Electronics and Communication Engineering - Communication Systems - Discussion
Discussion Forum : Communication Systems - Section 2 (Q.No. 28)
28.
The open loop gain of an amplifier is 50 but is likely to change by 20%. If negative feedback with β = 0.1 is used, the change in gain will be about
Answer: Option
Explanation:
.
Change in gain is about 4%.
Discussion:
3 comments Page 1 of 1.
Rohini sharma said:
9 years ago
How the change in gain is 4%? Explain it.
Raji said:
9 years ago
It is negative feedback,
So,
Af = Af/(1 + b * Af).
= 8.333.
Similarly for 20%;
50 * 20/100 = 10.
So, gain = 10/(1 + 10 * .1) = 5.
So 8.33 - 5 = 3.33,
Nearly 4.
So,
Af = Af/(1 + b * Af).
= 8.333.
Similarly for 20%;
50 * 20/100 = 10.
So, gain = 10/(1 + 10 * .1) = 5.
So 8.33 - 5 = 3.33,
Nearly 4.
Shadan said:
7 years ago
@ALL.
Av1=50(Given)
Beta(B)=0.1(Given)
so 20% of 50 is 10 then new Av2=50-10=40.
Now,
Avf1=Av/1+BAv=8.33.
Avf2= Av/BAv =8.
So change in gain is;
8.33-8/8,
0.0412 *100.
=4.1200%.
=4%.
So, the answer C is right.
Av1=50(Given)
Beta(B)=0.1(Given)
so 20% of 50 is 10 then new Av2=50-10=40.
Now,
Avf1=Av/1+BAv=8.33.
Avf2= Av/BAv =8.
So change in gain is;
8.33-8/8,
0.0412 *100.
=4.1200%.
=4%.
So, the answer C is right.
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