Question 8.6: An amplifier has an open-loop voltage amplification of A = 1...

Gain of the feedback amplifier, A_{f}, is calculated as

Let the open-loop gain, A change to A_{1} due to ageing. This change has been expressed in dB.

Therefore,                 20\log _{10}\left|\frac{A}{A_{1}} \right|=6

or,                  \frac{A}{A_{1}}=2

or,                    A_{1}=\frac{A}{2}=\frac{1,000}{2}=500

Now,                let A_{f_{1}}=\frac{A_{1}}{1+A_{1}β}=\frac{500}{1+500 ×0.01}=\frac{500}{6}=83.33

The percentage change in gain of the feedback amplifier is calculated as

Percentage change = \left\lgroup\frac {A_{f}-A_{f_{1}}}{A_{f}} \right\rgroup\times 100

= 8.4.