Signals and Systems - Electrical Engineering

5.8 Additional Properties 349

and a second derivative gives

d^2 x(t)

dt^2

=δ(t)− 2 δ(t− 1 )+δ(t− 2 )

Using the time-shift and the derivative properties, we get from the expression for the second

derivative and lettingX()be the Fourier transform ofx(t):

(j)^2 X()= 1 − 2 e−j+e−j^2 

=e−j[ej− 2 +e−j]

so that

X()=

2 e−j

^2

[1−cos()]

n

nExample 5.23

Consider the integral

y(t)=

∫t

−∞

x(τ)dτ−∞<t<∞

wherex(t)=u(t+ 1 )−u(t− 1 ). Find the Fourier transformY()directly and from the integration

property.

Solution

The integral is

y(t)=







0 t<− 1

t+ 1 − 1 ≤t< 1

2 t≥ 1

or

y(t)=[r(t+ 1 )−r(t− 1 )− 2 u(t− 1 )]

︸ ︷︷ ︸

y 1 (t)

+ 2 u(t− 1 )

The Fourier transform ofy 1 (t)is

Y 1 ()=

[

es−e−s

s^2

−

2 e−s

s

]

s=j

=

− 2 jsin()

^2

+j

2 e−j