The lsim command
Consider the plant
𝑃(𝑠) =
1
(𝑠 + 1)(𝑠 + 5)
1) What is the plant’s type?
2) Let C(s) = K (a proportional controller). Find the closed-loop transfer function
from reference to output using unity feedback.
3) Choose different gains for K within the range 1 to 100. Plot the unit step
response for the different gains. What happens with the transient response of the
closed-loop as K increases?
4) For 𝐾 = 20 find the maximum value attained by the output y(t) and the settling
time Ts
for a unit step input (the time it takes the output to settle within a band
of 2% around its final value). Also find what is the steady-state value of y(t)?
What is the steady-state error equal to?
5) Design a controller that will increase the system’s type by 1 and that will yield
the smallest settling time you can obtain for a step input. What is the settling
time? What is the steady-state error?
6) Plot the response of the closed-loop system to a unit ramp using the controller
you designed in part (5).
There are several ways to do this. One is to use the sawtooth command together
with the lsim command to obtain the time response of the system. Another way
is to implement the whole thing in Simulink, using a signal generator block to
produce the ramp input. And a third way still is to use the step command yet
again, noting that the Laplace transform of a unit step is 1
𝑠
, and the Laplace
transform of a unit ramp is 1
𝑠
2 = (
1
𝑠
) (
1
𝑠
).
Record your observations. Is there a steady-state error? If so, what is its
magnitude?
What would happen with the steady-state error if the input were, instead, a train
of steps? You do not need to give a plot for this question, just answer based on
the type number of the plant together with your controller design.
7) Comment on the performance limitations you found in part (5). Do this by
observing what happens if you make 𝐾 very small, or very large.
