control system basics-model based design-first order equation stability-poles-matlab

1. CONTROL SYSTEMS
12-Apr-17
1
Eng. Mahmoud Hussein
RTECS

2. Control Objectives3

3. Regulatory Control /
Load Disturbance Rejection4
 Maintains a parameter at or near a set-point despite
disturbance
 Is hence called
“Load Disturbance rejection problem”
 Example
 Cruise control
 Siphon Water Level Control

4. Regulatory Control
Siphon Water Level Control5
 Principle of Operation
 OPEN
 If water level becomes low in the reservoir, the lever comes
down by the weight of floating ball and the ball of valve-
room rotating by the lever opens the flow way.
 CLOSE
 If water level becomes high in the reservoir, the lever comes
up by the buoyancy of floating ball and the ball of valve-
room rotating by the lever inside of valve closes the flow
way.

5. Regulatory Control
Siphon Water Level Control6
 Set point
 Normalizing the scale, the set point can be taken as a reference
 Desired signal = 0
 Output signal @ steady state = 0
 Control signal @ steady state = 0
 Disturbance
 Flushing the toilet

6. Servo Control / Setpoint Tracking Control /
Trajectory Control7
 Cause the plant output to follow the changing input command
as closely as possible
 Examples
 Remote control car
 Automatic parking control

7. Servo Control / Tracking Control Automatic
Parking Control8

8. Control Structures9

9. Open Loop Control
10

10. Open Loop Control
11
 Characteristics
 Highly sensitive to extraneous disturbances
 Highly sensitive to parametrical variation in the process
 TypicalApplications
 Stepper motor
 Motor with a worm gear (very high reduction ratio)
 Electric toaster

11. Closed Loop Control
(Feedback Control)12
Gc(s)
Controller
  n
sensor
noise

w load disturbance

Gprocess(s)
u
control
y
output
r
reference
input, or
set-point
e
 sensed
error
GActuator(s)
Actuator Process

12. Closed Loop Control
(Feedback Control)13
 Closed loop control strategy involves
 Measurement of the actual output,
 Comparison with the required reference value and
 Employing a suitable correction compensating for any deviation
Gc(s)
Controller
  n
sensor
noise

w load disturbance

u
control
y
output


r
reference
input, or
set-point
e
sensed
error
GActuator(s) Gprocess(s)
Actuator Process

13. Five Signals of Feedback Control Loop
14
 Set Point
 Desired output to be performed by the system
 constant for regulation control and
 variable for trajectory control
 Error signal
 The deviation between the desired behaviour and the system sensed output.
 Command/Control signal
 The signal generated from the controller
Gc(s)
Controller
  n
sensor
noise

w load disturbance
u
control
y
output

reference
input, or
set-point
r e
sensed
error
GActuator(s Gprocess(s))
Actuator Process

14. Five Signals of Feedback Control Loop
15
Gc(s)
  n
sensor
noise

 Disturbance
 A temporary change in environmental conditions that causes a
pronounced change in the system behaviour
 Low frequency deterministic measurable change
 Noise
 Random change of the signal at a high frequency
 Only statistically measurable
w load disturbance

Gprocess(s)
Process
u
control
y
output

reference
input, or
set-point
r e
sensed
error
GActuator(s)
Controller Actuator

15. Advantages of Feedback Control
16
 Speeds up the response
 Reduces steady-state error
 Reduces the sensitivity of a system to
 External disturbances
 Variations in its individual elements & components
 Improves stability

16. The Cost of Feedback
17
 Increased number of components and complexity
 Possibility of instability
 Whereas the open-loop system is stable, the closed-loop system
may not be always stable.

17. One DOF Feedback Controller18

18. One DOF Feedback Controller Block Diagram
19
Gc(s)
Controller
  n
sensor
noise

w load disturbance

Gp(s)
Plant
u
control
y
output
r
reference
input, or
set-point
e
 sensed
error
W(s)
Gp (s)
1GcGp (s)
N(s) 
GcGp (s)
1GcGp (s)
R(s) 
GcGp (s)
1GcGp (s)
Y (s)

19. Tracking /Servo Control Problem
20
R(s)
GcGp (s)
1GcGp (s)
Y (s)

Gc(s)
Controller
  n
sensor
noise

w load disturbance

Gp(s)
Plant
u
control
y
output
r
reference
input, or
set-point
e
 sensed
error
1 if GcGp (s)  1

20. Disturbance Rejection Problem
21
Gc(s)
Controller
  n
sensor
noise

w load disturbance

Gp(s)
Plant
u
control
y
output
r
reference
input, or
set-point
e
 sensed
error
W(s)
Gp (s)
1GcGp (s)
Y(s)


21. Noise Problem
22
Gc(s)
Controller
  n
sensor
noise

w load disturbance

Gp(s)
Plant
u
control
y
output
r
reference
input, or
set-point
e
 sensed
error
N(s) c p
GcGp (s)
1 G G (s)
Y (s)
 1if GcGp (s)  1

22. One DOF Feedback Controller
23
 To address all aspects of the system, only one degree of
freedom is available, namely, Gc(s)
W(s)
Gp (s)GcGp (s) GcGp (s)
1GcGp (s)
R(s)  N(s) 
1 GcGp (s) 1GcGp (s)
Y (s)
Gc(s)
Controller
 n
sensor
noise

w load disturbance

Gp(s)
Plant
u
control
y
output
r
reference
input, or
set-point
e
sensed
error

23.  Series Connection
 G=G2*G1
 where G1 and G2 are LTI objects (tf, ss, or zpk)
 Parallel Connection
 G=G1+G2
 where G1 and G2 are LTI objects (tf, ss, or zpk)
 Feedback Connection
 G=feedback(G,H,Sign)
 If Sign=-1, the negative feedback structure is indicated; negative feedbackis
the default
Modeling of Interconnected Block Diagrams in
Matlab24

24. Modeling of Interconnected Block Diagrams in
Matlab Example
 Where H(s) is a feeedback
block representing a first order
low pass filter with the aim of
sensor noise cancellation
 tf
 Create transfer function model, convert
to transfer function model
 feedback
 Feedback connection of two models
 Gp = tf( [1 7 24 24], [1 10 35 50 24] )
 Gc = tf( [10 5], [1 0] )
 H = tf( 1, [0.01, 1] )
 G_cl_loop = feedback( Gp*Gc, H, -1)
25
0.01s1
H (s)
s
1
10s 5
s3
 7s2
 24s 24
10s3
 35s2
 50s 24
Gc (s) 
Gp (s) 
s4

25. Time Domain Analysis in Matlab26

26. Step response evaluations with MATLAB
27
 step(G)
 %automatic draw of step response curves
 [y,t]=step(G)
 %evaluate the responses, but not drawn
 [y,t]=step(G,t_f)
 % final simulation time t_f setting
 y=step(G,t)
 %simulation on user defined time vector t

27. Unit Step Response in MATLAB
28
 OL_Sys = zpk([],[0 -6],25)
 CL_Sys = feedback(OL_Sys,1)
 figure
 step(CL_Sys);
 grid on
 stepinfo(CL_Sys)
R(s) + Y(s)25
s(s 6)

28. Unit Step Response in MATLAB
29
 RiseTime: 0.3711
 SettlingTime: 1.1887
 SettlingMin: 0.9083
 SettlingMax: 1.0948
 Overshoot: 9.4773
 Undershoot: 0
 Peak: 1.0948
 PeakTime: 0.7829
Step Response
Amplitude
0.2 0.4 0.6 0.8
Time(sec)
1 1.2 1.4 1.6
0
0
0.2
0.4
0.6
0.8
1
1.2
1.4
System: CL_Sys
Rise Time(sec ): 0.372
System: CL_Sys
Peak amplitude: 1.09
Overshoot (%): 9.47
At time (sec): 0.777 System: CL_Sys
Settling Time (sec): 1.19

29. Biological Feedback Control Example
30
 Regulation of glucose in the bloodstream through the
production of insulin and glucagon by the pancreas.

30. RTECS 2015 12-Apr-17

31. Biological Feedback Control Example
32
 Can you find out the components of the feedback control loop?
Control Device
Pancreas
Actuator
Insulin OR
Glucagon
Process
Liver
Actual Blood
Glucose Conc.
error
Sensor
Glucoreceptors
In Pancreas
+
-
Measured Blood
Glucose
Desired Blood
GlucoseConc.
(90 mg per100 mL ofblood)

32. Stability33
RTECS 2015

33. Simple Real Pole p = -
 The natural response is
of the form:
y(t )  K e t
K is given by the
initial condition
 Assuming K > 0:
34
if   0y(t)
if   0
t
 Location in s-domain:
Im(s)
if   0 if   0
Re(s)
Note: since  = 0, y(t) is located on
the real axis.

34. Simple Real Pole y(t )  Ke t
35
Note: Since  = 0, y(t) are located
on the real axis.
 Location in s-domain:
Im(s)
t
stable
Re(s)
stable
 Assuming K >0, if  >0:
y(t)
y(t) decays
t
• Effect of the value of  >0:
y(t) y(t) decay


35. Simple Real Pole y(t )  Ke t
 Assuming K >0, if  <0:
36
Note: Since  = 0, y(t) are located
on the real axis.
 Location in s-domain:
Im(s)
t
y(t)
unstable unstable
Re(s)
y(t) grows
t
• Effect of the value of  <0:
y(t) y(t) grow


36. Simple Real Pole p = 0
37
RTECS 2015
 The natural response is
of the form:
y(t)  B
B is given by the
initial condition
 Assuming B > 0:
y(t)
t
Note: since  = 0 and  = 0, y(t) is
located on the origin.
 Location in s-domain:
Im(s)
unstable
Re(s)

37. Simple Pair of Imaginary Poles
p = ± jd38
RTECS 2015
Re(s)
 The natural response is
of the form:
y(t )  Acos(d t ) Bsin( dt )
A and B are given by
the initial conditions
 Assuming A > 0, B > 0:
y(t)
t Note: since  = 0, y(t) is located on
the imaginary axis.
 Location in s-domain:
Im(s)
  0

38.  Assuming A > 0, B >0:
39
Re(s)
t
y(t)

y(t) oscillates
forever
t
Simple Pair of Imaginary Poles
y(t )  Acos(d t ) Bsin(dt )
• Effect of the value of  < 0:
y(t) y(t) oscillate
 Location in s-domain:
Im(s)
unstable
unstable
2

unstable
unstable
Note: In the case  = 0, the systemis
said unstable or critically stable.
RTECS 2015

39. Simple Pair of Complex Poles
p = - ±jd
d dAcos( t ) Bsin(  t )y(t )  et
 The natural response is
of the form:
A and B are given by
the initial conditions
40
RTECS 2015
if   0
 Location in s-domain:
Im(s)
if   0
if   0
 Assuming A >0, B > 0:
y(t)
if   0
t
stable unstable
Re(s)
stable unstable

40. Effects of Zeros
41
 A zero in the left half-plane
 Increases the overshoot if the zero is within a factor of 4 of the real
part of the complex poles. whereas it has very little influence on the
settling time.
 A zero in the right half-plane
 Depress the overshoot (and may causes an initial undershoot (=the
step response starts out in the wrong direction).

41. Effects of Additional Poles
42
 An additional pole in the left half plane
 increases the rise time significantly (=slows down the response) if
the extra pole is within a factor 4 of the real part of the complex
poles.
 If the extra real pole  is more than 6 times Re{p}, the effect is
negligible.

42. RTECS 201543

43. Well Behaved Signals44

44. How To Specify Controller Requirements ?
45
 Step Input
 Specify the requirements on the response to a step reference signal
 Transient Response
 First part of system response when the output is still changing
 Steady State Response
 The final state for the system output

45. RTECS 201546
Settling time
OvershootControlled variable
Reference
Steady State
Transient State
Steady state error

46. Delay time, Td : the time needed for the
output response to reach 50% of its final
value.

47. Rise time, Tr : the time taken for the output
response to go from 10% to 90% of its final
value.

48. Settling time, Ts : the time taken for the
output response to reach, and stay within
2% of its final output value
dn
s
 
4

4
T 

49. Peak time, Tp : the time required for the
output response to reach the first, or
maximum peak
dn
pT


 

 1 2

50. Percentage overshoot, %OS : the amount
of the output response overshoots the final
value at the peak time, expressed in
percentage of steady-state value
cfinal
c final
100%%OS 
cmax
1 2
%OS  e 100%
1 2
)
cmax
( /
 c(tp ) 1 e


51. • The steady-state value of the system G can be
evaluated using the function
▫ K=dcgain(G)
Steady-state Value y(∞)
52
 The steady-state value of the system under the
step response is the output when t → ∞.
 Using the final value theorem
a0
s
y()  lim sG(s)
1
s0
 G(0) 
b0

52. Laplace Transform Method
53
 Inverse Laplace transformation is used to obtain the output
signal in the time domain
 Symbolic Toolbox of MATLAB

53. Laplace Transform Method Example
54
s4
u(t)  2 2e3t
sin(2t)
 For the transfer function G(s) and the input u(t),
obtain the output y(t)
 Hint
 Laplace command: laplace
 Inverse Laplace command: i laplace
 7s3
17s2
17s 6
s3
 7s2
3s  4
G(s) 

54. RTECS