Signals and Systems - Electrical Engineering

318 CHAPTER 5: Frequency Analysis: The Fourier Transform

Remarks

n When plotting|X()|versus, which we call the Fourier magnitude spectrum, for a periodic signal

x(t), we notice it is analogous to its line spectrum discussed before. Both indicate that the signal power is

concentrated in multiples of the fundamental frequency, the only difference being in how the information

is provided at each of the frequencies. The line spectrum displays the Fourier series coefficients at their

corresponding frequencies, while the spectrum from the Fourier transform displays the concentration of the

power at the harmonic frequencies by means of delta functions with amplitudes of 2 πtimes the Fourier

series coefficients. Thus, there is a clear relation between these two spectra, showing exactly the same

information in slightly different form.

n The Fourier transform of a cosine signal can now be computed directly as

F[cos( 0 t)]=F[0.5ej^0 t+0.5e−j^0 t]

=πδ(− 0 )+πδ(+ 0 )

and for a sine (compare this result with the one obtained before),

F[sin( 0 t)]=F

[

0.5

j

ej^0 t−

0.5

j

e−j^0 t

]

=

π

j

δ(− 0 )−

π

j

δ(+ 0 )

=πe−jπ/^2 δ(− 0 )+πejπ/^2 δ(+ 0 )

The magnitude spectra of the two signals coincide, but the cosine has a zero-phase spectrum, while the

phase spectrum for the sine displays a phase of±π/ 2 at frequencies± 0.

nExample 5.8

Consider a periodic signalx(t)with a period

x 1 (t)=r(t)− 2 r(t−0.5)+r(t− 1 )

If the fundamental frequency is 0 = 2 π, determine the Fourier transformX()analytically and

using MATLAB. Plot several periods of the signal and its Fourier transform.

Solution

The given periodx 1 (t)corresponds to a triangular signal. Its Laplace transform is

X 1 (s)=

1

s^2

(

1 − 2 e−0.5s+e−s

)

=

e−0.5s

s^2

(

e0.5s− 2 +e−0.5s

)