apunte advanced control
TRANSCRIPT
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Advanced ControlModeling Problem by Using MATLAB
Solved By
Abdulrahman Yossri Mahmoud
Presented To
Ass. Prof. Dr. Saber M Abdrabbo
Benha University
Faculty of Engineering Shoubra
Mechanical Design & Production
Department
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Assignment #1Modeling Problem by Using Laplace Transforms
Plotting Graph bet. Displacement & Time
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Case1:
Due to F.B.D
=
= + +
By using Laplace transform
( ) = [ + + ] ( )
Assume that;
M = 1 C = 2 K = 1 F = 1
= [ + 2 + 1] ( )
( ) = 1[ + 2 + 1]
= 1( + 1)
By using partial fraction
( ) = ++ 1
+( + 1)
1
( + 1) =
( + 1) + ( + 1) +
( + 1)
Sub S = 0 A = 1
Sub S = -1 C = -1
Sub S = 1 B = -1
( ) =1
1
+ 1
1
( + 1)
By using Inverse Laplace
( ) = 1
Solving Using MATLAB
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>> syms x s
>> x = 1/(s*((s^2)+ (2*s)+1));
>> ilaplace (x)
ans =
1+(-1-t)*exp(-t)
Then plot a graph between X and t Using MATLAB
>> t = 0:0.1:exp(2);
>> x = 1+(-1-t).*exp(-t);
>> Plot (t,x)
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Case2:
Due to F.B.D
=
= + +
By using Laplace transform
( ) = [ + + ] ( )
Assume that;
M = 1 C = 7 K = 12 F = 1
= [S + 7S+ 12] X(s)
X(s) = 1S[S + 7S+ 12]
=1
S( S+ 3)( S+ 4)
By using partial fraction
X(s) =A
S+
B
( S+ 3)+
C
( S+ 4)
1S( S+ 3)( S+ 4)
=A( S+ 3)( S+ 4) + BS( S+ 4) + CS( S+ 3)
S( S+ 3)( S+ 4)
Sub S = 0 A =
Sub S = -3 B =
Sub S = -4 C =
X(s) = 112
1S
13
1S+ 3
+ 14
1S+ 4
By using Inverse Laplace
X(t) =1
12
1
3e +
1
4e
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Solving Using MATLAB
>> syms x s
>> x = 1/(s*((s^2)+ (7*s)+12);
>> ilaplace (x)
ans =
1/4*exp(-4*t)+1/12-1/3*exp(-3*t)
Then plot a graph between X and t Using MATLAB
>> t = 0:0.1:exp(2_;
>> x = 1/4*exp(-4*t)+1/12-1/3*exp(-3*t));
>> Plot (t,x)
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Case3:
Due to F.B.D
= + +
By using Laplace transform
( ) = [ + + ] ( )
Assume that;
M = 1 C = 1 K = 2 F = 1
= [S + S + 2] X(s) X(s) =[ ]
Solving Using MATLAB
>> syms x s
>> x = 1/(s*((s^2)+ (s)+2));
>> ilaplace (x)
ans =
-1/2*exp(-1/2*t)*cos(1/2*7^(1/2)*t)-
1/14*7^(1/2)*exp(-1/2*t)*sin(1/2*7^(1/2)*t)+1/2
Then plot a graph between X and t Using MATLAB
>> t = 0:0.1:exp(3);
>> x = -1/2.*exp(-1/2.*t).*cos(1/2*7^(1/2).*t)-
1/14*7^(1/2).*exp(-1/2.*t).*sin(1/2*7^(1/2).*t)+1/2;
>> Plot (t,x)
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Assignment #2Modeling Problem by Using SIMULINK
Plotting Graph bet. Displacement & Time
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Case1:
Assuming:
R=1 L=1 J=2 C=7 Km=1.5 Ka=1.5
The block diagram given by:
The simulation will be:
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Case2 (using T.F):
Assuming:
R=1 L=2 J=2 C=7 Km=1.5 Ka=1.5
The block diagram given by:
The simulation will be:
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Case3:
Assuming:
R=2 L=1 J=2 C=5 Km=0.5 Ka=3
The block diagram given by:
The simulation will be:
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Assignment #3Modeling Problem of a Pneumatic System
Getting Response by Using SIMULINK
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Pneumatic Modeling Problem
System Analysis
System Equations Laplace Transform
=+
( ) =+
( )
=+
( ) =+
( )
= ( ) = ( ) ( )
= ( ) = ( )
= ( ) = ( + 1) ( )
= ( ) = ( + 1) ( )
= ( ) ( ) = ( ( ) ( ))
= ( ) =( )
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The Block Diagram Given By
Assuming That
a = 2 b = 5 k =3 A = 6 C = 4 R1=R2=9
By Using SIMULINLK, The Block Diagram Given By
The Time response of The System
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Assignment #4Derivative of LOGARITHMIC DECREMENT
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From the general equaon of 2nd
order system
( )
( )=
+ 2 +
Assume Impulse Input
( ) = ( )
( ) = 1
So,
( ) =+ 2 +
( ) =+ + 1 + 1
By using partial fraction
( ) =
+ + 1
+
+ 1
= = 2 1
Inverse Laplace
( ) =
=
But,
= 1
( ) =
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From trigonometric
sin =
2
So,
( ) = sin( )
Where
=
1
= 1
To get the LOGARITHMIC DECREMENT()
Let the response equation be
= sin( )
=sin( )
( )sin( ( + ))
Since the sin may be equal for tiny increase (d) so,
sin( ( + )) sin( )
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=( )
ln = ln ( )
ln = ln
ln = ln =
But, from the graph of response
= 2 = 2
1
ln =2
1
= ln =2
1
The relation between the logarithmic decrement and damping factor is
given by:
=4 +
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Assignment #5Solving a STEADY STATE ERROR problem
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Problem
For the shown system and corresponding response graph find: M, C
and K
Given
Mp = 0.095 Tp = 5 sec F = 2N
Solution
From the given data, the maximum peak = 0.095
= 0.095 =
ln0.095 =
1
ln 0.095=
1 = 1.33
1 = 1.33
1 = 2.78
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So, the damping factor is
= 0.599
By substituting at the peak time
= 5 = =1
5 =1 0.599
=51 0.599
= 0.784
So, the natural frequency is
= 0.784
From the mechanical motion of the system
Due to F.B.D
=
= + +
By using Laplace transform
( ) = [ + + ] ( )
( )
( )=
1
[ + + ]
But
( ) = 2
( ) =2
So, the transfer function given by
( ) =2
[ + + ]
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From the final value theorem
() = lim
( ) = lim
( )
So,
lim
( ) =2
[ + + ]=
2
But
() = 0.1
0.1 =2
= 20
From The general Formula of 2nd
order response equation
+ 2 +
2 =
=
So,
= =20
0.784= 32.5
And
= 2 = 2 32.5 0.599 0.784 = 30.56
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Assignment #6Solving a NYQUIST PLOTTING problem
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Problem:
=( + )( + )
Draw the Nyquist diagram, then:1) Find value of K for stability.
2) For K = 4 the phase margin and gain margin.
3) Find K at phase margin = 30.
4) Find K at gain margin =10 dB.
Solution:
Assume K = 1, So
( ) ( ) =
1
( + 1)( + 2)Along C1, convert from S-domain to frequency
domain substitute with S j
( ) ( ) =1
( + 1)( + 2)
The amplitude M is given by:
=1
( + 1)( + 4)
The angle is given by:
= 0 90 tan1
tan2
For any assumed get and M
M
0 -90
0.1 4.966 -98.57
0.25 1.924 -111.1
0.5 0.867 -130.6
1 0.316 -161.5
2 0.079 -198.4
5 0.007 -236.8
10 0.001 -252.9
0 -270
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Along C2 is the mirror of C1 around X-axis
Along C3 tends to be zero
= lim
= 0
Along C4 is given by:
= lim
So,
=1
2lim
lim
+ 112 lim
+ 1
But
lim
+ 1 =12
lim
+ 1 = 1
=1
2lim
=
So, it is an arc of radius and ploe d from 90 -90. Then draw a
polar plot of Nyquist criterion.
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The problem may plotted by using MATLAB as following
>> Num = [1];
>> Dum = [1 3 2 0];
>> T = tf(Num,Dum);
>> Nyquist(T)
1) To get the value of K for stability, find the intersection at = -180
= 0 90 tan1
tan2
= 180
By using trail and error method, we get
= 1.415
The stability occurred at M = 1. So,
=( + 1)( + 4)
= ( + 1)( + 4)
= 1 1.415 (1.415 + 1)(1.415 + 4)
= 6
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2) The gain margin at K = 4, subst ut e in the equaon of phas e angl e at
condition of = -180
= 0 90 tan1
tan2
= 180
So, as result before using trail and error method
= 1.415
Subst ut e at K = 4 to get the amp l itude of the gai n ma r gi n
=4
1.415 (1.415 + 1)(1.415 + 4)= 0.67
The gain margin is given by
. = 20 log 1
. = 20 log1
0.67= 3.48
The phase margin at K = 4, subst ut e in the equaon of ampl i tude at
condition of M = 1
=4
( + 1)( + 4)= 1
So,
( + 1)( + 4) = 4
( + 1)( + 4) = 16
Using trail and error method
= 1.145
Subst ut e at K = 4 to get the angl e of the phas e ma r gi n
= 0 90 tan1.145
1 tan
1.145
2= 168.65
The phase margin is given by
. = 180
. = 168.65 180 = 348.65
. = 11.35
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3) To get K at phase margin = 30= -330
. = 180
330 = 180
= 150
The corresponding frequency is given by
= 150 = 0 90 tan1
tan2
Using trail and error method
= 0.79
Substitute in the equation of amplitude at condition of M = 1
=( + 1)( + 4)
= 1
= 1 0.79 (.079 + 1)(0.79 + 4)
= 2.16
4) To get K at gain margin = 10 dB
. = 20 log 1
10
20= log
1
1= 3.18
So,
0.314
The corresponding frequency is given by
= 180 = 0 90 tan1
tan2
Using trail and error method
= 1.415
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Substitute in the equation of amplitude at condition of = -180
=( + 1)( + 4)
So,
0.314 =1.415 (1.415 + 1)(1.415 + 4)
= 0.314 1.415 (1.415 + 1)(1.415 + 4)
= 1.88