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

% 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');

Answer 1

[Wayne King]

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.

[Joe]

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

Answer 2

[Wayne King]

 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]

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

[Wayne King]

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);

I've added it above.

[Joe]

Just to clarify... What is t(2) and t(1)? I know that dt is the time separation, but I'm not sure why you used t(2) and t(1). Also, is there another way of getting dt?