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When we add a compensator(either Lead or Lag) to a plant in series, what is the effect in terms of addition of poles and zeros?

A pole-zero pair is added(meaning one pole and one zero)?

or

a pair of closed loop poles is added?(meaning two poles)

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  • \$\begingroup\$ You add a compensator (as far as I'm aware) to make the loop stable hence you are cancelling an undesired pole or zero in the raw plant. "Adding" poles or zeros seems to be missing the whole point. Just my opinion of course. \$\endgroup\$ – Andy aka Dec 17 '20 at 17:20
  • \$\begingroup\$ A pole-zero pair is added to the open loop and closed loop. \$\endgroup\$ – Chu Dec 17 '20 at 17:29
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Suppose we have a plant, whose feed-forward transfer function is

\$G(s)=\frac{1}{s^2-2s+2}=\frac{1}{(s-1+j)(s-1-j)}\$

Without feedback, this plant cannot be "controlled" because it has poles in the positive half-plane (of the complex plane). i.e. \$s=1-j\$ and \$s=1+j\$.

If we add feedback with a transfer function of H(s), the transfer function of the closed loop system will be

\$F(s) = \frac{G(s)}{1+G(s)H(s)}\$

Expanding that out gives

\$F(s) = \frac{\frac{1}{s^2-2s+2}}{1 + \frac{H(s)}{s^2-2s+2}} = \frac{1}{s^2-2s+2+H(s)}\$

To ensure stability we want \$s^2-2s+2+H(s)\$ to have no zeros in the positive half-plane.

There are a number of transfer functions \$H(s)\$ that will do the job. However, one of these is the high pass filter (lead compensator)

\$H(s) = \frac{14s}{s+3}\$

which has a zero at s=0 and a pole at s=-3.

To see how this compensator works, we calculate

\$P(s) = s^2-2s+2+\frac{14s}{s+3}= \frac{(s^2-2s+2)(s+3) + 14s}{s+3}\$

\$P(s) = \frac{s^3 + s^2 + 10s + 6}{s+3}\$

which has zeros at -0.61456, -0.1927+3.1187j, -0.1927-3.1187j

So, \$F(s)\$ has no poles with positive real parts, and so is stable. By adding a feedback compensator (which had a pole), we moved the poles that were in the positive half-plane without feedback, to the negative half plane. This is just one example, and the same compensator will not work for different \$G(s)\$ functions. But the general idea is to choose and \$H(s)\$ that "moves" problematic poles from the positive half-plane to the negative.

If the compensator is in the feed-forward path, but there is still feedback, the analysis is similar. The closed loop transfer function becomes

\$F(s) = \frac{G(s)H(s)}{1+G(s)H(s)}\$ or \$F(s) = \frac{G(s)H(s)}{1-G(s)H(s)}\$ depending upon whether one has positive or negative feedback.

Again, one wants to choose H(s) so that none of the zeros of \$P(s) = 1+G(s)H(s)\$ (alternatively \$P(s)=1-G(s)H(s)\$) lie on the positive half-plane.

!!Warning, don't rely on my math here because I make mistakes!!

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The compensator goes in series to the forward path of the loop.

Here is a random diagram from google showing it.

"Lead" and "Lag" refer to a compensator of the form (s+a)/(s+b). Alternatively (1+s/a)/(1+s/b), but this form changes the resulting K. A constant K must be included somewhere, in the diagram it is included with the integrator. It is a bit arbitrary as you can rearrange the terms in different ways.

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