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Nội dung Text: Lecture Fundamentals of control systems: Chapter 5 - TS. Huỳnh Thái Hoàng
- Lecture Notes
Fundamentals of Control Systems
Instructor: Assoc. Prof. Dr. Huynh Thai Hoang
Department of Automatic Control
Faculty of Electrical & Electronics Engineering
Ho Chi Minh City University of Technology
Email: hthoang@hcmut.edu.vn
huynhthaihoang@yahoo.com
Homepage: www4.hcmut.edu.vn/~hthoang/
6 December 2013 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 1
- Chapter 5
ANALYSIS OF CONTROL SYSTEM
PERFORMANCE
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 2
- Content
Performance criteria
Steady state error
Transient response
The optimal performance index
Relationship between frequency domain performances and
time domain performances.
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 3
- Performance criteria
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 4
- Performance criteria: Steady state error
yfb(t)
Y(s) ess
R(s) E(s) r(t)
+_ G(s)
Yfb(s) e(t) ess
H(s) t
0
Error: is the difference between the set-point
set point (input) and the
feedback signal.
e(t ) r (t ) y fb (t ) E ( s ) R ( s ) Y fb ( s )
Steady-state error: is the error when time approaching infinity.
ess lim
li e(t ) ess lim
li sE
E ( s)
t s 0
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 5
- Performance criteria – Percent of Overshoot (POT)
Overshoot: refers to an output exceeding its steady-state
steady state value.
value
y(t) y(t)
overshoot
ymax
ymax yss
yss yss
yss
No overshoot t t
0 0
Percentage of Overshoot (POT) is an index to quantify the
overshoot of a system, POT is calculated as:
y max y ss
POT 100%
y ss
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 6
- Performance criteria – Settling time and rise time
Settling time (ts): is the time required for the response of a
system to reach and stay within a range about the steady-
state value of size specified by absolute percentage of the
steady-state value (usually 2% or 5%)
Rise time (tr): is the time required for the response of a
system to rise from 10% to 90% of its steady-state
steady state value.
value
y(t) y(t)
(1+)yss
yss yss
(1)yss 0.9yss
0.1yss t
t
0
0
ts tr
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 7
- Steady--state error
Steady
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 8
- Steady--state error
Steady
R( )
R(s) E( )
E(s) Y( )
Y(s)
+_ G(s)
Yfb(s)
( )
H(s)
R( s)
Error expression: E (s)
1 G(s) H (s)
sR( s )
Steady-state error: ess lim sE ( s ) lim
s 0 s 0 1 G ( s ) H ( s )
Remark: Steady-state error not only depends on the structure
and p
parameters of the system
y but also depends
p on the input
p
signal.
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 9
- Steady--state error to step input
Steady
Step input: R( s ) 1 / s
1
Steady-state error: ess
1 Kp
with K p lim G ( s ) H ( s ) (position constant)
s 0
yfb(t) yfb(t)
1 1
t t
0 0
G(s)H(s) does not have G(s)H(s) has at least 1
any deal integral factor ideal integral factor
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 10
- Steady--state error to ramp input
Steady
Ramp input: R( s ) 1 / s 2
1
ess with K v lim sG ( s ) H ( s ) (velocity constant)
Kv s 0
s
yfb(t) yfb(t) yfb(t)
r(t) r(t) r(t)
ess 0 ess = 0
e(t)
t t t
0 0 0
G(s)H(s) does not have G(s)H(s) has 1 ideal G(s)H(s) has at least
deal integral factor integral factor 2 ideal integral factors
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 11
- Steady--state error to parabolic input
Steady
t R( s) 1 / s 3
P b li iinput:
Parabolic
1
ess with K a lim s G ( s ) H ( s )
2 (acceleration
Ka s 0 constant )
yfb(t) yfb(t) yfb(t)
r(t) r(t) r(t)
ess0 ess= 0
e(t)
t t t
0 0 0
G(s)H(s) has less than G(s)H(s) has 2 ideal G(s)H(s) has more than
2 ideal integral factors integral factors 2 ideal integral factors
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 12
- Transient response
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 13
- First--order system
First
R(s) K Y(s)
Ts 1
K
Transfer function: G (s)
T 1
Ts
1
First order system has 1 real pole: p1
T
1 K
Transient response: Y ( s ) R ( s )G ( s ) .
s Ts 1
y (t ) K (1 e t /T )
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 14
- First--order system (cont’)
First
Im s y(t)
(1+).K
K
(1).K
Re s
0 0.63K
1/T
t
0
T ts
Pole – zero plot Transient response
of a first order system of the first order
y (t ) K (1 e t /T )
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 15
- First--order system – Remarks
First
First order system has only one real pole at (1/T), its
transient response doesn’t have overshoot.
Time constant
Ti t t T:
T is
i the
th time
ti required
i d for
f the
th step
t response off
the system to reach 63% its steady-state value.
The further the pole (1/T) of the system is from the
imaginary axis, the smaller the time constant and the faster
the time response of the system.
Settling time of the first order system is:
1
ts T ln
l
where = 0.02
0 02 (2% criterion) or = 0.05
0 05 (5% criterion)
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 16
- First--order system
First
The relationship between the pole and the time response
The further the pole of the system is from the imaginary axis,
the smaller the time constant and the faster the time
response of the system.
Im s y(t)
K
Re s
0
t
0
Pole – zero plot Transient response
of a first order system of the first order
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 17
- Second--order oscillating system
Second
R(s) K Y(s)
T 2 s 2 2Ts 1
The transfer function of the second-order oscillating system:
K Kn2 1
G( s) 2 2 2 (n , 0 1)
T s 2Ts 1 s 2n s n
2
T
The system has two complex conjugate poles:
p1, 2 n jn 1 2
1 Kn2
Transient response: Y ( s ) R ( s )G ( s ) . 2
s s 2n s n2
y (t ) K 1
e nt
1 2
sin (n 1 2 )t
(cos )
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 18
- Second--order oscillating system (cont’)
Second
y(t)
Im s
cos =
j n 1 2 (1+).K
n K
((1)).K
R s
Re
n 0
j n 1 2
t
0
ts
Pole – zero plot of a second Transient response of a second
order oscillating system order oscillating system
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 19
- Second--order oscillating system – Remark
Second
A second order oscillation system has two conjugated
complex poles, its transient response is a oscillation signal.
If = 0,
0 transient
t i t response
is a stable oscillation signal =0
at the frequency n n = 0.2
is called natural oscillation = 0.4
frequency.
If 0
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