Control sys asg

Question1

Solution:-

Given differential equation can be written as

( ) − 2 ( ) = ( )…………

(1)

Let us consider the following state variables for the above differential equation ( )=

( )

( )=

( )

( )=

( )

Now substituting above considered y(t) and ( ) values in equation (1) we get ( )−2 ( )= ( )

In a vector matrix form above equation is represented by ( )=

Where ( )=

( )+

( )=

( ) ( ) ( )

0 1 0 = 0 0 1 2 0 0

( )=

( ) ( ) ( )

( )

( )

0 = 0 And 1

= [1 0 0]

Therefore state space representation of equation (1) can be written as ( ) 0 ( ) = 0 ( ) 2

1 0 0 1 0 0

Answer for Question 1 is

( ) 0 ( ) + 0 ( ) 1

( ) 0 ( ) = 0 ( ) 2

1 0 0

( ) = [1

( )

0 1 0 0

( ) = [1 0 0]

( ) 0 ( ) + 0 ( ) 1 ( ) ( ) 0] ( )

( )

( ) ( ) ( )

Question 2

Give data

Figure (a)

Consider

( ( ( (

)= )= )= )=

( ( ( (

) ) ) )

Solution:-

To derive a state-space model for the system, let us first consider the first car, i.e. From Newton’s second law states we know that

∝

From newton’s second law we can write an equation for all the forces acting on it: For

Similarly; for

( )= ( )−

( )= ( )−

( )+

( )=

( )=

Now let us substitute the given values of

( )−

( )−

( )+

( )− ( ) +

( )−

( )+

( )− ( )

( )…………… ( )− ( )

( ) − ( )…………..

( ) ( ) ( ) ( ) in equation (1)

( )= ( )−

( )=

( )− ( ) −

( )+

( )+

( )−

( )−

( )+

( )+

( )

( )

(1)

(2)

( )=

( )

( )+

−

( )−

( )= ( )=

( )−

( )……………

(3)

( ) ( ) ( ) ( ) in equation (2)

Similarly let us substitute the given values of ( )=

( )+

( )−

( )+

( )−

( )−

( )+

( )+

( )−

( )−

( )

( )

( )………………

(4)

From equations 3 and 4 we can obtain ( )=

( ( ( (

) ) ) )

0

=

0

1

0

0

0

0

( ) ( ) ( )

( )=

1

0

=

0

=

And

1 0 0 0

Therefore state space representation) can be written as ( ( ( (

0 − ) ) = ) 0 )

1 −

0 0

0

0 1 − −

( )=

1 0 0 0

( ) ( ) + ( )

0 ) ) = ) 0 ) ( )=

1 0

1

0 0

0

0 1 0 0 0 1

0

( )

( ) ( ) ( )

0 0 1 0

Answer for Question 2 is ( ( ( (

0 1

0 0

0 ( ) ( ) + ( ) 0

( ) ( ) ( )

( )

0 0 1 0

Question 3

Given dada

1 ( )= 0 0

2 2 −2 0 −1 1

1 ( )+ 1 1

( )

( ) = [ 1 1 0] ( )

From the above state space equation we can consider 1 = 0 0

Solution:-

2 2 −2 0 −1 1

1 = 1 1

= [1 1 0]

3a) Controllability

We know by the definition, controllability of a matrix can be known by Now let us find AB

CM = [

]

1 = 0 0

2 2 1 −2 0 1 −1 1 1

1+2+2 = 0 + (−2) + 0 0 + (−1) + 1

Similarly

5 = −2 0

can be find

= (

By substituting matrix A and AB value in above equation

)

1 2 2 5 = 0 −2 0 −2 0 −1 1 0 =

5 + (−4) + 0 0+4+0 0+2+0 1 = 4 2

By substituting matrix B AB and

values in CM we get 1 = 1 1

Rank = 3

5 1 −2 4 0 2

By using mat lab we can find rank of a matrix Since rank of CM is 3 it is controllable

3b) Transforming the system into canonical form To find the canonical form we have ℎ

P can be obtained by |SI − A| Let us find

−

−

( )

=

.

1 0 0 1 2 2 0 1 0 − 0 −2 0 0 0 1 0 −1 1

= −

( )=

−1 0 0

=

−2 +2 1

−2 0 −1

|SI − A| = (S − 1)(S + 2)(S − 1) |SI − A| = ( − 1)(

|SI − A| = Therefore matrix P can be obtained from

−2

|SI − A| = |SI − A| = =

By substituting the value of P in T we get

− 2 + 1)

+ +2

−4 +2

−3 +2 −3 +2

−3 0 1 0 1 0 1 0 0

=

.

1 5 1 −3 0 1 = 1 −2 4 0 1 0 1 0 2 1 0 0

−3 + 0 + 1 0 + 5 + 0 1 + 0 + 0 = −3 + 0 + 4 0 − 2 + 0 1 + 0 + 0 −3 + 0 + 2 0 + 0 + 0 1 + 0 + 0 = Determinant of the T is

−2 1 −1

5 1 −2 1 0 1

| | = −8 =

1

| |

( )

−2 −5 7 ( ) = −2 −1 3 −2 −5 −1 = Canconical form is a system equation is

( )=

Let us find AT first

Now we can obtain

1 = 0 0

1 −2 −5 7 −2 −1 3 −8 −2 −5 −1

( )=

( )+

( )

( )

2 2 −2 −2 0 1 −1 1 −1

5 1 −2 1 0 1

−2 1 5 = −2 4 −2 −2 2 0

by substituting the value of AT =

1 −2 −5 7 −2 1 5 −2 −1 3 −2 4 −2 −8 −2 −5 −1 −2 2 0 =

Similarly we can obtain =

0 1 0 0 0 1 −2 3 0

1 −2 −5 7 1 −2 −1 3 1 −8 −2 −5 −1 1 =

1 0 0 −8 −8

0 = 0 1

We can also find CT similarly

−2 5 1 = [1 1 0] 1 −2 1 −1 0 1 = [−1 3

2]

Therefore by substitution the obtained values in standard canonical equation we get ( )=

( )=

0 1 0 0 0 1 −2 3 0

( )=

( )+

0 ( )+ 0 1

( )

( ) ( )

………

( ) = [−1 3 2] ( )

(1)

3c) hence we had obtained canonical form above, so we can design full state feedback controller with eigenvalues of -6, -7 and -8 From the eigenvalues characteristic equation can be formed as

( + 8)( + 6)( + 7)

( + 8)(

+ 13

=

+ 104 + 336

+ 146 + 336 … … … . . |

From equation 1 we can get

Now let us find ( −

+ 42 + 8

+ 21

Now let us solve

+ 13 + 42)

−( −

)| = 0

0 1 0 0 0 1 −2 3 0

) first ( −

)=

( −

)=

0 1 0 0 0 1 −2 3 0 0 0 − −2

(2)

0 = 0 1

0 0 1 0 3−

0 0

0 0 0 1 −

Substitute matrix ( −

Therefore |

−( −

) value in

−( −

−( −

)=

−( −

)=

)|

|

−( − +

1 0 0

0 1 0 0 0 0 1 0 − − −2 3− 0 1

0 +2 )| = (

+(

) we get

−1

0 −1 +

−3 + +

− 3) + (

Comparing above equation 3 with equation 2 we get

+

− 3) +

0 1 −

+2=0

+ 2) = 0 … … … .

(3)

= 21

= 149 = 334 Therefore controller is obtained from 3b with eigenvalues at -6, -7 and -8 is ( )=

( )+ ( )

( ) = [334 149 21] ( ) + ( )

Then the state feedback controller is

( )=

( )=

( )+ ( )

−2 −5 7 1 [334 149 21] −2 −1 3 −8 −2 −5 −1

( )+ ( )

( ) = [126 240.5 −345.5] ( ) + ( )

Answers for question number 3 3a

Controllable since its rank is 3

3b

( )=

0 1 0 0 0 1 −2 3 0

0 ( )+ 0 1

( ) = [−1 3 2] ( )

3c ( )

state feedback controller

( ) = [126 240.5 −345.5] ( ) + ( )