b_lecture12 frequency response and plotautomatic control system
DESCRIPTION
Automatic control SystemTRANSCRIPT
-
Frequency Domain Analysis
-
Introduction
In practice, the performance of a control system is
more realistically measured by its time-domain
characteristics. The reason is that the performance of
most control systems is judged based on the time
responses due to certain test signals.
This is in contrast to the analysis and design of
communication systems for which the frequency
response is of more importance, since most of the
signals to be processed are either sinusoidal or
composed of sinusoidal components.
-
Introduction
We learned in time-domain that the time response of a control system is usually more difficult to determine analytically, especially for high-order systems.
In the frequency domain, there is a wealth of graphical methods available that are not limited to low-order systems.
There are correlating relations between the frequency-domain and the time-domain performances in a linear system, so the time-domain properties of the system can be predicted based on the frequency-domain characteristics.
-
Frequency Response
r
0
0
t
t
tAt 1sin
tAt 2sin
t
t0
0
c
)sin( 222 tAc
)sin( 111 tAc
tr tcsystem
-
1
1s
Ts
sinr ru A t 2 2r
r
AU s
s
2 21
1
rc r
AU s s U s
Ts s
1
2 2 2 2sin( arctan )
1 1
r rTc
A T Au t e t T
T T
2 2
lim sin( arctan )1
rc
t
Au t t T
T
2 2 ( )
1
rc
AA arctg T
T
Frequency Response
First-order system
Input
Transient response Steady state response
If the system is stable, then the transient response0 when t
The steady state response
-
22)(
s
AsR
n
i
ips
sm
sn
sms
1
)(
)(
)(
)()( tAtr sin)(
js
a
js
a
ps
k
ps
k
ps
k
s
A
ps
smsC
n
n
n
i
i
21
2
2
1
1
22
1
)(
)()(
R(s) C(s) )(sGeneral system
Where pi is assumed to be distinct poles (i=1,2,3n). Then, in partial fraction form, we have
Frequency Response
-
js
a
js
a
ps
k
ps
k
ps
ksC
n
n
21
2
2
1
1)(
Taking the inverse Laplace transform yields
For the stable system all poles (pi) have negative real parts, then the transient response
the steady state response:
0)(1
n
i
tp
itiektc
js
a
js
aLtcs
211)(
js
a
js
aLekekektc
tp
n
tptp n 211
2121)(
Transient response )(tct Steady state response )(tcs
Frequency Response
-
)90)((
12
)(
2)( )(
))(()(
ojj
js
ejA
j
Ajjs
jsjs
Asa
)90)((*
122
)( ojjejA
aa
js
a
js
aLtcs
211)(
))(sin()(
2)(
)(
)90)(()90)((
21
jtjA
eejA
eaeatc
oo jtjjtj
tjtj
s
Frequency Response
sincos
formula
je
Euler
j
j
ee
ee
jj
jj
2sin
2cos
-
The amplitude ratio of the steady-state output cs(t) versus sinusoid input r(t):
istic charactermagnitude jR
jC(j
A
jA
)(
)()
)(
The phase difference between the steady-state output and sinusoid input:
acteristicphase charjRjCjtjt )()( )()]([
Then we have :
jss
jR
jCj
)(
)(
)()(
Compare with the sinusoid input tAtr sin)( , we have:
)sin()()( 211
tjA
js
a
js
aLtcs
where )( j
Frequency Response
-
Definition : frequency response (or characteristic) the ratio of the complex vector of the steady-state output versus sinusoid
input for a linear system, that is:
Here: input sinusoid theoftion representactor complex ve the)( jR
output the of tionrepresentavector complex the )( jC
stic)characterir response(o frequency )( j
jss
jR
jCj
)(
)(
)()(
)] (sin[)()( output state-steady the
)sin()(
jtAjtc
tAtrwhen
rrs
rr
Frequency Response
Thus the steady-state response depends only on the magnitude
and phase of , at a specific frequency . )( j
-
A unity feedback control system, the open-loop transfer function:
ssG
5.0
1)(
Determine the steady-state response of the system.
)6028.6sin(10)( : ottrIf
Solution:
The closed-loop transfer function : 15.0
1
5.011
5.01
)(1
)(
)(
)()(
ss
s
sG
sG
sR
sCs
5.0 10 28.6 : TAknown r
2 2
1( ) ( )
1arctg T
T
2
0
1 6.28 ( ) 0.3
(0.5 6.28) 1
( ) (0.5 6.28) 72.4
when
arctg
)12.4-3sin(6.28t
)4.726028.6sin(103.0)] (sin[)()(
0
00
tjtAjtc rrs
Example
-
Graphic expression of the frequency response
1. Rectangular coordinates plot
2. Polar plot (Nyquist curve)
3. Bode diagram(logarithmic plot)
Frequency Response Plot
-
1. Rectangular coordinates plot
Example )2()2(1
10
12
10)(
12
10)(
1
2
tgj
jGs
sG
o
o
o
o
o
o
o
. .
jGjG
29849950 5
875.8224.14
538.8064.13
964.754.22
435.6347.41
4507.75.0
0100
)()(
-90o
0.5 1 2 3 4 50
)( jG
1
5
10
)( jG
Frequency Response Plot
-
2. Polar plot (Nyquist curve )
The magnitude and phase response:
Calculate and for different . )(G
The polar plot is easily useful for investigating system stability.
Frequency Response Plot
)}(Im{)}(Re{)()( GjGsGjG js
)()()()( )( jGjGejGjG jj
It is done in polar coordinates as varies from 0 to .
)( jGr )( jG
)(G
0
1
10
)( 1jG
Im
0
or
-
3. Bode diagram(logarithmic plot) Plot the frequency characteristic in a semilog coordinate:
Magnitude response Y-coordinate in decibels: )(log20 10 jG
X-coordinate in logarithm of : 10log
Phase response Y-coordinate in radian or degree: )( jG
X-coordinate in logarithm of : 10log
Frequency Response Plot
Note:
The logarithm of the magnitude is normally expressed in terms
of the logarithm to the base 10, multiplying by 20,where the units are
decibels (dB).
A unit change in in the rectangular coordinates is
equivalent to one decades of variation in ,that is from 1 to 10, 10 to 100, ,and so on.
)(log20 10 jG
10log 10log
-
Frequency Response Plot
Note:
The logarithm of the magnitude is normally expressed in terms
of the logarithm to the base 10, multiplying by 20,where the units are
decibels (dB).
A unit change in in the rectangular coordinates is
equivalent to one decades of variation in ,that is from 1 to 10, 10 to 100, ,and so on.
)(log20 10 jG
10log 10log
)dB( )(L
)/( srad
20
40
0 2 4 6 10 1 8 20 40 60
60
2
2
3
Bode Plot
)(lg20)( jGL
)()( jG
80 100
-
1. Inertial element
Frequency Response of The Typical Elements
1
1
TssG
1
1
TjjG
)(1
1
22
TarctgGT
Gjs
bandwidth- 3
3
1
b
sb
s
b t
TtT
TG
G 1
4
707.0
Characteristic point
1
707.0
0
4
2
b
T1
G
G
Rectangular coordinates plot
-
Polar plot (Nyquist curve )
Frequency Response of The Typical Elements
1. Inertial element
0
T/1
1 045
Im
0
GjtgTj eGeT
jG
arg
22 1
1
1
1
TssG
1
1
TjjG
-
4db 3707.0lg20 20lg
, 1
arctgTG
G
T
arctgTG
TG
TG
1lg20lg20
1
1
22
22
Frequency Response of The Typical Elements
1. Inertial element
Bode diagram(logarithmic plot)
T/1 Break frequency or Corner frequency
1
1
TssG
1
1
TjjG
Characteristic point
-
Asymptotic plot 1lg20lg2022 TG
1 with compare when neglected T)( because 0lg20 1 1 2 isGTT
.a
.b lg20lg20lg20 1 1 22 TTGTT
dBnTT
dBTT
dBTT
dBTT
nn 2010lg20lg2010
4010lg20lg2010
2010lg20lg2010
01lg20lg201
22
Blue curve: exact plot
Red curve: asymptotic plot(corner plot)
Frequency Response of The Typical Elements
1. Inertial element
[-20]
-
22
2
2 nn
n
sssG
22
2
2 nn
n
jjjG
2
22
21
1
nn
jG
2
1
2
n
narctgjG
Frequency Response of The Typical Elements
2. Oscillating element
-
eakresonant pG
requencyresonant f
Gd
jGd
m
m
m
nm
12
1
21
02
2
2
2
1
G
G
n
bandwidth-
707.0|2
1
707.0
b
707.0
nb
nG
Frequency Response of The Typical Elements
Rectangular coordinates plot
Polar plot
2. Oscillating element
-
[-40]
22
2
21lg20lg20
nn
G
2
1
2
n
narctgG
Frequency Response of The Typical Elements
Bode diagram
2. Oscillating element
dB0||lg20 1or Gn
n
)lg(40)lg(20lg20 1or 2
nnn
n |G|
n Break frequency or Corner frequency
Asymptotic plot
-
Frequency Response of The Typical Elements
Transfer function: KsG )(
Frequency response:
ojG
KjGLKjGKjG
0)()(
lg20)(lg20)()()(
3. Proportional element
Re
Im
K
0dB, 0o
100 10 1 0.1 )(lg
)( ),( L
dB log20)( KL o0)(
Polar plot Bode diagram
0
)(L
-
4. Integrating element
Transfer function: s
sG1
)(
Frequency response:
ojG
jGLjG
jjG
90)()(
lg20)(lg20)(1
)(1)(
Polar plot
0
Bode diagram
decdB /20
o90)(
Frequency Response of The Typical Elements
Re
Im
0dB, 0o
10 1 0.1 0.01 )(lg
)( ),( L
0
-
5. Differentiating element
Transfer function
aldifferentiordersecondss
aldifferentiorderfirsts
aldifferentis
sG
nn 1)/(2)/(
1)(2
Polar plot
Re
Im
Re
Im
1
Re
Im
1
differential first-order differential second-order differential
Frequency Response of The Typical Elements
-
Because of the transfer functions of the differentiating elements are
the reciprocal of the transfer functions of Integrating element, Inertial
element and Oscillating element respectively
]1)/(2)//[(11)/(2)/(
111
1
22
nn
inverse
nn
inverse
inverse
ssss
Tss
ss
Bode curves of the differentiating elements are symmetrical to the lg-axis with the Bode curves of the Integrating element, Inertial element and
Oscillating element respectively.
Then we have the Bode diagrams of the differentiating elements:
Frequency Response of The Typical Elements
5. Differentiating element
-
0dB, 0o
100 10 1 0.1
)(lg
)( ),( L
decdB /20 o90)(
differential
)( ),( L
1th-order differential
0dB, 0o
100 10 1 0.1 )(lg
decdB /20o45
o90
Frequency Response of The Typical Elements
0dB, 0o
100 10 1 0.1 )(lg
)( ),( L
decdB /40
o180
o90
n
2th-order differential
-
6. Delay element
Transfer function: sesG )(
)()(
0)(lg20)(1)()(
jG
jGLjGejG j
Polar plot
Re
Im R=1
0dB, 0o 0
)(lg
)(L
Bode diagram
)(
Frequency Response of The Typical Elements
0
)(L