Unformatted text preview: ear t he
p oint z = 1 (Fig. 12.4a) would result in a lowpass response. T he c orresponding
a mplitude a nd p hase response a ppears in Fig. 12.4a. For smaller values of w , t he
p oint e jwT ( a p oint o n t he u nit circle a t a n angle w T) is closer t o t he pole, a nd consequently t he g ain is higher. As w increases, t he d istance of t he p oint e jwT from t he
pole increases. Consequently t he g ain decreases, resulting i n a lowpass c haracteristic. Placing a zero a t t he o rigin does n ot c hange t he a mplitude r esponse b ut i t does
modify t he p hase response, as i llustrated i n Fig. 12.4b. Placing a zero a t z =  1,
however, changes b oth t he a mplitude a nd p hase response (Fig. 12.4c). T he p oint
z =  1 c orresponds t o f requency w = 1 f IT (z = e jwT = e j1r =  1). Consequently,
t he a mplitude r esponse now becomes more a ttenuated a t h igher frequencies, w ith
a zero gain a t w T = 1 f. W e c an a pproach ideal lowpass characteristics b y using
more poles staggered n ear z = 1 ( but w ithin t he u nit circle). Figure 12.4d shows
a t hirdorder lowpass filter w ith t hree poles n ear z = 1 a nd a t hirdorder zero a t
z =  1, w ith c orresponding a mplitude a nd p hase response. For a n i deal lowpass
filter we need a n e nhanced g ain a t every frequency i n t he b and (0, we). T his c an
b e achieved by placing a continuous wall of poles (requiring a n i nfinite n umber o f
poles) opposite t his b and.
e jOT Highpass Filters
A h igh p ass filter has a small gain a t lower frequencies a nd a h igh gain a t
h igher frequencies. S uch a c haracteristic c an b e r ealized b y p lacing a pole or poles
n ear z =  1 b ecause we w ant t he g ain a t w T = 1 f t o b e t he h ighest. Placing a
zero a t z = 1 f urther e nhances suppression of gain a t lower frequencies. F igure
12.4e shows a possible polezero configuration o f t he t hirdorder h ighpass filter w ith
c orresponding a mplitude a nd p hase response.
• E xample 1 2.2: B andpass F ilter Using trialanderror, design a tuned (bandpass) filter with zero transmission a t 0
Hz and also a t 500 Hz. T he resonant frequency is required to be 125 Hz. T he highest
frequency to be processed is Fh = 500 Hz.
Because Fh = 500, we require T :0: lO~O [see Eq. (8.17)]. Let us select T = 1 0 3 .
Since the amplitude response is zero a t w = 0 and w = 10001f, we need to place zeros at
eiwT corresponding to w = 0 and w = 10001f. For w = 0, z = e jwT = 1; for w = 10001f
(with T = 1 0 3 ), e jwT =  1. Hence, there must be zeros a t z = ± l. Moreover, we need
enhanced frequency response a t w = 2501f. This frequency (with w T = 1f/4) corresponds
to z = e jwT = ei1r / 4 . Therefore, to enhance the frequency response a t this frequency, we 726 12 F requency R esponse a nd D igital F ilters 111 12.2 727 F requency R esponse F rom P oleZero L ocation F igure 1 2.5b shows t he a mplitude r esponse for values o f bl = 0.83, 0.96, a nd 1. As
expected, t he g ain is zero a t w = 0 a nd a t 500 Hz (w = 100011"). T he g ain p eaks a t a bout
125 Hz (w = 25011"). T h...
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This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.
 Spring '13
 Bayliss
 Signal Processing, The Land

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