Signals and Systems - Electrical Engineering

302 CHAPTER 5: Frequency Analysis: The Fourier Transform

5.3 Existence of the Fourier Transform

For the Fourier transform to exist,x(t)must beabsolutely integrable—that is,

|X()|≤

∫∞

−∞

|x(t)e−jt|dt=

∫∞

−∞

|x(t)|dt<∞

Moreover,x(t)must have only a finite number of discontinuities and a finite number of minima and

maxima in any finite interval. (Given the limiting connection between the Fourier transform and the

Fourier series, it is not surprising that the above conditions coincide with the existence conditions for

the Fourier series.)

The Fourier transform

X()=

∫∞

−∞

x(t)e−jtdt

of a signalx(t)exists (i.e., we can calculate its Fourier transform via this integral) provided

n x(t)is absolutely integrable or the area under|x(t)|is finite.

n x(t)has only a finite number of discontinuites as well as maxima and minima.

From the definitions of the direct and the inverse Fourier transforms—both being infinite integrals—

one wonders whether they exist in general, and if so how to most efficiently compute them.

Commenting on the existence conditions, Professor E. Craig [17] wrote:

It appears that almost nothing has a Fourier transform—nothing except practical communi-

cation signals. No signal amplitude goes to infinity and no signal lasts forever; therefore, no

practical signal can have infinite area under it, and hence all have Fourier transforms.

Indeed, signals of practical interest have Fourier transforms and their spectra can be displayed using

aspectrum analyzer(or better yet, any signal for which we can display its spectrum will have a

Fourier transform). A spectrum analyzer is a device that displays the energy or the power of a signal

distributed over frequencies.

5.4 Fourier Transforms from the Laplace Transform

The region of convergence of the Laplace transformX(s)indicates the region in thes-plane whereX(s)

is defined. The following applies to signals whether they are causal, anti-causal, or noncausal.

If the region of convergence (ROC) ofX(s)=L[x(t)]contains thejaxis, so thatX(s)can be defined fors=j,

then

F[x(t)]=L[x(t)]|s=j=

∫∞

−∞

x(t)e−jtdt

=X(s)

∣∣

s=j (5.3)