5.8 Additional Properties 347
sinceu(t−τ)as a function ofτequals
u(t−τ)=
{
1 τ <t
0 τ >t
We thus have that
F
∫t
−∞
x(τ)dτ
=X()F[u(t)] (5.29)
- Since the unit-step signal is not absolutely integrable its Fourier transform cannot be found from
the integral definition, and we cannot use its Laplace transform either because its ROC does not
include thejaxis. Let’s transform it into an absolutely integrable signal by subtracting 1/2 and
multiplying the result by 2. This gives thesignsignal:
sgn(t)=2[u(t)−0.5]=
{
1 t> 0
− 1 t< 0
The derivative of sgn(t)is
dsgn(t)
dt
= 2 δ(t)
and thus ifS()=F[sgn(t)],
S()=
2
j
using the derivative property. The linearity of the Fourier transform applied to the definition of
sgn(t)gives
F[sgn(t)]= 2 F[u(t)]− 2 πδ() ⇒ F[u(t)]=
1
j
+πδ() (5.30)
Replacing the Fourier transform ofu(t)in Equation (5.29), we get
F
∫t
−∞
x(τ)dτ
=X()
[
1
j
+πδ()
]
=
X()
j
+πX( 0 )δ() (5.31)
Remarks
n Just like in the Laplace transform where the operator s corresponds to the derivative operation in time of
the signal, in the Fourier transform jbecomes the corresponding operator for the derivative operation in
time of the signal.
