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Hi: I am reading a book called "lectures on wiener and kalman filtering" by Professor Thomas Kailath. On page 18, it says the following:

To carry through this approach, let us first note that a smooth random process $ \{ Y(\tau), a \le \tau \le b \}$ can be approximately represented as

$$Y(\tau) = \sum_{i=0}^{n-1} Y(\tau_{i}) \times \sqrt \triangle \times p(\tau - i \triangle) = \sum_{i=0}^{n-1} Y_{i} \times p(\tau - i \triangle)$$

where

$ p(\tau) = \begin{cases}\frac{1}{\sqrt \triangle}, & 0 \le \tau \le \triangle \\ 0, & \text{otherwise}\end{cases}$

There is a picture included with the text:

enter image description here

$\triangle$ is the length of the space between the discrete $\tau_{i}$. The $\tau_{i}$ are the discrete versions of the time and are equally spaced.

I have some questions about this formula but I can send someone the pages from the book if they need them since, without a picture, it may be difficult to understand. ( note that I don't even follow it with the picture !!!!! ).

  1. I don't have any intuition for this smoothing formula. This seems like some kind of windowing process that uses weights that are smaller for further away values and larger for closer values ? Is there a name for it ?

  2. The second equality in the smoothing relation implies that $Y(\tau_{i}) \times \sqrt \triangle = Y_{i} $. I don't follow that ?

  3. It looks like the weights are going to be negative and positive depending on the value of $\tau$. Is this correct ?

  4. Like I said, I can send the 3 pages ( what I wrote is a part of a longer derivation of a functional relation for least squares ) to anyone who is interested. Dr. Kalaith makes this approximation sound trivial but it's not to me.

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  • $\begingroup$ i dunno if there is a good name for this within the context of Kalman or Wiener filtering. but it appears to me to be filtering through a Zero-order hold that has DC gain of $\frac{1}{\sqrt{\triangle}}$. (naw, now that i look more closely at it, it appears to have a DC gain of 1.) $\endgroup$ – robert bristow-johnson Jun 3 '16 at 17:41
  • $\begingroup$ thanks but I don't come from an EE background ( my background is statistics ) so I don't know what "zero order hold" or "DC gain" means. I'll click on the link to zero-order hold to see what that gives but I doubt that it will help my understanding. still, your help is much appreciated. $\endgroup$ – mark leeds Jun 3 '16 at 17:54
  • $\begingroup$ HI Robert: I clicked on the link and don't follow it but that link has other links so I'll make an attempt at understanding. The language is totally different for me and probably makes things seem more difficult than they actually are. Like I said, it's very much appreciated but if anyone has a statistical explanation that would still be appreciated also. I'll let you know how my understanding goes but it will be a while :). All the best. $\endgroup$ – mark leeds Jun 3 '16 at 18:13
  • $\begingroup$ Mark, "zero order hold" just means that the continuous-time signal is approximated by piecewise constant ("zero order") values taken at the sample instants. So if we sample $p(t)$ at $t=nT$ to get $p[n]$ and then try to reconstruct $p(t)$ as $\hat{p}(t)$ then $\hat{p}(t) = p[k]$ for $kT \le t \le (k+1)T$. $\endgroup$ – Peter K. Jun 3 '16 at 18:46
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1) I don't have any intuition for this smoothing formula. This seems like some kind of windowing process that uses weights that are smaller for further away values and larger for closer values ? Is there a name for it ?

The first misconception is that this is not a smoothing formula. It's a way of representing an already smooth continuous-time signal in discrete samples.

All it's doing is as @robert says in the comments: it's assuming that the value of the signal between $\tau_i$ and $\tau_{i+1}$ is just $Y(\tau_i)$ --- it's piecewise constant.

I believe that $p(t)$ is just a "pulse".

2) The second equality in the smoothing relation implies that $Y(\tau_{i}) \times \sqrt \triangle = Y_{i} $. I don't follow that ?

This is just allowing us to simplify $Y(\tau_{i}) \times \sqrt \triangle$ and allow us to talk about $Y_i$ as shorthand notation.

edit from r b-j: it's just an alternative shorthand notation for $Y(\tau)\bigg|_{\tau = \tau_i}$ if $Y_i$ was anything else (like coefficients from some other set), i can't say that it's a ZOH with a DC gain of 0 dB. and that "$\sqrt{\triangle}$" is chaff.

3) It looks like the weights are going to be negative and positive depending on the value of $\tau$. Is this correct ?

The "weights" are either $0$ or $\frac{1}{\sqrt{\triangle}}$. For any value of $\tau$ outside $0 \le \tau - i\triangle \le \triangle$ the "weight" (pulse) is zero.

4) Like I said, I can send the 3 pages ( what I wrote is a part of a longer derivation of a functional relation for least squares ) to anyone who is interested. Dr. Kalaith makes this approximation sound trivial but it's not to me..

Please give me feedback on the above answers.

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    $\begingroup$ hey, it might be a smoothing formula if you consider the samples to be attached to dirac impulses. "smoothing filters" and ZOH both share the property of being LPFs. $\endgroup$ – robert bristow-johnson Jun 3 '16 at 19:58
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    $\begingroup$ @robertbristow-johnson yes, but the text does not imply that's what is meant by "smooth"... YMMV $\endgroup$ – Peter K. Jun 3 '16 at 20:02
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    $\begingroup$ @Peter K: I printed out the long answers at the link and will read them sloooooooowly and look in references for whatever I need to understand. There are some good and detailed answers in there but it will take me a while so don't wait up :). I think the $\triangle$ in Kalaith's notation is analogous to the $\tau$ that Robert defines at the beginning when he's defining the dirac impulse. So, there's probably an answer in there somewhere. Thanks. $\endgroup$ – mark leeds Apr 19 at 2:43
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    $\begingroup$ Peter K: The answers at the link regarding zero order hold ( by Robert and Timo ) are probably better than those in any textbook. They even mention that most texts gloss over the issue or explain it incorrectly. I've been going through those answers slowly and will have to look up some things in textbooks but the issue of the scale factor in the formula is discussed in detail. So, I found the right place to go for sure. Thanks for your help. $\endgroup$ – mark leeds Apr 19 at 18:20
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    $\begingroup$ no problem peter K. just as a last comment for anyone who comes to this thread, if you want to understand the math-fourier transform theory behind zero-order hold, go to that link. There are really masterful answers over there. I'm still working on understanding them but atleast I'm on the right track. $\endgroup$ – mark leeds Apr 21 at 0:22

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