Transcript Slide 1

‫بسم ا‪ ...‬الرحمن الرحيم‬
‫سیستمهای کنترل خطی‬
‫پاییز ‪1389‬‬
‫دکتر حسین بلندي‪ -‬دکتر سید مجید اسما عیل زاده‬
Frequency Response
Frequency Response
Frequency Response
method to plot the magnitude response of the Bode diagram
s + 10)
10
(
G (s) =
s( s + 5)(s + 2)




Bode Form:
10( jw + 10)
10(1 + jw /10)
=
G ( jw ) =
jw ( jw + 5)( jw + 2) jw (1 + jw /5)(1 + jw /2)
Plot the asymptotic approximations for each
term separately, for both magnitude and angle.
Then add them together to get the system
asymptotic approximation.
Sketch in the Bode plot curve.
10 (1 + j w /10 )
j w (1 + j w /5 )(1 + j w /2 )
60
Magnitude (dB)
40
1 + jw /10
20
1
1 + jw / 5
0
1
jw
-20
-40
1
1 + jw / 2
-60
Phase (deg)
-90
-100
-120
-140
-160
-180
-2
10
-1
10
.2
.5 100 2
Frequency (rad/sec)
5
1
10
2
10
method to plot the magnitude response of the Bode diagram
method to plot the magnitude response of the Bode diagram
G ( s) =

Bode Form:
G ( jw ) =
=

5
100
 2
s + 1 s + 2 s + 100
5
100

jw + 1 ( jw )2 + 2 jw + 100
5
1

1 + jw /1 1 + 0.2 jw /10 - (w /10 )2
The damping ratio for the second-order
term is z = 0.1 and the natural frequency is
10 rad./s .
5 
100
jw +1 ( jw )2 + 2 jw +100
1
= 5 
1+ jw /1 1+ 0.2 jw /10- (w /10)2
G( jw) =
Magnitude (dB)
14
-20dB/dec.
0
14dB
-20
-60dB/dec.
-40
w = wr = wn 1 - 2z 2
-60
= 10 1 - 2  0.12  9.9
Phase (deg)
0
-50
-100
-150
Mr =
-45/dec.
-135/dec.
-200
-90/dec.
-250
-270
10 -1
10 0
p=1
10 1
wn = 10
Frequency (rad/sec)
10 2
1
2z 1 - 
2
 5.05 = 14.6 dB
method to plot the magnitude response of the Bode diagram
L(w ),  (w )
G ( s) =
10( s + 1)
40dB, 90o －20dB/dec
2 2
s(0.1s + 1)( 0.01 s + 0.01s + 1)
20dB,
0dB, 0o
45o
-40dB,
-90o
0.1
-20dB, -45o
1
10
－20dB/dec
1.25dB
100
wr
w (logw )
－60dB/dec
-60dB.-135o
-80dB,-180o
-100dB,-225o
-120dB,-270o
There is a resonant peak Mr at:
Mr =
1
2z 1 -  2
 1.154 = 1.25 dB
w = w r = w n 1 - 2z 2
= 100 1 - 2  0.5 2  70.7
method to plot the magnitude response of the Bode diagram
50, 000( jw + 10)
G( jw) =
( jw + 1)( jw + 500)
Given:
First: Always, always, always get the poles and zeros in a form such that
the constants are associated with the jw terms. In the above example
we do this by factoring out the 10 in the numerator and the 500 in the
denominator.
G( jw) =
Second:
50, 000 x10( jw /10 + 1)
100( jw /10 + 1)
=
500( jw + 1)( jw / 500 + 1)
( jw + 1)( jw / 500 + 1)
When you have neither poles nor zeros at 0, start the Bode
at 20log10K = 20log10100 = 40 dB in this case.
wlg
Bode Plot Magnitude for 100(1 + jw/10)/(1 + jw/1)(1 + jw/500)
1
60
1
1
1
1
1
1
1
1
1
1
1
40
20
dB Mag
dB Mag
0
-20
-60
-60
0.1
1
10
100 w (rad/sec)
1000
w (rad/sec)
10000
Bode Plots
Example :
The completed plot is shown below.
G( s) =
60
1
1
10(1 + s / 10)
s(1 + s / 100) 2
1
1
1
1
-20db/dec
40
20
dB Mag
-40 db/dec
0
-20
-40
-60
0.1
1
10
w (rad/sec)
100
1000
Bode Plots
Example:
80(1 + jw)3
G( s ) =
( jw)3 (1 + jw / 20)2
Given:
1
1
1
1
20 log80 = 38 dB
1
1
-60 dB/dec
60
dB Mag 40
-40 dB/dec
20
0
-20
.
0.1
1
w (rad/sec)
10
100
Bode Plots
Example:
G ( jw) =
Given:
1
1
10(1 - jw / 2)
(1 + j 0.025w)(1 + jw / 500)2
1
1
1
1
60
40
+ 20 dB/dec
dB Mag
-40 dB/dec
20
0
-20
-40
-60
0.1
1
2
10
w (rad/sec)
100
1000
Bode Plots
Example
Given:
1
(1 + jw / 30) (1+ jw / 100)
G ( jw) =
(1+ jw / 2) (1+ jw / 1700)
2
2
1
1
1
1
2
2
1
60
40
20
dB Mag
0
-40 dB/dec
-20
+ 40 dB/dec
-40
-60
0.1
1
10
w (rad/sec)
100
1000
Determine the transfer function in terms of the Bode diagram
The minimum phase system(or transfer function)
Compare following transfer functions:
K (s + 1)
G1 ( s ) =
(Ts + 1)
K (s - 1)
G2 ( s ) =
(Ts + 1)
K ( -s + 1)
K (s + 1)
G3 ( s ) =
G4(s) =
(Ts + 1)
(Ts - 1)

We have:
T 
G1 (w ) = G2 (w ) = G3 (w ) = G4 (w )
=K
(w ) 2 + 1
(Tw ) 2 + 1
The magnitude responses are the same.
But the net phase shifts are different when ω vary from zero to infinite. It can be
illustrated as following:
Sketch the polar plot:

G1 ( jw ) = - tg -1 (Tw ) + tg -1 (w ), G2 ( jw ) = - tg -1 (Tw ) + 180o - tg -1 (w )



G3 ( jw ) = - tg -1 (Tw ) - tg -1 (w ), G4 ( jw ) = - 180o - tg -1 (Tw ) + tg -1 (w )
Determine the transfer function in terms of the Bode diagram
Compare following transfer functions:
K (.1s + 1)
K (.1s - 1)
G1 ( s) =
G2 ( s) =
(1s + 1)
(1s + 1)
T 
K (-.1s + 1)
K (.1s + 1)
G3 ( s) =
G4(s) =
(1s + 1)
(1s - 1)

Bode Diagram
0
Magnitude (dB)
-5
-10
-15
-20
360
G1
Phase (deg)
G2
180
G3
G4
0
-180
-2
10
-1
10
0
10
1
10
2
10
3
10
Frequency (rad/sec)
Only for the minimum phase systems we can affirmatively determine the
relevant transfer function from the magnitude response of the Bode diagram .
Determine the transfer function from the magnitude response of
the Bode diagram
Example : suppose the system is minimum phase
－
we can get the G(s) from
the Bode diagram :
K ( 0.5s + 1)
G ( s) =
2
s
( 0.005s + 1)
and :
L(w )
40dB/dec
－
20dB/dec
w (logw )
0dB, 0o
0.12
120
200 100
10
40dB/dec
－
L(w ) = 2o log K - 20 log w 2 + 20 log(0.5w ) - 20 log(0.005w )
w = 20
= 0  K = 40
Bode Diagram
100
System: sys
50
Magnitude (dB)
Frequency (rad/sec): 20
Magnitude (dB): 0.000989
0
-50
Phase (deg)
-100
-90
-120
-150
-180
-1
10
0
10
1
10
2
10
3
10
4
10
Determine the transfer function from the magnitude response of
the Bode diagram
Example
L(w ),  (w )
－20dB/dec
10( 0.1s + 1)
10( 0.1s - 1) 0dB,
G1 =
G2 =
( s + 1)
( s + 1)
10( -0,1s + 1)
10( 0.1s + 1)
G3 =
G4 ( s ) =
( s + 1)
( s - 1)
All satisfy the magnitude response
10
1
100
w (logw )
Bode Diagram
0
-10
-15
10(0.1s + 1)
only G4 ( s ) =
( s - 1)
-20
360
G1
G2
Phase (deg)
satisfy the phase response sim ultaneously.
10(0.1s + 1)
So, we have :
G( s) =
(s - 1)
0.1
－90o
－180o
-5
Magnitude (dB)
But
0o
180
G3
G4
0
-180
-2
10
-1
10
0
10
1
10
Frequency (rad/sec)
2
10
3
10
Transfer Function Identification
• Frequency response characteristics can be be
obtained experimentally by applying sinusoidal
inputs of various frequencies, and measuring the
gain and phase relationships between input and
output.
• By fitting asymptotic approximations to the
frequency response characteristics obtain from
experimental measurements, an approximate
transfer function model of the system can be
obtained.
Transfer Function Identification:
Major Steps
1. Determine the initial slope  order of poles at the
origin.
2. Determine the final slope  difference in order
between the denominator and numerator (n-m).
3. Determine the initial and final angle  confirm
the results from above or detect the presence of a
non-minimum phase system (delays or zeroes in
the RHS).
4. Determine the low frequency gain (Bode gain).
Transfer Function Identification:
Major Steps
5. Detect the number and approximate location
of corner frequencies and fit asymptotes.
– examine expected -3dB points.
– try to separate second-order terms and use the
standard responses to estimate damping.
6. Sketch the phase plot for the identified
transfer function as a check of accuracy.
Transfer Function Identification:
Major Steps
7. Use the phase plot to check for non-minimum
phase terms and to calculate the time delay
value if one is present.
8. Calculate the frequency response for the
identified model and check against the
experimental data (MatLab or a few points by
hand calculation).
9. Iterate and refine the pole/zero locations and
damping of second-order terms.
Example I:
G ( s) =
10( s + 10)
s ( s + 5)(s + 2)
Magnitude (dB)
Transfer Function Identification Example I
40
20
0
-20
-40
-60
-80
-100
-120
Phase (deg)
-80
-100
-120
-140
-160
-180
-200
-220
10 -1
10 0
10 1
Frequency (rad/sec)
10 2
10 3





Initial slope = -20dB/dec.  a 1/jw term.
Final slope = -40dB/dec.  (n-m) = 2 .
Initial angle = -90 and the final angle is -180
which checks with the gain curve.
Low frequency gain is found to be |KB/w|dB = 35dB,
where w = 0.1  KB = 5.62 .
Through asymptotic fitting there are two poles
found at wc = 4 and 25, and one zero at wc = 70 .
Transfer Function Identification Example I
40
Magnitude (dB)
20
-20dB/dec.
0
-20
-40dB/dec.
-40
-60
-40dB/dec.
-80
-60dB/dec.
-100
-120
4
25
70
-80
Phase (deg)
-100
-120
-140
-90/dec.
-160
-45/dec.
-180
+45/dec.
-200
-220
10 -1
10 0
10 1
Frequency (rad/sec)
10 2
10 3

The estimated transfer function in Bode
form is
5.62(1+ jw /70)
G ( jw ) =
jw (1 + jw /4)(1 + jw /25)
The final form is
Bode Diagram
Magnitude (dB)
8( s + 70)
G ( s) =
s ( s + 4)( s + 25)
0
-50
-100
-150
-90
Actual
Approximate
Phase (deg)

50
-135
-180
-225
-1
10
0
10
1
10
Frequency (rad/sec)
2
10
3
10
Example II:
Magnitude (dB)
Transfer Function identification Example II
40
20
0
-20
-40
-60
Phase (deg)
0
-50
-100
-150
-200
10 -2
10 -1
10 0
Frequency (rad/sec)
10 1
10 2
 Initial slope = 0dB/dec.  no pole at the origin.
 Final slope = -40dB/dec.  (n-m) = 2 .
 Initial angle = 0 and the final angle is -180 which checks
with the gain curve.
 Low frequency gain is found to be
|KB|dB = 20dB  KB = 10 .
 Through asymptotic fitting a simple pole is found at wc = 0.2,
a simple zero at wc = 1.0 and a complex pole at wn = 5.0 .
Transfer Function Identification Example II
Magnitude (dB)
40
20
-20dB/dec.
15dB.
0
-20
-40dB/dec.
-40
-60
0.2
1.0
5.0
0
Phase (deg)
-45/dec.
-50
-90/dec.
-100
-90/dec.
-150
-200
10 -2
10 -1
10 0
Frequency (rad/sec)
10 1
10 2
 The peak at wn = 5.0 is  15dB which corresponds to a
damping ratio of z 0.1 .
 The estimated transfer function in Bode form is
+ jw /1 )
10
(
1
G ( jw ) =
(1 + jw /0.2)(1 + 0.2 jw /5 - w 2 /25)
 The final form is
+1)
50
(
s
G (s) =
(s + 0.2)(s2 + s + 25)
Bode Diagram
40
Magnitude (dB)
20
0
-20
-40
-60
0
Phase (deg)
-45
-90
-135
-180
-2
10
-1
10
0
10
Frequency (rad/sec)
1
10
2
10
Bode Plots
Design Problem:
Design a G(s) that has the following Bode plot.
Example
40
30 dB
20
-40dB/dec
+40 dB/dec
0
dB mag
?
0.1
1
?
10
30
w rad/sec
100
900
1000
Bode Plots
Procedure: The two break frequencies need to be found.
Recall:
#dec = log10[w2/w1]
Then we have:
(#dec)( 40dB/dec)
= 30 dB
log10[30/w1] = 0.75
Also:
log10[w2/900] (-40dB/dec) = - 30dB
This gives w2 = 5060 rad/sec
w1 = 5.33 rad/sec
Bode Plots
Procedure:
(1 + s / 5.3)2 (1 + s / 5060)2
G( s ) =
(1 + s / 30)2 (1 + s / 900)2
Clearing:
( s + 5.3)2 ( s + 5060)2
G( s ) =
( s + 30)2 ( s + 900)2
Use Matlab and conv:
N1 = ( s2 + 10.6s + 28.1)
N1 = [1 10.6 28.1]
N 2 = ( s2 + 10120s + 2.56 xe7 )
N2 = [1 10120 2.56e+7]
N = conv(N1,N2)
1
1.86e+3
2.58e+7
s4
s3
s2
2.73e+8 7.222e+8
s1
s0
Bode Plots
Procedure:
The final G(s) is given by;
( s + 10130.6 s + 2.571e s + 2.716e s + 7.194e )
G(s) =
( s + 1860s + 9.189e s + 5.022e s + 7.29e )
4
3
4
Testing:
3
8
2
2
2
8
7
8
8
We now want to test the filter. We will check it at w = 5.3 rad/sec
And w = 164. At w = 5.3 the filter has a gain of 6 dB or about 2.
At w = 164 the filter has a gain of 30 dB or about 31.6.
We will check this out using MATLAB and particularly, Simulink.
Matlab (Simulink) Model:
Filter Output at w = 5.3 rad/sec
Produced from Matlab Simulink
Filter Output at w = 70 rad/sec
vvv
Produced from Matlab Simulink

Consider the definitions of the gain and phase
margins in relation to the Bode plot of GH(jw) .

Gain Margin: the additional gain required to make
| GH(jw) | = 1 when /GH(jw) = -180 . On the Bode plot
this is the distance, in dB, from the magnitude curve up
to 0dB when the angle curve crosses -180 .

Phase Margin: the additional phase lag required to make
/GH(jw) = -180 when | GH(jw) | = 1 . On the Bode plot
this is the distance in degrees from the phase curve to 180 when the gain curve crosses 0dB.
Gain and Phase Margin Example
40
20
Magnitude (dB)
0
GM = 19dB
-20
-40
-60
-80
-100
8( s + 70)
G ( s) =
s ( s + 4)( s + 25)
-120
-80
Phase (deg)
-100
-120
-140
-160
fm = 40
-180
-200
-220
10 -1
10 0
10 1
Frequency (rad/sec)
10 2
10 3
2