Professor Walter W. Olson
Department of Mechanical, Industrial and Manufacturing Engineering
University of Toledo
Lecture 27a: Problem Session
Exercise 1: 1st Order ZN PID Design
 Design a PID controller for the system with a step response
below: (lines on next slide)
Exercise 1: 1st Order ZN PID Design
 Design a PID controller for the system with a step response
below:
Exercise 1: 1st Order ZN PID Design
L  10
T  37
1.2T
 4.44
L
Ti  2 L  20
Kp 
Td  0.5 L  5
1


CPID ( s )  4.44 1 
 5s 
 20 s

Exercise 2: Oscillatory ZN PID Design
 Design a PI Controller for the following system (Kcr=10):
Exercise 2: Oscillatory ZN PID Design
 Design a PI Controller for the following system (Kcr=10):
P  19 / 9  2.11
K p  0.45 K cr  4.5
2.11
Ti 
 1.76
1.2
1 

C PI ( s )  4.5 1 

 1.76 s 
9 complete cycles in 19 sec
Exercise 3: Lead Design (Root Locus)
 Design a lead controller for the open loop system below with
unity feedback which will result in a damping ratio of 0.36
while reducing the 5% settling time by 50%
5
G (s) 
s  s  4  s  6 
Part 1: where would you like to see the closed loop poles?
Exercise 3: Lead Design (Root Locus)
Part 1: where would you like to see the closed loop poles?
5
s  s  4  s  6 
G (s) 
specifications:
 =0.36
t snew 
new 
tsold
3
3


 1.1sec
2
2n 2  0.758  1.84
3
 tsnew

3
 7.57 rps
0.36  1.1
n  .36 * 7.57  2.725
n 1   2  7.57 1   0.36   7.06
2
Desired placement: -2.725  7.06i
Part 2: Placing a zero and a pole
Exercise 3: Lead Design (Root Locus)
Part 2: Placing a zero and a pole
G (s) 
5
s  s  4  s  6 
specifications:
 =0.36
t snew  1.1sec
new  7.57 rps
Desired placement: -2.725  7.06i
Try a zero at -1 and a pole at -10:
Need to bend the curve up more
Exercise 3: Lead Design (Root Locus)
Part 2: Placing a zero and a pole
G (s) 
5
s  s  4  s  6 
specifications:
 =0.36
t snew  1.1sec
new  7.57 rps
Desired placement: -2.725  7.06i
Try a zero at -1 and a pole at -15:
Closer…
Exercise 3: Lead Design (Root Locus)
Part 2: Placing a zero and a pole
G (s) 
5
s  s  4  s  6 
specifications:
 =0.36
t snew  1.1sec
new  7.57 rps
Desired placement: -2.725  7.06i
Try a zero at -1 and a pole at -18:
Very close: Could fine adjust more
Accepting this controller:
Clead ( s )  218
s 1
s  18
Exercise 4: Lead Design (frequency)
 For the following system, increase the static velocity error
2.0/sec with a phase margin of 50o:
5
G( s) 
s  s  4  s  6 
Exercise 4: Lead Design (frequency)
 For the following system, increase the static velocity error
2.0/sec with a phase margin of 50o:
G (s) 
5
s  s  4  s  6 
5
 0.2083
46
Need to increase the gain by at least 10
Kv 
Try 12 for a safety factor
New gain  5*12 = 60
New phase Margin = 43
Next Step!
Exercise 4: Lead Design (frequency)
 For the following system, increase the static velocity error
2.0/sec with a phase margin of 50o:
G (s) 
5
s  s  4  s  6 
5
 0.2083
46
New gain  5*12 = 60
Kv 
New phase Margin = 43 at 2.09 rps
m  2.09
new  43  7  safety factor  55
1  sin  1  0.8192


 0.0994
1  sin  1  0.8192
a  m   2.09 0.0994  0.6590
a 0.6590
b 
 6.6286
 0.0994
 6.6286   s  0.6590 
Clead  12 * 


 0.6590   s  6.6286 
Exercise 4: Lead Design (frequency)
 For the following system, increase the static velocity error
2.0/sec with a phase margin of 50o:
G (s) 
5
s  s  4  s  6 
5
 0.2083
46
New gain  5*12 = 60
Kv 
New phase Margin = 43 at 2.09 rps
1  sin  1  0.8192

 0.0994
1  sin  1  0.8192
Adjusting the compensator to the right 4.5 rps

a  m   4.5  5.1590
a
 4.5  11.1286

 11.1286  s  5.1590 
Clead  12 * 


 5.1590  s  11.1286 
b