210 C H A P T E R 3: The Laplace Transform

FIGURE 3.16

Inverse Laplace

transform of

X(s)=( 3 s^2 + 2 s− 5 )/

(s^3 + 6 s^2 + 11 s+ 6 ).

(a) Poles and zeros of

X(s)are given with (b)

the corresponding

inversex(t).

−− (^14) − 2 0 2
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
σ
jΩ
(a)
− (^10510)
−0.5
0
0.5
1
1.5
2
2.5
3
t
(b)
x(t)
iff(t)is the inverse ofN(s)/D(s), then
x(t)=f(t)+f(t−α)+f(t− 2 α)+···
Another possibility is when the function is given as
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)
  −···
  Iff(t)is the inverse ofN(s)/D(s), we then have
  x(t)=f(t)−f(t−α)+f(t− 2 α)−···
  The time-shifting property of Laplace makes it possible for the numeratorN(s)or the denominator
  D(s)to havee−σsterms. The procedure for inverting such functions is to initially ignore these terms
  and do the partial fraction expansion on the rest and at the end consider them to do the necessary
  time shifting. For instance, the inverse of
  X(s)=
  es−e−s
  s

=

es

s

−

e−s

s