Discrete Time Rect Function(4B)

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Transcription:

Discrete Time Rect Function(4B) Discrete Time Rect Functions

Copyright (c) 29-23 Young W. Lim. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version.2 or any later version published by the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts. A copy of the license is included in the section entitled "GNU Free Documentation License". Please send corrections (or suggestions) to youngwlim@hotmail.com. This document was produced by using OpenOffice and Octave.

Fourier Transform Types Discrete Time Fourier Series DTFS X [ k ] = N n = x[n] e j (2π/ N ) k n x[n] = k = + j(2π/ N )k n X [ k ] e Discrete Fourier Transform DFT X [k] = n = x [n] e j 2 / N k n x [n] = N k = j 2 / N k n X [k ] e Discrete Time Fourier Transform DTFT X e j = n = x[n] e j n x[n] = +π 2 π π X (e j ω ) e + j ω n 3

DTFS and DTFT L = 2 N + N L/ N N (L ) zero crossings N +N k DTFS (Discrete Time Fourier Series) X [k ] = N sin(π L k / N ) sin(π k / N ) = L N drcl (k / N, L) L = 2 N + ( L ) zero crossings L 2 π L 2π N +N ω DTFT (Discrete Time Fourier Transform) X (e j ω ) = sin( ω L/2) sin( ω/2) = L diric( ω, L) = L D L (e j ω ) 4

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DTFS Discrete Time Fourier Series DTFS X [ k ] = N n = x[n] e j (2π/ N ) k n x[n] = k = + j(2π/ N )k n X [ k ] e L = 2 N + N L/ N N (L ) zero crossings N +N k DTFS (Discrete Time Fourier Series) X [k ] = N sin(π L k / N ) sin(π k / N ) = L N drcl (k / N, L) 6

Rect N [n] * δ N [n] DTFS () Discrete Time Fourier Series DTFS X [ k ] = N n = x[n] e j (2π/ N ) k n x[n] = k = + j(2π/ N )k n X [ k ] e X [k ] = N x[n]e j(2π/ N )k n N n= = +N N n= N x[n]e j(2π/ N ) k n N X [k ] = e + j(2 π N / N ) k + + e j(2 π N / N ) k = e + j (2π/N ) N k e j(2π/ N )( 2N +) k e j(2π/ N ) k L = 2 N + N j(m)(2 N +) k = e + j (m) N k e m = (2π/ N e j( m)k ) j(m)( 2N +) k /2 = e + j (m) N k e e+ j(m)(2 N +) k/2 e j(m)(2 N+) k /2 e j(m) k /2 e + j(m) k/2 j(m) k/2 e = sin((m)(2 N +)k /2) sin((m)k /2) X [k ] = N sin((2 π/ N )(2 N +) k /2) sin((2π/ N ) k /2) Dirichlet Function N +N drcl(t, L) = sin(π Lt ) Lsin(π t) 7

Rect N [n] * δ N [n] DTFS (2) Discrete Time Fourier Series DTFS X [ k ] = N n = x[n] e j (2π/ N ) k n x[n] = k = + j(2π/ N )k n X [ k ] e X [k ] = N sin((2 π/ N )(2 N +) k /2) sin((2π/ N ) k /2) drcl (k / N, (2 N +)) = sin (π k (2 N +)/ N ) (2 N +)sin (π k / N ) = N sin (π k (2 N +)/ N ) sin(π k / N ) X [k ] = (2 N +) N drcl (k / N, (2 N +)) X [k ] = N sin(π k L/ N ) sin(π k / N ) X [k ] = L N drcl (k / N, L) L = 2 N + Dirichlet Function N drcl(t, L) = sin(π Lt ) Lsin(π t) N +N D L (e j ω ) = sin( ω L/2) Lsin( ω/2) 8

Rect N [n] * δ N [n] DTFS (3) Discrete Time Fourier Series DTFS X [ k ] = N n = x[n] e j (2π/ N ) k n x[n] = k = + j(2π/ N )k n X [ k ] e X [k ] = N sin(π k L/ N ) sin(π k / N ) Period : N (odd L), 2N (even L) X [k ] = L N drcl (k / N, L) L = 2 N + (L-) zero crossings Dirichlet Function N drcl(t, L) = sin(π Lt ) Lsin(π t) N +N D L (e j ω ) = sin( ω L/2) Lsin( ω/2) 9

Rect N [n] * δ N [n] DTFS (4) t = 2 t = t = t =+ t =+2 t = 2 t = t = t =+ t =+2 odd L=9 even L= 9 zero crossings 8 zero crossings k= 32 k= 6 k= k=+6 k =+32 k= 32 k= 6 k= k=+6 k=+32 (L-) zero crossings (L-) zero crossings Dirichlet Function drcl (t, L) = sin(π L t) L sin(π t) X [k ] = 9 drcl (k /6, 9) 6 3, 2,,, +, +2, +3, L

Rect N [n] * δ N [n] DTFS (5) Period : N (odd L), 2N (even L) k= 32 k= 6 k= k=+6 k=+32 (L-) zero crossings X [k ] = 9 6 drcl (k /6, 9)

Rect 2 [n] * δ 8 [n] DTFS Example Discrete Time Fourier Series DTFS X [ k ] = N n = x[n] e j (2π/ N ) k n x[n] = k = + j(2π/ N )k n X [ k ] e X [k ] = N sin(π k (2 N +)/ N ) sin(π k / N ) X [k ] = 8 sin(π k 5/8) sin(π k /8) X [k ] = L N drcl (k / N, L) X [k ] = 5 drcl (k /8, 5) 8 N =8 L = 5 (N = 2) Period : N = 8 (odd L = 5) (L ) = 4 zero crossings Dirichlet Function L = 2 N + N =8 drcl(t, L) = sin(π Lt ) Lsin(π t) N +N D L (e j ω ) = sin( ω L/2) Lsin( ω/2) 2

Rect 3 [n] * δ 6 [n] DTFS Example Discrete Time Fourier Series DTFS X [ k ] = N n = x[n] e j (2π/ N ) k n x[n] = k = + j(2π/ N )k n X [ k ] e X [k ] = N sin(π k (2 N +)/ N ) sin(π k / N ) X [k ] = L N drcl (k / N, L) Rect 3 [n] δ 6 [n] X [k ] = 6 sin (π k 7/6) sin(π k /6) X [k ] = 7 drcl (k /6, 7) 6 Period : N = 6 (odd L = 7) (L ) = 6 zero crossings N =6 L = 7 (N = 3) Dirichlet Function L = 2 N + N =6 drcl(t, L) = sin(π Lt ) Lsin(π t) N +N D L (e j ω ) = sin( ω L/2) Lsin( ω/2) 3

7/6 drcl(k/6, 7).5 L N = 7 6 N = 6 6 sin(π7 k /6) sin(π k /6).4 = 7 6 sin(π7 k /6) 7sin(π k /6).3 L = 7 = 7 6 drcl ( k 6, 7 ).2. N L = 6 7 6 zero crossings 6 7 2 6 7 4 Zeros 7 6 k 6 = 7 m k = 6 7 m 6 -. 6 7 6 7 3 6 7 5 -.2-2 -5 - -5 5 5 2 k 4

Phase of 7/6 drcl(k/6, 7) 3.5 3 π π = π 6 = 7 6 sin(π7 k /6) sin(π k /6) sin(π7 k /6) 7sin(π k /6) 2.5 = 7 6 drcl ( k 6, 7 ) 2.5.5 -.5-2 -5 - -5 5 5 2 5

Magnitude Response N = 6.45.4.35.3.25 6 = 7 6 sin(π7 k /6) sin(π k /6) sin(π7 k /6) 7sin(π k /6) = 7 6 drcl ( k 6, 7 ).2.5..5-2 -5 - -5 5 5 2 k 6

Phase Response () N = 6 3.5 3 2.5 2 6 = 7 6 sin(π7 k /6) sin(π k /6) sin(π7 k /6) 7sin(π k /6) = 7 6 drcl ( k 6, 7 ).5.5-2 -5 - -5 5 5 2 k 7

Phase Response (2) N = 6 4 3 2 6 = 7 6 sin(π7 k /6) sin(π k /6) sin(π7 k /6) 7sin(π k /6) = 7 6 drcl ( k 6, 7 ) - -2-3 -4-2 -5 - -5 5 5 2 k 8

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Rect N [n-n] * δ N [n] DTFS () Discrete Time Fourier Series DTFS X [ k ] = N n = x[n] e j (2π/ N ) k n x[n] = k = + j(2π/ N )k n X [ k ] e L = 2 N + N Dirichlet Function N +N drcl(t, L) = sin(π Lt ) Lsin(π t) 2

2

DFT Discrete Fourier Transform DFT X [k] = n = x [n] e j 2 / N k n x [n] = N k = j 2 / N k n X [k ] e L = 2 N + N L N (L ) zero crossings N +N k DFT (Discrete Fourier Transform) X [k ] = sin(π L k / N ) sin(π k / N ) = L drcl (k / N, L) 22

Rect N [n] * δ N [n] DFT Discrete Fourier Transform X [k] = n = x [n] e j 2 / N k n x [n] = N k = j 2 / N k n X [k ] e X [k ] = sin((2π/ N )(2 N +)k /2) sin ((2π/ N )k /2) = sin(π k / N (2 N +)) sin(π k / N ) = sin(π k / N L) sin(π k / N ) drcl(k / N, (2 N +)) = sin (π k / N (2 N +)) (2 N +)sin (π k / N ) X [k ] = (2 N +) drcl (k / N, (2 N +)) = L drcl(k / N, L) Dirichlet Function L = 2 N + N drcl(t, L) = sin(π Lt ) Lsin(π t) N +N D L (e j ω ) = sin( ω L/2) Lsin( ω/2) 23

Rect N [n] * δ N [n] DTFS & DFT Discrete Time Fourier Series DTFS X [ k ] = N n = x[n] e j (2π/ N ) k n x[n] = k = + j(2π/ N )k n X [ k ] e X [k ] = N sin(π k L/ N ) sin(π k / N ) X [k ] = L N drcl (k / N, L) Discrete Fourier Transform DFT X [k] = n = x [n] e j 2 / N k n x [n] = N k = j 2 / N k n X [k ] e X [k ] = sin(π k L/ N ) sin(π k / N ) X [k ] = L drcl (k / N, L) 24

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DTFT Discrete Time Fourier Transform DTFT X e j = n = x[n] e j n x[n] = +π 2 π π X (e j ω ) e + j ω n L = 2 N + ( L ) zero crossings L 2 π L 2π N +N ω DTFT (Discrete Time Fourier Transform) X (e j ω ) = sin( ω L/2) sin( ω/2) = L diric( ω, L) = L D L (e j ω ) 26

Rect N [n] DTFT Discrete Time Fourier Transform DTFT X e j = n = x[n] e j n x[n] = +π 2 π π X (e j ω ) e + j ω n X (e j ω ) = +N n= N e j ωn x[n] = {e + j ω N + + e j ω N } = e + j ω N { + + e j ω 2 N } j ω(2 N+) = e + j ω N e e j ω j ω(2 N+)/2 = e + j ω N e e j ω/2 e + j ω( 2N +)/2 e j ω(2 N+)/2 e + j ω/2 e j ω/2 = e+ j ω(2 N +)/2 j ω(2n +)/2 e = e + j ω/2 e j ω/2 X (e j ω ) = sin( ω L/2) sin( ω/2) = L diric( ω, L) sin( ω(2 N +)/2) sin( ω/2) = L D L (e j ω ) L = 2 N + Dirichlet Function N +N D L (e j ω ) = sin( ω L/2) Lsin( ω/2) 27

Dirichlet Functions D 9 (e j ω ) 2 π D 9 (e j ω ) = sin( ω9/2) 9sin ( ω/2) 8 zero crossings D (e j ω ) = sin ( ω/2) sin( ω/2) zero crossings 8 zero crossings D 3 (e j ω ) = sin ( ω3/2) 3sin( ω/2) 2 zero crossings D (e j ω ) 2 π D (e j ω ) = sin( ω/2) sin ( ω/2) 9 zero crossings 9 zero crossings D 2 (e j ω ) = D 4 (e j ω ) = sin ( ω2/2) 2sin( ω/2) sin( ω4/2) 4 sin ( ω/2) zero crossings 3 zero crossings 28

L D L (e jω ) () 7 6 L = 7 2 π X (e j ω ) = sin( ω L/2) sin( ω/2) 5 = L D L (e j ω ) 4 = L diric( ω, L) 3 L = 7 2 2 π L = 2 π 7 6 zero crossings 2 π 7 2 2 π 7 4 2π 7 6-2 π 7 2π 7 3 2π 7 5-2 - -8-6 -4-2 2 4 6 8 29

Magnitude Response of L D L (e jω ) 7 6 X (e j ω ) = sin( ω L/2) sin( ω/2) = L D L (e j ω ) 5 = L diric( ω, L) 4 3 2 - -8-6 -4-2 2 4 6 8 3

Phase Response of L D L (e jω ) 3.5 3 X (e j ω ) = sin( ω L/2) sin( ω/2) = L D L (e j ω ) 2.5 = L diric( ω, L) 2.5.5 - -8-6 -4-2 2 4 6 8 3

32

Rect N [n-n] DTFT Discrete Time Fourier Transform DTFT X e j = n = x[n] e j n x[n] = +π 2 π π X (e j ω ) e + j ω n L = 2 N + Dirichlet Function N +N D L (e j ω ) = sin( ω L/2) Lsin( ω/2) 33

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References [] http://en.wikipedia.org/ [2] J.H. McClellan, et al., Signal Processing First, Pearson Prentice Hall, 23 [3] G. Beale, http://teal.gmu.edu/~gbeale/ece_22/fourier_series_2.html [4] C. Langton, http://www.complextoreal.com/chapters/fft.pdf [5] M. J. Roberts, Fundamentals of Signals and Systems