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Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition 5 15 Copyright © Orchard Publications Computation of the State Transition Matrix We will use the following MATLAB script for the solution of (5.60). B=sym('[1 1 1; 1 2 4; 1 3 9]'); b=sym('[exp(t); exp(2*t); exp(3*t)]'); a=B\b; fprintf(' \n');. .. disp('a0 = '); disp(a(1)); disp('a1 = '); disp(a(2)); disp('a2 = '); disp(a(3)) a0 = 3*exp(t)-3*exp(2*t)+exp(3*t) a1 = -5/2*exp(t)+4*exp(2*t)-3/2*exp(3*t) a2 = 1/2*exp(t)-exp(2*t)+1/2*exp(3*t) Thus, (5.61) 4. We also use MATLAB to perform the substitution into the state transition matrix, and to per- form the matrix multiplications. The script is shown below. syms t; a0 = 3*exp(t)+exp(3*t) 3*exp(2*t); a1 = 5/2*exp(t) 3/2*exp(3*t)+4*exp(2*t);. .. a2 = 1/2*exp(t)+1/2*exp(3*t) exp(2*t);. .. A = [5 7 5; 0 4 1; 2 8 -3]; eAt=a0*eye(3)+a1*A+a2*A^2 eAt = [-2*exp(t)+2*exp(2*t)+exp(3*t), -6*exp(t)+5*exp(2*t)+exp(3*t), 4*exp(t)-3*exp(2*t)-exp(3*t)] [-exp(t)+2*exp(2*t)-exp(3*t), -3*exp(t)+5*exp(2*t)-exp(3*t), 2*exp(t)-3*exp(2*t)+exp(3*t)] [-3*exp(t)+4*exp(2*t)-exp(3*t), -9*exp(t)+10*exp(2*t)-exp(3*t), 6*exp(t)-6*exp(2*t)+exp(3*t)] Thus, 5.4.2 Multiple (Repeated) Eigenvalues In this case, we will assume that the polynomial of (5.62) a 0 3e t 2t e 3t + = a 1 5 2 --e t 4e 3 2 + = a 2 1 2 t e 1 2 + = e At 2e t e ++ 6 e t 5e e t e e t e + t e + t e + t e + 9e t 10e e + 6e t e + = det A λ I [] 0 =

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Chapter 5 State Variables and State Equations 5 16 Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition Copyright © Orchard Publications has roots, and of these roots are equal. In other words, the roots are (5.63) The coefficients of the state transition matrix (5.64) are found from the simultaneous solution of the system of equations of (5.65) below. (5.65) Example 5.8 Compute the state transition matrix given that Solution: 1. We first find the eigenvalues of the matrix and these are found from the polynomial of . For this example, and thus, nm λ 1 λ 2 = λ 3 = …λ m , λ m1 + λ n = a i e At a 0 Ia 1 Aa 2 A 2 a n1 A ++ + + = a 0 a 1 λ 1 a 2 λ 1 2 a λ 1 +++ + e λ 1 t = d d λ 1 -------- a 0 a 1 λ 1 a 2 λ 1 2 a λ 1 + () d d λ 1 --------e λ 1 t = d 2 d λ 1 2 0 a 1 λ 1 a 2 λ 1 2 a λ 1 + d 2 d λ 1 2 λ 1 t = d d λ 1 --------------- a 0 a 1 λ 1 a 2 λ 1 2 a λ 1 + d d λ 1 ---------------e λ 1 t = a 0 a 1 λ + a 2 λ + 2 a λ + + e λ + t = a 0 a 1 λ n a 2 λ n 2 a λ n + e λ n t = e A 1 0 21
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