Signals and Systems - Electrical Engineering

3.3 The One-Sided Laplace Transform 195

nExample 3.12

Suppose we wish to find the Laplace transform of the causal sequence of pulsesx(t)shown in

Figure 3.12. Letx 1 (t)denote the first pulse (i.e., for 0≤t<1).

x 1 (t) x(t)

t

01

t

· · ·

(^0213)
FIGURE 3.12
Generic causal pulse signal.
Solution
We have fort≥0,
x(t)=x 1 (t)+x 1 (t− 1 )+x 1 (t− 2 )+···
and 0 fort<0. According to the shifting and linearity properties, we have
X(s)=X 1 (s)

[

1 +e−s+e−^2 s+···

]

=X 1 (s)

[

1

1 −e−s

]

Notice that 1+e−s+e−^2 s+···= 1 /( 1 −e−s), which is verified by cross-mutiplying:

[1+e−s+e−^2 s+···]( 1 −e−s)=( 1 +e−s+e−^2 s+···)−(e−s+e−^2 s+···)= 1

The poles ofX(s)are the poles ofX 1 (s)and the roots of 1−e−s=0 (thesvalues such thate−s=1,

orsk=±j 2 πkfor any integerk≥0). Thus, there is an infinite number of poles forX(s), and the

partial fraction expansion method that uses poles to invert Laplace transforms, presented later,

will not be useful. The reason this example is presented here, ahead of the inverse Laplace, is to

illustrate that when we are finding the inverse of this type of Laplace function we need to con-

sider the time-shift property, otherwise we would need to consider an infinite partial fraction

expansion. n

nExample 3.13

Consider the causal full-wave rectified signal shown in Figure 3.13. Find its Laplace transform.