Part (e) of Problem 6.2 continues the trend of reviewing
material covered in recent lectures. (Here, specifically,
the October 26 lecture.)
[Only change Kd in part (e); use same G found in part(c).]
As mentioned in lecture (and on previous pages here), PD control
will generate an additional zero not seen in the closed-loop
transfer function for rate feedback control.
Problem 6.2 reviews this concept.
Compare your solutions with the examples in "Proportional,
derivative and integral control" (hand-out of Oct-26-00).
Note that a pole and zero which have identical
values are 'co-located'.
A co-located pole and zero therefore factor out of the polynomial
(in s) of that transfer function.
Thus, a pole and zero of identical value completely cancel out
one another.
Suggestion: Examine the effect on the system of making a pole and zero
closer and closer. (...not exactly 'co-located,' but approachingly closer.)
You can use MATLAB to experiment with systems, as one way of
looking at the effects of making the zero closer to the dominant pole.
You may wish to play with a variety of values for Kd (in part e)
to develop more intution.