Principles of Control Systems
Lecture#24
PID Controllers and Cascade
Compensators
Improving Transient Response
 Reduce overshoot
 Reduce settling time
Among CL poles A, B, and C
B has lowest setttling time (from the real part, sigma)
C has lowest overshoot percentage (from the angle, or zeta)
In this lecture, we will learn how to improve the settling
time of a system response. In this example, we will learn how
to move the CL pole from A to B.
Improving Transient Response via PD controller
We can speed up the response by dragging the root locus to the
left. This can be achieved by adding zero to the left of the dominant
pole.
Gc ( s )  s  zc
See root locus of compensated system with different zeros.
Notes:
 All systems (a,b,c,d) have the same percentage overshoot,
  0.4
 Compensated systems have lower settling time (Ts), due to
bigger  , or n of the dominant poles
 Higher static constant, Kp, means lower steady-state error
Phase Lead Compensator
Design the phase lead controller to
1. Increase the damping ratio,  by increase the phase margin
and hence reduce the overshoot
2. Increase the gain crossover frequency resulting in larger
bandwidth and hence reduce the settling time
Phase lead compensator
1
1
T
Gc ( s ) 
 s 1
T
s
where   1
For example,
Characteristics of the lead compensator
 The DC gain is 1. This means the gain in the lower frequency
range does not change.
 Smaller  means higher compensated gain, and phase
1  
 Compensator’s Maximum phase is max  sin 1 

1  
1
 Compensator’s Maximum phase occurs at max 
T 
 Compensator’s Magnitude at max is Gc (max ) 
1

Phase Lead Design
Given the forward transfer function of the uncompensated system
KG p ( s)  K
N p ( s)
D p ( s)
The compensated system is of the form

1 
 1 s
 N p ( s)
T
KGc ( s )G p ( s )  K 

1

 s
 D p ( s)

 T 

We have to find, K ,  , and T
Phase Lead Design Procedure
1. Compute K: Phase lead design does not affect the system at
low frequency range, find gain K such that the steady-state error
requirement is met. This could be the requirement on
N ( s)
N ( s)
, or Kv  lim sK
[Forced Response
K p  lim K
D( s )
s 0 D( s)
s 0
Error]
2. Compute the required damping ratio,  : Given the required
 %OS 
 ln 

100 

%OS,  can be calculated.  
[Chapter 4]
%
OS


 2  ln 2 

 100 
From the damping ratio, PM can be computed [Equation 10.73
in Chapter 10]
3. Compute Bandwidth (BW):
(i), If the peak-time ( Tp ) is specified, we can find the
required bandwidth (BW) from
BW 

Tp 1   2
1  2 2  
4 4  4 2  2 [Chapter 10]
(ii) Given the required settling-time ( Ts ), we can find the
required bandwidth (BW) from
BW 
4
Ts
1  2 2  
4 4  4 2  2 [Chapter 10]
4. Make bode plot of the uncompensated system to find its current
phase margin
5. Compute max : compute the difference between desired PM
(step2) and current phase margin (step 3) to obtain the required
PM from the compensator ( max )
1  
6. Compute  from max  sin 1 

1  
1
7. Compute Gc (max ) 

8. Find max . We will make max the new phase margin
frequency. So, from the uncompensated bode plot, find the
frequency where the magnitude is
9. Find T from max 
1
T 
.
.
10.
Combine your result,

1 
s

 N p ( s)
 1 
T
KGc ( s )G p ( s )   K  

   s  1  D p ( s )

 T 

11.
Check your design with the simulation.
12.
Redesign if necessary.
For example,
(i) If % OS is still too high, increase your damping ratio, and repeat
the steps (2-11)
(ii) If the step response is too slow, lower your settling time, and
recalculate (3-11)
Example 11.3: Lead Compensation Design
Given the system in the figure shown below, design a phase lead
compensator to obtain 20% overshoot, and Kv = 40 with a peak
time of 0.1 sec.
0. First, write down the open-loop gain of the uncompensated
system
G( s) 
100 K
s  s  36  s  100 
1. Compute gain K, From Kv  lim sK
s 0
N ( s)
D( s )
s100 K
K  1440
s 0 s  s  36  s  100  ,
40  lim
From the equation
G( s) 
So,
144,000
s  s  36  s  100 
2. Compute the required damping ratio,  :
 %OS 
 20 
 ln 
 ln 


100 
100 


 
 
 0.456
%
OS
20




 2  ln 2 
 2  ln 2 


 100  ,
 100 
This is equivalent to PM of 48.1 degree
3. Compute bandwidth: with Tp  0.1 sec.
BW 
1  2 0.456  

2
0.1 1   0.456 
2
4  0.456   4  0.456   2
4
2
 BW  46.6 rad/s
4. Bode plot of the uncompensated system:
5. Currently, PM is 34 at 29.6 rad/s
Since the compensated system will move gain cross over frequency
to the right (and lower phase margin), add extra 10 degree to the
design. So, Phase lead must add (48.1-34+10) = 24.1 degree
1 
6. for max  24.1, max  sin 1 
1 
7.
Gc (max ) 
1



 and   0.42 ,
,
1
 3.76 dB , at max
0.42
8.
Find the frequency max , this is where the uncompensated gain
is -3.76 dB. From the plot, max  39rad / s
1
1
 25.3 , and
 60.2 these are the
T
T
break frequencies (zeros and poles)
9.
Find T, T  0.0395 , So,
10. Hence, the open-loop gain of the compensated system is

1 
s

1

144,000
T
G(s) 


s  s  36  s  100    s  1 

 T 

G (s) 
144,000
s  25.3 

2.38


s  s  36  s  100  
s  60.2 
The final form is
G( s) 
342,600  s  25.3
s  s  36  s  100  s  60.2 
11. Check the bandwidth: The magnitude response is -7 dB (the
BW check point), at 68.8 rad/s. So the peak time spec Tp is met.
12. Do the simulation to check if all the specifications are met.