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Assume the observation of the original signal $s(t)$ is $x(t) = s(t) +n(t)$, where the signal and noise are independent.

Then we need to design the Wiener filter $g(t)$ to estimate $\frac{d s(t)}{dt}$ which is the derivative of the original signal $s(t)$.

The design of the Wiener filter is from the MMSE: $E[e]$. But how to formulate the error function here? $e(t) = \frac{d s(t)}{dt} - x(t)$ or $e(t) = \frac{d s(t)}{dt} - \frac{d x(t)}{dt}$?

I think this can be generalized to any linear operators like estimating $y = Hs$ (deconv, deblur etc.). But I have no idea the begining step. Could anyone give me some hints? Thanks!

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The key point here is the derivation / integration is a Linear Operator.
Since the Wiener Filter is also a linear operator it makes things easier.

Write the problem using the Wiener Filter Model:

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Neglecting the integrator we know the Wiener Filter for $ \hat{H} \left( w \right) $ (See Wikipedia - Wiener Deconvolution):

$$ G(f) = \frac{1}{\hat{H}(f)} \left[ \frac{ 1 }{ 1 + 1/(|\hat{H}(f)|^2 \mathrm{SNR}(f))} \right] $$

By defining $ H \left( w \right) = \frac{1}{j w} \hat{H} \left( w \right) $ as the fourier operation of integration (This is not exactly accurate, see The Integration Property of the Fourier Transform)

$$ G(f) = \frac{1}{{H}(f)} \left[ \frac{ 1 }{ 1 + 1/(|H(f)|^2 \mathrm{SNR}(f))} \right] $$

Basically, we defined the integration as a filter and used the known building blocks.

The interpretation is, with respect to the non integral case, is apply the derivative operator on the estimator and have some regularization on the SNR estimation as we estimate a derivative.

Resources

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  • $\begingroup$ Hi Royi, thanks for your answer. The wiener deconvolution is for engineering application, which treats the realization (in stochastic process term) as the approximate distribution. In true case, the estimation involves cross-correlation of $h(t)$ and $x(t)$, do you have idea of this term? Ref: dsp.stackexchange.com/a/51660/20519 $\endgroup$ – stander Qiu Nov 17 '20 at 10:38
  • $\begingroup$ Sure, I have created. Thanks. dsp.stackexchange.com/questions/71489/… $\endgroup$ – stander Qiu Nov 17 '20 at 10:55

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