Signals and Systems - Electrical Engineering

306 CHAPTER 5: Frequency Analysis: The Fourier Transform

or a sinc function where Aτ corresponds to the area under x 3 (t). The Fourier trans-

form X 3 () is an even function of . At =0 using L’Hopital’s rule we find thatˆ

X 3 ( 0 )=Aτ. Finally, the Fourier transform of the pulse becomes zero when = 2 kπ/τ,

k=±1,±2,....

If we letA= 1 /τ(so that the area of the pulse is unity), and letτ→0, the pulsex 3 (t)becomes

a delta functionδ(t)in the limit and the sinc function expands (forτ→0,X 3 ()is not zero

for any finite value) to become unity. On the other hand, if we letτ→∞, the pulse becomes

a constant signalAextending from−∞to∞, and the Fourier transform gets closer and closer

toδ()(the sinc function becomes zero at values very close to zero and the amplitude at= 0

becomes larger and larger, although the area under the curve remains constant). As shown above,

X 3 ()= 2 πAδ()is the transform ofx 3 (t)=A,−∞<t<∞.

n To illustrate the transition in the Fourier transform as the time support increases, we used the

following MATLAB script to compute the Fourier transform of pulses of the same amplitudeA= 1

but different time supports 1 and 4. The script below shows the case when the support is 1, but

it can be easily changed to get the support of 4. The symbolic MATLAB functionfouriercomputes

the Fourier transform. The results are shown in Figure 5.1.

%%%%%%%%%%%%%%%%%%%%%

% Time-frequency relation

%%%%%%%%%%%%%%%%%%%%%

syms t w

x = heaviside(t + 0.5)−heaviside(t−0.5);

subplot(211)

ezplot(x, [−3]);axis([−3 3−0.1 1.1]);grid

X = fourier(x) % Fourier transform

subplot(212)

ezplot(X, [−50 50]); axis([−50 50−1 5]);grid

In summary, the support ofX()is inversely proportional to the support ofx(t). Ifx(t)has a Fourier transform

X()andα6= 0 is a real number, thenx(αt)is an

n Contracted(α > 1 ),

n Contracted and reflected(α <− 1 ),

n Expanded( 0 < α < 1 ),

n Expanded and reflected(− 1 < α < 0 ), or

n Simply reflected(α=− 1 )

signal, and we have the pair

x(αt) ⇔

1

|α|

X

(



α

)

(5.6)

First let us mention that the symbol⇔means that to a signalx(t)in the time domain (on the left)

there corresponds a Fourier transformX()in the frequency domain (on the right). This isnotan

equality—far from it!