Feedback systems & oscillations

Question

The transfer function of feedback system is: $$ \frac{V_{\rm out}}{V_{\rm in}} = \frac{A}{1+Af} $$ Where $A$ is the open loop gain, and $f$ is the feedback gain.

Now for oscillation to happen, $Af$ term (loop transmission) must be $-1$, which means the system is configured to be on a pole, because the denominator goes to zero. I have got couple of questions on this one.

1. For oscillation to happen (steady state), the pole must be on $y$-axis (frequency axis) in the $s$-plane. I believe this is an assumption in the oscillation theory not reflected by the above condition that loop transmission is $-1$. Because loop transmission is $-1$ in all poles in the system. So it is up to the designer to make sure that the circuit operates on a pole on $y$-axis. Is this correct?

2. If I ignore all the Laplace transform thing, and just look at the transfer function $$\frac{V_{\rm out}}{V_{\rm in}} = \frac{A}{1+Af}$$ On the pole, the output is supposed to be very high (infinity). If I only consider this equation on standalone basis, then how come we have a steady state oscillation, the output voltage is supposed to be very high as denominator is zero. My explanation is that it does not happen, because the supply voltage to the Op-Amp/MOSFET (or whatever we use to implement the circuit) restricts that output voltage. Is this correct?

Answer 1

I think that the problem is that you are considering that $$ T = \frac{A}{1+Af} $$ is the actual transfer function of your system (an analog amplifier I guess) and this can be miss leading because that expression is just and approximation for the region of frequencys were the amplifier response is flat. If you want to make assumptions regarding the stability of the system you should work with the full Laplace transform, this is: $$ T(s) = \frac{A(s)}{1+A(s)f(s)} $$ In order to oscillate at a given frequency, the system needs to have conjugate poles on the imaginary-axis at that frequency, but in a practical case, if the system has poles on the right half-plane, as you said, the supply voltage of the amplifiers will move this poles into the j-axis

I don't have enough reputation to comment, but I think that this question that I asked a few weeks ago may be helpful. https://electronics.stackexchange.com/questions/279365/positive-feedback-with-constant-transfer-functions

Answer 2

I think you may be confusing the poles of the open-loop system with the poles of the closed-loop system. Or you may be a little confused on something else. In particular, I believe this statement is incorrect:

Now for oscillation to happen, Af term (loop transmission) must be -1,

If $Af=-1$, then the closed loop transfer function of the system is just $A/0$. That does not yield an oscillatory impulse response. It may be a little to hard see this, but it is easy enough to take the limit as Af approaches -1. i.e. try this in matlab

A=1;
f=-0.9;
b1 = A;
a1 = 1+A*f;
sys = tf(b1, a1);
[Y,t] = impulse(sys);
figure;
plot(t, Y);

Repeat for f=-0.99, -0.999, -0.9999 (i.e. approaching -1). You can get as close as you would like to f=-1, and the system will never oscillate. (It may also be instructive for you to investigate the step response and frequency response at the same time, with matlab functions step() and freqs())

Now, about this statement

for oscillation to happen (steady state), the pole must be on Y-axis (frequency axis) in the s-plane.

That is correct, but perhaps you are not interpreting it correct. What this means is that your closed loop transfer function needs to look something like this

$$\frac{1}{s^2+a^2}$$

where $a$ is a real number. Then your system has a pair of complex conjugate poles at $s=+ia$ and $s=-ia$.

For example, if you take $a=1$, and then try b1 = 1;a1 = [1,0,1]; in the above matlab code, you should get an undamped oscillation.

So, given that the closed loop transfer function is indeed $A/(1+Af)$, if you want an oscillation, you don't want $Af=-1$, you want $Af=s^2$ (or something similar).

Once you understand the above, you should hopefully be able to see the answer to your second question. If not, then perhaps ask it again.