SOLUTION OF STATE EQUATION
LAPLACE TRANSFORM SOLUTION
The standard form of state equation is
x (t )  Ax(t )  Bu(t )
y(t )  C x(t )
(1)
Consider the first equation in (1)
x1  a11 x1  a12 x2  ... a1n xn  b11u1  ... b1r ur
Its LAPLACE transform is
sX 1  x1 (0)  a11 X 1  a12 X 2  ...  a1n X n
 b11U1  ...  b1rU r
The second equation in (1) yields
sX 2  x1 (0)  a21 X1  a22 X 2  ... a2n X n
 b21U1  ...  b2rU r
The remaining equations in (1) yield equation
of the same form. These equations may be
written in matrix form as
sX( s)  x(0)  AX(s)  BU(s)
where
x(0)  x1 (0) x2 (0)...xn (0)
T
We will solve this equation for X(s), so
we collect all terms containing X(s) on
the left side
sX( s)  AX(s)  x(0)  BU(s)
To solve this we factorized X(s)
sIX(s)  AX(s)  (sI  A)X(s)  x(0)  BU(s)
This equation now be solved for X(s)
X(s)  (sI  A)1 x(0)  (sI  A)1 BU(s)
And the state vector x(t) is the inverse
LAPLACE transform of this equation
SOLUTION OF STATE EQUATION
LAPLACE TRANSFORM SOLUTION
To obtain a general relationship for the solution we define the state
transition matrix as
Φ(t )  L 1[(sI  A)1 ]
This Matrix is also called the fundamental matrix.
The matrix (sI – A)-1 is called resolvant of A.
Finding the inverse Laplace transform of resolvant is difficult, time consuming and prone to
error
More practical procedure is computer simulation.
Example finding transition matrix
Consider the system described by
Y (s)
1
G( s) 
 2
s 3 s  2
U (s)
Using observer canonical form we write
the state equation as:
  3 1
0
x (t )  
x(t )   u(t )

 2 0
1
y(t )  1 0x(t )
To find the state transition matrix, first we
calculate the matrix (sI-A)
1 0   3 1 s  3 1
sI  A  s 




s 
0 1  2 0  2
The next step is that we have to find the
inverse of the matrix (sI-A). First we find
the adjoint of (sI-A)
1 
 s
Adj( sI  A)  

 2 s  3
Its determinant is
det (sI-A)= s2+3s+2=(s+1)(s+2)
The inverse is then the adjoint matrix
divided by the determinant
s
1


 ( s  1)(s  2) ( s  1)(s  2) 
1
( sI  A )  

2
s3


 ( s  1)(s  2) ( s  1)(s  2) 
The state transition matrix is the inverse
Laplace transform of this matrix
  e t  2e 2t
Φ(t )  
t
 2t

2
e

2
e

e t  e 2t 
t
 2t 
2e  e 
With the definition of state transition
matrix , the equation for complete solution
of the state equation can be found
Example finding solution of the state equation
We are going to solve the following equation
X(s)= (sI-A)1x(0)+(sI-A)1BU(s)
(1)
Consider the same system as before
  3 1
0
x (t )  
x
(
t
)


1u(t )

2
0


 
 1 t 1  2 t 
 2  e  2 2e 
-1
1
L ((sI  A) BU ( s)  
3
1  2t 
t
  2e  e 
2
2

The state transition matrix is the inverse
Laplace transform of (1)
y(t )  1 0x(t )
With (s) the Laplace transform of
transition matrix is given by
s

 ( s  1)(s  2)
1
Φ( s )  ( sI  A)  
2

 ( s  1)(s  2)
The inverse Laplace transform of this terms is
1

( s  1)(s  2) 

s3

( s  1)(s  2) 
Suppose that the input is a unit step. Then
U(s)=1/s. And the second term of (1)
becomes
s
1


 ( s  1)(s  2) ( s  1)(s  2)  0 1
1
( sI  A) BU ( s)  
 

2
s

3

 1 s
 ( s  1)(s  2) ( s  1)(s  2) 
  e  t  2e 2t
e  t  e 2t   x1 (0) 
x(t )  


t
 2t
2e t  e  2t   x2 (0)
 2e  2e
 1 t 1  2 t 
 2  e  2 2e 

3
1  2t 
t
  2e  e 
2
2

Finding the complete solution is long even
for a second order system. The necessity
for reliable machine solutions, such as a
digital computer simulation is evident
Convolution solution of the state equation
We are going to find the inverse L.T of
X(s)= (sI-A)1x(0)+(sI-A)1BU(s)
(1)
Using Convolution theorem we find that:
t
x(t )  Φ(t )x(0)   Φ(t  τ)Bu(τ)dτ
0
or
t
x(t )  Φ(t )x(0)   Φ(τ)Bu(t  τ)dτ
0
• Note that the solution is composed of to terms
• The first term is referred to
– the zero input part or the initial condition part of the
solution
• The second term is called
– the zero-state part or the forced part.
Example convolution solution
Consider the same system as previous

  e  (t  τ  2e 2(t  τ)
Φ(t )Bu( τ)dτ   
0  2e ( t  τ)  2e  2 ( t  τ)

t
e  (t  τ)  e 2(t  τ)  0
dτ
( t  τ)
 2 ( t - τ)   
2e
e
 1
 t e (t  τ)  e 2(t  τ) dτ 


  t0
 2e (t  τ)  e 2(t - τ) dτ 
 0

 1   e t   1 e  2t 
 
 


2
2
  
   2t 
t
3
1

e



 e
 2 

2
Only the force part is derived here the initial condition part is derived
the same way as the previous example
Infinite series solution
One method of differential eq. is to assume as
a solution an infinite series with unknown
coefficient. This method is know used to find
the transition matrix
The state equation may be written as
x (t )  Ax(t )
(1)
with the solution
x(t )  Φ(t )x(0)
(2)
The solution is assumed to be of the form
x(t )  (K 0  K1t  K 2t 2  )x(0) (3)

  K i t i x(0)  Φ(t )x(0)
i 0
where the (nn) matrices Ki are unknown
and t is the scalar time. Differentiating this
expression yields:
x (t )  (K1  2K 2t  3K 2t 2  )x(0) (4)
Substituting (3) and (4) into (1) yield
(K1  2K 2t  3K 3t 2  )x(0) 
A(K 0  K1t  K 2t 2  )x(0) (5)
Next perform the following operations:
1. Evaluate (5) at t=0
2. Differentiate (5) and evaluate the result at t=0
3. Repeat step 2 again and again
The result is the following equations
K1 = AK0
(6)
2K2 = AK1
3K3 = AK2
Evaluating (3) at t = 0 shows that K0=I, then
the other matrices are evaluated from (6)
K1  A
2
A
K2 
2!
...
Hence from (3) the state transition matrix is:
2
t
Φ(t )  I  At  A
  e At
2!
2
Example
A satellite system shown below is assume to be
rigid and in frictionless environment and to
rotate about an axis perpendicular to the page.
Torque is applied to the satellite by firing
thrustors. Thrustors can be fired left or right to
increase or decrease angle . Torque (t) is the
input and angle (t) is the output.
The state model is then
0
0 1
x (t)  Ax(t)  Bu (t )  
x(t)   1 u(t )

 j 
0 0
Thus we have
(t)
(t)
Thrustors
0 1
0 1 0 1 0 0
2
A
,
A








0 0
0 0 0 0 0 0
0 0
A 
,
0
0


n
for n  2
The state transition matrix is then:
Then
2
( s )
d
 12
τ (t )  J 2θ and G ( s ) 
( s ) Js
dt
0 1 0 1 1 t 
Φ(t )  I  At  

t



0 0 0 0 0 1
Transfer function
We are going to find the transfer function if the
state representation is known. The standard
form of state equation when D=0 is
x (t )  Ax(t )  Bu(t )
y(t )  C x(t )
(1)
(2)
X(s) can easily be solve
X(s)= (sI-A)1BU(s)
(3)
The L.T of the output equation in (1) yields
Y(s)= CX(s)
(4)
Eq. (3) and (4) gives
Y(s)= C(sI-A)1BU(s)=G(s)U(s)
(5)
The transfer function G(s) is then
G(s) = C(sI-A)1B = C(s)B
G(s) = C(sI-A)1B = C(s)B + D
(7)
Example:
The state representation is as follows
Ignoring the initial condition, its L. T. is
sX(s)= A X(s)+BU(s)
For the case that D  0, the T.F. is
(6)
  3 1
0
x (t )  
x
(
t
)

u(t )



 2 0
1
y(t )  1 0x(t )
The transfer function is given by
1
 s  3  1 0
G ( s )  C( sI  A) B  1 0 
s  1
 2
1

 2 1
( s  1)( s  2) s  3s  2
1
MATLAB: A=[-3 1];-2 0];B=[0;1];C=[1 0];
D=0; [n,d]=ss2tf(A,B,C,D)
Result: n 0 0 1; d 1 2 3
Similarity transformations

Finding the S.S. model from Diff. Eq. or T.F.
has been presented.

It has been shown that a unique state model
does not exist.

Two general state model control canonical
form and observer canonical form can always
be found



v2(t) = q21x1(t) + q22x2(t) ++ q2nxn(t)
•
•
•
•
The number of internal models (state model) is
unbounded
The state model of SISO system is
x (t )  Ax(t )  Bu(t )
y(t )  C x(t )  Du(t )
(1)
vn(t) = qn1x1(t) + qn2x2(t) ++ qnnxn(t)
In matrix form
v(t) = Qx(t)= P1x(t)
x(t) = Pv(t)
And the transfer function G(s) is given by
G(s) = C(sI-A)1B = C(s)B
(2)
There are many combination of matrices A, B, C,
and D that will satisfy (1) for a given G(s). We will
show that this combination is unbounded
Suppose that we are given a ss model as in (1).
Now define state vector v(t) that is the same
order of x(t), such that the elements of v(t) is a
linear combination of the elements of x(t), that
is
v1(t) = q11x1(t) + q12x2(t) ++ q1nxn(t)




Matrix P is called the transformation matrix
this will transform one set of state vector to a
different state vector
This transformation alter the internal model
but not the input output relationship
This typo of transformation is called similarity
transformation