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

Accepted answer

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.

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

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);

I've added it above.

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