ECE - Questions and solutions set-4

Electronics and communication Engineering Questions and solutions set-4
Question:-
Option (A)
3.0 sec
Option (B)
4.0 sec
Option (C)
5.0 sec
Option (D)
6.0 sec
Correct Option:
C
Question Solution:
H(S) = (K/1+τS)
For Step i/p Y(S) = K/[S(1+τS)] = K[(1/S)- τ/(1+τS)]
Þ y(t) = K(1- e-t/τ) u(t)
At t = ∞, y(t) = 2 Þ K = 2
Now 1.1 = 2 (1 – e-4/τ) (at t=4 y(t) is 1.1)
Þ τ= 5.0 sec
Question:-
Option (A)
27 dB
Option (B)
30 dB
Option (C)
33 dB
Option (D)
34 dB
Correct Option:
D
Question Solution:
If B.W. is increased, 4 times then effect of noise will also become 4 times. So,
S/N ratio will also decrease by 4 times.
Earlier SNR = 104, Now SNR = 2500
So, SNR in dB = 10 log 2500 = 34dB
Question:-
Option (A)
0.04 m2 v-1 sec-1 and 1.74×1022 m-3
Option (B)
0.4 m2 v-1 sec-1 and 1.74×1022 m-3
Option (C)
0.04 m2 v-1 sec-1 and 1.74×1018 m-3
Option (D)
4.0 m2 v-1 sec-1 and 1.74×1018 m-3
Correct Option:
A
Question Solution:
RH=(1/nq)
\ n= 1 / (3.6x10-4x1.6x10-19)
N = 10 / (3.6x1.6)=1.7x1022/m3

Mobility (μ) = σ. RH
=(RH/ρ)=(3.6x10-4/9x10-4)=0.04
Question:-
Option (A)
Low pass filter
Option (B)
High pass filter
Option (C)
Band pass filter
Option (D)
Band stop filter
Correct Option:
B
Question Solution:
When w = 0, v0 = 0
w = ¥, v0 = finite
Question:-
Option (A)
0.125 v
Option (B)
0.250 v
Option (C)
0.525 v
Option (D)
0.625 v
Correct Option:
D
Question Solution:
For step size i/p is 001
But v0 =(VR/2N)(bN-12N-1+-------+b0.1)
Here b2=b1=0 and b0=0
=(5/8)(0.22+0.21+1.20)
=0.625 volt
Question:-
Option (A)
12.5 m sec
Option (B)
25 m sec
Option (C)
37.5 m sec
Option (D)
75 m sec
Correct Option:
A
Question Solution:
Vx=mE=(mv/d)=(d/t)
Þ t=(d2/mv)={(0.5x10-3)2m2/((0.2m2/vsec)x100x10-3)}
T=12.5 msec