solved16

# solved16 - For x4 =[s z4 X4 =[X[0 X[2 X[4 X[6 X[8 X[10 X[12...

This preview shows pages 1–2. Sign up to view the full content.

S 16.1 (P 3.26) _______________ (i) x is obtained from y by zero-padding to double its length. Thus Y = [ X0 X2 X4 X6 X8 X10 ].' (ii) x is also obtained from [a b c].' by zero-padding to four times its length. Thus [ a b c ].' <--> [ X0 X4 X8 ].' S is a periodic extension of [a b c].' to four times its length. Thus S = 4*[ X0 0 0 0 X4 0 0 0 X8 0 0 0 ].' S 16.2 (P 3.27) _______________ For x1=s : X1 = [X[0] X[4] X[8] X[12]].' (since the zero-padded vector has length = 4*length of s). For x2 = [s ; s] : X2 = [2*X[0] 0 2*X[4] 0 2*X[8] 0 2*X[12] 0].' (periodic extension of s to twice its length) For x3 = [s ; s ; s ; s ; s] : X3 = [5*X[0]; z4 ; 5*X[4] ; z4 ; 5*X[8] ; z4 ; 5*X[12] ; z4]

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: For x4 = [s; z4] : X4 = [X[0] X[2] X[4] X[6] X[8] X[10] X[12] X[14]].' (since x has exactly twice as many entries as [s;z4]) For x5 = [z4 ; s] : X4 = [X[0] -X[2] X[4] -X[6] X[8] -X[10] X[12] -X[14]].' (circular shift of x4 by 4=8/2 units, so multiplication by (-1)^k in the frequency domain) For x6 = [s ; z4 ; s ; z4] : X6 = [2*X[0] 0 2*X[2] 0 2*X[4] 0 2*X[6] 0 . .. 2*X[8] 0 2*X[10] 0 2*X[12] 0 2*X[14] 0 ].' (periodic extension of [s;z4] to twice its length) For x7 = [s ; s ; z4 ; z4] : x7 = x + (P^4)*x and thus X7 = X + F^(-4)*X Since 4*(2*pi/16) = pi/2, we have X7[k] = (1+(-j)^k)*X[k] for every k (k=0:15) ....
View Full Document

## This note was uploaded on 10/18/2011 for the course ENEE 241 taught by Professor Staff during the Spring '08 term at Maryland.

### Page1 / 2

solved16 - For x4 =[s z4 X4 =[X[0 X[2 X[4 X[6 X[8 X[10 X[12...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online