Signals and Systems - Electrical Engineering

324 CHAPTER 5: Frequency Analysis: The Fourier Transform

function by lettings=j, (2) by using the symbolic functionfourier, and (3) samplex(t)and find

its Fourier transform (this requires sampling theory—see Chapter 7).

The following script is used to compute and plot the signalx 2 (t)=e−tu(t)and the magnitude and

phase of its Fourier transform using symbolic MATLAB. A similar script is used forx 1 (t).

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% Example 5.11

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symm t

x2 = exp(−t)∗heaviside(t);

X2 = fourier(x2)

X2m = sqrt((real(X2))ˆ2 + (imag(X2))ˆ2; % magnitude

X2p = imag(log(X2); % phase

Notice the way that the magnitude and the phase are computed. To compute the magnitude we

use

|X 2 ()|=

√

Re[X 2 ()]^2 +Im[X 2 ()]^2.

The computation of the phase is complicated by the lack of the functionatan2in symbolic MAT-

LAB, which extends the principal values of the inverse tangent to(−π,π] by considering the sign

of the real part. The phase computation can be done by using thelogfunction; indeed

log(X 2 ())=log

(

|X 2 ()|ej∠X^2 ()

)

=log(|X 2 ()|)+j∠X 2 ()

so that

∠X 2 ()=Im[log(X 2 ())]

Analytically, the phase of the Fourier transform ofx 1 (t)=u(t)−u(t− 1 ) can be found by

considering the advanced signalz(t)=x 1 (t+0.5)=u(t+0.5)−u(t−0.5)with Fourier transform

Z()=

sin(/ 2 )

/ 2

Given thatZ()is real, its phase is either zero whenZ()≥0 and±πwhenZ() <0 (using these

values so that the phase is an odd function of). Sincez(t)=x 1 (t+0.5), thenZ()=X 1 ()ej0.5,

so that

X 1 ()=e−j0.5Z()

and as such

∠X 1 ()=∠Z()−0.5=

{

−0.5 Z()≥ 0

±π−0.5 Z() < 0