3.4 Inverse Laplace Transform 209
function pfeLaplace(num,den)
%
disp(’>>>>>Zeros<<<<<’)
z = roots(num)
[r,p,k]=residue(num,den);
disp(’>>>>>Poles<<<<<’)
p
disp(’>>>>>Residues<<<<<’)
r
splane(num,den)
The functionpfeLaplaceuses the MATLAB functionrootsto find the zeros ofX(s)defined by the coef-
ficients of its numerator and denominator given in descending order ofs. For the partial fraction
expansion,pfeLaplaceuses the MATLAB functionresidue, which finds coefficients of the expansion
as well as the poles ofX(s). (The residuer(i)in the vectorrcorresponds to the expansion term for
the polep(i); for instance, the residuer( 1 )=8 corresponds to the expansion term corresponding
to the polep( 1 )=−3.) The symbolic functionilaplaceis then used to find the inversex(t); as input
toilaplacethe functionX(s)is described in a symbolic way. The MATLAB functionezplotis used for
the plotting of the symbolic computations.
The analytic results are shown in the following, and the plot ofx(t)is given in Figure 3.16.
>>>>>Zeros<<<<<
z = -1.6667
1.0000
>>>>>Poles<<<<<
p = -3.0000
-2.0000
-1.0000
>>>>>Residues<<<<<
r = 8.0000
-3.0000
-2.0000
>>>>>Inverse Laplace<<<<<
x = 8 * exp(-3 * t)-3 * exp(-2 * t)-2 * exp(-t) n
3.4.2 Inverse of Functions Containinge−ρsTerms
WhenX(s)has exponentialse−ρsin the numerator or denominator, ignore these terms and perform partial
fraction expansion on the rest, and at the end consider the exponentials to get the correct time shifting.
In particular, when
X(s)=
N(s)
D(s)( 1 −e−αs)
=
N(s)
D(s)
+
N(s)e−αs
D(s)
+
N(s)e−^2 αs
D(s)
+···
