On 9/25/2014 2:31 PM, Evgeny Filatov wrote:
> On 9/24/2014 9:34 PM, robert bristow-johnson wrote:
>>
>> i have to confess, Bob, that i am not persuaded.  i set up the question
>> a bit more with the start of an answer (in two modes, one using the
>> language of Hilbert Spaces and the other not).  please take another look
>> at the page, if you want.
>>
>> L8r,
>>
>
> Hi,
>
> To borrow a quote from "Monday Starts on Saturday", "It's nonsense to
> look for a solution if it already exists. We are talking about how to
> deal with a problem that has no solution."
>
> The problem is that your problem in the general form as expressed in
> your initial post has no solution which is proved by a few
> counterexamples provided here.
>
> May be, it's prudent to provide some additional constraints. For
> example, simulations I've run indicate that your theorem seem to work IF
> the signals X, Y and Z are both even AND convex curves (within a
> period). It doesn't work if X,Y, and Z are only even but not convex. It
> doesn't work if X,Y, and Z are only convex but not even.
>
> Hope that helps.
>
> Evgeny.
>

Oops. I've meant to say that a reasonable guess about the sufficient 
conditions for Robert's theorem is that the shapes of signals X, Y, and 
Z are both symmetrical and rise monotonously from their minimum to 
maximum values (within a period, that is).

Even then you would have to deal with the ambiguity related to the fact 
that there might be more than one maximum in cross-correlation functions 
per period, so that's assuming there is only one.

Evgeny.

Robert

I've only spent a few minutes thinking about this so I'm likely not correct. I have one question though. The cross- correlation between any 2 of your 3 periodic signals could have a negative peak instead of a positive peak. In that case your formula for finding the location of the max Rxy would fail ( well your formula will give some answer but should you be looking for the location of the max abs(Rxy)  instead ?)

Bob

On 9/24/2014 9:34 PM, robert bristow-johnson wrote:
>
> i have to confess, Bob, that i am not persuaded.  i set up the question
> a bit more with the start of an answer (in two modes, one using the
> language of Hilbert Spaces and the other not).  please take another look
> at the page, if you want.
>
> L8r,
>

Hi,

To borrow a quote from "Monday Starts on Saturday", "It's nonsense to 
look for a solution if it already exists. We are talking about how to 
deal with a problem that has no solution."

The problem is that your problem in the general form as expressed in 
your initial post has no solution which is proved by a few 
counterexamples provided here.

May be, it's prudent to provide some additional constraints. For 
example, simulations I've run indicate that your theorem seem to work IF 
the signals X, Y and Z are both even AND convex curves (within a 
period). It doesn't work if X,Y, and Z are only even but not convex. It 
doesn't work if X,Y, and Z are only convex but not even.

Hope that helps.

Evgeny.

On 9/24/14 7:06 AM, radams2000@gmail.com wrote:
> Oops, I mixed up the value of the peak correlation with the position of the peak. I'll think about it some more.
>
> Here's a counter- example.
>
>   Let the transfer function between X and Y be 1 + a1* Z^-k where a1 = 1.1.
>
> Let the transfer function between Y and Z be 1 - a2*Z^-k where a2 = 0.5.
>
> The peak correlation between X and Y will occur at a lag of K samples (because a1>  1). The peak correlation between Y and Z will occur at a lag of zero samples (because a2<  1). The peak correlation between X and Z will occur at a lag of zero samples.  That's because some cancellation of the delayed impulse at sample k took place, reducing its value to the point where it no longer dominates.
>

i have to confess, Bob, that i am not persuaded.  i set up the question 
a bit more with the start of an answer (in two modes, one using the 
language of Hilbert Spaces and the other not).  please take another look 
at the page, if you want.

L8r,

-- 

r b-j                  rbj@audioimagination.com

"Imagination is more important than knowledge."

On 9/24/14 3:44 AM, rickman wrote:
> On 9/23/2014 9:52 AM, robert bristow-johnson wrote:
>>
>> ain't ya just proud of me?
>>
>> it's at
>>
>> http://dsp.stackexchange.com/questions/18316/cascaded-cross-correlation
>>
>>
>> doesn't look like anyone is taking it up. would anyone here at comp.dsp
>> be willing to look at it? it's about cross-correlation and using it to
>> sorta measure the delay between one signal and another. i would like to
>> know if that delay measure adds when the sorta delay elements (there's
>> potentially filtering besides the delay) are cascaded.
>>
>> if someone can take a whack at it, i would appreciate it. i'll be
>> whacking at it myself.
>
> Uh, are you asking people here to go there to answer your question? Why
> not ask it here?
>

i was thinking about that.  i'll admit that i've grown tired of doing 
ASCII math.

in a sense, i *am* asking here.  it's why i posted this.  just asking 
that you go to the link to see the expression of the question.

-- 

r b-j                  rbj@audioimagination.com

"Imagination is more important than knowledge."

Have a free upvote on the grounds I dont like stackexchange and I have a peripheral interest in autocorrelation

On 9/24/2014 3:06 PM, radams2000@gmail.com wrote:
> Oops, I mixed up the value of the peak correlation with the position of the peak. I'll think about it some more.
>
> Here's a counter- example.
>
>   Let the transfer function between X and Y be 1 + a1* Z^-k where a1 = 1.1.
>
> Let the transfer function between Y and Z be 1 - a2*Z^-k where a2 = 0.5.
>
> The peak correlation between X and Y will occur at a lag of K samples (because a1> 1). The peak correlation between Y and Z will occur at a lag of zero samples (because a2 < 1). The peak correlation between X and Z will occur at a lag of zero samples.  That's because some cancellation of the delayed impulse at sample k took place, reducing its value to the point where it no longer dominates.
>
> Bob
>

It's a nice example. Should I trust my gut feeling that (strictly 
speaking) the above reasoning holds correct only in cases involving zero 
ISI?

Evgeny.

Oops, I mixed up the value of the peak correlation with the position of the peak. I'll think about it some more. 

Here's a counter- example.

 Let the transfer function between X and Y be 1 + a1* Z^-k where a1 = 1.1. 

Let the transfer function between Y and Z be 1 - a2*Z^-k where a2 = 0.5. 

The peak correlation between X and Y will occur at a lag of K samples (because a1> 1). The peak correlation between Y and Z will occur at a lag of zero samples (because a2 < 1). The peak correlation between X and Z will occur at a lag of zero samples.  That's because some cancellation of the delayed impulse at sample k took place, reducing its value to the point where it no longer dominates. 

Bob

On 9/23/2014 9:52 AM, robert bristow-johnson wrote:
>
>   ain't ya just proud of me?
>
> it's at
>
>   http://dsp.stackexchange.com/questions/18316/cascaded-cross-correlation
>
>
> doesn't look like anyone is taking it up.  would anyone here at comp.dsp
> be willing to look at it?  it's about cross-correlation and using it to
> sorta measure the delay between one signal and another.  i would like to
> know if that delay measure adds when the sorta delay elements (there's
> potentially filtering besides the delay) are cascaded.
>
> if someone can take a whack at it, i would appreciate it.  i'll be
> whacking at it myself.

Uh, are you asking people here to go there to answer your question?  Why 
not ask it here?

-- 

Rick