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SOLUTION OF STATE EQUATION 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 LAPLACE TRANSFORM SOLUTION The standard form of state equation is (1) Its LAPLACE transform is The second equation in (1) yields The remaining equations in (1) yield equation of the same form. These equations may be written in matrix form as where To solve this we factorized X(s) This equation now be solved for X(s) And the state vector x(t) is the inverse LAPLACE transform of this equation

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SOLUTION OF STATE EQUATION LAPLACE TRANSFORM SOLUTION To obtain a general relationship for the solution we define the state transition matrix as The matrix (sI – A) -1 is called resolvant of A. This Matrix is also called the fundamental matrix. More practical procedure is computer simulation. Finding the inverse Laplace transform of resolvant is difficult, time consuming and prone to error

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

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Example finding solution of the state equation Consider the same system as before 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 The inverse Laplace transform of this terms is The state transition matrix is the inverse Laplace transform of (1) We are going to solve the following equation X(s)= (sI-A) 1 x(0)+ (sI-A) 1 BU(s) (1) 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

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Convolution solution of the state equation 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. Using Convolution theorem we find that: We are going to find the inverse L.T of X(s)= (sI-A) 1 x(0)+ (sI-A) 1 BU(s) (1) or

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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

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Infinite series solution The state equation may be written as with the solution The solution is assumed to be of the form where the (n n) matrices K i are unknown and t is the scalar time. Differentiating this expression yields: Substituting (3) and (4) into (1) yield 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: 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 K 1 = AK 0 2K 2 = AK 1 3K 3 = AK 2 Evaluating (3) at t = 0 shows that K 0 =I, then the other matrices are evaluated from (6)... Hence from (3) the state transition matrix is: (6)

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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. The state transition matrix is then: (t) (t)(t) Thrustors Then and Thus we have

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Transfer function Ignoring the initial condition, its L. T. is The L.T of the output equation in (1) yields sX(s)= A X(s)+ BU(s) (2) X(s) can easily be solve We are going to find the transfer function if the state representation is known. The standard form of state equation when D=0 is (1) X(s)= (sI-A) 1 BU(s) (3) Y(s)= C(sI-A) 1 BU(s)=G(s)U(s) (5) Y(s)= CX (s) (4) Eq. (3) and (4) gives The transfer function G(s) is then G(s) = C(sI-A) 1 B = C (s)B (6) G(s) = C(sI-A) 1 B = C (s)B + D (7) For the case that D 0, the T.F. is Example: The state representation is as follows The transfer function is given by 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

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Similarity transformations 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 In matrix form 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 (1) v(t) = Qx(t)= P 1 x(t) And the transfer function G(s) is given by G(s) = C(sI-A) 1 B = 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 v 1 (t) = q 11 x 1 (t) + q 12 x 2 (t) + + q 1n x n (t) v 2 (t) = q 21 x 1 (t) + q 22 x 2 (t) + + q 2n x n (t) v n (t) = q n1 x 1 (t) + q n2 x 2 (t) + + q nn x n (t) x(t) = Pv(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

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