Ps7soln - MIT OpenCourseWare http/ocw.mit.edu 2.161 Signal Processing Continuous and Discrete Fall 2008 For information about citing these

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 . 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 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...
View Full Document

This note was uploaded on 02/27/2012 for the course MECHANICAL 2.161 taught by Professor Derekrowell during the Fall '08 term at MIT.

Page1 / 15

Ps7soln - MIT OpenCourseWare http/ocw.mit.edu 2.161 Signal Processing Continuous and Discrete Fall 2008 For information about citing these

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online