Transcript Slide 1

Chapter 3
DYNAMIC RESPONSE
Feedback Control System
Dr.-Ing. Erwin Sitompul
http://zitompul.wordpress.com
The Block Diagram
• Block diagram is a graphical tool to visualize the model of a
system and evaluate the mathematical relationships between
its components, using their transfer functions.
• In many control systems, the system equations can be written
so that their components do not interact except by having the
input of one part be the output of another part.
• The transfer function of each components is placed in a box,
and the input-output relationships between components are
indicated by lines and arrows.
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The Block Diagram
U1 (s) G1 (s)  Y1 (s)
• Using block diagram, the system equations can be
simplified graphically, which is often easier and more
informative than algebraic manipulation.
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Elementary Block Diagrams
Blocks in series
Y2 ( s)
 G1G2
U1 ( s)
Blocks in parallel with
their outputs added
Pickoff
point
Summing
point
Y ( s)
 G1  G2
U ( s)
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Elementary Block Diagrams
Single loop
negative feedback
Y (s)  R(s)  Y (s)  G2  G1
Y (s)  G1R(s)  G1G2Y (s)
Y (s)1  G1G2  G1R(s)
G1
Y ( s)

R( s) 1  G1G2
Negative
feedback
G1
Y ( s)

R( s) 1  G1G2
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?
What about single loop
with positive feedback?
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Block Diagram Algebra
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Block Diagram Algebra
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Transfer Function from Block Diagram
Example:
Find the transfer function of the system shown below.
2s  4
2
Y (s)
2s  4
s

T (s) 
 2
R ( s ) 1  2 s  4 s  2s  4
s2
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Transfer Function from Block Diagram
Example:
Find the transfer function of the system shown below.
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Transfer Function from Block Diagram
G1G2
1G1G3
G1G2G4
1G1G3
Y ( s)
T (s) 

R( s ) 1 
 G6


 G5 
 G2

G1G2G5  G1G6

1  G1G3  G1G2G4
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Transfer Function from Block Diagram
Example:
Find the response of the system Y(s) to simultaneous
application of the reference input R(s) and disturbance D(s).
When D(s)  0,
YR ( s)
G1G2

R( s) 1  G1G2 H
When R(s)  0,
?
YD ( s)
G2
G2


D( s) 1  (G1 )G2 H 1  G1G2 H
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Transfer Function from Block Diagram
Y (s)  YR (s)  YD (s)
G1G2
G2
Y ( s) 
R( s ) 
D( s )
1  G1G2 H
1  G1G2 H
G2
Y ( s) 
 G1R(s)  D(s) 
1  G1G2 H
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Definition of Pole and Zero
Consider the transfer function F(s):
Y (s)
B( s)
 F ( s)  F ( s) 
U (s)
A( s)
Numerator polynomial
Denominator polynomial
The system response is given by:
B( s )
Y ( s)  F ( s)U ( s) 
U ( s)
A( s)
• The poles are the values of s for which
the denominator A(s)  0.
• The zeros are the values of s for which
the numerator B(s)  0.
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Effect of Pole Locations
Consider the transfer function:
1
Y ( s)
1

Y
(
s
)

U (s)
H (s) 

s 
U (s) s  
A form of first-order
transfer function
The impulse response will be an
exponential function:
y(t )  e t 1(t ) •How?
• When  > 0, the pole is located at s < 0,
 The exponential expression y(t) decays.
 Impulse response is stable.
• When  < 0, the pole is located at s > 0,
 The exponential expression y(t) grows with time.
 Impulse response is referred to as unstable.
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Effect of Pole Locations
Example:
Find the impulse response of H(s),
2s  1
2s  1
1
3
H ( s)  2
•PFE


s  3s  2 ( s  1)( s  2)
s 1 s  2
 1 
1  1 
h(t )  L 
  3L 

 s  1
s  2
h(t )  (et  3e2t ) 1(t )
2
1
1.5
h(t )
The terms e–t and e–2t, which are stable,
are determined by the poles at s  –1
and –2
1
0.5
0
-0.5
0
1
2
3
4
Time (sec)
• This is true for more complicated cases as well.
• In general, the response of a transfer function is determined
by the locations of its poles.
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Effect of Pole Locations
Time function of impulse response assosiated
with the pole location in s-plane
LHP
RHP
LHP
RHP
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: left half-plane
: right half-plane
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Representation of a Pole in s-Domain
• The position of a pole (or a zero)
in s-domain is defined by its real
and imaginary parts, Re(s) and
Im(s).
• In rectangular coordinates, the
complex poles are defined as
(– ± jωd).
• Complex poles always come in
conjugate pairs.
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A pair of
complex
poles
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Representation of a Pole in s-Domain
• The denominator corresponding to a complex pair will be:
A( s)  ( s    jd )( s    jd )
 ( s   )2  d2
• On the other hand, the typical polynomial form of a secondorder transfer function is:
n2
H ( s)  2
s  2n s  n2
• Comparing A(s) and denominator of H(s), the
correspondence between the parameters can be found:
  n and d  n 1   2
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ζ : damping ratio
ωn : undamped natural frequency
ωd : damped frequency
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Representation of a Pole in s-Domain
• Previously, in rectangular coordinates,
the complex poles are at (– ± jωd).
• In polar coordinates, the poles are at
(ωn, sin–1ζ), as can be examined from
the figure.
  n
d  n 1   2
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

 
d
 n   2  d2
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Unit Step Resonses of Second-Order System
y(t )
n t
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Effect of Pole Locations
Example:
Find the correlation between the poles and the impulse
response of the following system, and further find the exact
impulse response.
2s  1
H ( s)  2
s  2s  5
n2
2
Since H ( s)  2
,

n  5  n  5  2.24 rad sec
2
s  2n s  n
2n  2    0.447
The exact response can be otained from:
2s  1
2s  1

H ( s)  2
 poles at s  1  j 2
2
2
s  2s  5 ( s  1)  2
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Effect of Pole Locations
To find the inverse Laplace transform, the righthand side of
the last equation is broken into two parts:
2s  1
H ( s) 
( s  1) 2  22
s 1
1
2
2

2
2
( s  1)  2 2 ( s  1) 2  22
h(t )  L1  H (s) 
1 t
 t

  2e cos 2t  e sin 2t  1(t )
2


Damped
sinusoidal
oscillation
2
1.5
1
h(t ) 0.5
0
-0.5
-1
0
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2
3
4
Time (sec)
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6
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Time Domain Specifications
Specification for a control system design often involve certain
requirements associated with the step response of the system:
1. Delay time, td, is the time required for the response to reach
half the final value for the very first time.
2. Rise time, tr, is the time needed by the system to reach the
vicinity of its new set point.
3. Settling time, ts, is the time required for the response curve
to reach and stay within a range about the final value, of
size specified by absolute percentage of the final value.
4. Overshoot, Mp, is the maximum peak value of the response
measured from the final steady-state value of the
response (often expressed as a percentage).
5. Peak time, tp, is the time required for the response to reach
the first peak of the overshoot.
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Time Domain Specifications
tr ,10%90% , tr ,5%95% , tr ,0%100%
ts,1% , ts,2% , ts,5%
%M p 
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y(t p )  y()
y ( )
100%
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First-Order System
The step response of the first-order system in the typical form:
1
H ( s) 
 s 1
is given by:
1 1
Y (s) 

 s 1 s
1
1
 
s s  (1  )
1

1
y (t )  L  

 s s  (1  ) 
1
y(t )  (1  et  ) 1(t )
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•  : time constant
• For first order system,
Mp and tp do not apply
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Second-Order System
The step response of the second-order system in the
typical form:
n2
H ( s)  2
s  2n s  n2
is given by:
n2
1
Y ( s)  2

2
s  2n s  n s
s  n
n
1
 

2
2
s (s  n )  d (s  n )2  d2

 t
1
cos d t 
e t sin d t
y(t )  L Y (s)  1  e
1  2



 t
 cos d t 
y(t )  1  e
sin d t 
2


1




n
n
n
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Second-Order System
• Time domain specification
parameters apply for most
second-order systems.
• Exception: overdamped
systems, where ζ > 1
(system response similar to
first-order system).
• Desirable response of a
second-order system is
usually acquired with
0.4 < ζ < 0.8.
y(t )  1  e
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n t



 cos d t 
sin d t 
2


1 


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Rise Time
The step response expression of the second order system
is now used to calculate the rise time, tr,0%–100%:
y(tr )  1  1  e
Since e
n tr
cos d tr 
n tr



 cos d tr 
sin d tr 
2


1




 0 , this condition will be fulfilled if:

1 
2
sin d tr  0
tan   
d

φ
Or:
1  2
d
tan d tr  



 d   (   )
tr 
tan   
d
d
  
1
1
tr is a function of ωd
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  n
d  n 1   2
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Settling Time
Using the following rule:
A sin   B cos   C cos(   ),
with:
C
 A
A  B ,   tan  
B
The step response expression
can be rewritten as:
en t
y (t )  1 
  cos(d t   ) 
2
1 
where:
 

1

  tan 
2
 1  


2
2
1
ts is a function of ζ

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n
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Settling Time
The time constant of the envelope curves shown previously is
1/ζωd , so that the settling time corresponding to a certain
tolerance band may be measured in term of this time constant.
ts ,5%  3 
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n
ts ,2%  4 
4
n
ts ,1%  5 
5
n
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Peak Time
When the step response y(t) reaches its maximum value
(maximum overshoot), its derivative will be zero:



 cos d t 
y(t )  1  e
sin d t 
2


1







n t
 cos d t 
y(t )  n e
sin d t  
2


1






d
n t
 d sin d t 
e
cos d t 
2


1




n t
2



n
n t

y(t )  e
 d  sin d t
 1  2



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Peak Time
At the peak time,
y(t p )  e
n t p
  2

n

 d  sin d t p  0
2
 1 



d t p  0,  , 2 ,3 ,
≡0
Since the peak time corresponds to the first peak overshoot,

tp 
d
tp is a function of ωd
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Maximum Overshoot
Substituting the value of tp into the expression for y(t),



 cos d t p 
y(t p )  1  e
sin d t p 
2


1







n  d
 cos  
y (t p )  1  e
sin    1  e
2


1




n t p
%M p 
M p  y (t p )  y ()
 (1  e
Mp e

1 2
1
2
y(t p )  y()
y ( )
1 2
100%
) 1
%M p  e

1 2
100%
if y(∞) = 1
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Example 1: Time Domain Specifications
Example:
Consider a system shown below with ζ  0.6 and ωn  5 rad/s.
Obtain the rise time, peak time, maximum overshoot, and
settling time of the system when it is subjected to a unit step
input.
2

Y
(
s
)
n
After block diagram simplification,
 2
R(s) s  2n s  n2
Standard form of
second-order system
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Example 1: Time Domain Specifications
  0.6, n  5 rad/s  d  1   2 n  1  0.62  5  4 rad/s
   n  0.6  5  3 rad/s
 d 
tr 
tan   
d
  
1
1
In second quadrant
1
4 1
1 
 tan     (  0.927)  0.554 s
4
 3 4
 
tp 
  0.785 s
d 4
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Example 1: Time Domain Specifications
M p  y (t p )  y ()  (1  e

 0.6  0.8
1 2
Mp e

1 2
e
%M p  e

1 2
100%  9.48%
ts ,2%
ts ,5%
) 1
 0.0948
4
4


 1.333 s
n 0.6  5
3


1 s
n 0.6  5
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Check y () for unit step
input, if
n2
Y ( s)
 2
R(s) s  2n s  n2
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Example 1: Time Domain Specifications
tr  0.554 s, t p  0.785 s
M p  9.48%, ts  1.333 s
1.2
1
Amplitude
0.8
0.6
0.4
0.2
0
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Time (sec)
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Example 2: Time Domain Specifications
Example:
For the unity feedback system shown below, specify the
gain K of the proportional controller so that the output y(t)
has an overshoot of no more than 10% in response to a
unit step.
K
s ( s  2)
K
s ( s  2)
Y (s)

R(s) 1 

%M p  10%  e
K
s 2  2s  K

1 2
0  K  2.853
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 2n  2
 n2  K
 0.1    0.592
1
1
 n  
 1.689
 0.592
 K  n2  1.6892  2.853
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Example 2: Time Domain Specifications
1.2
1
:K=2
: K = 2.8
:K=3
Amplitude
0.8
0.6
0.4
0.2
0
0
1
2
3
4
5
6
Time (sec)
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Feedback Control System
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Homework 2
 No.1
Obtain the overall transfer function of the system given below.
+
+
 No.2, FPE (5th Ed.), 3.20.
Note: Verify your design using MATLAB and submit also the
printout of the unit-step response.
 Deadline: 17.05.2011, 7:30 am.
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Feedback Control System
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