What is the effect of addition of a compensator(either Lead or Lag) to a system?in terms of addition of poles and zeros?

Question

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)

Answer 1

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!!

Answer 2

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.