I need to approximate a function f, but I cannot do so with frequencies that exceed 1kHz

What is the best approximation I can get? Is taking the Fourier transform then zeroing any term above 1kHz the best approximation? Or can I fiddle with the lower order terms and get a better fit?

This is a made up scenario, but I have to prove the same concept with Walsh transforms. I am fairly certain that the lower order terms form the best approximation from random twiddling and hill climbing searches, but I need proof.

I believe the proof is something very similar to a least squares regression proof, but I can't get it. Has this problem been solved before? At least in the Fourier domain?


1 Answer 1


Depends on what you mean by 'best'. In terms of least squares, then yes, just knocking off the terms above 1kHz will give the closest approximation. I guess one way to prove it would be: $$ x(t) = \int_{-\infty}^{\infty} X(f)\exp\left( j2\pi ft \right) dt\\ \hat x(t) = \int_{-1k}^{1k} X(f) \exp\left( j2\pi ft \right) dt $$ leaving a residual error of $$ e(t) = \int_{|f| > 1k} X(f)\exp\left( j2\pi ft \right) dt . $$

Compare this to a 'fiddled' version, using $\tilde X(f)$ for the coefficients: $$ \hat x(t) = \int_{-1k}^{1k} \tilde X(f) \exp\left( j2\pi ft \right) dt, $$ which has a residual error of $$ e(t) = \int_{|f| \le 1k} \left(X(f) - \tilde X(f)\right) \exp\left( j2\pi ft \right) dt + \int_{|f| > 1k} X(f)\exp\left( j2\pi ft \right) dt , $$ which is greater than the error in the first instance unless $\tilde X(f) = X(f) \; \forall f$.

Something like that ought to do it, anyway.

EDIT: It is possible to prove that a projection onto any orthogonal set of functions (doesn't need to be the Fourier set) gives the least squares fit:

$$ \int f_k(t) f_m(t) dt = \cases{ 0, & $k \ne m$ \\ 1/\lambda_k, & $k = m$} $$ for any orthogonal basis set $\{f_k\}$ (I'm not bothering with weighting functions), where $$ \lambda_k = \frac{1}{\int f^2(t) dt} $$ The projection of $g(t)$ onto the set is given by $$ c_k = \lambda_k \int g(t)f_k(t) dt. $$ The least squares fit of a set of orthogonal basis function is the minimisation of $$ \int \left[g(t) - \sum_k d_k f_k(t)\right]^2 dt. $$ Minimise by setting the derivative (wrt $d_m$) to zero: $$ 2\int\left(g(t) - \sum_kd_kf_k(t)\right)\left(-f_m(t)\right) dt = 0 \\ -\int f_m(t)g(t)dt + \int f_m(t)\sum_kd_kf_k(t)dt = 0 \\ -\int f_m(t)g(t)dt + \sum_kd_k \int f_m(t)f_k(t)dt = 0 \\ -\int f_m(t)g(t)dt + \sum_kd_k \frac{1}{\lambda_m} = 0 $$ which rearranges to give the same value as for $c_k$, thus the least squares fit is exactly the straightforward projection onto the orthogonal set.

  • $\begingroup$ Thank you. I had something like this floating in pieces. I had an idea that the orthogonality was going to be an important part of it. This answer is awesome. Thank you so much for the extra detail $\endgroup$
    – MaxRunFast
    Mar 21, 2013 at 16:32

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