Signals and Systems - Electrical Engineering

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

The above results can be explained by looking for a dual of the derivative property. Multiplying by

−tthe signalx(t)corresponds to differentiatingX(s)with respect tos. Indeed for an integerN>1,

dNX(s)

dsN

=

∫∞

0

x(t)

dNe−st

dsN

dt

=

∫∞

0

x(t)(−t)Ne−stdt

Thus, ifx(t)=u(t),X(s)= 1 /s, then−tx(t)has Laplace transformdX(s)/ds=− 1 /s^2 , ortu(t)and

1 /s^2 are Laplace transform pairs. In general, the Laplace transform oftN−^1 u(t), forN≥1, hasN

poles at the origin.

What about multiple real (different from zero) and multiple complex poles? What are the

corresponding inverse Laplace transforms? The inverse Laplace transform of

2  0 s/(s^2 +^20 )^2

having double complex poles at±j 0 , is

tsin( 0 t)u(t)

Likewise,

te−atu(t)

has as Laplace transform 1/(s+a)^2. So multiple poles correspond to multiplication bytin the

time domain. n

nExample 3.10

Obtain from the Laplace transform ofx(t)=cos( 0 t)u(t)the Laplace transform of sin(t)u(t)using

the derivative property.

Solution

The causal sinusoid

x(t)=cos( 0 t)u(t)

has a Laplace transform

X(s)=

s

s^2 +^20

Then,

dx(t)

dt

=u(t)

dcos( 0 t)

dt

+cos( 0 t)

du(t)

dt