# Amplifier Feedback Gain

I'm stuck on the part b of this tutorial question. I thought I understood the theory, but I can't get the right answer. For part a I got: Af(jw) = 10^3 * jw / (3jw + 200)

From my understanding, the formula I should use for part b is:

Af = s*Am / (s + wL), where Am is the mid-high frequency gain and wL is the lower 3dB frequency.

However this gives me an Am of 10^3 and a wL of 200, which doesn't match the answers.

If anyone could show me where I've gone wrong I'd be extremely grateful.

Edit: I've realized part of my mistake. I'm looking for the NEW Am and wL, meaning the values for when I include feedback. According to my notes I should find the original values from A(s) using the formula: A(s) = sAm / (s + wL). So this gives me Am = 10^3 and wL = 200.

Now I need to find the new Am using this formula: Midband gain = Am / (1 + Am*B). This gives me 333.3, which is the correct answer.

Next formula I am supposed to use: wfL = wL / (1 + AmB), where wfL is the new value for wL now that feedback is included. I used 200 as my value for wL and 0.002 as my value for B. I tried both 10^3 and 333.3 as values for Am, giving me wfL = 66.7 and wfL = 120. I still cannot get the right answer for wfL; the answer is supposed to be 10.61Hz.

I've finally figured out what I was doing wrong. To see how I got the correct answer for the midband gain, look at the edit at the bottom of my question. As for finding wfL, it turns out 66.7 was the correct answer, I just hadn't converted it from radians/second to hertz. 66.7rad/s = 10.61Hz, which is the correct answer.

• +1 for providing an answer. Don't forget you can accept your own answer (if you think it's correct!). – Transistor Jul 29 '16 at 1:29

Part (a): For a block A(s) with negativ feedback (factor 0.002) we apply the classical formula from H. Black (closed loop gain Acl):

Acl(s)=A(s)/[1+0.002*A(s)]=1/[1/A(s)+0.002].

Now you can insert the expression for A(s), set s=jw and re-arrange the resulting expression to get the classical form similar to the form of A(jw).

(I think, your result is not correct).

Part (b): The 3dB frequency for the open-loop block can simply be identified as wo=200 (rad/sec). The same principle can be used for the closed-loop system (if the transfer function with feedback has the proper normalized form).

I must admit that I don`t know the meaning of the term "mid-high frequency gain". Perhaps the high-frequency gain is wanted (where the gain has settled) knowing that for very large frequencies the gain will drop again (however, not according to the transfer function which is not realistic for practical systems).

• Thanks for your help! However I'm still getting the same answer for part a. I'll have another go at part b. – The Impossible Squish Jul 28 '16 at 23:02