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

on 28 Jun 2013

### Wayne King (view profile)

% 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');

### Wayne King (view profile)

on 29 Jun 2013
Edited by Wayne King

### Wayne King (view profile)

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.

Joe

### Joe (view profile)

on 29 Jun 2013

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

### Wayne King (view profile)

on 28 Jun 2013
Edited by Wayne King

### Wayne King (view profile)

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

### Joe (view profile)

on 29 Jun 2013

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

Wayne King

### Wayne King (view profile)

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);

Joe

### Joe (view profile)

on 5 Jul 2013

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?