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# ps7soln - MIT OpenCourseWare http/ocw.mit.edu 2.161 Signal...

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MIT OpenCourseWare http://ocw.mit.edu 2.161 Signal Processing: Continuous and Discrete Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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summationdisplay summationdisplay parenleftBig parenrightBig parenleftBig parenrightBig parenleftBig parenrightBig MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING 2.161 Signal Processing - Continuous and Discrete Fall Term 2008 Solution of Problem Set 7 Assigned: October 30, 2008 Due: November 6, 2008 Problem 1: There are a number of ways to prove this – here are two: (a) Start with an acausal filter of the same length ( N 1) / 2 H a ( z ) = h n z n n = ( N 1) / 2 with N odd, and with real, odd symmetric coeﬃcients h n = h n so that the impulse response is a real, odd function. Since DTFT { h n } ⇐⇒ H ( e j Ω ) , from the properties of the DTFT, namely that a real, odd function has a imaginary, odd DTFT, then H a ( e j Ω ) is imaginary and odd. H a ( z ) may be made causal by adding a delay of ( N 1) / 2, that is H ( z ) = z ( N 1) / 2 H a ( z ) so that H ( e j Ω ) = e j Ω( N 1) / 2 H a ( e j Ω ) Since H a ( e j Ω ) is imaginary negationslash H ( e j Ω ) = negationslash je j Ω( N 1) / 2 = ( N 1) Ω ± π/ 2 2 which is linear with frequency Ω = ω Δ T (apart from possible jump discontinuities where H a ( e j Ω ) changes sign. Note that in practical filters, any jump in the phase response will occur outside the pass-band, and is generally ignored.) The equivalent delay is found from the slope, and represents a delay of ( N 1) / 2 samples. (b) Assume the causal filter and let N 1 H ( z ) = h n z n n =0 = h 0 z 0 + h 1 z 1 + . . . + h N 1 z ( N 1) with N odd, and with odd symmetric coeﬃcients about the mid-point, h n = h N 1 n . Group the symmetric components together H ( z ) = h 0 z 0 z ( N 1) + h 1 z 1 z ( N 2) + h 2 z 2 z ( N 2) + . . . + h ( N 1) / 2 z ( N 1) / 2 1
summationdisplay parenleftBig parenrightBig summationdisplay parenleftBig parenrightBig summationdisplay parenleftBig parenrightBig summationdisplay ( N 1) / 2 1 = h n z n z ( N 1 n ) + h ( N 1) / 2 z ( N 1) / 2 n =0 ( N 1) / 2 1 = z ( N 1) / 2 h n z ( N 1) / 2 n z (( N 1 / 2) n ) n =0 since, as noted in the problem statement, for an odd-symmetric function about the mid-point h ( N 1) / 2 = 0. The frequency response is H ( e j Ω ) = H ( z ) | z = e j Ω ( N 1) / 2 1 H ( e j Ω ) = e j Ω( N 1) / 2 h n e j Ω(( N 1) / 2 n ) e j Ω(( N 1 / 2) n ) n =0 ( N 1) / 2 1 = je j Ω( N 1) / 2 2 h n sin (Ω(( N 1) / 2 + n )) n =0 = je j Ω( N 1) / 2 H a ( e j Ω ) , The function H a ( e j Ω ) is purely real, therefore negationslash H ( e j Ω ) = negationslash je j Ω( N 1) / 2 = ( N 1) Ω ± π/ 2 2 which is linear with frequency Ω = ω Δ T (apart from possible jump discontinuities of π at sign changes in H a ( e j Ω ), which as noted above occur outside the pass-band) and represents a delay of ( N 1) / 2 samples.

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