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# How do I convert the x-axis of an FFT from frequency to wavelength?

Asked by Joe on 28 Jun 2013
% speed of light m/s
c = 299792458;

% pulse duration (T=tau)
FWHM = 30e-15;
T = FWHM/(2*(sqrt(log(2))));

% central wavelngth and central angular frequency
lambda0 = 800e-9;
w0 = (2*pi*c)/lambda0;

% chirp
eta = 2;

% time interval
t = -55e-15:.1e-15:55e-15;

% spectral phase
phi_t = 0;

% length of FFT # of sampling points
nfft = 200;

% This is an evenly spaced frequency vector
f = [0:nfft - 1]/nfft;

% wavelength interval and angular frequency conversion
lambda = (740e-9:20e-9:860e-9);
w = (2.*pi.*c)./lambda;

% electric field
E_t = exp((-t.^2/(2*T.^2)) + (-i*w0*t-(1/2)*i*eta*t.^2)); %*phi_t));

% Take fft, padding with zeros so that length(E_w) is equal to nfft
E_w = abs(fftshift(fft(E_t,nfft)));
I_w = ((abs(E_w)).^2);
I_lambda = I_w.'*((2.*pi.*c)./lambda); (This is where I'm trying to convert to wavelength)

% Plot
subplot(2, 1, 1);
plot(t, real(E_t));
title('Gaussian Pulse Signal');
xlabel('time (s)');
ylabel('E_t');

subplot(2, 1, 2);
plot(lambda, E_w);
xlabel('\lambda');
ylabel('E_\omega');


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Answer by Wayne King on 29 Jun 2013
Edited by Wayne King on 29 Jun 2013

This runs perfectly for me:

    c = 299792458;
% pulse duration (T=tau)
FWHM = 30e-15;
T = FWHM/(2*(sqrt(log(2))));
% central wavelngth and central angular frequency
lambda0 = 800e-9;
w0 = (2*pi*c)/lambda0;
% chirp
eta = 2;
% time interval
t = -55e-15:.1e-15:55e-15;
% spectral phase
phi_t = 0;
% wavelength interval and angular frequency conversion
lambda = (740e-9:20e-9:860e-9);
w = (2.*pi.*c)./lambda;
% electric field
E_t = exp((-t.^2/(2*T.^2)) + (i*w0*t-(1/2)*i*eta*t.^2)); %*phi_t)); 
   Edft = fftshift(fft(E_t));
dt = t(2)-t(1);
Fs = 1/dt;
df = Fs/length(E_t);
freq = -Fs/2+df:df:Fs/2;
lambda = c./freq;
plot(lambda,abs(Edft).^2);
set(gca,'xlim',[0 10e-7])

And it shows the correct wavelength.

## 1 Comment

Joe on 29 Jun 2013

Awesome Thanks! Now I just got to get the y-axis right... haha.

Answer by Wayne King on 28 Jun 2013
Edited by Wayne King on 29 Jun 2013
 E_t = exp((-t.^2/(2*T.^2)) + (i*w0*t-(1/2)*i*eta*t.^2));
Edft = fftshift(fft(E_t));
dt = t(2)-t(1);
Fs = 1/dt;
df = Fs/length(E_t);
freq = -Fs/2+df:df:Fs/2;
lambda = c./freq;
plot(lambda,abs(Edft).^2);

Now the wavelength you expect is

c/(w0/(2*pi))


so zoom in on that

set(gca,'xlim',[0 10e-7])


You see the peak at 8x10^{-7} as you expect

Joe on 29 Jun 2013

What is "df" that you have for the "freq"?

Wayne King on 29 Jun 2013

That is the frequency separation between the DFT bins. Sorry I forgot that line of code:

    df = 1/(dt*length(E_t));

or equivalently

    df = Fs/length(E_t);