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Fourier transform of simple sin wave in matrix

I am trying to show the spectrum of a simple sine wave, as we know that one strong wave with a fixed frequency must have a peak in its spectrum. I am writing this code, but I cannot get this one peak, what is wrong in my code:

clc
nsteps=200;%number of signal elements in time domain
i=sqrt(-1);
NFREQS=100;%number of elements in frequency domain
ddx=1e-9;
dt=ddx/(6e8);%separation between each time domain elements
lambdai=150e-9;
lambdaf=500e-9;
freqi=3e8/lambdai;
freqf=3e8/lambdaf;
freq=zeros(1,NFREQS);
for j=1:NFREQS
freq(j)=freqi-j*(freqi-freqf)/NFREQS;%desired frequency domain
end
arg=2*pi*freq*dt;
et=zeros(nsteps,1);
    for j=1:nsteps
       et(j)=sin(2*pi*3e15*j*dt);%sin wave in time domain
    end
e=zeros(NFREQS,1);
for n=1:NFREQS
for j=1:nsteps  
 e(n)=e(n)+et(j)*exp(-i*arg(n)*n);%sin wave in frequency domain
end
end
lambda=linspace(lambdai,lambdaf,NFREQS);
plot(lambda,abs(e))

The result is here

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2 answers


The key point here is using the window function. The resulting code:

clc
clear
nsteps=40000;
i=sqrt(-1);
Sc=0.5;
ddx=1e-9;
dt=ddx*Sc/(3e8);
lambdai=100e-9;
lambdaf=700e-9;
lambda=lambdai:1e-9:lambdaf;
[m,NFREQS]=size(lambda);

    freq=3e8./lambda;
et=zeros(nsteps,1);
    for j=1:nsteps
       et(j)=sin(2*pi*3e8/(300e-9)*dt*j);%sin wave in time domain
    end
    w=blackman(nsteps);%window function for periodic functions 
    for j=1:nsteps
        et(j)=et(j)*w(j);
    end
    e=zeros(1,NFREQS);
    for n=1:NFREQS
        for j=1:nsteps
            e(n)=e(n)+et(j)*exp(-i*2*pi*freq(n)*dt*j);
        end
    end
   plot(lambda,abs(e))


And the result:

enter image description here

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You can use the built-in Fourier transform that MATLAB provides instead of writing your own. See http://www.mathworks.se/help/techdoc/math/brentm1-1.html

Update

The feature fft

you should be aware of has a few quirks. First, the array it generates must be normalized with a factor 1/N

, where N

is the size of the array (this is due to the implementation of the MATLAB function). If you do not do this, all amplitudes in the frequency domain will be N

times greater than they "actually".



Second, you need to somehow find the frequencies that each element in the output array corresponds to. The third line in the code below converts the array of samples to frequencies that match the array fourier

.

This function takes a signal in the time domain and displays it in the frequency domain. t

is the time array, and y

are the signal amplitudes.

function plotSpectrum(t, y)
freqUnit = 1 / (abs(t(2) - t(1)) * length(t));
f = -length(t) * freqUnit : freqUnit : (length(t) - 1) * freqUnit;
oneSidedFFT = fft(y);
fourier = horzcat(oneSidedFFT, oneSidedFFT);
plot(f, abs(fourier));
xlabel('Frequency');
ylabel('Magnitude');
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