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ECE 486 Control Systems Spring 2009
Solution to Homework 9
Problem 23 A delay of length T can be expressed as an LTI system with transfer function e−T s. Using the approximation ex^ ≈ 1 + x, x ∼ 0, we obtain the low-frequency approximation,
e−T s^ = e−T s/^2 eT s/^2
1 − T s/ 2 1 + T s/ 2
Consider the DC motor model with transfer function [(s + 1)s]−^1 , and a one second delay. The overall transfer function is approximated by
Gp(s) =
2 − s 2 + s
s(s + 1)
(a) Explain using a root locus plot that a stabilizing PI compensator can be devised. Note that only a 0 ◦-locus will do. Also, explain why you might use a PI compensator for a type I plant. A PI compensator has the form Gc = Kp + K sI. It is used for a type 1 plant to achieve perfect ramp input tracking. By placing a zero close to the origin (Kp/KI small), we can place the closed loop poles in the open left-hand plane.
(b) Explain using a Bode plot that a stabilizing PI compensator can be devised. Try Gc = 15 s+1 s /^2 and a stabilizing control can be achieved.
(c) Explain using a Bode plot that a stabilizing PI compensator can be devised. On setting the numerator and denominator of the plant,
nump = [0 0.0000 -1.0000 2.0000 ] denp = [1 3 2 0]
the plant transfer function is given in matlab by
Gp=tf(nump,denp)
Transfer function: 2.665e-15 s^2 - s + 2
s^3 + 3 s^2 + 2 s
The margin command is useful for looking at its Bode plot:
−
−
−
−
0
20
40
Magnitude (dB)
10 −1^100 101
0
45
90
135
180
225
270
Phase (deg)
Bode Diagram Gm = 1.58 dB (at 0.894 rad/sec) , Pm = 8.91 deg (at 0.786 rad/sec)
Frequency (rad/sec)
With proportional control K = 1/10 we obtain almost 90◦^ PM, with ωc ≈ 0 .1, which takes care of “steps 1 and 2” in our three-step approach to design. We can then slap-in a PI with a zero at 0.01, GP I (s) = (s + 0.01)/s to obtain the overall compensator Gc = 0.1(s + 0.01)/s. Here is the resulting Bode plot for the compensator:
−
−
−
−
0
20
40
Magnitude (dB)
10 −1^100 101
0
45
90
135
180
225
270
Phase (deg)
Bode Diagram Gm = 1.58 dB (at 0.894 rad/sec) , Pm = 8.91 deg (at 0.786 rad/sec)
Frequency (rad/sec)
Using the matlab command [K,S,E]=lqr(A,B,eye(4),1,0) gives K = [1. 5 , 5. 1 , 6. 5 , 1]. u = −Kx is the feedback law minimizing ∫ (^) ∞
0
‖xA(t)‖^2 + u^2 (t) dt
This results in a closed loop bandwidth near 1 rad/sec:
eig(A-B*K) ans = -2. -1. -0.5648 + 0.6894i -0.5648 - 0.6894i
That is, the dominant poles have natural frequency near unity. While not well motivated at this point in the course (take ECE 515 or 555 for motivation), we can use the LQR command to place the observer poles: [L,S,E]=lqr(Ap’,Cp’,10*eye(3),10,0) and then L=L’ gives
L =
The resulting observer poles have natural frequency only slightly larger than the state feedback poles:
eig(Ap-Cp*L) reveals observer poles at {− 9. 2548 , − 0 .8806 + 0. 9782 i, − 0. 8806 − 0. 9782 i}. Alternatively, we can use the place command. For example, placing the observer poles at the consequentive values − 10 , − 11 , −12 is done as follows:
L=place(Ap’,Cp’,[-10,-11,-12]); L=L’ L = 30.0000 150.0000 90.
Observe the relatively high gain required to place the observer poles at these locations. The block diagram is the same as in FPE Figure 7.
