Assume we have $N$ measurements $z_1, ..., z_N \in \mathbb{R}^{n_z} $ that generated by

$$ z_i = M v_i + e_i $$

where $v_i \in \mathbb{R}^{n_v}$, $n_v < n_z$ and $e_i$ an error sampled from normal distribution $\mathcal{N}(0, \sigma^2 I) $. Hereby $M$ should have full rank but be unknown.

Given only the measurements $z_i$ and no information about $M$, $v_i$ and $e_i$, my task is to find a matrix $B$ that estimates

$$ \hat v_i \approx B \hat z_i $$

for new samples $(\hat v_i, \hat z_i)$ generated by same distributions as before. This is in general referred to as blind signal separation (BSS).

Now my question is the following: I dont know the distribution of $v_i$ at all, but i have the information, that the entries $v_{i,j}$ will always be around $0$ for all indices $j=1,...,n_v$ except a small number of indices, up to k many (its around $k/n_v \leq 0.2$ ...). So $v_i$ is always "approximately sparse".

So is there a good way to use this information to improve certain BSS algorithms to deliver a $B$ which will more probably deliver solutions $\hat v_i$ which tend to have the above sparsity property? Or are there certain algorithms for BSS that are especially well suited for this special case? Is there maybe a clever way to reformulate the BSS problem to contain the sparsity property?

To explain a bit about my application ... I want to build a software which is filled with piano music and should extract the notes being played. Herefor, i take several mp3s and genrate samples $z_i$ from wavelet transforming the sound signal at different time stamps. So a $z_i$ is basically the vector of amplitudes to different frequencies at a certain time. Then the $v_i$ is the vector of the intensities of the different notes being played (obviously unknown resp. not practicable to look up). Keep in mind, that if u play one note $v_{i,j}=\delta_{ij}$ u get all its harmonic frequencies in $z_i$ which makes it difficult to filter out.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.