Talk:Matched filter

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[edit] Derivations of matched filter

On July 29, 2006, I made significant changes to the (formerly) "Problem statement" section of this article. The derivation that I found there was of great help. I added in a few missing lines of derivation noted by a previous contributor and clarified some points, such as the covariance matrix. I also added another section with an alternate derivation of the matched filter. I believe that it is best to derive the matched filter in the context of the inner conjugate product of the filter and the observed signal, since this requires only the use of vectors, matrices, and their conjugate transposes -- without a need to use transposes and complex conjugation alone. I believe that this significantly reduces the mathematical complexity. I have edited the Lagrangian derivation accordingly. --Rabbanis 04:46, 30 July 2006 (UTC)

I found the derivations to be very helpful, but I was thrown off by the notation for conjugate transpose. Perhaps the superscript 'H' notation should be explained in the article? In my background, I've never come across the 'H' (I think a dagger is what I saw in my textbooks). Cheers. Unexpect 07:04, 10 February 2007 (UTC)

I added a brief explanation of the H notation. Steve8675309 15:46, 10 February 2007 (UTC)
I agree that this is a good idea. Thanks for adding that explanation about the 'H'. Rabbanis 21:56, 20 February 2007 (UTC)

[edit] Notational Problem in definition of y, the conjugate inner product of the filter and the observed signal

The definition for y, the conjugate inner product of the filter and the observed signal, is given as a discrete convolution

y = \sum_{k=-\infty}^{+\infty} h^{*}[k] x[h] = h^{H}x = ... .

This looks wrong to me: how can h, *the filter*, be the index parameter to x? Shouldn't it actually be something like

y = \sum_{k=-\infty}^{+\infty} h^{*}[k] x[n-k] = h^{H}x = ... ?

69.121.100.249 22:48, 27 October 2006 (UTC)

You are absolutely right. This is a typo. I intended to use the index k. I did not use the discrete convolution, however, because I find that the derivation of the matched filter is more intuitive with the conjugate inner product. I realize that this is slightly confusing, as it is inconsistent with the use of h earlier, where it is the impulse response of an LTI system. Thanks for pointing this out. I have changed the index to k. Rabbanis 15:31, 31 October 2006 (UTC)


[edit] Changes on 15 Jan 2007 Regarding i.i.d. Noise

I believe that the recent changes to the derivation section on this date are not consistent with the body of the article. Specifically, I refer to the second line of the section stating that x is the sum of a signal h and i.i.d. noise. My concerns are:

  • We have been using s as the signal, while h is the impulse response of the filter.
  • The noise is not necessarily i.i.d. If it is, then the impulse response of the filter is the time-reversed complex-conjugate of the signal, but farther down we allow for correlated noise by generalizing the derivation for an arbitrary covariance matrix Rn = E{nnH}.

I propose that we revert back to the version prior to 15 Jan 2007. --Rabbanis 19:08, 16 January 2007 (UTC)

You're right. I didn't read far enough to see that h was not the signal. I removed my edit.
I have a question, though. Isn't conjugation assumed in the inner product operation? (http://planetmath.org/encyclopedia/InnerProduct.html) If so, shouldn't the term "conjugate inner product" (used a few places in the text) be replaced with just "inner product"?
Steve8675309 21:26, 16 January 2007 (UTC)
That's a good question. I've looked around and it seems that "dot product" and "inner product" are not exactly the same thing. I am not sure how "inner product" is defined for complex-valued vectors. Perhaps someone with a rigorous mathematics background can illuminate this point and make corrections as necessary. Rabbanis 00:22, 18 January 2007 (UTC)
I looked at Wiki's page on this. It says that inner products have to satisfy a few axioms including conjugate symmetry (\langle x,y\rangle =\overline{\langle y,x\rangle}). That axiom wouldn't be satisfied unless one of the vectors in an inner product is automatically conjugated. I'm pretty confident that the phrase "conjugate inner product" should just be "inner product". Steve8675309 01:48, 27 January 2007 (UTC)
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