Apunte Advanced Control

8/22/2019 Apunte Advanced Control

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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

=

= + +


( ) = [ + + ] ( )

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)


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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>> 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)


>> 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

= + +


( ) = [ + + ] ( )

Assume that;

M = 1 C = 1 K = 2 F = 1

= [S + S + 2] X(s) X(s) =[ ]


>> 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


>> 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




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Case3:

Assuming:

R=2 L=1 J=2 C=5 Km=0.5 Ka=3




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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


( ) =

+ + 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

=

= + +


( ) = [ + + ] ( )

( )

( )=

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


= 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


= 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

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