# Getting stability information from Bode plots

Discussion in 'Math' started by marathonman, Mar 4, 2015.

1. ### marathonman Thread Starter New Member

Mar 4, 2015
3
0
Hi.

I am trying to understand the theory behind gain and phase margins from Bode Plots for a system with negative feedback:

For the above system, the transfer function is:

$\frac{KG(s)}{1+KG(s)}$

The poles of this equation determine the stability, and these poles occur at any s such that KG(s) = -1.

I understand that when all poles are in the left-hand plane (negative real part) the system is stable, when any pole is in the right-hand plane the system is unstable, and when any pole is on the imaginary axis the system is at best marginally stable.

If we let s=jω (i.e. confine ourselves to s along the imaginary axis), we can draw the Bode plot. If there exists ω=φ such that |KG(jφ)| = 1 and ∠(KG(jφ)) = -180°, then we know that s=jφ must be a pole. Since this pole lies on the imaginary axis, we can say that the system is (at best) marginally stable.

I think what I have written above is correct, but if there is anything I am misunderstanding please correct me

Now what I do not understand is what happens if the Bode plot does not pass through 0dB when the phase is -180°? How can you get any information about the poles in this situation (other than knowing that they're not on the imaginary axis)? Surely all you have is information about KG(s) for s along this axis?

I've read the usual stuff on gain and phase margins, but can't find any proper justification for these rules in terms of the positions of the poles.

I would really appreciate any help!
Thanks!

2. ### marathonman Thread Starter New Member

Mar 4, 2015
3
0
Edit: aplogies, I realised I drew the wrong diagram! What I meant to post was this:

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
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783
Are you familiar with the concepts of gain and phase margin?

4. ### marathonman Thread Starter New Member

Mar 4, 2015
3
0
I am familiar with them, but I don't understand the theory behind why the rules work. That is what I'm trying to understand. I cannot see how you go from "Bode Plot" back to "Positions of Poles in the s-Plane" in order to prove that a positive gain margin means the system must be stable, etc.

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
The Bode plot does not explicity inform you as to the location of the system poles. Rather, one normally starts with the known open loop poles and zeros as the basis for constructing the Bode plot. That said, it can of course be constructed by direct measurement of the open loop frequency response.
If one is specifically interested in how the pole positions change with loop gain in closed loop form, then one would look to the root locus method.

Last edited: Mar 9, 2015