Electronics and Communication Engineering - Exam Questions Papers

46. 

The transfer function of a system is . The characteristic equation of the system is :

A. 2s2 + 6s + 5 = 0
B. (s + l)2 (s + 2) = 0
C. 2s2 + 6s + 5 + (s + 1 )2(s + 2) = 0
D. 2s2 + 6s + 5 - (s + l)2(s + 2) = 0

Answer: Option C

Explanation:

Characteristic equation = 1 + GH = 0

(s + 1)2(s + 2) + (2s2 + 6s + s) = 0

Apply R-H.


47. 

Given f = GEF + KHIJ + LMON + TUVWXYZ and

  1. E + F + G = LMN
  2. V + W = U . Z . Y . X . T
  3. NAND B
  4. (UVW ⊕ WVU) = KJ(XY ⊕ XY)

Then f is equivalent to

A. 0
B. 1
C. EF + UZ + HI
D. None of these

Answer: Option B

Explanation:

(UVW ⊕ WVU) = 1 = KJ(XY ⊕ XY)

KJ = 1

NAND B

HI = 1

Since f = GEF + 1 + LMON + TUVWXYZ = 1.


48. 

If V = 4 in the figure, the value of IS is given by,

A. 6 A
B. 5/2 A
C. 12 A
D. None of these

Answer: Option A

Explanation:

Using Nodal analysis

At node V1

At node V2

2V2 - V1 = 0

V1 = 2V2 Put V2 = V = 4

V1 = 8 V

Put in equation (i), IS = x 8 - 4 = 10 - 4 = 6 A.


49. 

Choose correct option for a stable system :

  1. Roots of the characteristic equation of the system are real and negative
  2. Area within the impulse response is finite

A. I is true, II is true
B. I is true, II is false
C. I is false, II is true
D. I is false, II is false

Answer: Option A

Explanation:

From Statement 1: Roots of the characteristics equation of the system are real and negative i.e. the poles are on the left half plane.

Hence the system is stable.

From Statement 2: Area within the impulse response is finite i.e. it is finite duration signal. It produces a bounded output. Hence system is stable.

So both the statements are true.


50. 

The number of distinct Boolean expressions of 3 variables is

A. 8
B. 256
C. 64
D. 7

Answer: Option B

Explanation:

22n = 223 ⇒ 28 = 256.


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