Electronics and Communication Engineering - Exam Questions Papers

16. 

A conductor in x-y plane and having length 1 m is moving with a velocity V = (2i + 3j + k) m/sec. A magnetic induction field B = (i + 2j) Wb/m2 is applied to the conductor. The potential difference between the ends of the conductor is

A. 0
B. 4.88 V
C. 6 V
D. None of these

Answer: Option C

Explanation:

= - (V x B) = - (2i + 3j + k) x (i + 2j)

|| = (E2x + E2y + E2z) = 6 v/m

Potential Difference across 1 m length = 6.


17. 

A 1000 kHz carrier wave modulated 40% at 4000 Hz is applied to a resonant circuit tuned to a carrier frequency and having Q = 140. What is the degree of modulation after passing the wave through this circuit?

A. 0.4
B. 0.2
C. 0.27
D. 0.554

Answer: Option C

Explanation:

Resulting depth of modulation is given by :

when δ =

fc = 1000 x 103 Hz

fm = 4 x 103 Hz

δ =

m0 =

= = 0.27 .


18. 

The VTH at terminals A and B is equal to,

A. 0 V
B. 5 V
C. - 5 V
D. None of these

Answer: Option B

Explanation:

Apply KVL to loop 1

5V - 3I1 - 2(I1 - I2) - 5V = 0

I2 = 0

5 - 3I1 - 2I1 - 5V = 0

I1 = 0

VAB = 5V.


19. 

If , then y has a

A. maximum at x = e
B. minimum at x = e
C. maximum at x = e-1
D. minimum at x = e-l

Answer: Option A

Explanation:

.


20. 

If . dt, then F[y(t)] is

A. - jsgn(f). x(f)
B. j2pf . e-jptf . x(f)
C.
D.

Answer: Option A

Explanation:

⇒ F{Y(t)} = Y(f) = X(f) x

sin (t) ⇌

By duality :⇌ sgn + (- f)

jsgn (- f)

⇌ - j sgn (f)

∵ sgn(- f) = - sgn(f)

⇒ Y(f) = X(f). [- j sgn(f)].