# ECE - Questions and solutions set-7

Electronics and communication Engineering Questions and solutions set-7
Question:-
Option (A)
25
Option (B)
100
Option (C)
200
Option (D)
400
Correct Option:
B
Question Solution:
Fm = 500 Hz, fs = 1000 Hz
Frequency resolution =(fs/N)
10 = (1000/N) → N = 100
Question:-
Option (A)
12.5 ohm
Option (B)
50 ohm
Option (C)
25 ohm
Option (D)
250 ohm
Correct Option:
C
Question Solution:
if ZL = 0 → (K = -1)
So angle of reflection will be 180°
Now VSWR = (1+|K|)/(1-|K|) =3
|K| = (1/2) → K = ± (1/2)
but here negative sign of K is taken because position of Minima is not Changed.
So K = -(1/2) = (RL-75)/(RL+75)
→ RL + 75 =- 2RL + 2 x 75
→ 3RL = 75
RL = 25 ohm
Question:-
Option (A)
5 & 5.03
Option (B)
6 & 5.83
Option (C)
5 & 6.64
Option (D)
6 & 5.32
Correct Option:
D
Question Solution:
Minm bits are 2n = 40 → n = 6
Entropy = log2 40 = 5.32
Question:-
Option (A)
(t2+t)
Option (B)
(2t+t)
Option (C)
2t + t2 + t + 1
Option (D)
Data insufficient
Correct Option:
A
Question Solution:
x(t) = t3 + 2t + 4t2
X(S) = (6/S4) + (2/S2) + (4.2/S3) = {(6+ 2S2 + 8S) / S3}
Y(t) = t2 + t
Y(S) = (2/S3) + (1/S2) = {(2+S) /S3}
So H(S) = (Y(S)/X(S) ) = {(S+2) / (2S2 + 8S +6)}
New input x(t) is x1(t) = (3t2 + 8t + 2)
X1(S) = (3.2/S3) + (8/S2) +(2/S) = {(6+ 8S+ 2S2) / S3}
New output will be
Y1(S) = {(S+2) /S3} = (1/S2) + (2/S3)
Y1(S) =(1/S2) + (2/S3)
Y1 (t) = t+ t2