3.3 The One-Sided Laplace Transform 193
=− 0 sin( 0 t)u(t)+cos( 0 t)δ(t)
=− 0 sin( 0 t)u(t)+δ(t)
so that the Laplace transform ofdx(t)/dtis given by
sX(s)−x( 0 −)=− 0 L[sin( 0 t)u(t)]+L[δ(t)]
Thus, the Laplace transform of the sine is
L[sin( 0 t)u(t)]=−
sX(s)−x( 0 −)− 1
0
=
1 −sX(s)
0
=
0
s^2 +^20
sincex( 0 −)=0 andX(s)=L[cos(oT)] given above. n
Notice that whenever the signal is discontinuous att=0, as in the case ofx(t)=cos( 0 t)u(t), its
derivative will include aδ(t)signal due to the discontinuity. On the other hand, whenever the signal
is continuous att=0, for instancey(t)=sin( 0 t)u(t), its derivative does not containδ(t)signals. In
fact,
dy(t)
dt
= 0 cos( 0 t)u(t)+sin( 0 t)δ(t)
= 0 cos( 0 t)u(t)
since the sine is zero att=0.
3.3.3 Integration
The Laplace transform of the integral of a causal signaly(t)is given by
L
∫t
0
y(τ)dτu(t)
=Y(s)
s
(3.14)
This property can be shown by using the derivative property. Call the integral
f(t)=
∫t
0
y(τ)dτu(t)
Using the fundamental theorem of calculus, we then have that
df(t)
dt
=y(t)u(t)
