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Electronics and Communication Engineering - Exam Questions Papers

Exercise : Exam Questions Papers - Exam Paper 8
31.
The logic circuit given below converts a binary code y1 y2, y3 into
Excess - 3 code
Gray code
BCD code
Hamming
Answer: Option
Explanation:

Let y1 = 1, y2 = 0, y3 = 1

then x1 = 1

x2 = y1 ⊕ y2 ⇒ 1 ⊕ 0 ⇒ 1

x3 = x2 ⊕ y3 = 1 ⊕ 0 = 1

which is gray code conversion.

32.
A system described by the following differential equation is initially at rest. For input x(t) - 2u(t), the output y(t) is :
(1 - 2e-t + e -2t)
(1 + 2e-t - 2e -2t)u(t)
(0.5 + e-t + 1.5e -2t)u(t)
(0.5 + 2e-t + 2e -2t)u(t)
Answer: Option
Explanation:

Taking L.T.

s2y(s) + 3sy(s) + 2y(s) = x(s)

I.L.T = (1 - 2e-t + e-2t)U(t).

33.
A transmission line terminates in two branches, each of length λ/4, as shown. The branches are terminated by 50Ω loads. The lines are lossless and have the characteristics impedances shown. Determine the impedance Z1 as seen by the source
200Ω
100Ω
50Ω
25Ω
Answer: Option
Explanation:

.

34.
The Nyquist plot of
will start from (ω = ∞) in the first quadrant and will terminate (ω = 0) in the second quadrant
will start from (ω = ∞) in the fourth quadrant and will terminate (ω = 0) in the second quadrant
will start form (ω = ∞) in the second quadrant and will terminate (ω = 0) in the fourth quadrant
will start from (ω = ∞) in the first quadrant and will terminate (ω = 0) in the fourth quadrant
Answer: Option
Explanation:

35.
The counter shown in figure is built using 4 negative edge triggered toggle FFs. The FFs can be set synchronously when R = 0. The combinational logic required to realize a modulo 13 counter is
F= Q4 Q3 Q2 Q1
F= Q4 + Q3 + Q2 + Q1
F = Q4 + Q3 + Q2 + Q1
F = Q4 Q3 Q2 Q1
Answer: Option
Explanation:

According to figure, output of combinational logic is applied to R of all FF

N = 13 = (1101)2

The logic required is Q4 Q3 Q2 Q1 The gate used is a NAND.

Y = Q4 + Q3 + Q2 + Q1 .

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