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FREQUENCY RESPONSE, FILTERS, AND RESONANCE
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EXAMPLE 12.5 Three network functions H1 , H2 , and H3 are given by a H1 1 s 1 b H2 s2 1 p 2s 1 c H3 1 1 s3 2s2 2s 1 s 1 s2 s 1
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Find the magnitudes of their frequency responses. frequency at !0 1. a b c jH1 j2
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Show that all three functions are low-pass with half-power
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1 1 1 j! 1 j! 1 !2 1 1 p p jH2 j2 1 !2 j 2! 1 !2 j 2! 1 !4 1 1 jH3 j2 1 !2 1 !2 j! 1 !2 j! 1 !6
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For all three functions, at ! 0, 1, and 1, we have jHj2 1, 1/2, and 0, respectively. Therefore, the three network functions are low-pass with the same half-power frequency of !0 1. They are rst-, second-, and third-order Butterworth lters, respectively. The higher the order of the lter, the sharper is the cuto region in the frequency response.
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PASSIVE AND ACTIVE FILTERS
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Filters which contain only resistors, inductors, and capacitors are called passive. Those containing additional dependent sources are called active. Passive lters do not require external energy sources and they can last longer. Active lters are generally made of RC circuits and ampli ers. The circuit in Fig. 12-16(a) shows a second-order low-pass passive lter. The circuit in Fig. 12-16(b) shows an active lter with a frequency response V2 =V1 equivalent to that of the circuit in Fig. 12-16(a).
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EXAMPLE 12.6 Find the network function V2 =V1 in the circuits shown in (a) Fig. 12-16(a) and (b) Fig. 12-16(b). (a) In Fig. 12-16(a), we nd V2 from V1 by voltage division. 1 V1 V1 1 V1 Cs R Ls 1=Cs LCs2 RCs 1 LC s2 R=L s 1=LC p p Substituting for R 1, L 1= 2, and C 2, and dividing by V1 , we get V2 V2 1 2 p V1 s 2s 1 (b) In Fig. 12-16(b), we apply KCL at nodes A and B with VB V2 .
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Fig. 12-16
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FREQUENCY RESPONSE, FILTERS, AND RESONANCE p p VA V1 2 VA V2 2 VA V2 2s 0 p V2 s V2 VA 2 0 Thus,
Node A: Node B:
6a 6b
By eliminating VA in (6a) and (6b), the network function H s V2 =V1 is obtained. V2 1 p V1 s2 2s 1 Note that the circuits of Figs. 12-16(a) and (b) have identical network functions. worth low-pass lters with half-power frequencies at ! 1 rad/s.
They are second-order Butter-
BANDPASS FILTERS AND RESONANCE The following network function is called a bandpass function. H s ks s as b
where a > 0; b > 0; k > 0
The name is especially appropriate when the poles are complex, close to the j! axis, and away from the origin in the s-domain. The frequency response of the bandpass function is k2 !2 k2 2 8 2 b !2 a2 !2 a b !2 2 =!2 p The maximum of jHj occurs when b !2 0 or ! b, which is called the center frequency !0 . At the center frequency, we have jHjmax jH !0 j k=a. The half-power frequencies are at !l and !h , where H j! kj! b !2 aj! jHj2 jH !l j2 jH !h j2 1 jH !0 j2 2 By applying (8) to (9a), !l and !h are found to be roots of the following equation: b !2 2 a2 !2 p !l a2 =4 b a=2 q !h a2 =4 b a=2 9b (9c) 9d 9a
Solving,
From (9c) and (9d) we have !h !l a The bandwidth is de ned by !h !l a The quality factor Q is de ned by Q !0 = p b=a 10c 10b and !h !l b !2 0 10a
The quality factor measures the sharpness of the frequency response around the center frequency. This behavior is also called resonance (see Sections 12.11 to 12.15). When the quality factor is high, !l and !h may be approximated by !0 =2 and !0 =2, respectively.
EXAMPLE 12.7 Consider the network function H s 10s= s2 300s 106 . and upper half-power frequencies, the bandwidth, and the quality factor. Since !2 106 , the center frequency !0 1000 rad/s. 0 The lower and upper half-power frequencies are, respectively, Find the center frequency, lower
FREQUENCY RESPONSE, FILTERS, AND RESONANCE q q a2 =4 b a=2 3002 =4 106 300=2 861:2 rad=s q q !h a2 =4 b a=2 3002 =4 106 300=2 1161:2 rad=s !l The bandwidth !h !l 1161:2 861:2 300 rad/s. The quality factor Q 1000=300 3:3.
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