Signals and Systems - Electrical Engineering

3.3 The One-Sided Laplace Transform 189

L

[

d^2 f(t)

dt^2

u(t)

]

=s^2 F(s)−sf( 0 −)−

df(t)

dt

∣

∣t= 0 − (3.12)

In general, iff(N)(t)denotes anNth-order derivative of a functionf(t)that has a Laplace transformF(s), we

have

L[f(N)(t)u(t)]=sNF(s)−

N∑− 1

k= 0

f(k)( 0 −)sN−^1 −k (3.13)

wheref(m)(t)=dmf(t)/dtmis themth-order derivative,m> 0 , andf(^0 )(t),f(t).

The Laplace transform of the derivative of a causal signal is

L

[

df(t)

dt

u(t)

]

=

∫∞

0 −

df(t)

dt

e−stdt

This integral is evaluated by parts. Letw=e−st, thendw=−se−stdt, and letv=f(t)so thatdv=
[df(t)/dt]dt, and
∫
wdv=wv−

∫

vdw

We would then have

∫∞

0 −

df(t)

dt

e−stdt=e−stf(t)

∣∣∞

0 −−

∫∞

0 −

f(t)(−se−st)dt

=s

∫∞

0 −

f(t)e−stdt−f( 0 −)

=sF(s)−f( 0 −)

wheree−stf(t)|t= 0 −=f( 0 −)ande−stf(t)|t→∞=0 since the region of convergence guarantees that

lim

t→∞

f(t)e−σt= 0

For a second-order derivative we have that

L

[

d^2 f(t)

dt^2

u(t)

]

=L

[

df(^1 )(t)

dt

u(t)

]

=sL[f(^1 )(t)]−f(^1 )( 0 −)

=s^2 F(s)−sf( 0 −)−

df(t)

dt

|t= 0 −

where we used the notationf(^1 )(t)=df(t)/dt. This approach can be extended to any higher order to
obtain the general result shown above.