# How to complete the fourier Analysis using Matlab ?

1 Ansicht (letzte 30 Tage)
Megh am 11 Jan. 2015
Kommentiert: Katherine Zheng am 18 Mär. 2022
Hello !!
I have tried using the Matlab tutorial for FFT and DFT but I'm having extreme difficulty understanding the code and how I can use it in my question. My experience with matlab is only in data manipulation and plotting, so I'm struggling with the concepts.
So here is the question...
To compute the Cn coefficient given by
• Cn = 1/T * ∫ f(t)*e^(-2*pi*n*t/T) dt,
in which T is the period, and then Amplitude is
• sqrt(Re(Cn)^2 + Im(Cn)^2)
• F = n/T
then... how can I plot Amplitude vs Frequency using matlab functions. So far I have this function and req as mentioned above.
1. f(t) = e^-t from -3 to 3
2. T = 6
3. F = -10:1:10
I would like to plot using matlab, but so far I'm doing integral manually and then just iterating values, and plotting them in matlab, can I use fft function or any other function to speed up the process ??
Many thanks
Megh
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dpb am 11 Jan. 2015
There's a complete example of computing/plotting the one-sided PSD at
doc fft
that should be pretty easy to follow what it's doing.

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### Akzeptierte Antwort

Rick Rosson am 11 Jan. 2015
Bearbeitet: Rick Rosson am 11 Jan. 2015
doc fft
doc fftshift
doc abs
doc angle
doc plot
doc stem
doc xlabel
doc ylabel
doc grid
doc xlim
doc ylim
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Megh am 13 Jan. 2015
Thanks !!

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### Weitere Antworten (2)

Youssef Khmou am 11 Jan. 2015
Try to study and alter this example :
Fs=10;
t=-3:1/Fs:3;
x=exp(-t);
plot(t,x)
N=1000; % N points for frequency computation
fx=fftshift(fft(x,N))/sqrt(N);
fx=fx.*conj(fx);
% frequency axis
f=(-N/2:N/2-1)*Fs/(2*N);
figure; plot(f,fx);
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Megh am 13 Jan. 2015
Thanks !!

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Megh am 13 Jan. 2015
Bearbeitet: Megh am 13 Jan. 2015
Ughhh I tried using above answers but It felt unsure about lots of operation (Since I have yet to understand many of the FFT properties and theorems) ... Well I chose the following code which I'm more confident about the mathematical operations.
This code works for now, well I still need to figure out the better way.
% clear the memory and workspace
clc, clear, close all
syms t fr
g_t = sin(2*pi*t); % function
T = 10; % period of the function
a = -5; b = 5; % to integrate from a to b
n = -100:1:100; % n is an integer by DEFINATION
freq = n./T;
% Integrate function
f = g_t*exp(-2*pi*fr*i*t);
Cn = 1/T * int(f,t,a,b);
% get the amplitude and phase spectra function
Amp = sqrt(real(Cn)^2 + imag(Cn)^2);
Pha = atan(imag(Cn)/real(Cn));
% Do this to combat datapoints/frequency in which function yields
% infinty or division by zero error, take the limit
% NEED TO FIND BETTER Way
AmpV = zeros(1, length(freq));
for n = 1:length(AmpV)
try AmpV(n) = subs(Amp, 'fr', freq(n));
catch AmpV(n) = limit(Amp, fr, freq(n));
end
end
% Plot the discrete Amplitude spectrum
figure(1)
stem(freq,AmpV)
xlim([-10 10])
xlabel('\bfFrequency in Hz')
ylabel('\bfAmplitude')
title('\bfAmplitude spectra of F_{t}')
Note this image is for function exp(-t)*sin(2*pi*t) from 0 to 5, with T of 10s.
##### 1 Kommentar-1 ältere Kommentare anzeigen-1 ältere Kommentare ausblenden
Katherine Zheng am 18 Mär. 2022
I just really appreciate such poster who figure out their issue and give a detailed answer about it!!! Thanks!!

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