Signals and Systems - Electrical Engineering

334 CHAPTER 5: Frequency Analysis: The Fourier Transform

and stable filter with frequency response H(j)should satisfy the condition

∫∞

−∞

|log(H(j))|

1 +^2

d <∞ (5.22)

To satisfy this condition, H(j)cannot be zero in any band of frequencies, because in such cases the

numerator of the integrand would be infinite.The Paley-Wiener integral condition is clearly not

satisfied by ideal filters.So they cannot be implemented or used in actual situations, but they can be

used as models for designing filters.

n That ideal filters are not realizable can be understood also by considering what it means to make the

magnitude response of a filter zero in some frequency bands. A measure of attenuation is given by theloss

functionin decibels, defined as

α()=−10 log 10 |H(j)|^2

=−20 log 10 |H(j)|dB

Thus, when|H(j)|= 1 and there is no attenuation the loss is 0 dB, and when|H(j)|= 10 −^5 for a

large attenuation the loss is 100 dB. You quickly convince yourself that if a filter achieves a magnitude

response of 0 at any frequency this would mean a loss or attenuation at that frequency of∞dBs! Values

of 60 to 100 dB attenuation are considered extremely good, and to obtain that the signal needs to be

attenuated by a factor of 10 −^3 to 10 −^5. A curious termJNDor“just noticeable difference”is used by

experts in human hearing to characterize the smallest sound intensity that can be judged by a human

as different. Such a value varies from0.25to 1 dB. To illustrate what is loud in the dB scale, consider

the following cases: A sound pressure level higher than 130 dB causes pain; 110 dB is generated by an

amplified rock band performance [65].

nExample 5.15

The Gibb’s phenomenon, which we mentioned when discussing the Fourier series of periodic

signals with discontinuities, consists in ringing around these discontinuities. To see this, consider

a periodic train of square pulsesx(t)of periodT 0 displaying discontinuities atkT 0 /2, fork=

±1,±2,.... Show how the Gibb’s phenomenon is due to ideal low-pass filtering.

Solution

Choosing 2N+1 of the Fourier series coefficients to approximate the signalx(t)is equivalent to

passingx(t)through an ideal low-pass filter,

H(j)=

{

1 −c≤≤c

0 otherwise

having as impulse response a sinc functionh(t). If the Fourier transform of the periodic signalx(t)

of fundamental frequency 0 = 2 π/T 0 is

X()=

∑∞

k=−∞

2 πXkδ(−k 0 )