1. SOLUTION OF STATE EQUATION
The remaining equations in (1) yield equation
of the same form. These equations may be
written in matrix form as
LAPLACE TRANSFORM SOLUTION
The standard form of state equation is
(1)
where
Consider the first equation in (1)
We will solve this equation for X(s), so
we collect all terms containing X(s) on
the left side
Its LAPLACE transform is
To solve this we factorized X(s)
This equation now be solved for X(s)
The second equation in (1) yields
And the state vector x(t) is the inverse
LAPLACE transform of this equation

2. SOLUTION OF STATE EQUATION
LAPLACE TRANSFORM SOLUTION
To obtain a general relationship for the solution we define the state transition matrix as
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.

3. Example finding transition matrix
Consider the system described by
Its determinant is
det (sI-A)= s2+3s+2=(s+1)(s+2)
Using observer canonical form we write
the state equation as:
The inverse is then the adjoint matrix
divided by the determinant
The state transition matrix is the inverse
Laplace transform of this matrix
To find the state transition matrix, first we
calculate the matrix (sI-A)
With the definition of state transition
matrix , the equation for complete solution
of the state equation can be found
The next step is that we have to find the
inverse of the matrix (sI-A). First we find
the adjoint of (sI-A)

4. Example finding solution of the state equation
We are going to solve the following equation
The inverse Laplace transform of this terms is
X(s)= (sI-A)1x(0)+(sI-A)1BU(s) (1)
Consider the same system as before
The state transition matrix is the inverse
Laplace transform of (1)
With (s) the Laplace transform of
transition matrix is given by
Suppose that the input is a unit step. Then
U(s)=1/s. And the second term of (1)
becomes
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

5. 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: or
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.

6. Example convolution solution
Consider the same system as previous
Only the force part is derived here the initial condition part is derived
the same way as the previous example

7. 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
Next perform the following operations:
Evaluate (5) at t=0
Differentiate (5) and evaluate the result at t=0
Repeat step 2 again and again
The result is the following equations
K1 = AK0
2K2 = AK1
3K3 = AK2
The state equation may be written as
with the solution
(6)
The solution is assumed to be of the form
Evaluating (3) at t = 0 shows that K0=I, then the other matrices are evaluated from (6)
...
where the (nn) matrices Ki are unknown
and t is the scalar time. Differentiating this
expression yields:
Hence from (3) the state transition matrix is:
Substituting (3) and (4) into (1) yield

8. Example The state model is then
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.
Thus we have
(t)
(t)
Thrustors
The state transition matrix is then:
Then
and

9. 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
For the case that D  0, the T.F. is
G(s) = C(sI-A)1B = C(s)B + D (7)
(1)
Example:
The state representation is as follows
Ignoring the initial condition, its L. T. is
sX(s)= A X(s)+BU(s) (2)
X(s) can easily be solve
X(s)= (sI-A)1BU(s) (3)
The L.T of the output equation in (1) yields
The transfer function is given by
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
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
G(s) = C(sI-A)1B = C(s)B (6)

10. 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
The number of internal models (state model) is unbounded
The state model of SISO system is
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)
v2(t) = q21x1(t) + q22x2(t) ++ q2nxn(t)
vn(t) = qn1x1(t) + qn2x2(t) ++ qnnxn(t)
In matrix form
(1)
v(t) = Qx(t)= P1x(t)
x(t) = Pv(t)
And the transfer function G(s) is given by
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
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