Signal Processing and Linear Systems-B.P.Lathi copy

# 1 22 sinusoidal input and t he corresponding o utput

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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 hird-order lowpass filter w ith t hree poles n ear z = 1 a nd a t hird-order 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 pole-zero configuration o f t he t hird-order h ighpass filter w ith c orresponding a mplitude a nd p hase response. • E xample 1 2.2: B andpass F ilter Using trial-and-error, 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 ole-Zero 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.

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