FPGA Central

On 2009-08-05 15:33:29 +0200, Jerry Avins <[email protected]> said:

> Alex wrote:
>> On 2009-08-05 13:56:15 +0200, Jerry Avins <[email protected]> said:
>>
>>> Alex wrote:
>>>
>>> ...
>>>
>>>> What kind of signal are these? Here is some extra information:
>>>>
>>>> All my signals are a 40kHz.
>>>> X, Y, X2 and Y2 have values that can go from -inf to +inf.
>>>
>>> Ah! I see the problem! How are these limits represented in your data?
>>> Can you distinguish between inf and inf/2?
>>>
>>> Get real.
>>>
>>> ...
>>>
>>> Jerry
>>
>> +inf and -inf was only a way of speaking to say that the signals are
>> not feeted in a range.
>
> "Feeted"? I can't decode that.

I meant fitted... sorry for that. :-)

>
>> The signal is comming in real time from a sensor that could give really
>> high values, but in practice it rarelly go over +10 and -10.
>>
>> That helps?
>
> It helps to know that when you write one thing, you might well mean
> another. All sensors have bounded outputs.
>
>> I just want to know if there are any techniques to filter a signal
>> using knowledge of the original sources, other than a range of
>> frequencies.
>
> I'm sure there are, but those methods must be specific to
> characteristics of the signal that you haven't yet made clear. What do
> you mean by "the norm of every couple of values"? That condition
> appears to apply to the sources conjointly. How so?

What I mean by "the norm of every couple of values" is that:

X1 and Y1 are a the couple of source signals I'm mainly looking for.
After the source separation X1 is a series of values {x11, x12, x13,
etc... } and Y1 is another series of values {y11, y12, y13, etc... }

If we take a sample of X1 and another of Y1 at the same time, lets say
x13 and y13, these two values represent a vector (x13,y13) whose norm
is 1. In other words:

SquareRoot(x13 * x13 + y13 * y13) = 1

That also means that we can find the ideal value of x13 using a suposed
rigth y13 in this way:

x13 = SquareRoot(1 - y13*y13)

and vice versa.

>
>> I have a very strong condition for the output signal and the dependency
>> of X and Y and I know I could have a better result if my filter could
>> take it into account.
>
>
>> I'm not asking for a solution, but for some tips about where to look.
>
> You haven't given enough specifics about what the situation really is.

What else I can say? All the signals are 1D, the sensor gives float
values but I do not know the precision of each sample.

I'm a PhD in Computer Graphics, but just a beginner in DSP, so please
tell me what else you will need to know and I'll do my best to be
specific considering that I cannot talk a lot about the device I'm
using. :-/

>
> Jerry

In any case, thaks a lot for your help.

Alex

Risto Ekholm wrote:

...

> SquareRoot(x13 * x13 + y13 * y13) = 1

I think you mean sqrt(x1[3]^2 + y1[3]^2 = 1, as if x1 and y1 were sine
and cosine. How then can either of them exceed 1?

> That also means that we can find the ideal value of x13 using a suposed
> rigth y13 in this way:
>
> x13 = SquareRoot(1 - y13*y13)
>
> and vice versa.
>
>>
>>> I have a very strong condition for the output signal and the dependency
>>> of X and Y and I know I could have a better result if my filter could
>>> take it into account.
>>
>>
>>> I'm not asking for a solution, but for some tips about where to look.
>>
>> You haven't given enough specifics about what the situation really is.
>
> What else I can say? All the signals are 1D, the sensor gives float
> values but I do not know the precision of each sample.
>
> I'm a PhD in Computer Graphics, but just a beginner in DSP, so please
> tell me what else you will need to know and I'll do my best to be
> specific considering that I cannot talk a lot about the device I'm
> using. :-/

The signals are noisy and you want to remove as much of the noise as you
can, right? How fast can they change?

Jerry
--
Engineering is the art of making what you want from things you can get.
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