Phase Margin Example
A Detailed Look At Using/Setting Phase Margin

        We will assume that you want to control a system with a transfer function:

G(s) = 2/[(.5s + 1)(s2 + 4s + 1)]

That is the transfer function for G(s) in the block diagram below.  You will probably need to design a compensator for the controller - and determine the value for the gain.

We also assume that you have the following specifications:

We start by examining the Bode' plot for G(s)

In this plot we have drawn lines at -140o and -180o.  We can see that the phase plot intersects -140o at a frequency of about 0.25 Hz.  If we set the zero db crossing there, the phase margin will be 40o.  To get the magnitude plot to zero db at f = -.25 requires a gain of about 12 db.  Since we already have a DC gain of 6 db that will move the overall magnitude plot up so that the DC gain will be 18db - or a gain of 8.  That gives a SSE of 1/[1 + 8] or 11%.  Here's the plot with a vertical line drawn at the frequency where the phase is -140o so you can see how it fits together.

Now, for this system, we find the following.

        Now, let's see what we can do with a compensator.

        Here is what we will do.

        We can begin to see what the problem really is in this system.  As we get up around f = .2 or .3, the magnitude really starts to drop off quickly, and the phase heads toward -180o fairly quickly.  We also see that we have an open loop pole at s = 2, so there is a corner frequency at f = .318.  There is also a resonant frequency at jw = 1, or f = -.159.  So, around f = .3, the magnitude is moving to drop off at -60 db/decade, and the phase is heading south to -270o.  To get a good system, we need to correct that problem, and we may well need two lead networks!