It is known that Fourier Transform has an "accumulation property". If we define:

$$X(\omega) = F[x(n)]$$ and $$y(k) = \sum\limits_{n=-\infty}^k x(n)$$ i can write: $$y(k)-y(k-1)=\sum\limits_{n=-\infty}^k x(n)-\sum\limits_{n=-\infty}^{k-1} x(n)=x(k)$$ Where both $x(n)$ and $y(n)$ are discrete functions. Now, if i'm not mistaken, the accumulation property states that the Fourier Transform of $y(n)$ can be written as $$\frac{X(\omega)}{(1-e^{-i\omega})}+\pi \sum\limits_{n=-\infty}^{+\infty} X(2\pi n) \delta(\omega-2\pi n)$$

Now, I understand the second term exists to yield the result of $Y(\omega)$ when $x(n)$, but I don't understand how this writing works, because when $\omega=0$ the first term goes to infinity anyway.

My question is, shouldn't the definition be:

$$ \begin{cases} \frac{X(\omega)}{(1-e^{-i\omega})}, \omega\neq0 \\ \pi \sum\limits_{n=-\infty}^{+\infty} X(2\pi n) \delta(\omega-2\pi n), \omega=0 \end{cases} $$

because i think the other expression will always be infinite when $\omega=2 \pi n$.

Thanks in advance.


Consider the discrete fourier transform of the step function $u(t)$

$$\frac{1}{1-e^{i\omega}}+\pi\sum\limits_{n=-\infty}^{+\infty} \delta(\omega-2\pi n)$$

And the continuous fourier transform


It is supposed to be like this. The formula is correct. It is important to note that the step function is not actually an absolutely integrate function, so its fourier transform is more like a principal value.

  • $\begingroup$ Thanks for your help. With your answer I noted my problem was with the FT of the step function. After reading this article I understood why it is like that: cs.uaf.edu/~bueler/M611heaviside.pdf Thanks :) $\endgroup$
    – Dayman75
    Jun 11 '15 at 18:34

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