Determine the constraint on r = \z\ for each of the following sums to converge: 10.2. (a) ~ (~)n+1 -n (b) ~(~)-n+lzn n=-1 n=l (c) ~{ l+(~l)"}z-n (d) ~ (4)1n1 cos(in)z-n
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Textbook Solutions for Signals and Systems
Question
The bilinear transformation, introduced in the previous problem, may also be used to obtain a discrete-time filter, the magnitude of whose frequency response is similar to the magnitude of the frequency response of a given continuous-time lowpass filter. In this problem, we illustrate the similarity through the example of a continuous-time second-order Butterworth filter with system function Hc(s). (a) Let Hd(Z) = Hc(s)Js= 1-z-1 l+z-1 Show that (b) Given that 1 Hc(s) = 14 14 (s + e/rr )(s + e- j1r ) and that the corresponding filter is causal, verify that Hc(O) = 1, that IHc(Jw )J decreases monotonically with increasing positive values of w, that IHc(j)J 2 = 112 (i.e., that We = 1 is the half-power frequency), and that Hc(oo) = 0. (c) Show that if the bilinear transformation is applied to Hc(s) of part (b) in order to obtain Hd(Z), then the following may be asserted about Hd(Z) and Hd(ejw): 1. Hd(Z) has only two poles, both of which are inside the unit circle. 2. Hd(ej0 ) = 1. 3. IHd(ejw)l decreases monotonically as w goes from 0 to TT. 4. The half-power frequency of Hd(ejw) is TT/2.
Solution
The first step in solving 10 problem number 66 trying to solve the problem we have to refer to the textbook question: The bilinear transformation, introduced in the previous problem, may also be used to obtain a discrete-time filter, the magnitude of whose frequency response is similar to the magnitude of the frequency response of a given continuous-time lowpass filter. In this problem, we illustrate the similarity through the example of a continuous-time second-order Butterworth filter with system function Hc(s). (a) Let Hd(Z) = Hc(s)Js= 1-z-1 l+z-1 Show that (b) Given that 1 Hc(s) = 14 14 (s + e/rr )(s + e- j1r ) and that the corresponding filter is causal, verify that Hc(O) = 1, that IHc(Jw )J decreases monotonically with increasing positive values of w, that IHc(j)J 2 = 112 (i.e., that We = 1 is the half-power frequency), and that Hc(oo) = 0. (c) Show that if the bilinear transformation is applied to Hc(s) of part (b) in order to obtain Hd(Z), then the following may be asserted about Hd(Z) and Hd(ejw): 1. Hd(Z) has only two poles, both of which are inside the unit circle. 2. Hd(ej0 ) = 1. 3. IHd(ejw)l decreases monotonically as w goes from 0 to TT. 4. The half-power frequency of Hd(ejw) is TT/2.
From the textbook chapter The z-Transform you will find a few key concepts needed to solve this.
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