Control Systems Final

Control Systems Final - Problem 5 - Solution

Comments for next time:
This problem caused untold grief - much greater than previous years. They hadn't assimilated a number of points.

They got mixed up with Kp and DC gain, and didn't correct for the gain in G(s). Thus, the error of around 10% got translated into 30-40%, even though they got Kp right.
They didn't realize what the unit circle was.
They couldn't scale the plot for a gain half of the original.
There was all kinds of confusion with dbs when they weren't even part of the problem. We did a fair amount of work getting phase margin on Bode' plots, and they were so unfamiliar with dbs that they were unable to separate the issues. This problem was largely due to problems with linear systems for this group.

5. Here is the block diagram for a system.

Dr. Abner Mallity has a Nyquist plot for G(jw). In other words, the Bode' plot does not account for K_p.

Mallity has to have the answer to the following two questions.

(10 pts) What is the smallest SSE (Step Input) that can be achieved while remaining stable? What is the gain for the smallest SSE?

SSE = 9.1%
K_p = 2.0

(5 pts) What is the smallest number of poles this system can have? Be careful answering this question and be sure to take into account the odd behavior in the frequency response.

Minimum number of poles is 4.

(5 pts) For a gain, K_p = 0.5, estimate the phase margin for the closed loop system.

f_m = 59.5 degrees

Solution

To determine the smallest SSE, we need to determine the largest gain for stability. From the Nyquist plot it appears that the plot intersects the negative real axis (indicating -180^o) at around -0.5. That means that the system gain can be doubled by making the proportional gain equal to 2.0. Since the original DC gain was 5, it will be increased to 10 when we increase the proportional gain. That's going to give an SSE of 1/(1 + 10) = .0909 -> 9.1%.

The smallest number of poles that the system can have is figured as follows.

The phase goes to -270^o.
However, there is an increase in phase as the frequency increases.
The increase in phase is less than 90^o, so there is probably one zero.
To get to =-270^o we will have to add -360^o, so there are probably four poles in the system.
If we are wrong about the number of zeros (and there has to be at least one!), then there could be more poles, but four is the minimum.
And, by the way, it's the same system as in Problem 4!

Now, if you decrease the gain by a factor of 2, the plot will contract. To contract and be on the unit circle requires that the point originally have a magnitude of 2. Here a plot where we have estimated where the plot has a gain of 2.

Here is the same plot shrunk by a factor of 2, with the unit circle drawn on the same plot

It looks like the tangent of the angle from -180^o is about 1.7 since the line intersects the vertical line from the -1 point at -1.7j. That gives an angle of 59.5^o, so that's the phase margin.