File:Variations of the Fourier transform.tif

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DescriptionVariations of the Fourier transform.tif	English: Illustration of using Dirac comb functions and the convolution theorem to model the effects of sampling and/or periodic summation. At lower left is a DTFT, the spectral result of sampling s(t) at intervals of T. The spectral sequences at (a) upper right and (b) lower right are respectively computed from (a) one cycle of the periodic summation of s(t) and (b) one cycle of the periodic summation of the s(nT) sequence. The respective formulas are (a) the Fourier series integral and (b) the DFT summation. The relative computational ease of the DFT sequence and the insight it gives into S(f) make it a popular analysis tool.
Date	13 December 2011
Source	Own work
Author	Bob K
Permission (Reusing this file)	I, the copyright holder of this work, hereby publish it under the following license:   This file is made available under the Creative Commons CC0 1.0 Universal Public Domain Dedication. The person who associated a work with this deed has dedicated the work to the public domain by waiving all of their rights to the work worldwide under copyright law, including all related and neighboring rights, to the extent allowed by law. You can copy, modify, distribute and perform the work, even for commercial purposes, all without asking permission. http://creativecommons.org/publicdomain/zero/1.0/deed.enCC0Creative Commons Zero, Public Domain Dedicationfalsefalse
Other versions	File:Fourier_transform,_Fourier_series,_DTFT,_DFT.svg, File:Fourier_transform,_Fourier_series,_DTFT,_DFT.gif
Source code InfoField	This TIF graphic was created with LibreOffice by Bob K.
Source code InfoField	LibreOffice code Source code pkg load signal graphics_toolkit gnuplot %======================================================= % Consider the Gaussian function e^{-B (nT)^2}, where B is proportional to bandwidth. T = 1; % Choose a relatively small bandwidth, so that one cycle of the DTFT approximates a true Fourier transform. B = 0.1; N = 1024; t = T*(-N/2 : N/2-1); % 1xN y = exp(-B*t.^2); % 1xN % The DTFT has a periodicity of 1/T=1. Sample it at intervals of 1/8N, and compute one full cycle. % Y = fftshift(abs(fft([y zeros(1,7*N)]))); % Or do it this way, for comparison with the sequel: X = [-4*N:4*N-1]; % 1x8N xlimits = [min(X) max(X)]; f = X/(8*N); W = exp(-j*2*pi * t' * f); % Nx1 × 1x8N = Nx8N Y = abs(y * W); % 1xN × Nx8N = 1x8N % Y(1) = SUM(n=1,2,...,N): { e^(-B × t(n)^2) × e^(-j2π ×-4096/8N × t(n)) } % Y(2) = SUM(n=1,2,...,N): { e^(-B × t(n)^2) × e^(-j2π ×-4095/8N × t(n)) } % Y(8N) = SUM(n=1,2,...,N): { e^(-B × t(n)^2) × e^(-j2π × 4095/8N × t(n)) } Y = Y/max(Y); % Resample the function to reduce the DTFT periodicity from 1 to 3/8. T = 8/3; t = T*(-N/2 : N/2-1); % 1xN z = exp(-B*t.^2); % 1xN % Resample the DTFT. W = exp(-j*2*pi * t' * f); % Nx1 × 1x8N = Nx8N Z = abs(z * W); % 1xN × Nx8N = 1x8N Z = Z/max(Z); %======================================================= hfig = figure("position", [1 1 1200 900]); x1 = .08; % left margin for annotation x2 = .02; % right margin dx = .05; % whitespace between plots y1 = .08; % bottom margin y2 = .08; % top margin dy = .12; % vertical space between rows height = (1-y1-y2-dy)/2; % space allocated for each of 2 rows width = (1-x1-dx-x2)/2; % space allocated for each of 2 columns x_origin1 = x1; y_origin1 = 1 -y2 -height; % position of top row y_origin2 = y_origin1 -dy -height; x_origin2 = x_origin1 +dx +width; %======================================================= % Plot the Fourier transform, S(f) subplot("position",[x_origin1 y_origin1 width height]) area(X, Y, "FaceColor", [0 .4 .6]) xlim(xlimits); ylim([0 1.05]); set(gca,"XTick", [0]) set(gca,"YTick", []) ylabel("amplitude") %xlabel("frequency") %======================================================= % Plot the DTFT subplot("position",[x_origin1 y_origin2 width height]) area(X, Z, "FaceColor", [0 .4 .6]) xlim(xlimits); ylim([0 1.05]); set(gca,"XTick", [0]) set(gca,"YTick", []) ylabel("amplitude") xlabel("frequency") %======================================================= % Sample S(f) to portray Fourier series coefficients subplot("position",[x_origin2 y_origin1 width height]) stem(X(1:128:end), Y(1:128:end), "-", "Color",[0 .4 .6]); set(findobj("Type","line"),"Marker","none") xlim(xlimits); ylim([0 1.05]); set(gca,"XTick", [0]) set(gca,"YTick", []) ylabel("amplitude") %xlabel("frequency") box on %======================================================= % Sample the DTFT to portray a DFT FFT_indices = [32:55]*128+1; DFT_indices = [0:31 56:63]*128+1; subplot("position",[x_origin2 y_origin2 width height]) stem(X(DFT_indices), Z(DFT_indices), "-", "Color",[0 .4 .6]); hold on stem(X(FFT_indices), Z(FFT_indices), "-", "Color","red"); set(findobj("Type","line"),"Marker","none") xlim(xlimits); ylim([0 1.05]); set(gca,"XTick", [0]) set(gca,"YTick", []) ylabel("amplitude") xlabel("frequency") box on %======================================================= % Output (or use the export function on the GNUPlot figure toolbar). print(hfig,"-dtif", "-S1200,900","-color", 'C:\Users\BobK\Fourier transform, Fourier series, DTFT, DFT.tif')

${\displaystyle s(t)\quad {\stackrel {\mathcal {F}}{\Longleftrightarrow }}\quad S(f)\,}$

${\displaystyle S[k]\quad S_{1/T}(f)\quad S_{N}[k]\,}$

${\displaystyle \underbrace {s(t)*\sum _{k=-\infty }^{\infty }\delta (t-kP)} _{\color {BrickRed}\operatorname {rept} _{P}(s(t))}\quad {\stackrel {\mathcal {F}}{\Longleftrightarrow }}\quad {\color {OliveGreen}{\frac {1}{P}}}\underbrace {S(f)\cdot \sum _{k=-\infty }^{\infty }\delta (f-k/P)} _{\color {OliveGreen}\operatorname {comb} _{1/P}(S(f))}\,}$

${\displaystyle \underbrace {s(t)\cdot \sum _{n=-\infty }^{\infty }\delta (t-nT)} _{\operatorname {comb} _{T}(s(t))}\quad {\stackrel {\mathcal {F}}{\Longleftrightarrow }}\quad {\frac {1}{T}}\underbrace {S(f)*\sum _{n=-\infty }^{\infty }\delta (f-n/T)} _{\operatorname {rept} _{1/T}(S(f))}\,}$

${\displaystyle s(t)\cdot \sum _{n=-\infty }^{\infty }\delta (t-nT)=\color {Red}\operatorname {comb} _{T}(s(t))\,}$

${\displaystyle {\frac {1}{T}}S(f)*\sum _{n=-\infty }^{\infty }\delta (f-n/T)=\color {Blue}{\frac {1}{T}}\operatorname {rept} _{1/T}(S(f))\,}$

${\displaystyle \underbrace {{\color {BrickRed}\operatorname {rept} _{P}(s(t))}\cdot \sum _{n=-\infty }^{\infty }\delta (t-nT)} _{\operatorname {comb} _{T}({\color {BrickRed}\operatorname {rept} _{P}(s(t))})}\quad {\stackrel {\mathcal {F}}{\Longleftrightarrow }}\quad {\frac {1}{T}}{\color {OliveGreen}{\frac {1}{P}}}\underbrace {{\color {OliveGreen}\operatorname {comb} _{1/P}(S(f))}*\sum _{n=-\infty }^{\infty }\delta (f-n/T)} _{\operatorname {rept} _{1/T}({\color {OliveGreen}\operatorname {comb} _{1/P}(S(f))})}}$

${\displaystyle \underbrace {\operatorname {comb} _{T}(s(t))*\sum _{k=-\infty }^{\infty }\delta (t-kP)} _{\operatorname {rept} _{P}(\operatorname {comb} _{T}(s(t)))}\quad {\stackrel {\mathcal {F}}{\Longleftrightarrow }}\quad {\frac {1}{P}}{\frac {1}{T}}\underbrace {\operatorname {rept} _{1/T}(S(f))\cdot \sum _{k=-\infty }^{\infty }\delta (f-k/P)} _{\operatorname {comb} _{1/P}(\operatorname {rept} _{1/T}(S(f)))}}$

${\displaystyle {\color {Red}\operatorname {comb} _{T}(s(t))}*\sum _{k=-\infty }^{\infty }\delta (t-kP)=\operatorname {rept} _{P}({\color {Red}\operatorname {comb} _{T}(s(t))})}$

${\displaystyle {\frac {1}{P}}{\color {Blue}{\frac {1}{T}}\operatorname {rept} _{1/T}(S(f))}\cdot \sum _{k=-\infty }^{\infty }\delta (f-k/P)={\frac {1}{P}}{\color {Blue}{\frac {1}{T}}}\operatorname {comb} _{1/P}({\color {Blue}\operatorname {rept} _{1/T}(S(f))})}$