Presentation on theme: "SOLUTION OF STATE EQUATION"— Presentation transcript:
1SOLUTION OF STATE EQUATION The remaining equations in (1) yield equationof the same form. These equations may bewritten in matrix form asLAPLACE TRANSFORM SOLUTIONThe standard form of state equation is(1)whereConsider the first equation in (1)We will solve this equation for X(s), sowe collect all terms containing X(s) onthe left sideIts LAPLACE transform isTo solve this we factorized X(s)This equation now be solved for X(s)The second equation in (1) yieldsAnd the state vector x(t) is the inverseLAPLACE transform of this equation
2SOLUTION OF STATE EQUATION LAPLACE TRANSFORM SOLUTIONTo obtain a general relationship for the solution we define the state transition matrix asThis 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 errorMore practical procedure is computer simulation.
3Example finding transition matrix Consider the system described byIts determinant isdet (sI-A)= s2+3s+2=(s+1)(s+2)Using observer canonical form we writethe state equation as:The inverse is then the adjoint matrixdivided by the determinantThe state transition matrix is the inverseLaplace transform of this matrixTo find the state transition matrix, first wecalculate the matrix (sI-A)With the definition of state transitionmatrix , the equation for complete solutionof the state equation can be foundThe next step is that we have to find theinverse of the matrix (sI-A). First we findthe adjoint of (sI-A)
4Example finding solution of the state equation We are going to solve the following equationThe inverse Laplace transform of this terms isX(s)= (sI-A)1x(0)+(sI-A)1BU(s) (1)Consider the same system as beforeThe state transition matrix is the inverseLaplace transform of (1)With (s) the Laplace transform oftransition matrix is given bySuppose that the input is a unit step. ThenU(s)=1/s. And the second term of (1)becomesFinding the complete solution is long evenfor a second order system. The necessityfor reliable machine solutions, such as adigital computer simulation is evident
5Convolution solution of the state equation We are going to find the inverse L.T ofX(s)= (sI-A)1x(0)+(sI-A)1BU(s) (1)Using Convolution theorem we find that:orNote that the solution is composed of to termsThe first term is referred tothe zero input part or the initial condition part of the solutionThe second term is calledthe zero-state part or the forced part.
6Example convolution solution Consider the same system as previousOnly the force part is derived here the initial condition part is derivedthe same way as the previous example
7Infinite series solution One method of differential eq. is to assume asa solution an infinite series with unknowncoefficient. This method is know used to findthe transition matrixNext perform the following operations:Evaluate (5) at t=0Differentiate (5) and evaluate the result at t=0Repeat step 2 again and againThe result is the following equationsK1 = AK02K2 = AK13K3 = AK2The state equation may be written aswith the solution(6)The solution is assumed to be of the formEvaluating (3) at t = 0 shows that K0=I, then the other matrices are evaluated from (6)...where the (nn) matrices Ki are unknownand t is the scalar time. Differentiating thisexpression yields:Hence from (3) the state transition matrix is:Substituting (3) and (4) into (1) yield
8Example 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)ThrustorsThe state transition matrix is then:Thenand
9Transfer functionWe are going to find the transfer function if the state representation is known. The standard form of state equation when D=0 isFor the case that D 0, the T.F. isG(s) = C(sI-A)1B = C(s)B + D (7)(1)Example:The state representation is as followsIgnoring the initial condition, its L. T. issX(s)= A X(s)+BU(s) (2)X(s) can easily be solveX(s)= (sI-A)1BU(s) (3)The L.T of the output equation in (1) yieldsThe transfer function is given byY(s)= CX(s) (4)Eq. (3) and (4) givesY(s)= C(sI-A)1BU(s)=G(s)U(s) (5)The transfer function G(s) is thenMATLAB: 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 3G(s) = C(sI-A)1B = C(s)B (6)
10Similarity 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 foundThe number of internal models (state model) is unboundedThe state model of SISO system isSuppose 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 isv1(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)= P1x(t)x(t) = Pv(t)And the transfer function G(s) is given byMatrix P is called the transformation matrixthis will transform one set of state vector to a different state vectorThis transformation alter the internal model but not the input output relationshipThis typo of transformation is called similarity transformationG(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