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Signals.and.Systems.with.MATLAB.Computing.and.Simulink.Modeling.4th_Part17

Signals.and.Systems.with.MATLAB.Computing.and.Simulink.Modeling.4th_Part17

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Chapter 4 Circuit Analysis with Laplace Transforms 4 26 Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition Copyright © Orchard Publications and in matrix form We use the MATLAB script below we obtain the values of the currents. Z=[z1+z2 z2; z2 z2+z3]; Vs=[1/s 2/s]'; Is=Z\Vs; fprintf(' \n');... disp('Is1 = '); pretty(Is(1)); disp('Is2 = '); pretty(Is(2)) Is1 = 2 2 s - 1 + s ------------------------------- 2 3 (6 s + 3 + 9 s + 2 s ) Is2 = 2 4 s + s + 1 - ------------------------------- 2 3 (6 s + 3 + 9 s + 2 s ) conj(s) Therefore, (1) (2) We use MATLAB to express the denominators of (1) and (2) as a product of a linear and a quadratic term. p=[2 9 6 3]; r=roots(p); fprintf(' \n'); disp('root1 ='); disp(r(1));... disp('root2 ='); disp(r(2)); disp('root3 ='); disp(r(3)); disp('root2 + root3 ='); disp(r(2)+r(3));... disp('root2 * root3 ='); disp(r(2)*r(3)) root1 = -3.8170 root2 = -0.3415 + 0.5257i z 1 z 2 + ( ) I 1 s ( ) z 2 I 2 s ( ) 1 s = z 2 I 1 s ( ) z 2 z 3 + ( ) I 2 s ( ) + 2 s = z 1 z 2 + ( ) z 2 z 2 z 2 z 3 + ( ) I 1 s ( ) I 2 s ( ) 1 s 2 s = I 1 s ( ) s 2 2s 1 + 2s 3 9s 2 6s 3 + + + ------------------------------------------- = I 2 s ( ) 4s 2 s 1 + + 2s 3 9s 2 6s 3 + + + ------------------------------------------- =
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Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition 4 27 Copyright © Orchard Publications Solutions to End of Chapter Exercises root3 = -0.3415 - 0.5257i root2 + root3 = -0.6830 root2 * root3 = 0.3930 and with these values (1) is written as (3) Multiplying every term by the denominator and equating numerators we obtain Equating , , and constant terms we obtain We will use MATLAB to find these residues. A=[1 1 0; 0.683 3.817 1; 0.393 0 3.817]; B=[1 2 1]'; r=A\B; fprintf(' \n');... fprintf('r1 = %5.2f \t',r(1)); fprintf('r2 = %5.2f \t',r(2)); fprintf('r3 = %5.2f',r(3)) r1 = 0.48 r2 = 0.52 r3 = -0.31 By substitution of these values into (3) we obtain (4) By inspection, the Inverse Laplace of first term on the right side of (4) is (5) The second term on the right side of (4) requires some manipulation. Therefore, we will use the MATLAB ilaplace(s) function to find the Inverse Laplace as shown below.
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