Signals and Systems - Electrical Engineering

686 C H A P T E R 11: Introduction to the Design of Discrete Filters

The desired impulse response is thus

hd[n]=

1

2 π

∫π

−π

Hd(ejω)ejωndω=

1

2 π

∫π/^4

−π/ 4

ejωndω

=

{

sin(πn/ 4 )/(πn) n6= 0

0.25 n= 0

Using a rectangular window, the FIR filter is then of the form (the delay is(N− 1 )/ 2 =10)

Hˆ(z)=Hw(z)z−^10 =

∑^20

n= 0

hd[n−10]z−n

=0.25z−^10 +

∑^20

n=0,n6= 10

sin(π(n− 10 )/ 4 )

π(n− 10 )

z−n

The magnitude and the phase of this filter are shown in Figure 11.23 when we use a rectangular

and a Hamming window.

The magnitude and the phase responses of the filter designed using the Hamming window is much

improved over the one obtained using the rectangular window. Notice that the second lobe in the

stopband for the Hamming window design is at about−50 dB while for the rectangular window

design it is at about−20 dB, a significant difference. In both cases, the phase response is linear

in the passband of the filter, corresponding to the impulse responseh[n] being symmetric with

respect to the(N− 1 )/ 2 =10 sample. n

nExample 11.13

Design a high-pass filter of order 14 and a cut-off frequency of 0.2πusing the Kaiser window. Use

MATLAB.

Solution

Lethlp[n] be the impulse response of an ideal low-pass filter:

Hlp(ejω)=

{

1 −ωc≤ω≤ωc

0 otherwise in [−π,π)

According to the modulation property of the DTFT, we have that

2 hlp[n] cos(ω 0 n) ⇔ Hlp(ej(ω+ω^0 ))+Hlp(ej(ω−ω^0 ))

If we letω 0 =π, then the right term gives a high-pass filter, and sohhp[n]= 2 hlp[n] cos(πn)=

2 (− 1 )nhlp[n] is the desired impulse response of the high-pass filter. The following script shows

how to use the functionfirto design the high-pass filter.