2
$\begingroup$

As I know the Fourier transform of $\delta(t)$ is equal to $1$, on the other hand, shift with $a$ in time domain is equal to multiplying with $e^{-jaw}$ so the Fourier transform of $\delta(t-a)$ is equal to $e^{-jaw}$, but on the other hand the Fourier transform of Dirac comb ($\sum_{n=-\infty}^{+\infty}\delta(t-nT)$which is equal to sum of Dirac functions) is also another Dirac comb, This doesn't make sense to me. The Fourier transform of Dirac comb should be sum of the $e^{-jnTw}$.

$\endgroup$
2
$\begingroup$

Your argumentation is perfectly right. The only thing you are missing is that the sum of complex exponentials becomes a comb function in the limit. See also this link and this matlab code:

N = 2048;
n = (0:(N-1))';

k = 4*(0:20);

signals = zeros(N,length(k));
for kk = 1:length(k)
    signals(:,kk) = exp(-1j*2*pi*k(kk).*n/N);
end

plot(abs(sum(signals, 2)));

Code output diagram

$\endgroup$
  • $\begingroup$ Is it okay to say that "The sum of complex exponentials is Fourier series expansion of Fourier transform of Dirac comb" $\endgroup$ – Navin Prashath Nov 12 '16 at 13:02
1
$\begingroup$

I wrote about this exact problem - an impulse train in time has a Fourier Transform that is another impulse train, although signals narrow in time should have a wide spectral representation - in my recent article on DSPrelated.

$\endgroup$
  • 1
    $\begingroup$ Your article is amazing! $\endgroup$ – user137927 Nov 17 '16 at 20:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.