Convolution relation

I have a question about a particular relationship with convolutions.
We want to transmit a particular waveform X, but through that process
the input signal X gets convolved with some annoyance term H and we
actually send Y. I.e.

Y = H * X (where * = convolution)

By knowing the X we want and the Y we actually end up with, our
problem is to make a different input signal Z that ends up being equal
to X. I.e.

H * Z = X

A solution can be found by deconvolution z = x^2/y (where lowercase
means the Fourier transform). However noise usually renders this
solution unusable.

My colleague came up with an approach that "works": Z = 2X - Y.
However I don't really know what this means because substituting that
expression into the two equations above leads to Y = X, which is...
wrong.

On 2008-07-11, spasmous <spasmous@gmail.com> wrote:
>
> My colleague came up with an approach that "works": Z = 2X - Y.
> However I don't really know what this means because substituting that
> expression into the two equations above leads to Y = X, which is...
> wrong.

Are you sure that the "annoying H" is really linear?

-- 
Ben Jackson AD7GD
<ben@ben.com>
http://www.ben.com/

On 11 Jul, 23:29, spasmous <spasm...@gmail.com> wrote:
> I have a question about a particular relationship with convolutions.
> We want to transmit a particular waveform X, but through that process
> the input signal X gets convolved with some annoyance term H and we
> actually send Y. I.e.
>
> Y = H * X (where * = convolution)
>
> By knowing the X we want and the Y we actually end up with, our
> problem is to make a different input signal Z that ends up being equal
> to X. I.e.
>
> H * Z = X

How did you find this expression? What about

Y = X * H

and instead of finding H try and estimate X from Y:

Y * G  = X'

where G is some filter to be determined and X' is an
estimator for the original X.

If you minimize the estimation error

E = | X' - X |^2

the problem all of a sudden looks deceptively similar
to the Wiener filter problem...

> A solution can be found by deconvolution z = x^2/y (where lowercase
> means the Fourier transform). However noise usually renders this
> solution unusable.
>
> My colleague came up with an approach that "works": Z = 2X - Y.
> However I don't really know what this means because substituting that
> expression into the two equations above leads to Y = X, which is...
> wrong.

No. The solution fits straight into the argument above with the
simplest G possible, i.e. G = 1. While your colleague might have
found that answer by an unconventional line of reasoning, the
solution is perfectly valid. The question is whether there are
other solutions on the same form that in some sense are better.

Rune

>I have a question about a particular relationship with convolutions.
>We want to transmit a particular waveform X, but through that process
>the input signal X gets convolved with some annoyance term H and we
>actually send Y. I.e.
>
>Y = H * X (where * = convolution)
>
>By knowing the X we want and the Y we actually end up with, our
>problem is to make a different input signal Z that ends up being equal
>to X. I.e.
>
>H * Z = X
>
>A solution can be found by deconvolution z = x^2/y (where lowercase
>means the Fourier transform). However noise usually renders this
>solution unusable.
>
>My colleague came up with an approach that "works": Z = 2X - Y.
>However I don't really know what this means because substituting that
>expression into the two equations above leads to Y = X, which is...
>wrong.
>

%%%%

Is this not the same problem as the transmission of data thru linear
channel?

where H is channel, X is transmitted symbol, Y is received symbol

Now at receiver we need to perform two taks: Design a filter and estimate
X.

Two options:
1) if you assume that H is known, then use MMSE criteria and get a filter

say A and X'=A*Y, where X' is estimate of data.

2) if you do not know H, then u can train your filter using LMS or RLS
algorithm, and estimate X.

Chintan