Control
Systems Final - Problem 4
Name_____________________________
4.
A lunar lander must land at the correct angle. Using variable thrust,
it is possible to produce a torque on the lander to change the vertical
angle. That torque turns the lander. To control the angle,
the system in the figure needs to be designed.
The transfer function
for the lander (the "plant") is:
Gp(s)
= 1/[100s2]
This system is particularly
obnoxious to control, and it has been suggested to try a compensator of
the form:
Gc(s)
= Kc(s - z)/(s - p)
Answer the following questions.
-
Is it possible to control
the system (i.e., make it stable) if the zero is to the left of the pole?
No
-
Is it possible to control
the system (i.e., make it stable) if the zero is to the right of the pole?
Yes
-
(5 pts) Explain
both answers above for full credit. See
below
-
(5 pts) If we choose
z = -1 (i.e. a zero at s = -1, and a (s + 1) factor in the numerator),
where can the pole be located for stability?
-
(10 pts) Where can
the pole be located to put closed loop poles at s = - 1 + 1.5j and s =
-1 - 1.5j?
-
Sketch the root locus
for your compensated system on the plot given. Show the open loop
poles and zeroes - including for the compensator - on your root locus plot.
Solution
If you add a pole and a zero in a compensator, then the center of gravity
will shift
-
A pole in the LHP shifts
the CG to the left.
-
A zero in the LHP shifts
the CG to the right.
We can't afford to shift
the CG to the right, so we need to have the pole to the left of the zero.
If we place the zero at s = -1, then, any pole to the left of that will
make the system stable for all gains. Here is the root locus
for a pole at s = -1.1. Careful examination of the pole locations
(in Matlab, for example) shows that the closed loop poles are just barely
into the LHP. Clearly, to get the locus further into the LHP will
require a pole much further to the left.
Now, lets's consider trying to get a closed loop pole at -1 +1.5j.
Here is the root locus (actually, for the solution) that shows a pole at
some point on the negative real axis.
-
Clearly, the angle from
the zero is +90o.
-
We can get the angle off
the horizontal for the poles at the origin by taking the inverse tangent
of 1.5. That's 56.3o. That means the angle
from a pole at the origin (s = 0) is 180o - 56.3o
= 123.7o. Then, we have to remember that there
are two poles at the origin.
-
Finally, we write the
equation for the angle criterion:
-
90o
- a
- 2(123.7o) = -180o.
-
a
= 180o + 90o - 2(123.7o)
= 270o - 247.7o = 22.3o
-
That means that we can
determine how far to the left of the zero the pole is located.
-
1.5/x = tan(22.3o)
or x = 1.5/tan(22.3o) = 3.66
-
Then, the pole should
be placed at -4.66.