1. CONTROLLER DESIGN Objective ~ Type of compensator
~ Design of compensator in time response
~ Design of compensator in frequency response

2. Type of compensator Basic block diagram + - Actuator Sensor Input
Output
Compensator
Actuator
Plant + -
Controller
Sensor
Actuator
Takes low-energy signal to transform and amplify before going to the plant
Sensor
Takes high-energy signal from plant to transform to low-energy signal for further action

3. Forward path compensator
Commonly used due to easy implementation + + + -

4. Feedback path compensator
To avoid time delay and lag + + + -

5. Feed-forward path compensator
To absorb for disturbance + + + + + -

6. Inner feedback loop Used to eliminate any minor disturbance + + + + -
Also known as cascade control
Used to eliminate any minor disturbance + + + + - -

7. Velocity feedback Also known as rate feedback + + + + - -
To overcome the problem of feedback instantaneous change + + + + - -

8. Selection of compensator
(1) Flow
Very noisy, thus need derivative-action
Overall gain less than, integral-action will ensure no steady-state error.
(2) Level
Normally is of type 1, thus just required proportional action
(3) Temperature
Thermal delay, thus normally required PID-action
(4) Pressure
Characteristic can be fast or slow depending on application, thus required PI or P.

9. PID compensator Proportional compensator (P) + -
Use to improve steady state error type +
R(s)
E(s)
A(s)
Y(s) -
B(s)
Consider P-compensator transfer function as
and
where A(s) is the actuator signal.
gives better steady state but poor transient response
High
Too high
can cause instability.

10. Example: + -
For a unit step input
Final value

11. Integral compensator (I)
Use to eliminate error for type 0
Consider the I-compensator
and actuating signal of
E(s)
A(s) +
R(s)
Y(s) -
B(s)
Slow response, can be used with P-compensator to remedy this problem

12. + - Example: Closed-loop transfer function
For a unit step input, the response is

13. Derivative compensator (D)
Consider the D-compenator as
and actuating signal as
R(s) +
E(s)
A(s)
Y(s) -
B(s)
Quick response
No effect at steady state because no error signal
Useful for controlling type 2 together with a P-controller
Response only to rate of change and no effect to steady state

14. , , e Example: where and Compensator Determine
, if damping ratio 0.6 and
Error detector
Amplifier
Servo motor e

15. , , , , New open loop transfer function Give closed loop For and .
C.f. nominal second order
dan
Hence

16. Proportional-integral compensator (PI)
Use for combination of fast convergence zero steady state error or

17. + - Example: Consider a spped control system employing a PI-controller
Closed-loop transfer function
For unit step input

18. Proportional-integral-derivative compensator (PID)

19. PID Tuning
There are many techniques for tuning PID

20. Zeigler-Nichols Step Response
Suitable for process control where time lag is significant k
0.63k T a L
The above response can be approximated by
where k is static gain, T time constant, L time delay and

21. Based rule-of-thumb, for the controller gains can be appoximated by
PID PI P
1/a
0.9/a
1.2/a 3L 2L
L/2
Controller

22. Zeigler-Nichols Closed-Loop Method
Using Routh-Hurwitz criteria, we can determined the margainally stable’s gain, and its period
of oscilation, for a controller with a gain of K
With rule-of-thumb of thumb
Controller
PID PI P
0.833T
0.5T
0.125T

23. Example
Design a P, PI and PID controller for a plant with an open loop transfer function as follows
With a gain of K connected in cascade to the system, its characteristic equaton becomes
Forming the Routh array 1 11 6
6(1+K)
1+K

24. and its frequency of osccilation
Thus
and its frequency of osccilation
which gives T= s
This gives
Controller P PI
PID 5
4.5 6
0.947
0.237
1.57
Their step response are

26. Effect of adding zero and pole
Example
Consider an open-loop transfer function
By introducing a zero and normalizing the response
The step response for uncompensated, a=4 and a=8.

30. Reduce rise time, peak time Increases overshoot
The effect of introducing zero
Reduce rise time, peak time
Increases overshoot
As the zero approach the orgin its contribution is more significant
Bandwidth increase
Improve gain margin
For a compensator with a pole and normalizing the output
The step response for uncompensated, a=4 and a=8.

33. The effect of introducing pole
Reduce oscillation, and as its more dominant the response becomes sluggish

34. Design in time response
Lag compensator
Use for steady state improvement without affecting the transient response
Consider a lag compensator transfer function
where .

35. Dc gain without compensator
Consider a plant
Dc gain without compensator
where
and
zeros and poles position from test point.
With compensator As
, thus
with that the steady state error is reduced.

36. Lead compensator For transient response where Uncompensated.
Uncompensated
Compensated
Thus

37. + - Example: Use a lead compensator
so that the overall system has undamped frequency
of 4 rad.s-1 and rate of decay 0.5 s-1, if + -
Open loop transfer function

38. Third pole is determined from
Dominant poles
and
that give a damping ratio of
and
Third pole is determined from

39. s-plane
c j3.97
 j
c
-11-10
 -j
c -j3.97

40. Hence, the dc gain
Which give the closed loop transfer function

41. Location of zero and pole of the system together with the compensator.
Satah s s
 j
  O 
-12
-10
 j

42. Using the angle condition on the closed loop poles of
where
the angle of
from
i.e. the angle contribution required by
Thus the compensator transfer function

43. Design in frequency response
Basic concepts in the design criteria
Increase of phase margin, the overshoot will be reduced
Increase of bandwith, the response will be faster
Increase of low frequency magnitude, the steady state will be reduced

44. Gain adjustment
Use to reduce the overshoot during transeint period by incresing the phase margin dB
GM
 (rad.s-1) 
Plot LM PM o
-180
PM’
Plot 
Design Procedure
(i) With a chosen gain, obtain the Bode plot and the gain crossover frequency,
(ii) Determine the required phase margin
(iii) Obtain the gain that is required for the new cossover frequency,
This gain is the extension of the gain in (i).

45. + - Example: Consider a type-1 system.
Determine the uncompensated phase margin.
If a P-compensator is cascaded to the system, determine the required gain so that the
phase margin is 30. o + -

46. Solution:
By replacing
, the frequency response in corner frequency form is
and
and
and

47. (rad.s-1) 1 10
500
(dB/dekad)
-20
(dB/dec)
Total
-40
-60
Pole at origin
Real pole
Real pole
slope
and

48.  (rad.s-1) 1 50
100
5000
(deg/dec)
-45
Total slope 45
-90
Real pole
Real pole
and

49. (rad.s-1) 1 10
500
10000
Total slope
(dB/dec)
-20
-40
-60
LM (dB) 12 -8
-76
-154

50.  (rad.s-1) 1 50
100
5000
10000
Total slope
(deg/dec)
-45
-90
(deg)
-166
-193
-269
>> w1=[ ];LM=[ ];
>> w2=[ ];ph=[ ];
>> subplot(2,1,1);semilogx(w1,LM);
subplot(2,1,2);semilogx(w2,ph);
From the Bode plot, and for uncompensated system

51. That gives the compensator gain as,
For the new phase margin,
d an increase of gain
That gives the compensator gain as

52. Lag compensator The compensator can also be presented as where
. The pole is chosen to be closed with the origin, example
As the zero is far away from the pole, no obvious transient response will be affected
Increased in the open-loop gain will improve the steady state error
Asymtote approximation of the Bode plot

53. Example of actual plot for and 5,dB
 (rad.s1) 
 (rad.s-1)
Example of actual plot for
and 5, ,
>> alpha1=2;T=0.1;alpha2=5
>> sys1=tf([1 1/T],[1 1/(alpha1*T)]);
>> sys2=tf([1 1/T],[1 1/(alpha2*T)]);
>> bode(sys1,sys2)

54. Gambarajah 6.47: Plot Bode bagi pemampas mengekor dengan penghampiran asimtot

55. gain so we can incresed the gain for steady state improvement
Observation from the plot:
At low frequency, the compensator will not have influenced on the open-loop
gain so we can incresed the gain for steady state improvement
Design Procedure
Prosedur rekabentuk adalah seperti berikut:
Determine the gain K for the required steady state error and draw the Bode plot
(ii) Obtained the required phase margin and add, compensator phase angle, =5 - 12 o o
(iii) Compensator’s zero will be a decade below of the new gain crossover frequency of (ii).
Compensator’s pole is obtained by drawing a slope of -20dB/dec from the compensator’s zero
until the line touched the 0dB.

56. Example
If the new gain crossover frequency is,
, and the LM at that frequency is
dB.
Compensator’s zero
While, compensator’s pole is obtained by finding the frequency where the slope of -20dB/dec which
begin from the compensator’s zero cross the 0dB-axis. dB
 (rad.s-1)
20dB/dekad -Y
(iv) Adjust the gain K for the specified error coefficient

57. Example:
Design a lag compensator so that the ramp error coefficient is 10 ses-1, gain margin of at
least 20 dB and phase margin of at least 45o. Assume the compensator’s phase angle is 5o. + -
Open loop transfer function
Consider
and K=50, which give an open loop transfer function of

58. and
>> w1=[ ];LM=[ ];
>> w1a=[ ];LMa=[0 0];
>> subplot(2,1,1);semilogx(w1,LM,w1a,LMa,'-');
>> w2=[ ];ph=[ ];
>> w2a=[ ];pha=[ ];
>> subplot(2,1,2);semilogx(w2,ph,w2a,pha,'-');
From the Bode plot

60. Given
Open loop transfer function without compensator
with initial value of
Thus shift 20dB upward from original plot.
New phase margim with =5o, PM’=45o+5o=50o. thus
’=0.65 rad.s-1.
Hence the compensator must provide –20.

62. >> w1=[0.1 1 5 20 100 1000];LM=[40 20 -8 -44 -100 - 180];
>> w1a=[ ];LMa=[0 0];
>> subplot(2,1,1);semilogx(w1,LM,w1a,LMa,'-');title('PLOT BODE');ylabel('Magnitud (dB)')
>> w2=[ ];ph=[ ];
>> w2a=[ ];pha=[ ];
>> w2b=[ ];phb=[ ];
>>subplot(2,1,2);semilogx(w2,ph,w2a,pha,w2b,phb,'-');xlabel('Frekuensi (rad/s)');ylabel('Sudut fasa (Darjah)') At ’, At ’,

63. The compensator Hence pole of the compensator dB 0.0065 0.065
 (rad.s-1)
-20dB/dekad
-20
Gain of the compensator
The compensator
>> syms s;
>> den=expand((s )*s*(s+5)*(s+20)*(s+1))
den = s^ /2000*s^ /1000*s^3+1613/16*s^2+13/20*s
>> num=expand(0.1*(s+0.065)*2)
num = 1/5*s+13/1000
>> num=[1/5 13/1000];den=[ / / /16 13/20 0];
>> bode(num,den)

65. >> syms s;
>> den=expand((s+0.065)
*s*(s+5)*(s+20)*(s+1))
den = s^5+5213/200*s^ /100*s^3+865/8*s^2+13/2*s
>> num=expand(0.1*(s+0.65)*2)
num = 1/5*s+13/100
>> num=[1/5 13/100];den=[1 5213/ / /8 13/2 0];
>> bode(num,den)
>> num=[1/5 13/100];den=[1 5213/ / /8 13/2 0];
bode(num,den)
sys=tf(num,den);
[GM,PM,Wg,Wp] = margin(sys)
GM_dB = 20*log10(Gm)
GM =
PM =
Wg =
Wp =
GM_dB =