Transcript Slide 1
Feedback Control Systems
Lecture 2 Dr.-Ing. Erwin Sitompul President University
http://zitompul.wordpress.com
President University Erwin Sitompul FCS 2/1
Chapter 3 Dynamic Response
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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Chapter 3 Dynamic Response
The Block Diagram
U s G s
1 ( ) 1 ( )
Y s
1 ( ) Using block diagram, the system equations can be simplified graphically, which is often easier and more informative than algebraic manipulation.
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Chapter 3 Dynamic Response
Elementary Block Diagrams
Blocks in series Blocks in parallel with their outputs added
Y s
2 ( )
U s
1 ( )
G G
1 2 President University
Pickoff point
Erwin Sitompul
Summing point
G
1
G
2 FCS 2/4
Chapter 3 Dynamic Response
Elementary Block Diagrams
Single loop negative feedback ( ) 2
G
1
G R s
1 ( )
G G Y s
1 2 ( )
G G
1 2
G R s
1 ( )
Negative feedback
1
G
1
G G
1 2 1
G
1
G G
1 2
What about single loop with positive feedback?
?
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Chapter 3 Dynamic Response
Block Diagram Algebra
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Chapter 3 Dynamic Response
Block Diagram Algebra
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Chapter 3 Dynamic Response
Transfer Function from Block Diagram
Example: Find the transfer function of the system shown below.
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s
4 1 2
s s
2 4
s
2
s
2 2
s
2
s
4 4 Erwin Sitompul FCS 2/8
Chapter 3 Dynamic Response
Transfer Function from Block Diagram
Example: Find the transfer function of the system shown below.
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Chapter 3 Dynamic Response
Transfer Function from Block Diagram
President University 1 2 1
G G
1 2 4 1
G G G
1 2 5
G G
1 3
G G
1 6
G G G
1 2 4
G
6
G
2
G
5 Erwin Sitompul FCS 2/10
Chapter 3 Dynamic Response
Transfer Function from Block Diagram
Example R(s) and : Find the response of the system Y(s) to simultaneous application of the reference input disturbance D(s).
When D(s) 0,
R
1
G G
1 2
G G H
1 2 When R(s) 0,
D
G
2
G G H
1 ) 2
G
2 1
G G H
1 2 President University Erwin Sitompul
?
FCS 2/11
Chapter 3 Dynamic Response
Transfer Function from Block Diagram
R
D
1
G G
1 2
G G H
1 2 1
G
2
G G H
1 2
G
2 1
G G H
1 2
G R s
1 ( ) President University Erwin Sitompul FCS 2/12
Chapter 3 Dynamic Response
Definition of Pole and Zero
Consider the transfer function F(s): The system response is given by: 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.
Numerator polynomial Denominator polynomial
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Chapter 3 Dynamic Response
Effect of Pole Locations
Consider the transfer function F(s):
s
1
s
A form of first-order transfer function
1 The impulse response will be an exponential function:
e
t
1( ) •
How?
When
σ
>0, the pole is located at s < 0, The exponential expression y(t) decays.
Impulse response is stable.
When
σ
<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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Chapter 3 Dynamic Response
Effect of Pole Locations
Example : Find the impulse response of H(s),
s
2 2
s
1 3
s
2 •
PFE
(
s
2
s
1)(
s
1 2)
s
1 1 2
s
3 2
L
1
s
1 1
e
t
3
e
2
t
3
L
1
s
1 2 1.5
1 0.5
0 ● The terms e
–t stable, are determined by the poles at s = –1 and –2.
● This is true for more complicated
cases as well.
and e –2t , which are
● In general, the response of a
transfer function is determined by the locations of its poles.
-0.5
0 1 2 Time (sec) 3 4 President University Erwin Sitompul FCS 2/15
LHP
Chapter 3 Dynamic Response
Effect of Pole Locations
Time function of impulse response assosiated with the pole location in s-plane
RHP
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LHP : left half-plane RHP : right half-plane
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Chapter 3 Dynamic Response
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 (–s ± jω
d
).
Complex poles always come in conjugate pairs.
A pair of complex poles
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Chapter 3 Dynamic Response
Representation of a Pole in s-Domain
The denominator corresponding to a complex pair will be:
s s
) 2
j
d d
2 )(
s j
d
) On the other hand, the typical polynomial form of a second order transfer function is:
s
2 2 2
n n s
2
n
Comparing A(s) and denominator of H(s), the correspondence between the parameters can be found:
n
and
d
n
1 2
ζ ω ω n d
: damping ratio : undamped natural frequency : damped frequency
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Chapter 3 Dynamic Response
Representation of a Pole in s-Domain
Previously, in rectangular coordinates, the complex poles are at (–
± jω d
).
In polar coordinates, the poles are at (ω
n
, sin –1
ζ
the figure.
), as can be examined from
n
d
n
1 2
n
d
2
d
2 President University Erwin Sitompul FCS 2/19
Chapter 3 Dynamic Response
Unit Step Resonses of Second-Order System
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n t
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Chapter 3 Dynamic Response
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.
s
2 2
s
2 1
s
5 Since
s
2 2 2
n n s
n
2 ,
n
2 2
n
n
5 2.24 rad sec 0.447
The exact response can be otained from:
s
2 2
s
2
s
1 5 (
s
2
s
1) 2 1 2 2 poles at
s j
2 President University Erwin Sitompul FCS 2/21
Chapter 3 Dynamic Response
Effect of Pole Locations
To find the inverse Laplace transform, the righthand side of the last equation is broken into two parts: 2
s
1 (
s
1) 2
s
1 2 2 2 (
s
L
1 1) 2 2 2 1 2 (
s
2 1) 2 2 2 2 2
e
t
cos 2
t
1 2
e
t
sin 2
t
1( ) 1.5
1
Damped sinusoidal oscillation
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0 -0.5
-1 0 President University Erwin Sitompul 1 2 3 Time (sec) 4 FCS 2/22 5 6
Chapter 3 Dynamic Response
Time Domain Specifications
Specification for a control system design often involve certain requirements associated with the step response of the system: 1. Delay time, t
d
, is the time required for the response to 2. Rise time, t
r
, is the time needed by the system to reach the 3. Settling time, t
s
, is the time required for the response of size specified by absolute percentage of the final value.
4. Overshoot, M
p
, is the maximum peak value of the response (often expressed as a percentage).
5. Peak time, t
p
, is the time required for the response to President University Erwin Sitompul FCS 2/23
Chapter 3 Dynamic Response
Time Domain Specifications
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t r t s
,
t s
,
t r
,
t s
,
t r
%
M p
y t p y
Erwin Sitompul
y
100% FCS 2/24
Chapter 3 Dynamic Response
First-Order System
The step response of first-order system in typical form:
s
1 1 is given by:
s
1 1 1
s s s
1
L
1 1
s s
1
e
t
•
: time constant
• For first order system, M
and t
p
do not apply
p
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Chapter 3 Dynamic Response
Second-Order System
The step response of second-order system in typical form:
s
2 2 2
n n s
2
n
is given by:
s s
2 2 2
n n
(
s
s
n
2
s
n
n
) 2
L
1
e
n t
cos 1
s
d
d
2
t
e
n t
(
s
cos
1
n
n
) 2
d
2
t
sin
d
2
d
1
t
2
e
n t
sin
d t
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Chapter 3 Dynamic Response
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.
e
n t
cos
d t
1 2 sin
d t
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Chapter 3 Dynamic Response
Rise Time
The step response expression of the second order system is now used to calculate the rise time, t r,0%–100% :
y t r
e
cos
t d r
1
2 sin
t d r
tan
d
e
fulfilled if: 0 cos
t d r
1 or, tan
t d r
1
2
sin
t d r
2
d t r
1
d
tan 1
d
0 (
d
)
φ
n
d
n
1 2
t r
is a function of ω
d
President University Erwin Sitompul FCS 2/28
Chapter 3 Dynamic Response Using the following rule:
A
sin
B
cos
C
cos( ), with:
C
A
2
B
2 , tan 1
A B
The step response expression can be rewritten as:
e
n t
where:
tan 1 1
2 1
2 cos(
d t
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)
Settling Time
Erwin Sitompul
t s
is a function of ζ
1
n
FCS 2/29
Chapter 3 Dynamic Response
Settling Time
The time constant of the envelope curves shown previously is 1/ζω
n
, so that the settling time corresponding to a certain constant.
t s
3 3
n t s
4 4
n t s
5 5
n
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Chapter 3 Dynamic Response
Peak Time
When the step response y(t) reaches its maximum value (maximum overshoot), its derivative will be zero:
e
n t
cos
d t
1
2 sin
d t
( )
n e
n t e
n t
d
cos
d t
sin
d t
1 2
d
1 2 sin
d t
cos
d t
( )
e
n t
1
n
2
d
sin
d t
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Chapter 3 Dynamic Response
Peak Time
At the peak time, ( )
p
e
1
n
2
d
sin
t d p
≡0
t d p
0 Since the peak time corresponds to the first peak overshoot,
t p
d
t p
is a function of ω
d
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Chapter 3 Dynamic Response
Maximum Overshoot
Substituting the value of t
p y t p
e
cos
t d p
into the expression for y(t),
1
2 sin
t d p
y t p
e
d
cos
1
2 sin
e
1 2
M p
y t p e
y
1 2
M p
e
1 2 %
M p
y t p y y
100% %
M p
e
1 2
if y(∞) = 1
100% President University Erwin Sitompul FCS 2/33
Chapter 3 Dynamic Response
Example 1: Time Domain Specifications
Example : Consider a system shown below with ζ input.
0.6 and ω
n
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 After block diagram simplification,
s
2 2 2
n n s
n
2
Standard form of second-order system
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Chapter 3 Dynamic Response
Example 1: Time Domain Specifications
0.6,
n
5 rad/s
d
1
n
n
t r
1
d
tan 1
d
1 4 tan 1 4 3
In second quadrant
1 4 ( 0.927) 0.554 s 3 rad/s 2 4 rad/s
t p
d
4 0.785 s President University Erwin Sitompul FCS 2/35
Chapter 3 Dynamic Response
Example 1: Time Domain Specifications
M p
y t p y
( )
e
1 2
M p
e
1 2
e
0.6
0.8
0.0948
%
M p
e
1 2 100% 9.48%
t s
4
n
4 1.333 s
t s
3
n
3 President University 1 s Check y(∞) for unit step input, if
s
2 2 2
n n s
2
n
Erwin Sitompul FCS 2/36
Chapter 3 Dynamic Response
Example 1: Time Domain Specifications
t r M
0.554 s,
t p p
9.48%,
t s
0.785 s 1.333 s 0.6
0.4
0.2
1.2
1 0.8
0 0 0.2
President University 0.4
0.6
0.8
1 Time (sec) 1.2
Erwin Sitompul 1.4
1.6
1.8
2 FCS 2/37
Chapter 3 Dynamic Response
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.
1
K
2)
K
2)
s
2
K
2
s
K
%
M p
10%
e
K
2.853
1 2 0.1
2
n
n
2 2
K
n K
0.592
1
n
2 1 0.592
1.689
2 1.689
2.853
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1.2
Chapter 3 Dynamic Response
Example 2: Time Domain Specifications
1 0.8
: K = 2 : K = 2.8
: K = 3 0.6
0.4
0.2
0 0 1 2 3 Time (sec) 4 5 6 President University Erwin Sitompul FCS 2/39
Chapter 3 Dynamic Response
Homework 2
No.1
Obtain the overall transfer function of the system given below.
+ + No.2, FPE (5 th Ed.), 3.20.
Note: Verify your design using MATLAB and submit also the printout of the unit-step response. Deadline: 18.05.2012, at 07:30.
President University Erwin Sitompul FCS 2/40