Quiz Electronics and Communication Engineering

Exam: GATE

Topic: Miscellaneous

Date: 05/08/2020

Each Question carries 2 Mark

Negative Marking: 1/3

Time: 20 Minutes

Q1. If X(z) = (z+z^(-3))/(z+z^(-1) ), then x(n) series has

(a) alternate 0’s

(b) alternate 1’s

(c) alternate 2’s

(d) alternate -1’s

Q2. Consider the following statements:

The frequency response of a control system has very sharp cut off characteristics. This implies that:

1. It has large peak resonance.

2. It has large bandwidth.

3. It is a less stable system.

Which of the statements given above is/are correct?

(a) 1 only

(b) 2 and 3

(c) 1 and 3

(d) 1, 2 and 3

Q3. A coil having a resistance of 10 Ω and inductance of 31.8 mH is connected to 230V, 50 Hz supply. The circuit current will be

(a) 16.20 A

(b) 1.62A

(c) 16.26A

(d) 62A

Q4. If (2.3)_4+(1.2)_4=(y)_4; what is the value of y?

(a) 10.1

(b) 10.01

(c) 10.2

(d) 1.02

Q5. Rectification efficiency of a full wave rectifier without filter is nearly equals to:

(a) 51%

(b) 61%

(c) 71%

(d) 81%

Q6. Assertion(A): The stability of the system is assured if the Region of Convergence(ROC) includes the unit circle in the z-plane.

Reason(R): For a causal stable system all the poles should be outside the unit circle in the z-plane.

(a) Both A and R are True and R is the correct explanation of A.

(b) Both A and R are True but R is not the correct explanation of A.

(c) A is True but R is False.

(d) A is False but R is True.

Q7. Application to Norton’s theorem to a circuit yields

(a) equivalent impedance source

(b) equivalent impedance

(c) equivalent current source and impedance in series

(d) equivalent current source and impedance in parallel

Q8. The extremely high input impedance of a MOSFET is primarily because of

(a) Absence of its channel

(b) Depletion of current carriers

(c) Extremely small leakage current of its gate capacitor

(d) Negative V_GS

Q9. The bandwidth required for the transmission of a PCM signal increases by a factor of ____________ when the number of quantization levels is increased from 4 to 64.

(a) 3

(b) 2

(c) 6

(d) 1

Q10. A combination of left-handed circular polarization and right-handed polarization leads to _______ Polarization.

(a) parabolic

(b) elliptical

(c) circular

(d) none of the above

SOLUTIONS

S1. Ans.(a)

Sol. Given:

⇒ X(z) = (z+z^(-3))/(z+z^(-1) )

⇒ X(z) = ((1+z^(-4)))/((1+z^(-2)))

⇒ X(z) = (1+z^(-4))〖(1+z^(-2))〗^(-1)

⇒ X(z) = (1+z^(-4))(1 – z^(-2) + z^(-4) – . . . . . . . . . )

⇒ X(z) = (1) (1 – z^(-2) + z^(-4) – . . . . . . . . . ) +

(z^(-4))(1 – z^(-2) + z^(-4) – . . . . . . . . . )

⇒ X(z) = (1 – z^(-2) + z^(-4) – . . . . . . . . . ) + (z^(-4) – z^(-6) + . . . . . . . )

⇒ X(z) = (1 – z^(-2) + 2z^(-4) – 2z^(-6). . . . . . . . . )

So, x(n) = 𝛿(n) – 𝛿(n – 2) + 2𝛿(n – 4) – 2𝛿(n – 6) + . . . . .

Hence, x(n) has alternate zeros i.e. at n = 0, 2, 4, 6, . . . .

S2. Ans.(c)

S3. Ans.(c)

Sol. Given:

Resistance of coil is R = 10 Ω

Inductive = 31.8 mH

frequency(f) = 50 Hz

V = 230 volts

Inductive reactance is

⇒ X_L = ωL = 2πfL

⇒ X_L = 2 × 3.14 × 50 × 31.8 × 10^(-3)

⇒ X_L = 2 × 50 × 10^(-3) × 99.582

∴ X_L = 10Ω

Net Coil impedance is

⇒ Z = √((10)^2+(10)^2 ) = √200

∴ Z = 14.14 Ω

Circuit current is

⇒ I =V/Z

⇒ I = 230/14.14

So, I = 16.26 A

S4. Ans.(a)

Sol. It is having base as 4 so there will be total 4 numbers in this i.e., 0, 1, 2 & 3.

S5. Ans.(d)

Sol. Efficiency(η) is the ratio of dc output power to ac input power.

The maximum efficiency of full wave rectifier is 81%.

S6. Ans.(c)

Sol. For a causal system all the poles should lie inside the unit circle in z-plane.

S7. Ans.(d)

Sol. From Norton Theorem, we get Norton current source and Norton equivalent resistance in parallel to it.

S8. Ans.(a)

Sol. Channel forming in the MOSFET will decrease the input impedance of the MOSFET and in absence of channel it achieves extremely high input impedance.

S9. Ans.(a)

Sol. It is given that the number of quantization levels is increased from 4 to 64.

We know that in PCM,

⇒ BW = nfS

And, n = log2L

For L= 4,

⇒ n1 = log24

So, n1 = 2 bits

For L = 64,

⇒ n2 = log264

So, n2 = 6 bits

Now,

⇒ (BW)1 = n1fS = 2fS

⇒ (BW)2 = n2fS = 6fS

So, BW_2/BW_1 = (6f_S)/〖2f〗_S = 3

Therefore, BW increases by 3 times.

S10. Ans (b)

Sol. A circularly polarized wave can be in one of two possible states, right circular polarization in which the electric field vector rotates in a right-hand sense with respect to the direction of propagation and left circular polarization in which the vector rotates in a left-hand sense. Circular polarization is a limiting case of the more general condition of elliptical polarization. The sum of a left-handed circular polarization and right-handed polarization leads to Elliptical Polarization.