Signals and Systems - Electrical Engineering

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

and so

L

[

df(t)

dt

]

=sF(s)−f( 0 )

=Y(s)

sincef( 0 )=0 (the integral over a point), then

F(s)=L





∫t

0

y(τ)dτ



=Y(s)

s

nExample 3.11

Suppose that

∫t

0

y(τ)dτ= 3 u(t)− 2 y(t)

Find the Laplace transform ofy(t), a causal signal.

Solution

Applying the integration property gives

Y(s)

s

=

3

s

− 2 Y(s)

so that solving forY(s)we obtain

Y(s)=

3

2 (s+0.5)

corresponding toy(t)=1.5e−0.5tu(t). n

3.3.4 Time Shifting...........................................................................

If the Laplace transform off(t)u(t)isF(s), the Laplace transform of the time-shifted signalf(t−τ)u(t−τ)is

L[f(t−τ)u(t−τ)]=e−τsF(s) (3.15)

This indicates that when we delay (advance) the signal to getf(t−τ)u(t−τ) (f(t+τ)u(t+τ))its

corresponding Laplace transform isF(s)multiplied bye−τs(eτs). This property is easily shown by a

change of variable when computing the Laplace transform of the shifted signals.