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I would like to derive the inverse discrete Fourier transform from the Fourier series. This should be possible since the DFT is just the Fourier series of a discrete time signal but I have run into an infinite series that I cannot figure out how to eliminate. Help is appreciated.

I start with the Fourier series:

\begin{aligned} x(t) & =\sum_{k=-\infty}^{+\infty}C_{k}e^{i2\pi kt/T},\\ C_{k} & =\frac{1}{T}\int_{-T/2}^{T/2}x(t)e^{-i2\pi kt/T}\textrm{d}t. \end{aligned}

I sample $x(t)$ N times during its period $T$ and arrive at the DFT coefficients, using a Riemann sum approximation:

\begin{aligned} C_{k}&=\frac{1}{T}\int_{-T/2}^{T/2}x(t)e^{-i2\pi kt/T}\textrm{d}t\\&=\frac{1}{T}\sum_{n=0}^{N-1}x(nT/N)e^{-i2\pi kn/N}\left(\frac{T}{N}\right)\\&=\frac{1}{N}\sum_{n=0}^{N-1}x(nT/N)e^{-i2\pi kn/N}. \end{aligned}

Where I am stuck is the inverse transform:

\begin{aligned} x(t) & =\sum_{k=-\infty}^{+\infty}C_{k}e^{i2\pi kt/T}\\ x\left(\frac{nT}{N}\right) & =\sum_{k=-\infty}^{+\infty}C_{k}e^{i2\pi kn/N}\\ & =\sum_{h=-\infty}^{+\infty}\sum_{k=0}^{N-1}C_{k+hN}e^{i2\pi(k+hN)n/N}\\ & =\sum_{h=-\infty}^{+\infty}\sum_{k=0}^{N-1}C_{k+hN}e^{i2\pi kn/N}\\ & =\ldots ? \end{aligned}

Since the equation should be:

\begin{aligned} x\left(\frac{nT}{N}\right)&=\sum_{k=0}^{N-1}C_{k}e^{i2\pi kn/N} \end{aligned}

It appears there must be a way to eliminate all values where $h\neq 0$. All textbooks I've read say that because it is periodic, there is no need to look at other periods, but I do not see how to arrive at this analytically.

Can someone please point out the key piece I'm missing?

Roxy
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  • I notice this was asked previously and had some discussion, but no answer. – Roxy Dec 04 '20 at 23:49
  • I think these are fundamentally different objects. Here when you do the Riemann sum approximation, notice that $C_k = C_{k+hN}$, so $$x(t) =\sum_{k=-\infty}^{+\infty}C_{k}e^{i2\pi kt/T}$$ does not even converge. This is known as aliasing in signal processing. – 61plus Dec 05 '20 at 00:10
  • @61plus I know this, hence the question. There is a way to get from the Fourier series to the IDFT since the difference between the two is simply that one is continuous and the other is discrete. I'm looking for the missing piece that connects the two in my derivation. – Roxy Dec 05 '20 at 00:14
  • Interestingly, Dr. Andrzej Borys has recently published a paper on almost exactly this, except that he derives the IDTFT from the Fourier Transform, while I am looking for the IDFT from the Fourier series. He arrives at a similar infinite summation and is able to resolve it by assuming the signal is band limited and using a bounds trick. Unfortunately, I do not see how the trick he used is able to help my scenario above. – Roxy Dec 05 '20 at 00:25

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