Monday, Due February 11
EE/MFS 599
HW#6
In our controls system lab, we have a DC Servo motor called the Motomatic. We would like to design a controller for this DC servo using the following block diagram:
I= F C onstant +
a
Ra + I a
L a V out + d V m K = b b dt + Vg 1/Na
+ V ax m
vin
V a
J T , m m F
out
Vi m n
-
-
We can operate the Motomatic in either velocity mode or position mode. If we operate in velocity mode, the Motomatic is approximately a first order system. The open-loop unit step response of the Motomatic in velocity mode is given below
Unit Step Response for Motomatic 14
12
10
Vg (Volts)
8
6
4
2
0 0
0.2
0.4
0.6
0.8
1 1.2 time (sec)
1.4
1.6
1.8
2
1.a) Find the transfer function, G(s), then find the value of Kp, Kv, and ess due to a step and ess due to a ramp Ans: From the step response, G(s) = K/(Ts+1) where K = 13 and T = 1.2 seconds. Thus, G(s) = 13/(1.2s+1). Kp = 13/1.2 = 10.833 and Kv = 0. Therefore, ess due to a step = 1/(1+Kp) = 0.0845 and ess due to a ramp is infinite. b) Design a PI compensator, GPI(s) to meet the following specs: 1) ts < 0.4 seconds 2) Mp < 5% 3) ess due to a step = 0
Ans: We need a PI compensator to meet the 3rd criterion (i.e., we need a Type 1 system). The first specification means that the dominant poles have to be to the left of the line = -0.4/4 = -10. The second means that the damping ratio should be less than 0.707 (which produces 4.32% overshoot according to Mohannad). Lets pick s1=-11+j11. Gpi = Kp + Ki/s = K(s+z)/s where K=Kp and z= Ki/Kp. To design the PI, we lump the 1/s term in with G(s) and solve K(s1+z) = -1/[G(s1)/s1] = 0.8462 +j21.4923. Thus, K = 21.49/11 = 1.9538 and z = 0.8462/K +11 = 11.4331. So Kp = K = 1.954 and Ki =
z*Kp = 22.339
c) Ans: Simulate your closed-loop design on Matlab
EE/MFS 599: Closed-Loop PI Compensated Step response for Motomatic 1.4
1.2
1
0.8 y(t) 0.6 0.4 0.2 0 0
0.1
0.2
0.3
0.4
0.5 0.6 time (sec)
0.7
0.8
0.9
1
The overshoot is much greater than expected (19% versus 4.32%) and the settling time is actually less than expected (0.31 seconds vers 4/11 = 0.364 seconds). d) If your step response does not meet all the specifications, comment on why Ans: Again, the presence of the zero at -11.431 causes extra overshoot to occur just as in HW#5 2. a) Design a PID compensator to meet the following specs: 1) ts < 0.2 seconds 2) Mp < 5% 3) ess due to a ramp = 1/50
Ans: The specification we address first in the PID design is the steady-state error spec (number 3). Recall, Gpid = Kp + Ki/s + Kds. Our desired Kv = 50 = Ki*13/1.2 . Thus, the necessary Ki = 50*1.2.13 = 4.6154. Next, we can address the transient specifications. The first specification means that the dominant poles have to be to the left of the line = -0.2/4 = -20. The second means that the damping ratio should be less than 0.707 (which produces 4.32% overshoot according to Mohannad). Lets pick s1=-21+j21. Thus, Gpid(s1) = Kp + Ki/s1 +Kds1 = -1/G(s1). When we lump the Ki/s1 term on the right hand side we have Kp+Kds1 = -Ki/s1 -1/G(s1) = 1.9714 - 1.8286. Equating real and imaginary parts we find that Kd = -1.8286/21 = -0.0871 and Kp = 1.9714 Kd(-21) = 0.1429. Thus, the entire PID compensator is given by: Gpid(s) = Kp + Ki/s +Kds = 0.1429 + 4.6154/s + -0.0871s b) Simulate your design on Matlab Ans:
EE/MFS 599: Closed-Loop Compensated PID Step response for Motomatic 5
0
-5 y(t) -10 -15 -20 0
0.1
0.2
0.3
0.4
0.5 0.6 time (sec)
0.7
0.8
0.9
1
EE/MFS 599: Closed-Loop PID Compensated Ramp response for Motomatic 1
0.8
0.6
0.4 y(t) 0.2 0 -0.2 -0.4 0
0.1
0.2
0.3
0.4
0.5 0.6 time (sec)
0.7
0.8
0.9
1
c)
If your step response does not meet all the specifications, comment on why
Ans: While the closed loop poles are definitely at -21+j21, the PID has added two zeroes to the system. If we take the roots of Kds2+Kps+Ki for our values we find that we have added a zero at 8.1468 and one at -6.5062. The zero in the RHP at 8.1468 is causing the initial instability that we see. There is significant undershoot and the overshoot is about 400%. The settling time is very close to 0.2 seconds as expected. In terms of the ramp response, the steady-state error is about 0.017 which is very close to our theoretical value of 1/50 = 0.02 The Bode Plot for the Motomatic in Position mode is given on the next page.
number is q=1 because the slope at low frequencies is -20 dB/decade. Finally, at frequencies below the breakpoints, G(s) = K/s. If we pick =1, then the magnitude is about 33 dB or 44.67. Thus K/(1.0) = 44.67 or K = 44.67. Thus, in position mode, our transfer function for the Motomatic is G(s) = 44.67/[s(s/5+1) (s/1100+1)]. Kp is infinite and Kv = 44.67. Thus, ess for a step is zero and ess for a ramp is 1/44.67 = 0.0224
e) Use the 2nd Ziegler-Nichols method described in class to tune a PID controller for this system (i.e., find the gain K0 that makes the closed-loop system marginally stable then record this gain and the period of oscillation) Ans: For a K0 =1 we obtain the following closed-loop unity-feedback response:
d) Find the new transfer function, G(s), then find the value of Kp, Kv, and ess due to a step and ess due to a ramp Ans: G(s) = K/[sq(s/1+1) (s/2+1)]. From the intersection of the asymptotes, we see that 1=5 and 2=1100. We also see that the type
1.6 1.4 1.2 1 0.8 0.6 0.4 ...