Unformatted text preview: 0.7304 sin wT)2
V 1.533 - 1.4608 cos w T L H[eiWT]=_tan- 1 ( 0 .7304sinwT ) (12.52)
T herefore 1 - 0.7304 cos w T
S ubstituting T = 7r X IH a (J.W )1 1 0-6 i n t he a bove e quations, we o btain jH[eiWT]j = O .h
V 1.533 - 1.4608cos(7r LH[eiwT] = - tan- 1 X 1O- 6 w) ( 12.51a) 6 [ 0 .7304sin(7r X 1O- w) ]
1 - 0.7304 cos (7r X 1O- 6 w) (12.51b) = 105 Vw2 + 1010 a nd L Ha(jw) = _ tan- 1 1~5 F igures 1 2.11b a nd 12.11c show t he a mplitude a nd p hase r esponse o f t he a nalog a nd t he
( normalized)t d igital f ilter over t he f requency r ange 0 ::; w ::; 7r/T = 10 • O bserve t hat
t The frequency response H[e iwT ] is higher t han t he desired frequency response Ha(jw) because o f
aliasing. We c an p artially correct this difference b y mUltiplying H[z] w ith a normalizing constant,
forcing t he resulting H[eiwT ] t o b e equal t o Ha(jw) a t w = O. T he normalizing constant K is
defined as t he r atio of Ha(jO) t o H[e JO ] = H[I], which in this case is 1/1.653 = 0.858. T he
normalized amplitude response in Fig. 12.11 b is t hat of 0.858 ( z !!Ol;;04) . 4
740 12 Frequency Response a nd D igital F ilters t he behavior of the analog and the digital filter is very close over the range W :S W e = 105 .
However, for higher frequencies, there is considerable aliasing, especially in the phase
• o C omputer E xample C 12.3
Using MATLAB, find t he impulse invariance digital filter to realize the first-order
analog Butterworth filter in Example 12.4.
The analog filter transfer function is 105/(8 + 105) and the sampling interval T =
10 - 6".. A suitable MATLAB function to solve this problem is ' impinvar'. The input
d ata are the coefficients of the numerator and the denominator polynomials of H a(8)
[entered as (n + I)-element vectors num and den] a nd the sampling interval T. MATLAB
returns b and a, the numerator and the denominator polynomial coefficients of the desired
digital filter H [z].
In designing impulse invariance filter, we use t he criterion h[k] = T ha(kT) in Eq.
(12.39), whereas most books, including MATLAB, use the criterion h[k] = h a(kT). Hence,
o ut answer will be T times the answer returned by MATLAB. To correct this discrepancy
we mUltiply num by T .
T =pi/lO A n um=T*[O WAS]; d en=[110A5];
MATLAB returns b=O. 3142 and a=1 -0.7304. Therefore
H [z] = 12.6 Recursive F ilter design: T he B ilinear Transformation M ethod a lower s ampling r ate c ompared t o t he i mpulse invariance m ethod b ecause o f t he
a bsence o f aliasing. I n a ddition, t he filter rolloff characteristics a re s harper w ith
t his m ethod c ompared t o t hose o btained u sing t he i mpulse invariance m ethod. T he
a bsence o f a liasing is t he r esult of one-to-one m apping f rom S p lane t o z p lane
i nherent i n t his m ethod.
T he f requency-domain design criterion is [see Eq. (12.37)]
lim H[e sT ] = Ha(s) L et u s c onsider t he following power series for t he h yperbolic t angent (see Sec. B.7-3)
sT 2 Design a digital filter t o realize a n a nalog transfer function Therefore, as T -t 0 'V Limitations o f t he Impulse Invariance...
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