Question 31 to 40 Solutions

GATE-EC 2023 Paper Solutions | Authentic and Elaborative Question 31 to 40 Solutions

Q.31. The signal-to-noise ratio (SNR) of an ADC with a full-scale sinusoidal input is given to be 61.96 dB. The resolution of the ADC is __ bits (rounded off to the nearest integer).

Ans. 10

Explanation :- Signal-to-noise ratio (SNR) is a measure used to quantify the level of desired signal relative to the level of background noise or unwanted interference in a system. It is a ratio of the power or amplitude of the signal of interest to the power or amplitude of the noise.

For the sinusoidal input , the signal to noise ratio (SNR) is given as,

SNR = 1.76 + 6.02 n dB

61.96 dB = 1.76 +6.02 n dB

6.02 n = 61.96 – 1.76

n= 10 bits

The resolution of ADC is 10 bits.

Q.32. In the circuit shown below, the current i flowing through 200 Ω resistor is __ mA (rounded off to two decimal places).

Ans. 1.36

Video Explanation:

Explanation :- The given diagram is,

Applying source transformation,

Changing the current source into voltage source

By Nodal Analysis,

\frac{V_{a}-2}{2k\Omega }+\frac{V_{a}}{2k\Omega }+\frac{V_{a}+1}{1.2k\Omega }=1 \: mA

V_{a}\left [ \frac{1}{2}+\frac{1}{2} +\frac{1}{1.2} \right ]= 1+\frac{2}{2}-\left ( \frac{1}{1.2} \right )=0.636 V

V_A= 0.636 V

The current through 200 Ω resistor is given by,

i=\frac{V_{A}+1}{1.2}=\frac{0.636+1}{1.2}=1.36 mA

The current flowing through the 200 ohm resistor is 1.36 mA

Q.33. For the two port network shown below, the [Y]-parameters is given as

[Y]=\frac{1}{100}\begin{bmatrix}2 & -1\\-1 & 4/3\end{bmatrix}S

The value of load impedance Z_L, in Ω, for maximum power transfer will be _ (rounded off to the nearest integer).

Ans. 80

Explanation :- Given that $[Y]=\frac{1}{100}\begin{bmatrix}2 & -1\\-1 & 4/3\end{bmatrix}S$

For the given Y parameter the two port network will be

Y₁₁= Y_a+Y_b= 2/100

Y₁₂= Y₂₁=-Y_b = $-\frac{1}{100}$

Y₂₂ = Y_b + Y_c = $\frac{4}{300}$

Solving these, we get

Y_b= 1/100 s

Y_a= 1/100 s

Y_c= 1/300 s

The network becomes,

The Z_th is given as,

Z_TH = 60 + [(20+10)||60]

Z_TH= $60+\frac{30\times 60}{30+60}= 80 \; \Omega$

Now, for maximum power transfer ,

Z_L=Z_Th= 80 Ω

Q.34. For the circuit shown below, the propagation delay of each NAND gate is 1 ns. The critical path delay, in ns, is __ (rounded off to the nearest integer).

Ans. 2

Explanation :- In general, The critical path delay is the maximum amount of time it takes for a signal to propagate through the longest path in a circuit, from the input to the output.

The given circuit can be drawn as,

The critical path delay = 1ns+1ns = 2 ns

Q.35. In the circuit shown below, switch S was closed for a long time. If the switch is opened at ???? = 0, the maximum magnitude of the voltage V_R, in volts, is _ (rounded off to the nearest integer).

Ans. 4

Explanation :- Let us consider two cases,

Case-I :- At t=0-

We will close the switch S, and inductance is deactivated,

The current in the inductance is given as i_L

i_{L}(0^{-})=\frac{2}{1}=2 A

Case-II :- At t=0+

The voltage is given as VR. This voltage is present behind the resistance.

V_R= -2 x 2 = -4

Magnitude of Voltage VR.

|V_R| = 4

The maximum magnitude of the voltage V_R in volts is 4.

Q.36. A random variable X, distributed normally as ????(0,1), undergoes the transformation Y=h(X), given in the figure. The form of the probability density function of ???? is
(In the options given below, ????,????,???? are non-zero constants and ????(????) is piece-wise continuous function)

a) $a\delta (y-1)+b\delta (y+1)+g(y)$

b) $a\delta (y+1)+b\delta (y)+c\delta (y-1)+g(y)$

c) $a\delta (y+2)+b\delta (y)+c\delta (y-2)+g(y)$

d) $a\delta (y+2)+b\delta (y-2)+g(y)$

Ans. (b)

Explanation :- Given that X= N(0,1)

f_{x}(x)=\frac{1}{\sqrt{2\pi }}e^{-x^{2}/2}

From the graph,

Y=-1; x ≤ -2

0 ; -1 ≤ x ≤ 1

1; x ≥ 2

x+1 ; -2 ≤ x ≤ -1

x-1 ; 1 ≤ x ≤ 2

Y is a mixed random variable as it is taking both the discrete set of values and a continuous range of values.

From the given options, the probability density function of ‘Y’ will be,

f_y(y) = $a\delta (y+1)+b\delta (y)+c\delta (y-1)+g(y)$

Q.37. The value of line integral $\int_{P}^{Q}(z^{2}dx+3y^{2}dy+2xzdz)$ along the straight line joining the points P(1,1,2) and Q(2,3,1) is

a) 20

b) 24

c) 29

d) -5

Ans. (b)

Explanation :- Given that , $\int_{P}^{Q}(z^{2}dx+3y^{2}dy+2xzdz)$

The line integral is joining the points P(1,1,2) and Q(2,3,1) is

= \int_{P(1,2)}^{P(2,1)}z^{2}dx+2xydz+\int_{y=1}^{3}3y^{2}dy

Solving the integration we get,

(xz^{2})^{(2,1)}_{(1,2)}+(y^{3})^{3}_{1}

= (2 x 1² – 1 x 2²) +(3³-1³)

= -2+26 = 24

The correct option is (b)

Q.38. Let ???? be an ????×1 real column vector with length $l=\sqrt{x^{T}x}$ The trace of the matrix ????=????????^???? is

a) l²

b) $\frac{l^{2}}{4}$

c) l

d) $\frac{l^{2}}{2}$

Ans. (a)

Explanation :- To determine the trace of the matrix P = xx^T, where x is an n×1 real column vector, we need to consider the definition of the trace and properties of matrix multiplication.

Given that, $l=\sqrt{x^{T}x}$ , ????=????????^????

Trace of (p) = Σ P(i,i) from i=1 to n

= Σ x(i) x(i)) from i=1 to n

= Σ x(i)² from i=1 to n

P=xx^T

= $\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\;;\\x_{n}\end{bmatrix}[x_{1}x_{2}x_{3…x_{n}}]$

P= x₁² + x₂²+….+ x_n² = l²

Option (a) is the correct answer.

Q.39. The $\frac{V_{out}}{V_{in }}$ of the circuit shown below is ,

a) $-\frac{R_{4}}{R_{3}}$

b) $\frac{R_{4}}{R_{3}}$

c) $1+\frac{R_{4}}{R_{3}}$

d) $1-\frac{R_{4}}{R_{3}}$

Ans. (a)

Explanation :- The given circuit diagram is,

From the fig, A₁ is the inverting amplifier and A₂ is the non-inverting amplifier,

The first input voltage is given by, $V_{01} =\frac{-R_{2}}{R_{1}}V_{in}$

The second input voltage is given by,

V_{02} =\left ( 1+\frac{R_{2}}{R_{1}} \right )V_{in}

There is third inverting amplifier A3, which is an inverting summing amplifier

V_{out}=\frac{-R_{4}}{R_{3}}V_{01}-\frac{R_{4}}{R_{3}}V_{02}

Putting values of V₀₁and V₀₂

V_{out}=\frac{-R_{4}}{R_{3} }\left [ \frac{R_{2}}{R_{1}}V_{in} +\left ( 1+\frac{R_{2}}{R_{1}} \right )V_{in}\right ]

V_{out}=\frac{-R_{4}}{R_{3}}V_{in}

The gain is the ratio of output voltage to the input voltage . It is given by,

\frac{V_{out}}{V_{in}}=\frac{-R_{4}}{R_{3}}

Therefore, option (a) is the correct answer.

Q.40. In the circuit shown below, D₁ and D₂ are silicon diodes with cut-in voltage of 0.7 V. V_IN and V_OUT are input and output voltages in volts. The transfer characteristic is