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454 Chapter 8 IIR FILTER DESIGN
>> % Digital Highpass Filter Cutoff frequency:
>> wphp = 0.6*pi; % Passband edge frequency
>> % LP-to-HP frequency-band transformation:
>> alpha = -(cos((wplp+wphp)/2))/(cos((wplp-wphp)/2))
alpha = -0.3820
>> Nz = -[alpha,1]; Dz = [1,alpha];
>> [bhp,ahp] = zmapping(blp,alp,Nz,Dz); [C,B,A] = dir2cas(bhp,ahp)
C = 0.0243
B = 1.0000 -2.0000 1.0000
1.0000 -2.0000 1.0000
A = 1.0000 1.0416 0.4019
1.0000 0.5561 0.7647
The system function of the highpass filter is
H(z) =
0.0243(1 − z −1 ) 4
(1+0.5661z −1 +0.7647z −2 )(1+1.0416z −1 +0.4019z −2 )
which is essentially identical to that in Example 8.25.
□
8.6.1 DESIGN PROCEDURE
In Example 8.26 a lowpass prototype digital filter was available to transform
into a highpass filter so that a particular band-edge frequency was
properly mapped. In practice we have to first design a prototype lowpass
digital filter whose specifications should be obtained from specifications
of other frequency-selective filters as given in Figure 8.30. We will now
show that the lowpass prototype filter specifications can be obtained from
the transformation formulas given in Table 8.2.
Let us use the highpass filter of Example 8.25 as an example. The
passband-edge frequencies were transformed using the parameter α =
−0.38197 in (8.70). What is the stopband-edge frequency of the highpass
filter, say ω s , corresponding to the stopband edge ω ′ s =0.3π of the prototype
lowpass filter? This can be answered by (8.70). Since α is fixed for
the transformation, we set the equation
α = − cos[(0.3π + ω s)/2]
cos[(0.3π − ω s )/2] = −0.38197
This is a transcendental equation whose solution can be obtained iteratively
from an initial guess. It can be done using MATLAB, and the
solution is
ω s =0.4586π
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