Signals and Systems - Electrical Engineering

304 CHAPTER 5: Frequency Analysis: The Fourier Transform

(c) The Laplace transform ofx 3 (t)is

X 3 (s)=

1

s+ 1

+

1

−s+ 1

=

2

1 −s^2

with a region of convergence ROC={s=σ+j:− 1 < σ <1,−∞<  <∞}that contains

thejaxis. Then the Fourier transform ofx 3 (t)is

X 3 ()=X 3 (s)|s=j=

2

1 −(j)^2

=

2

1 +^2 n

5.5 Linearity, Inverse Proportionality, and Duality

Many of the properties of the Fourier transform are very similar to those of the Fourier series or of

the Laplace transform, which is to be expected given the strong connection among these transforma-

tions. The linearity and the duality between time and frequency of the Fourier transform will help us

determine the transform of signals that do not satisfy the existence conditions given before.

5.5.1 Linearity

Just like the Laplace transform, the Fourier transform is linear.

IfF[x(t)]=X()andF[y(t)]=Y(), for constantsαandβ, we have that

F[αx(t)+βy(t)]=αF[x(t)]+βF[y(t)]

=αX()+βY() (5.4)

nExample 5.2

Suppose you create a periodic sine

x(t)=sin( 0 t) −∞<t<∞

by adding a causal sinev(t)=sin( 0 t)u(t)and an anti-causal siney(t)=sin( 0 t)u(−t), for each

of which you can find Laplace transformsV(s)andY(s). Discuss what would be wrong with this

approach to find the Fourier transform ofx(t)by lettings=j.

Solution

The Laplace transforms ofv(t)andy(t)are

V(s)=

 0

s^2 +^20

ROC:Re[s]> 0

Y(s)=

− 0

(−s)^2 +^20

ROC:Re[s]< 0

givingX(s)=V(s)+Y(s)=0. Moreover, the region of convergence ofX(s)is the intersection of

the two given ROCs, which is null, so it is not possible to obtain the Fourier transform ofx(t)this