Signals and Systems - Electrical Engineering

3.4 Inverse Laplace Transform 211

is obtained by first considering the term 1/s, which hasu(t)as inverse, and then using the information
in the numerator to get the final response,

x(t)=u(t+ 1 )−u(t− 1 )

The two sums

∑∞

k= 0

e−αsk=

1

1 −e−αs

∑∞

k= 0

(−e−αs)k=

1

1 +e−αs

can be easily verified by cross-multiplying. So when the function is

X 1 (s)=

N(s)

D(s)( 1 −e−αs)

=

N(s)

D(s)

∑∞

k= 0

e−αsk

=

N(s)

D(s)

+

N(s)e−αs

D(s)

+

N(s)e−^2 αs

D(s)

+···

and iff(t)is the inverse ofN(s)/D(s), we then have

x 1 (t)=f(t)+f(t−α)+f(t− 2 α)+···

Likewise, when

X 2 (s)=

N(s)

D(s)( 1 +e−αs)

=

N(s)

D(s)

∑∞

k= 0

(− 1 )ke−αsk

=

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 2 (t)=f(t)−f(t−α)+f(t− 2 α)−···

nExample 3.19

We wish to find the causal inverse of

X(s)=

1 −e−s

(s+ 1 )( 1 −e−^2 s)

Solution

We let

X(s)=F(s)

∑∞

k= 0

(e−^2 s)k