Control Example
10
April 5, 2014 FAHEM_HW 06 Page 1 of 10 Problem -1- Solution: a) Transfer frequency function is : () b) Bode diagram : clc close s=tf('s') T=8; G=1/(T*s+1); bode(G) -40 -35 -30 -25 -20 -15 -10 -5 0 Magnitude (dB) System: G Frequency (rad/s): 0.125 Magnitude (dB): -3.01 10 -3 10 -2 10 -1 10 0 10 1 -90 -45 0 System: G Frequency (rad/s): 0.125 Phase (deg): -45 Phase (deg) Bode Diagram Frequency (rad/s) G
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good Control Example
Transcript of Control Example
-
April 5, 2014 FAHEM_HW 06
Page 1 of 10
Problem -1- Solution:
a) Transfer frequency function is :
( )
b) Bode diagram :
clc
close
s=tf('s')
T=8;
G=1/(T*s+1);
bode(G)
-40
-35
-30
-25
-20
-15
-10
-5
0
Ma
gn
itu
de
(d
B)
System: G
Frequency (rad/s): 0.125
Magnitude (dB): -3.01
10-3
10-2
10-1
100
101
-90
-45
0
System: G
Frequency (rad/s): 0.125
Phase (deg): -45
Ph
as
e (
de
g)
Bode Diagram
Frequency (rad/s)
G
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April 5, 2014 FAHEM_HW 06
Page 2 of 10
From the bode diagram the cutoff frequency is ( )
Recall to the lecture:
Discussion: I can get the from the figure by moving the cursor on the
intersection point between the 45 deg. And the phase curve. Also this is the
same result when I calculated this value theoretically. Also, at this point the low
and high frequencies are equal.
c)
Form the bode diagram we choose two point in high frequency:
Point -1- has:
Point -2- has:
Slope=
( )
(
)
-80
-70
-60
-50
-40
-30
-20
-10
0
Ma
gn
itud
e (
dB
)
System: G
Frequency (rad/s): 10
Magnitude (dB): -38.1
System: G
Frequency (rad/s): 100
Magnitude (dB): -58.1
10-1
100
101
102
103
-90
-60
-30
Ph
as
e (
de
g)
Bode Diagram
Frequency (rad/s)
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April 5, 2014 FAHEM_HW 06
Page 3 of 10
Theoretically:
The magnitude of first order system is :
| |
| |
| | ( )
| |
| | ( )
Then : | | | |
d)
Then: ( )
( )
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April 5, 2014 FAHEM_HW 06
Page 4 of 10
-30
-25
-20
-15
-10
-5
0M
ag
nit
ud
e (
dB
)
System: G
Frequency (rad/s): 50.1
Magnitude (dB): -3
100
101
102
103
-90
-45
0
System: G
Frequency (rad/s): 50.3
Phase (deg): -45
Ph
as
e (
de
g)
Bode Diagram
Frequency (rad/s)
G
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April 5, 2014 FAHEM_HW 06
Page 5 of 10
Problem -2- Solution:
a) Transfer frequency function is :
( )
Bode diagram :
clc
close
s=tf('s')
T=5;
G=2/(T*s+1);
bode(G)
From the figure the system bandwidth is : ( )
-30
-25
-20
-15
-10
-5
0
5
10
System: G
Frequency (rad/s): 0.2
Magnitude (dB): 3.01
Ma
gn
itu
de
(d
B)
10-2
10-1
100
101
-90
-45
0
System: G
Frequency (rad/s): 0.2
Phase (deg): -45Ph
as
e (
de
g)
Bode Diagram
Frequency (rad/s)
G
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April 5, 2014 FAHEM_HW 06
Page 6 of 10
b)
s =
RiseTime: 11.5109 SettlingTime: 19.5490
SettlingMin: 1.7999
SettlingMax: 1.9999
Overshoot: 0
Undershoot: 0
Peak: 1.9999
PeakTime: 50
clc
close
s=tf('s')
T1=5;
t=0:0.001:50;
figure(1)
G1=2/(T1*s+1);
step(G1,t);
x=step(G1,t);
S=stepinfo(x,t,'RiseTimeLimits',[0 0.9])
figure(2)
bode(G1)
hold on
T2=.5;
G2=2/(T*s+1);
bode(G2)
0 5 10 15 20 25 30 35 40 45 500
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Step Response
Time (seconds)
Amplitu
de
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April 5, 2014 FAHEM_HW 06
Page 7 of 10
c)
Then: ( )
( )
-60
-50
-40
-30
-20
-10
0
10
20
Ma
gn
itu
de
(d
B)
System: G2
Frequency (rad/s): 50.3
Magnitude (dB): 3.01
System: G1
Frequency (rad/s): 0.2
Magnitude (dB): 3.01
10-2
10-1
100
101
102
103
-90
-45
0
Ph
as
e (
de
g)
System: G2
Frequency (rad/s): 50.3
Phase (deg): -45
System: G1
Frequency (rad/s): 0.2
Phase (deg): -45
Bode Diagram
Frequency (rad/s)
G1
G2
T=0.5
T=5
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April 5, 2014 FAHEM_HW 06
Page 8 of 10
Problem -3- Solution :
a) ( )
( )
b) From the characteristic equation , we can find :
-80
-60
-40
-20
0
20
Ma
gn
itu
de
(d
B)
10-1
100
101
102
-180
-135
-90
-45
0
Ph
as
e (
de
g)
Bode Diagram
Frequency (rad/s)
G
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April 5, 2014 FAHEM_HW 06
Page 9 of 10
c)
From the bode diagram
d) the cutoff frequency from bode diagram is :
-15
-10
-5
0
5
10
Ma
gn
itu
de
(d
B)
System: G
Frequency (rad/s): 2.83
Magnitude (dB): 8.93
System: G
Frequency (rad/s): 2.74
Magnitude (dB): 9.17
System: G
Frequency (rad/s): 4.3
Magnitude (dB): -3
100.4
100.5
100.6
100.7
-180
-135
-90
-45
System: G
Frequency (rad/s): 2.83
Phase (deg): -90
Ph
as
e (
de
g)
System: G
Frequency (rad/s): 2.74
Phase (deg): -79.4
System: G
Frequency (rad/s): 4.3
Phase (deg): -157
Bode Diagram
Frequency (rad/s)
G
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April 5, 2014 FAHEM_HW 06
Page 10 of 10
e) From the characteristic equation , and use FRF, we can
calculate :
And:
( )
( )
Discussion:
The result I got from these two methods (c and e), are very close.
So, the frequency value and amplitude show me the real resonance ( )
accrued before theoretical resonance states, also the system at real
resonance state has bigger value of amplitude then theoretical value. As
a result, the designer should give the good tension, because the
systems maybe go to failure before it arrives to theoretical value of
resonance.
clc
close
s=tf('s')
z=0.1768;
wn=8^0.5
G=wn^2/(s^2+2*z*wn*s+wn^2);
bode(G)
