cm} and \hspace{0.2cm} \frac{20}{11} k \Omega$

  3. Infinity and 2 k$\Omega$

  4. Infinity and $\frac{20}{11} k \Omega$


Correct Option: B
Explanation:

Given, rd = 20 k$\Omega$, IDSS = 10 mA, Vp = -8 V

Transconductance in milli-Siemens (mS) and voltage gain of the amplifier are respectively

  1. 1.875 mS and 3.41

  2. 1.875 mS and - 3.41

  3. 3.3 mS and -6

  4. 3.3 mS and 6


Correct Option: B
Explanation:

The open loop transfer function of a unity feedback is given by $G(s) = \dfrac{3e^{-2s}}{s(s+2)}$

The gain and phase crossover frequencies in rad/sec are respectively

  1. 0.632 and 1.26

  2. 0.632 and 0.485

  3. 0.485 and 0.632

  4. 1.26 and 0.632


Correct Option: D
Explanation:

The open loop transfer function of a unity feedback is given by $G(s) = \dfrac{3e^{-2s}}{s(s+2)}$

Based on the above results, the gain and phase margins of the system will be

  1. -7.09 dB and 87.5

  2. 7.09 dB and 87.5

  3. 7.09 dB and - 87.5

  4. -7.09 dB and - 87.5


Correct Option: D
Explanation:

The band gap of Silicon at room temperature is

  1. 1.3 eV

  2. 0.7 eV

  3. 1.1 eV

  4. 1.4 eV


Correct Option: C
Explanation:

For Silicon at 0 K,
At any temperature
$E_{gT} = E_{g0} - 3.6 \times 10^{-4}T$
At T = 300 K,
$E_{g300} = 1.21 - 3.6 \times 10^{-4} \times 300 = 1.1 eV$
This is standard value, that must be remembered.

The cascode amplifier is a multistage configuration of

  1. CC-CB

  2. CE-CB

  3. CB-CC

  4. CE-CC


Correct Option: B
Explanation:

The CE configuration has high voltage gain as well as high current gain. It performs basic function of amplifications. The CB configuration has lowest $R_i$ and highest $R_0$. It is used as last step to match a very low impedance source and to drain a high impedance load. Thus cascade amplifier is a multistage configuration of CE-CB

Despite the presence of negative feedback, control systems still have problems of instability because the

  1. used components have nonlinearities

  2. dynamic equations of the subsystems are not known

  3. mathematical analysis involves approximations

  4. system has large negative phase angle at high frequencies


Correct Option: A
Explanation:

Despite the presence of negative feedback, control systems still have problems of instability because used components have nonlinearity. There are always some variations as compared to ideal characteristics.

The input resistance Ri of the amplifier shown in the figure is

  1. $\frac{30}{4} k \Omega$

  2. 10 k$\Omega$

  3. 40 k$\Omega$

  4. infinite


Correct Option: B
Explanation:

Since the inverting terminal is at $\pi$ virtual ground, the current flowing throught the voltage source is

$$ dkfhdkjfh$$

The region of convergence of Z-transform of the sequence $\bigg(\frac{5}{6}\bigg)^n u(n) - \bigg(\frac{6}{5}\bigg)^n u(n - 1)$ must be

  1. $|z| < \frac{5}{6}$

  2. $|z| > \frac{6}{5}$

  3. $\frac{5}{6} < |z| < \frac{6}{5}$

  4. $\frac{6}{5} < |z| < \infty$


Correct Option: C
Explanation:

Given the matrix $\begin{bmatrix} \ -4 & 2 \\ \ 4 & 3 \\ \end{bmatrix}$, the eigen vector is

  1. $\begin{bmatrix} \ 3 \\ \ 2 \\ \end{bmatrix}$

  2. $\begin{bmatrix} \ 4 \\ \ 3 \\ \end{bmatrix}$

  3. $\begin{bmatrix} \ 2 \\ \ -1 \\ \end{bmatrix}$

  4. $\begin{bmatrix} \ -1 \\ \ 2 \\ \end{bmatrix}$


Correct Option: C
Explanation:

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