# Orthogonal basis for waveform expansion

I have many signals where each signal has a different waveform f(x). One example of such a waveform could be this f(x) sampled at 11 x positions:

I am looking for a basis, Bi, for a series expansion of f(x): $\ f(x) \simeq \sum\limits_{i=0}^m c_i B_i(x)$

The expansion should have the following properties:

1. Give a good approximation with few terms (low m): low square error.

2. Given a expansion for one waveform with the following coefficient vector: $\mathbf{c}_i$; another waveform that is "similar" to the first waveform should have nearly the same coefficient vector. By similar I mean e.g. same shape as the first but slightly wider, same shape as the first but slightly higher peak etc. The basis should NOT be scale invariant; e.g. a similar waveform but much smaller should get a very different coefficient vector.

The expansion will be used on all signals to automatically classify them into clusters; signals that have close coefficients vector belong to one cluster.

Thanks in advance for any answers!

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From what I understand, you need to provide more information. For example, just take $B_1 = f$, then $f$ has a trivial expansion (ie, $c_1 = 1$, all others $0$). – copper.hat Jun 11 '12 at 8:11
Yes. I have updated my question to clarify. – Andy Jun 11 '12 at 8:32
Could you explain what you mean by "a similar waveform but much smaller should get a very different coefficient vector."?? If $f(x) \approx \sum_i c_iB_i(x)$, then $\alpha\cdot f(x) \approx \sum_i (\alpha\cdot c_i)B_i(x)$, that is, scaling $f(x)$ by $\alpha$ simply scales the coefficient vector by $\alpha$, and thus does not give a "very different coefficient vector" as you want it to. – Dilip Sarwate Jun 11 '12 at 11:51