In my recent studies of the Fourier Series, I came along to proof the properties of the Fourier Series (just to avoid confusion, not the fourier transform but the series itself in discrete time domain).

I also came into the following property:

The question

For any Discrete Fourier Series whose average is zero (condition for convergence), the following property is met:

$$y[n] = \sum_{r=-\infty}^{n}x[r]$$ $$\Longrightarrow Y_k = \frac{1}{1-e^{-i\frac{2\pi k}{N}}} X_k$$

so that:

$$y[n] = \sum_{k=0}^{N-1} \Big(\frac{1}{1-e^{-i\frac{2\pi k}{N}}} X_k\Big) e^{i\frac{2\pi k}{N} n}$$

Where $Y_k$ and $X_k$ are the respective Fourier Coefficients, and $N$ is the period.

The question is, ¿how can I prove that? or better, ¿what is wrong with what I have approached?

What I have been able to do

The first thing I did, is to define $y[n]$ as:

$$y[n] = \sum_{r=0}^{n} x[r]$$

Since the average per period is zero.

Then, using the linearity property and the displacement in time property, I got into:

$$y[n] = \sum_{r=0}^{n} \sum_{k=0}^{N-1} X_ke^{-i\frac{2\pi k}{N} r} e^{i\frac{2\pi k}{N} n} = \sum_{k=0}^{N-1} X_k\Big( \sum_{r=0}^{n}e^{-i\frac{2\pi k}{N} r} \Big) e^{i\frac{2\pi k}{N} n}$$

Using the fact that, from hypothesis, $X_0$ is zero (since $X_0$ represents the average of one period), we can use the geometric progression into the second summatory:

$$y[n] = \sum_{k=1}^{N-1}X_k \frac{1-(e^{-i\frac{2\pi k}{N} (n+1)})}{1-e^{-i\frac{2\pi k}{N}}} e^{i\frac{2\pi k}{N} n}$$

Which is, after splitting the summatory:

$$y[n] = \sum_{k=1}^{N-1} \underbrace{\Big(\frac{1}{1-e^{-\frac{2\pi k}{N}}}X_k \Big)}_{\text{Yay!}} e^{i\frac{2\pi k}{N} n} - \sum_{k=1}^{N-1} \Big(\frac{1}{1-e^{-\frac{2\pi k}{N}}}X_k \Big) e^{-i\frac{2\pi k}{N}}$$

The first expression is what I am looking for, but I haven't been able to work with the second expression/summatory, I suppose that there might be a way to prove that it is zero, but I am not very sure about it, so I don't know if either there is something wrong in the procedure or I am missing something.

Thanks for the help in advance.


Let $\alpha=e^{\frac{-2\pi ik}{N}}$ to ease the notation a little bit. Substituting into the definiton of DTFS yields:

$$\begin{align}Y_k &=\sum_{n=0}^{N-1}\sum_{r=0}^nx[r]\alpha^n=\sum_{r=0}^{N-1}x[r]\sum_{n=r}^{N-1}\alpha^n=-\sum_{r=0}^{N-1}x[r]\sum_{n=0}^{r-1}\alpha^n=\sum_{r=0}^{N-1}x[r]\frac{\alpha^r-1}{1-\alpha} \\ &=\frac{1}{1-\alpha}\left(\sum_{r=0}^{N-1}x[r]\alpha^r+\sum_{r=0}^{N-1}x[r]\right)=\frac{X_k}{1-\alpha}\end{align}$$

since $\sum_{r=0}^{N-1}x[r]=0$, and the other term is the definition of DTFS of $x[n]$.

  • $\begingroup$ Thanks for the answer. Can't even imagine the answer was this easy. $\endgroup$ – Zumbock Jan 26 at 20:29

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