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I am trying to derive the formula from continuous Fourier transform to discrete Fourier transform, however, I encountered a small problem. My derivations are the following.

For a periodic function $f(x)=f(x+a)$, the continuous Fourier transform read \begin{align} f(x) &= \sum_{n=-\infty}^{+\infty} c_n e^{i\frac{2\pi n}{a}x}\\ c_n &= \frac{1}{a} \int_0^a dx e^{-i\frac{2\pi n}{a}x}f(x) \end{align} In order to get the discrete Fourier tranform, I discretize the $x$-axis into $N$ pieces, and for the sampling $x$-points $x_j = \frac{a}{N}j$, they satisfy $f_j = f_{j+N}$, where I defined $f_j\equiv f(x_j)$. Therefore the first equation becomes \begin{equation} f_j = \sum_{n=-\infty}^{+\infty} c_n e^{i\frac{2\pi n}{a}x_j} = \sum_{n=-\infty}^{+\infty} c_n e^{i\frac{2\pi n}{N}j} \end{equation} and the second equation becomes \begin{equation} c_n = \frac{1}{a}\sum_{j=0}^{N-1} \frac{a}{N} e^{-i\frac{2\pi n}{a} j\frac{a}{N}} f_j = \frac{1}{N}\sum_{j=0}^{N-1} e^{-i\frac{2\pi n}{N}j} f_j \end{equation}

My question comes from in the third equation, the n-summation runs from $-\infty$ to $+\infty$ which differs from the standard DFT that from $0$ to $N-1$. What's wrong with my derivation? Thanks in advance.

jwyan1126
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  • Note that $e^{\frac{2\pi (N-j)}{N}} = e^{-\frac{2\pi j}{N}}$. Rewrite your summation. – fantasie Jul 05 '20 at 11:42
  • Sorry, could you give more details? I know that $e^{i\frac{2\pi n}{N}j}$ is a periodic function with respect to $n$. However, if I restrict the summation in the third equation from $0$ to $N-1$, an infinite pre-factor comes out. – jwyan1126 Jul 05 '20 at 11:56
  • Usually, what you call continuous Fourier transform is called Fourier series, since the Fourier transform is the case of non-periodic functions. – LL 3.14 Jul 06 '20 at 06:47
  • The way how it was called is not important, but what's wrong with my derivation? – jwyan1126 Jul 06 '20 at 18:50

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