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State-Space Method: EE 515 - Feedback Control Systems (Lab)

The document provides the steps to obtain the transfer function of two systems using state-space methods. For the first system, it derives the state-space representation and computes the inverse and determinant of the state matrix. For the second system, it similarly derives the state-space representation and computes the inverse of the state matrix to obtain the transfer function.

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Arie Emmanuel Liston
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33 views5 pages

State-Space Method: EE 515 - Feedback Control Systems (Lab)

The document provides the steps to obtain the transfer function of two systems using state-space methods. For the first system, it derives the state-space representation and computes the inverse and determinant of the state matrix. For the second system, it similarly derives the state-space representation and computes the inverse of the state matrix to obtain the transfer function.

Uploaded by

Arie Emmanuel Liston
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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EE 515 FEEDBACK CONTROL SYSTEMS (LAB)

ACTIVITY 5: State-Space Method

Name: _Arie Emmanuel Liston Date: September 27, 2017

Problem:

Obtain the analogous electric circuit of the following systems and


determine the transfer function and block diagram of the systems
shown using state-space method.

3 ()
a)
()

()
b) ()
Computation:

a.

Left loop

(4 2 + 2 + 6)1 () (2)2 () = 0

Middle loop

(1 )() + (4 2 + 4 + 6)2 () (6)3 () = ()

Right loop

(6)2 () + (4 2 + 2 + 6)3 () = 0

State Variables
1
1 = 1 ; 2 2 = 2 ; 3 3 = 3

1 2
1 = 1 ; 2 = 2

Left loop

1 + 1 + 21 = 0

Middle loop

() + 21 + 2 +22 + 2 = 0

Right loop

2 + 3 + 23 = 0

Outer loop

() + 1 + 1 + 2 + 22 + 3 + 23 = 0

Node left and middle loop

1 = 1 + 2

Node middle and right loop

2 = 2 + 3

1 + 2 2 0 1 0 1 0
2 2 0 0 1 2 1
3 = 0 0 +2 0 1 3 + 0 ()
1 0 1 0 1/2 1 0
[ 0]
[ 2 ] [ 0 1/3 1 0 ] [ ]
2

1
2
= [0 0 1 0 0] 3
1
[2 ]
+ 2 2 0 1 0
2 0 0 1
( ) = 0 0 +2 0 1
0 1 0 1/2
[ 0 1/3 1 0 ]

+

[
] +


[ ]
[ ]
( ) =
+

+




[

]

MATLAB Script:

syms s

A = [s+2 -2 0 1 0;-2 s 0 0 1;0 0 s+2 0 1;0 1 0 s 1/2;0 1/3


1 0 s];
inv0 = inv(A);

disp(inv0);

[ (- 3*s^4 - 6*s^3 + 4*s^2 + 2*s)/(- 3*s^5 - 12*s^4 +


4*s^3 + 40*s^2 + 3*s - 8), (- 12*s^3 - 30*s^2 + s +
8)/(2*(- 3*s^5 - 12*s^4 + 4*s^3 + 40*s^2 + 3*s - 8)),
-(3*(3*s + 2))/(2*(- 3*s^5 - 12*s^4 + 4*s^3 + 40*s^2 + 3*s
- 8)), -(- 3*s^3 - 6*s^2 + 4*s + 2)/(- 3*s^5 -
12*s^4 + 4*s^3 + 40*s^2 + 3*s - 8), (3*(3*s^2
+ 8*s + 4))/(2*(- 3*s^5 - 12*s^4 + 4*s^3 + 40*s^2 + 3*s -
8))]
[-(6*s)/(- 3*s^3 - 6*s^2 + 13*s + 8),
-(3*s*(s + 2))/(- 3*s^3 - 6*s^2 + 13*s + 8),
-3/(- 3*s^3 - 6*s^2 + 13*s + 8),
6/(- 3*s^3 - 6*s^2 + 13*s + 8),
(3*(s + 2))/(- 3*s^3 - 6*s^2 + 13*s + 8)]

[-(2*s)/(- 3*s^5 - 12*s^4 + 4*s^3 + 40*s^2 + 3*s - 8),


-(s*(s + 2))/(- 3*s^5 - 12*s^4 + 4*s^3 + 40*s^2 + 3*s -
8), -(3*s^4 + 6*s^3 - 13*s^2 - 8*s + 1)/(- 3*s^5 - 12*s^4
+ 4*s^3 + 40*s^2 + 3*s - 8),
2/(- 3*s^5 - 12*s^4 + 4*s^3 + 40*s^2 + 3*s - 8), -
(3*(- s^3 - 2*s^2 + 4*s + 2))/(- 3*s^5 - 12*s^4 + 4*s^3 +
40*s^2 + 3*s - 8)]

[ (6*s^2 + 11*s - 8)/(- 3*s^5 - 12*s^4 + 4*s^3 + 40*s^2 +


3*s - 8), ((s + 2)*(6*s^2 + 11*s - 8))/(2*(- 3*s^5 -
12*s^4 + 4*s^3 + 40*s^2 + 3*s - 8)), -(3*(s^2 - 8))/(2*(-
3*s^5 - 12*s^4 + 4*s^3 + 40*s^2 + 3*s - 8)), -(3*s^4 +
12*s^3 - 4*s^2 - 34*s + 8)/(- 3*s^5 - 12*s^4 + 4*s^3 +
40*s^2 + 3*s - 8), -(3*(- s^3 - 2*s^2 + 8*s + 16))/(2*(-
3*s^5 - 12*s^4 + 4*s^3 + 40*s^2 + 3*s - 8))]

[ (2*s*(s + 2))/(- 3*s^5 - 12*s^4 + 4*s^3 + 40*s^2 + 3*s -


8), (s*(s + 2)^2)/(- 3*s^5 - 12*s^4 + 4*s^3 + 40*s^2 + 3*s
- 8), -(3*(- s^3 - 2*s^2 + 4*s + 2))/(- 3*s^5 - 12*s^4 +
4*s^3 + 40*s^2 + 3*s - 8), -(2*(s + 2))/(- 3*s^5 - 12*s^4
+ 4*s^3 + 40*s^2 + 3*s - 8), (3*(s + 2)*(- s^3 - 2*s^2 +
4*s + 2))/(- 3*s^5 - 12*s^4 + 4*s^3 + 40*s^2 + 3*s - 8)]

dt0 = det(inv0);
disp(dt0);

-3/(- 3*s^5 - 12*s^4 + 4*s^3 + 40*s^2 + 3*s - 8)

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