What is the Fourier transform of $\delta(t-a)$

Question

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 $e^{-jnTw}$.

Answer 1

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)));

Answer 2

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.