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Lecture 10 A Design Problem
The block diagram of an antenna angular position command control
system at right includes the following components:
plant: with elements
(i) motorspeed: and (ii) antenna:
controller: .
(a)Obtain the motor positiontransfer function in units of degrees/volt.
Solution:
. Hence, the position t.f. is:
(b)One of the closed loop design specifications is that the steady state tracking error for a unit ramp command input should be no more than 0.025o. Show that a proportional controller, cannotsatisfy this specification.
Solution:
Assume that the closed loop system is stable. Then it is a Type-1 system. Because it is a unity feedback system, the error constant is: . Hence, for a unit ramp, . To satisfy this specification requires.
Well, Mr. President. At this point, you should know that in politics, as in many walks of life, the assumption of stability is, unfortunately, all too often taken for granted.
So, let’s analyze the closed loop stability for this value of K.
Method 1 Use the Root Locus:
> s=tf('s');
> Gm=120/(s*(s+1));
> Ga=1000/(s^2+50*s+1000);
> Gma=Gm*Ga;
> rlocus(Gma)
> grid
Method 2 Use the Routh Array The open loop is: .
Hence, the CL characteristic polynomial is: .
The associated Routh array is given at the right, where:
and .
Closed loop stability requires:
. Hence Kmax = .1639.
(c)The totality of specifications the closed loop should satisfy include:
(S1): unit ramp steady state error no greater than 0.025o.
(S2): All time constants no greater than 0.5 seconds
(S3) Dominant complex pole damping ratio at least 0.85.
In an attempt to achieve these, let’s use the root locus pole-placement method of controller design with a PD controller .
Step 1: Find the location of the desired CL pole, call it s*, that corresponds to the intersection of the and lines in the complex plane.
Solution: The real coordinate of the desired pole is . Let . Then the imaginary coordinate is . Hence, s* = -2 + i1.24.
Step 2: Determine the amount of angle (in degrees) that the controller must add to satisfy the angle criterion.
Solution: TO WORK OUT IN CLASS [Should be 100.72o]
Step 3: Find the location of the real-valued controller zero, zc.
Solution: , or zc = -1.765.
Step 4: Use the data cursor to find K. [See RL plot @ right.]
Solution:We require K = 0.0224
Step 5: Plot the CL (i) unit stepresponse and (ii) unit ramperror response over the time interval seconds.
Solution: . The plots are in BLUE.
Figure 1.Closed loop step and ramp error responses without lead in (d) (BLUE) and with lead in (d) (RED)
Step 6: Estimate the steady state ramp error in relation to (S1), and comment.
Solution: The error is ~0.21o, which is much larger than the 0.025o (S1).
(d)Augment the PD controller with a lagcontroller to increase the static gain of the open loop without appreciably changing the location of the placed pole. Then overlay plots of the CL step and ramp error responses on your plots in (c).
Solution:
The PD controller gain is . So the static gain is .
A ramp steady state error of .025o requires that, ultimately,. [See (b).]
Hence, we need an additional controller with static gain (from (a)): .
The controller form is:, and we must have.
QUESTION: But won’t the added pole and zero screw up the root locus (hence, other specs.)?
ANSWER: Not of the controller doesn’t mess with the angle criterion too much.
Just place the pole and zero sufficiently close to the origin.
Well, Jonathan? We’re waiting…
I will choose . Hence, (2pts). Hence, the augmenting
lag controller is . The plots are shown above. (You tell ME!)