Signals.and.Systems.with.MATLAB.Computing.and.Simulink.Modeling.4th_Part24

# Signals.and.Systems.with.MATLAB.Computing.and.Simulink.Modeling.4th_Part24

This preview shows pages 1–3. Sign up to view the full content.

Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition 5 47 Copyright © Orchard Publications Solutions to End of Chapter Exercises and thus b. and thus c. and it is given that . Then, and thus 7 . a. Matrix is the same as Matrix C in Exercise 6. Then, and since is a matrix the state transition matrix is (1) Then, 1 λ λλ 2 6 ++ 0 = λ 2 7 = λ 1 7 = λ 2 7 = B a0 a b = det B λ I () det a b λ 10 01    a λ 0 a b λ 0 == = a λ b λ 0 = λ 1 a = λ 2 b = C 010 00 1 6 11 6 = det C λ I 1 6 11 6 λ 100 001 = λ 0 λ 1 6 11 6 λ 0 = = λ 2 6 λ 6 11 λ λ 3 6 λ 2 11 λ 6 +++ 0 λ 1 1 = λ 3 6 λ 2 11 λ 6 λ 1 + ---------------------------------------------- λ 2 5 λ 6 λ 1 + λ 2 + λ 3 + 0 λ 1 1 = λ 2 2 = λ 1 3 = A λ 1 1 = λ 2 2 = λ 1 3 = A3 3 × e At a 0 Ia 1 Aa 2 A 2 =

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Chapter 5 State Variables and State Equations 5 48 Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition Copyright © Orchard Publications syms t; A=[1 1 1; 1 2 4; 1 3 9];. .. a=sym('[exp( t); exp( 2*t); exp( 3*t)]'); x=A\a; fprintf(' \n');. .. disp('a0 = '); disp(x(1)); disp('a1 = '); disp(x(2)); disp('a2 = '); disp(x(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, Now, we compute of (1) with the following MATLAB script: syms t; 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); A=[0 1 0; 0 0 1; 6 11 6]; fprintf(' \n');. .. eAt=a0*eye(3)+a1*A+a2*A^2 eAt = [3*exp(-t)-3*exp(-2*t)+exp(-3*t), 5/2*exp(-t)-4*exp(-2*t)+3/ 2*exp(-3*t), 1/2*exp(-t)-exp(-2*t)+1/2*exp(-3*t)] [-3*exp(-t)+6*exp(-2*t)-3*exp(-3*t), -5/2*exp(-t)+8*exp(-2*t)- 9/2*exp(-3*t), -1/2*exp(-t)+2*exp(-2*t)-3/2*exp(-3*t)] [3*exp(-t)-12*exp(-2*t)+9*exp(-3*t), 5/2*exp(-t)-16*exp(- 2*t)+27/2*exp(-3*t), 1/2*exp(-t)-4*exp(-2*t)+9/2*exp(-3*t)] Thus, a 0 a 1 λ 1 a 2 λ 1 2 ++ e λ 1 t = a 0 a 1 a 2 + e t = a 0 a 1 λ 2 a 2 λ 2 2 e λ 2 t = a 0 2a 1 4a 2 + e 2t = a 0 a 1 λ 3 a 2 λ 3 2 e λ 3 t = a 0 3a 1 9a 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 8

Signals.and.Systems.with.MATLAB.Computing.and.Simulink.Modeling.4th_Part24

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online