Signals.and.Systems.with.MATLAB.Computing.and.Simulink.Modeling.4th_Part12

Signals.and.Systems.with.MATLAB.Computing.and.Simulink.Modeling.4th_Part12

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Chapter 3 The Inverse Laplace Transformation 3 10 Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition Copyright © Orchard Publications (3.44) The residues are The value of the residue can also be found without differentiation as follows: Substitution of the already known values of and into (3.44), and letting * , we obtain or from which as before. Finally, (3.45) Check with MATLAB: syms s t; Fs=(s+3)/((s+2)*(s+1)^2); ft=ilaplace(Fs) ft = exp(-2*t)+2*t*exp(-t)-exp(-t) We can use the following script to check the partial fraction expansion. syms s Ns = [1 3]; % Coefficients of the numerator N(s) of F(s) expand((s + 1)^2); % Expands (s + 1)^2 to s^2 + 2*s + 1; d1 = [1 2 1]; % Coefficients of (s + 1)^2 = s^2 + 2*s + 1 term in D(s) d2 = [0 1 2]; % Coefficients of (s + 2) term in D(s) Ds=conv(d1,d2); % Multiplies polynomials d1 and d2 to express the % denominator D(s) of F(s) as a polynomial [r,p,k]=residue(Ns,Ds) * This is permissible since (3.44) is an identity. F 4 s () s3 + s2 + s1 + 2 ----------------------------------- r 1 + ---------------- r 21 + 2 ------------------ r 22 + ++ == r 1 + + 2 = 1 r 21 + + ----------- = 2 r 22 d ds ----- + +   = + + + 2 -------------------------------------- = 1 = r 22 r 1 r 21 s0 = + + 2 + = 1 + = 2 + 2 = r 22 + = = 3 2 -- 1 2 2r 22 = r 22 1 = F 4 s + + + 2 = 1 + 2 + 2 1 + = e 2t 2te t e t + f 4 t =
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Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition 3 11 Copyright © Orchard Publications Partial Fraction Expansion r = 1.0000 -1.0000 2.0000 p = -2.0000 -1.0000 -1.0000 k = [] Example 3.5 Use the partial fraction expansion method to simplify of (3.46) below, and find the time domain function corresponding to the given . (3.46) Solution: We observe that there is a pole of multiplicity at , and a pole of multiplicity at . Then, in partial fraction expansion form, is written as (3.47) The residues are F 5 s () f 5 t F 5 s F 5 s s 2 3 + s1 + + 3 s2 + 2 ------------------------------------- = 3 = 2s 2 = F 5 s F 5 s r 11 + 3 ------------------ r 12 + 2 r 13 + ---------------- r 21 + 2 r 22 + ++ = r 11 s 2 3 + + + 2 ------------------------- = 1 == r 12 d ds ----- s 2 3 + + + 2    = = + 2 3 + 2s 2 + s 2 3 + + + 4 --------------------------------------------------------------------------------------------- = s4 + + 3 = 3 =
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Chapter 3 The Inverse Laplace Transformation 3 12 Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition Copyright © Orchard Publications Next, for the pole at , and By substitution of the residues into (3.47), we obtain (3.48) We will check the values of these residues with the MATLAB script below.
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Signals.and.Systems.with.MATLAB.Computing.and.Simulink.Modeling.4th_Part12

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