Signals and Systems - Electrical Engineering

332 CHAPTER 5: Frequency Analysis: The Fourier Transform

multiplied by a constant. This is one more example of the inverse relationship between time and

frequency. In this case, the support of the result of the windowing is finite, while the convolu-

tion in the frequency domain gives an infinite support forY()given thatW()has an infinite

support. n

5.7.2 Ideal Filters

Frequency-discriminating filters that keep low-, middle-, and high-frequency components, or a

combination of these, are calledlow-pass, band-pass, high-pass, andmultibandfilters, respectively. A

band-eliminatingornotchfilter gets rid of middle-frequency components. It is also possible to have an

all-passfilter that although it does not filter out any of the input frequency components, it changes

the phase of the input signal.

The magnitude frequency response of an ideal low-pass filter is given by

|Hlp(j)|=

{

1 − 1 ≤≤ 1

0 otherwise

and the phase frequency response of this filter is

∠Hlp(j)=−α

which as a function ofis a straight line with slope−α, thus its termlinear phase. The frequency 1

is called thecut-off frequencyof the low-pass filter. The above magnitude and phase responses only

need to be given for positive frequencies, given that the magnitude and the phase responses are the

even and odd function of. The rest of the frequency response is obtained by symmetry.

An ideal band-pass filter has a magnitude response

|Hbp(j)|=

{

1  1 ≤≤ 2 and − 2 ≤≤− 1

0 otherwise

with cut-off frequencies 1 and 2. The magnitude response of an ideal high-pass filter is given by

|Hhp(j)|=

{

1 ≥ 2 and ≤− 2

0 otherwise

with a cut-off frequency of 2. For both of these filters, i is assumed the phase is linear in the pass-

band (band of frequencies where the magnitude is unity).

From these definitions, we have that the ideal band-stop filter has as magnitude response of

|Hbs(j)|= 1 −|Hbp(j)|

The sum of the magnitude responses of the given low-, band-, and high-pass filters gives the

magnitude response of an ideal all-pass filter

|Hap(j)|=|Hlp(j)|+|Hbp(j)|+|Hhp(j)|= 1