Signals and Systems - Electrical Engineering

308 CHAPTER 5: Frequency Analysis: The Fourier Transform

This property is shown by a change of variable in the integration,

F[x(αt)]=

∫∞

−∞

x(αt)e−jtdt=















1

α

∞∫

−∞

x(ρ)e−jρ/αdρ α > 0

−^1 α

∞∫

−∞

x(ρ)ejρ/αdρ α < 0

=

1

|α|

X

(



α

)

by change of variableρ=αt. If|α|>1, when compared withx(t)the signalx(αt)contracts while its

corresponding Fourier transform expands. Likewise, when 0<|α|<1, the signalx(αt)expands, as

compared withx(t), and its Fourier transform contracts. Ifα <0, the corresponding contraction or

expansion is accompanied by a reflection in time. In particular, ifα=−1, the reflected signalx(−t)

hasX(−)as its Fourier transform.

nExample 5.3

Consider a pulsex(t)=u(t)−u(t− 1 ). Find the Fourier transform ofx 1 (t)=x( 2 t).

Solution

The Laplace transform ofx(t)is

X(s)=

1 −e−s

s

with the wholes-plane as its region of convergence. Thus, its Fourier transform is

X()=

1 −e−j

j

=

e−j/^2 (ej/^2 −e−j/^2 )

2 j/ 2

=

sin(/ 2 )

/ 2

e−j/^2

To the finite-support signalx(t)correspondsX()of infinite support. Then,

x 1 (t)=x( 2 t)=u( 2 t)−u( 2 t− 1 )=u(t)−u(t−0.5)

and its Fourier transform is found, again using its Laplace transform, to be

X 1 ()=

1 −e−j/^2

j

=

e−j/^4 (ej/^4 −e−j/^4 )

j

=

1

2

sin(/ 4 )

/ 4

e−j/^4 =

1

2

X(/ 2 )

which is an expanded version ofX()in the frequency domain and coincides with the result from

the property. See Figure 5.2.