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I am working on an application for which a matched filter seems like the right concept. In the derivation for the matched filter which I went through (here), they define the SNR as the ratio of signal power to noise power as the basis for the derivation of the expression for the matched filter. I think that for my application, the "amplitude" SNR would be a better target for optimization. However, it looks like I reach a mathematical dead-end that is avoided by the use of the power SNR. Below is my work which roughly follows the derivation from wiki:

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Is there a manipulation I am missing that could get me past this obstacle or does someone know if this result has been solved for already? I have not found it anywhere.

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Power is proportional to the square of amplitude, so there's really no difference (since SNR is always a positive number) when maximizing: A filter that maximizes "amplitude SNR" maximizes "power SNR".

So, if your application requires you to maximize the amplitude ratio between signal and noise (be careful, amplitudes can be negative or complex, depending on how you define them!), then that's the same as requiring you to maximize the power ratio.

I.e.

$$h_\text{opt} = \arg\max_h E\left\{\left(h^H (vv^H)h\right)^{\frac12}\right\} = \arg\max_h E\left\{h^H (vv^H)h\right\} $$

since the quadratic form $h^H (vv^H)h$ (which is the power of $y$) can never be negative, and that makes your square root both unambiguously invertible and monotonous.

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  • $\begingroup$ Thank you! Now I just added a small proof of the monotonicity of f(x) = x^2 and now the original derivation does all the work! Thank you! $\endgroup$
    – ChateauDu
    May 27 '20 at 18:54
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You could use Cauchy's inequality for expectations, $E\left\{(h^H (vv^H)h)^{1/2}\right\}$, you would need to know $E\{\sqrt{v}\}$. This would give you the minimum amplitude ratio which could be used in the expression. In any scenario when you take this route, you would need to apply a min/max or in other words bound on the ratio. The reason is you are now trying to optimize oven non linear functions of statistical measure of the noise.

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  • $\begingroup$ Thank you for the comment, I was not aware of this form of the Cauchy inequality. Good to know. $\endgroup$
    – ChateauDu
    May 27 '20 at 18:53

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