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

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.

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)

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

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.

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

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

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

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)

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