# Decomposition of periodic spikes into frequencies

Let $a_1, a_2, ..., a_n$ be a sequence such that:

$$a_n = \begin{cases} 1 & \text{if K_a divides n} \\ 0 & \text{otherwise} \end{cases}$$

And similarly let other two sequences $b_1, b_2, ..., b_n$ and $c_1, c_2, ..., c_n$ be:

$$b_n = \begin{cases} 1 & \text{if K_b divides n} \\ 0 & \text{otherwise} \end{cases}$$ $$c_n = \begin{cases} 1 & \text{if K_c divides n} \\ 0 & \text{otherwise} \end{cases}$$

Suppose now we have $x_1, x_2, ..., x_n$ where $x_n = a_n + b_n + c_n + \epsilon_n$, where $\epsilon_n$ is an error term (white noise).

How would it be possible to analyse this sequence and obtain the "frequencies" $1/K_a, 1/K_b, 1/K_c$, given a large $n$?

Applying Discrete Fourier Transform to any of the original sequences, say $a_1, ..., a_n$, does not work as the fourier transform creates a series of spikes in $1/K_a, 2/K_a, 3/K_a$ and so on. Applying DFT directly onto the $x$ sequence therefore creates many periodic spikes, and finding the three frequencies in the transformed sequence is no easier than in the original one.

While the Fourier Series is the decomposition of a function into a sum of sinusoidal waves, here it looks like we need a decomposition of a function into a sum of periodic spikes (Dirac Combs). How can we solve such a problem? What if (1) there are many such sequences added together, not only three, and also (2) each sequence has a different magnitude (not always 1 when not zero)?

• What about an $\ell_1$ fit where the dictionary is all the possible pulse trains? Commented Jun 26, 2017 at 18:24

So, what would you do if you wanted to find the period of a single impulse train sequence $\{a_n\}$? The simplest approach I can imagine would be to search for the smallest $n>0$, say, $n_0$, such that $a_{n_0}>0$. Then you would declare $n_0$ as the period of $\{a_n\}$.
1. Find the smallest $n_0>0$ such that $x_n>0$.
2. Declare $n_0$ as the period of one of the superimposed signals
3. Sustract the found impulse train from $x_n$, i.e., set $x_n\leftarrow x_n-\sum_{k}\delta[n-k n_0]$