1. Lecture 9:
Compensator Design in
Frequency Domain

2. Objectives Recognize the relationship between: the overshoot (%OS),
the damping ratio (ζ) , and
the phase margin (ΦM ).
Use two approaches employing Bode diagrams to design compensators (controllers) which achieve a desired phase margin:
Simple gain adjustment.
Lag compensator.

3. Relationship between %OS, ζ , ΦM
Consider a unity feedback system with the following open-loop transfer function
The closed loop transfer function is the standard 2nd order system

4. The time response of the second order under-damped system

5. The percentage overshoot, %OS, is given by
Note that %OS is a function only of the damping ratio, ζ.
The inverse is given by

6. There is also a relationship between damping ratio and phase margin
There is also a relationship between damping ratio and phase margin. The phase margin is obtained by solving |G(jω)| = 1 to obtain the frequency as
The phase margin is
There is a simpler formula
Thus if we can vary the phase margin, we can vary the percent overshoot.

7. Transient response via gain adjustment
The phase margin, can be varied via a simple gain adjustment as shown.

8. Example
For the position control system shown below, find the value of preamplifier gain, K, to yield a 9.5% overshoot in the transient response for a step input. Use only frequency response methods.

9. Solution The open-loop transfer function (with K = 1 for now) is
We draw the Bode plot for this open-loop system.

10. Bode plots for the example

11. The value of ζ achieving %OS = 9.5% is
In the Bode diagram we locate the point at which the phase is -120°. At this point, the frequency is 14.3 rad/sec and the gain is about -55dB.
For this point to be the phase margin, the gain must be 0dB, so we need a gain of +55dB.

12. Lag compensation
The function of the lag compensator is to increase the phase margin of the system to yield the desired transient response without affecting the low-frequency gain and hence it does not reduce system stability or steady error constant.

13. Steady-state error constants
The steady error constants are:
position constant
velocity constant
acceleration constant
The value of the steady-state error decreases as the steady error constants increases.

14. Visualizing lag compensator
The transfer function of the lag compensator is
where z > p.
The gain of this lag compensator at low frequency is unity and at high frequency is

15. Visualizing lag compensator
For example, the frequency response of a lag compensator:

16. Visualizing lag compensator
In the figure below, the uncompensated system is unstable since the gain at -180° is greater than 0dB. The lag compensator, while not changing the low-frequency gain, does reduce the high frequency gain. The magnitude curve can be shaped to go through 0dB at the desired phase margin to obtain the desired transient response.

17. Example
Given the following open-loop transfer function of a position control system, use Bode diagram to design a lag compensator to yield a percent overshoot of 9.5%.

18. Solution The response is too oscillatory!
Before solving the problem, let us draw the step response of the closed-loop system whose open loop transfer function is given.
The response is too oscillatory!
What do you expect about the value of the damping ratio and phase margin of the system?

19. Solution
Using the command margin, we draw the Bode diagram and find the phase margin to be only 1.58°.
Too low value.
This system must be compensated!

20. Solution Now, let us start solving the problem.
We want to design a lag compensator of the following transfer function
First we need to plot the Bode diagram, see next.

21. Choose a point!

22. Solution
From the previous example, we know that 9.5 %OS corresponds to a phase margin 60°. So, we should look for the point at which the phase passes -120°.
However, as the lag compensator add some extra little phase lag, we will seek a phase margin of 70°. That is we look for the point at which the phase passes -110°.
From the figure, we can locate this point at frequency rad/sec at which the gain is 29.3dB.

23. Solution
At the frequency rad/sec, the open loop has a gain 29.3dB.
Therefore, it is required that the lag compensator, at frequency rad/sec, to have a gain of
Arbitrarily, locate the zero of the lag compensator at

24. Solution
Thus, the required lag compensator is

25. Checking the phase margin of the new system
As a check, we use the command margin, to draw the Bode diagram again. Now the phase margin is 64.6° which is more than enough!!

26. Checking the step response of the new system
It is interesting to plot the step response of the closed loop system after adding the lag compensator.
As shown, the response is highly stable which is very good (compared to the oscillatory response of the original system) although it is sluggish too.