# Stability of system with negative feedback

Discussion in 'General Electronics Chat' started by epsilonjon, May 10, 2013.

1. ### epsilonjon Thread Starter Member

Feb 15, 2011
65
1
Hello.

I have a question about the stability of systems with negative feedback. If we have the feedback amplifier in the diagram below, the transfer function is

$A_{f}(j\omega) = \frac{S_o(j\omega)}{S_i(j\omega)} = \frac{a(j\omega)}{1+a(j\omega)b}$.

I can see that if the magnitude of the loop gain, L = a(jω)b, is exactly 1 at the frequency for which the phase shift of a(jω) is 180°, then the closed-loop gain will go to infinity. As such, the system is unstable.

But what about if the magnitude of the loop gain is greater than 1 at the 180° frequency? The book I am reading says that if this is the case, and if the phase of the loop gain is a monotonic function of frequency, then the system will be unstable. It justifies it simply by saying that it will have poles in the right-half of the complex plane.

I've thought about it a lot, but I don't see how or why this is the case. Would someone be able to explain it to me please?

Thanks!

2. ### GopherT AAC Fanatic!

Nov 23, 2012
7,983
6,774
I don't think I can explain it any better than texas Instruments. See their application note, it might look familiar...
Here

3. ### ramancini8 Active Member

Jul 18, 2012
473
145
Since I wrote the referenced app note, maybe I should attempt to answer your question. When ab = -1 the denominator equals zero, and the numerator goes to infinity, or instability. At values more negative than -1, say -10, the inherent non-linarites in the amplifier reduce the value to -1. Try this, increase the input signal to a high gain amplifier and watch the output signal increase, then as the input signal increases, the output starts to get distorted (non-linear), gets more distorted as the input signal increases.