• Period: The period is the smallest value of T satisfying $$g(t + T) = g(t)\label{0}\tag{0}$$ for all t. The period is defined so because if g(t + T) = g(t) for all t, it can be verified that g(t + T') = g(t) for all t where T' = 2T, 3T, 4T, ... In essence, it's the smallest amount of time it takes for the function to repeat itself. If the period of a function is finite, the function is called "periodic". Functions that never repeat themselves have an infinite period, and are known as "aperiodic functions". The period of a periodic waveform will be denoted with a capital T. The period is measured in seconds.

  • Kronecker delta function: $$ \delta_{ij}= \begin{cases} 0&\text{if}\, i\neq j\\ 1&\text{if}\, i= j \end{cases} \label{1}\tag{1} $$

  • Discrete-time signal x: $x(n)=\sum_{k=-\infty}^{\infty}x(k)\delta(n-k)\label{2}\tag{2}$

  • Continuous-time signal x: $$x(t)=\int_{-\infty}^{\infty}x(\tau)\delta(t-\tau)\delta{\tau}\label{3}\tag{3}$$
  • Euler formula derivations:



$$\cos(\theta) = \frac{1}{2}(e^{-j\theta}+e^{j\theta})\label{6}\tag{6}$$

$$\sin(\theta) = \frac{1}{2j}(e^{-j\theta}-e^{j\theta})\label{7}\tag{7}$$


The fourier expansion of a periodic signal $x_T(t)=x_T(t+T)$ is


where $X[k]$ is the fourier coefficient

$$X[k]=F[x_T(t)]=\frac{1}{T}\int_Tx_T(t)e^{-jkw_0t}dt \label{9}\tag{9}$$

where $k=0, \pm1,\pm2,...$


I'd like to prove this:


I believe by considering the above definitions it should be possible to get a nice simple proof but I don't know how to proceed here.

  • $\begingroup$ This question is far too broad to have any reasonable answer. Want the series to converge to the function? You'll need some extra niceness condition on the function - and exactly what that is depends on what sort of convergence you're looking for. There are quite a few standard theorems on the matter. $\endgroup$ – jmerry Jan 28 '19 at 13:42
  • $\begingroup$ When you say I'll need some extra niceness conditions on the function I guess you're talking about the domain,range and smoothness specs? So I guess if we just consider the definition of periodic signal provided here that still wouldn't allow us to start proving that equality? Feel free to specify which are the minimum function requirements so we can create an easy and intuitive proof though to that very interesting result on the time domain. $\endgroup$ – BPL Jan 28 '19 at 17:32

I think the problem of convergence of the Fourier series of a periodic function $f$ to $f$ is very wide, when we don’t specify conditions on $f$, like continuity and smoothness and a type of convergence of the series, like pointwise or uniform. Thus I think that Wikipedia article on convergence of Fourier series provides an excellent brief survey of the state of the art.

In mathematics, the question of whether the Fourier series of a periodic function converges to the given function is researched by a field known as classical harmonic analysis, a branch of pure mathematics. Convergence is not necessarily given in the general case, and certain criteria must be met for convergence to occur.

Determination of convergence requires the comprehension of pointwise convergence, uniform convergence, absolute convergence, $L^p$ spaces, summability methods and the Cesàro mean.

Thus, when the problem conditions are not specified I think there is no much sense to argue about convergence proofs. On the other hand, I have a few theorems providing sufficient conditions for pointwise and uniform convergence of the Fourier series of a periodic function $f$ to $f$ in my student analysis book (in Ukrainian). But my working experience with application people like programmers or engineers shows that they usually don’t need a proof, but at most a reference to a theorem providing a needed result.

  • 1
    $\begingroup$ Thanks for the wikipedia link, really appreciated, I'll take a look. Thing is, I've taken few courses about signal processing many years ago and nowadays my skills about it have gotten pretty rustic as you can deduce from my "basic" question. And yeah, you're correct and usually software engineers we just care about the right abstractions as well as ready to be implemented results. That said, I do really care about these type of proofs so I can continue studying more abstract concepts like integral transforms, operators with some a solid foundation established $\endgroup$ – BPL Feb 3 '19 at 18:16

Very roughly (i.e. with little rigour) the identity follows from the fact that one can write the Dirac Delta function as $$ \delta(x)=\frac1{2\pi}\sum_{n=-\infty}^\infty e^{inx}. $$ (cf https://en.wikipedia.org/wiki/Dirac_delta_function#Fourier_kernels) This means that we have \begin{equation}\begin{aligned} \frac1T\sum_{k=-\infty}^\infty\int_0^Tdt'x(t')e^{2\pi ik\frac{t-t'}T}&=\frac1T\int_0^Tdt'x(t')\sum_{k=-\infty}^\infty e^{2\pi ik\frac{t-t'}T}\\ &=x(t). \end{aligned}\end{equation} Again, this is very sketchy and not rigorous at all. I'm swapping limits and messing around with Dirac deltas without much thought.

  • $\begingroup$ I think considering the dirac delta is a good idea. In fact, I believe by expressing a discrete or continuous signal as a sum of weighted dirac deltas first could lead to the fourier expansion very intuitively. In this answer, after you've defined the dirac delta those results are a little bit messy and I don't know where they coming from. When I'm back home I'll give it a shot myself. Thx anyway to provide the hint about the dirac delta, just because of that, you deserve +1 :) $\endgroup$ – BPL Feb 1 '19 at 18:39
  • $\begingroup$ I've edited the question... in case you're still interested $\endgroup$ – BPL Feb 2 '19 at 18:35
  • $\begingroup$ If your happy with swapping the infinite sum and the integral as well as the expression for the Dirac delta function then that is a "proof". The left hand side of the second expression is $\sum_k X[k]e^{jk\omega_0 t}$. I've written $i$ for $j$ and used $\omega_0=\frac{2\pi}{T}$. $\endgroup$ – Alec B-G Feb 2 '19 at 18:43
  • $\begingroup$ For rigorous proofs, one needs to worry about convergence and smoothness conditions, as jmerry already pointed out. $\endgroup$ – Alec B-G Feb 2 '19 at 18:44
  • $\begingroup$ If that's so, feel free to improve your answer by talking about those constraints, jmerry said those were necessary but he didn't provide any relevant information to help. I've edited again my question to make easier the task to possible readers... My intuition tells me the proof should be possible to achieve by consider $\ref{0}$ & $\ref{2}$ (discrete) or $\ref{0}$ & $\ref{3}$ (continuous) but I'm still thinking how. Btw, in my previous comment I meant "weighted kronecker deltas" instead dirac ones. $\endgroup$ – BPL Feb 2 '19 at 19:14

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