Signals and Systems - Electrical Engineering

330 CHAPTER 5: Frequency Analysis: The Fourier Transform

the desirable frequency band or bands, and let it be close to zero in those frequencies we would not

want in the output signal.

If the input signalx(t)is periodic of periodT 0 , or fundamental frequency 0 = 2 π/T 0 , the Fourier

transform of the output is

Y()=X()H(j)

= 2 π

∑

k

XkH(jk 0 )δ(−k 0 ) (5.21)

where the magnitude and the phase of each of the Fourier series coefficients are changed by the fre-

quency response of the filter at the harmonic frequencies. Indeed,Xkcorresponding to the frequency

k 0 is changed into

XkH(jk 0 )=|Xk||H(jk 0 )|ej(∠Xk+∠H(jk^0 ))

The filter outputy(t)is also periodic of periodT 0 but is missing the harmonics of the input that have

been filtered out.

The above shows that independent of whether the input signalx(t)is periodic or aperiodic, the output

signaly(t)has the frequency components allowed through by the filter.

nExample 5.13

Consider how to obtain a dc source using a full-wave rectifier and a low-pass filter (it keeps only

the low-frequency components). Let the full-wave rectified signalx(t)be the input of the filter

and let the output of the filter bey(t). We wanty(t)=1 volt. The rectifier and the low-pass filter

constitute a system that converts alternating into direct voltage.

Solution

We found in Chapter 4 that the Fourier series coefficients ofx(t)=|cos(πt)|,−∞<t<∞, are

given by

Xk=

2 (− 1 )k

π( 1 − 4 k^2 )

so that the average ofx(t)isX 0 = 2 /π. To filter out all the harmonics and leave only the average

component, we need an ideal low-pass filter with a magnitudeAand a cut-off frequency 0<

c<  0 where 0 = 2 π/T 0 = 2 πis the fundamental frequency ofx(t). Thus, the filter is given by

H(j)=Afor− 0 < c<  0 and zero otherwise. According to the convolution property, then

Y()=H(j)X()=H(j)



 2 πX 0 δ()+

∑

k6= 0

2 πXkδ(−k 0 )





= 2 πAX 0 δ()