Reply by Stephan M. Bernsee●October 26, 20042004-10-26
On 2004-10-25 23:53:14 +0200, Skint
<wayneREMOVECAPSmarsh@dsl.pipexREMOVE.cXXXXom> said:
> Okay, yes. I don't have any formal mathematical background, as you can
> tell. I know that what I'm suggesting isn't exactly standard
> deviation, but it seemed a fair enough description at the time.
> What I mean was that instead of multiplying and summing/integrating a
> window against a signal to produce the output, you could instead take
> the difference of each sample in comparison, and take the average of
> that. The inverse of this result is then used as the output signal
> (as when the windows are perfectly aligned the result would be
> smaller, when we actually want it to be bigger).
>
> Does that make sense at all to anybody? Anywhere? Somebody?
Yes, that is called the average (squared) magnitude difference function
(AMDF). It is sometimes used instead of the correlation function for
similar purposes, usually in speech recognition instead of the
auto-correlation to extract the fundamental frequency.
--
Stephan M. Bernsee
http://www.dspdimension.com
Reply by Bob Cain●October 26, 20042004-10-26
Kedi wrote:
> Hi Bob,
>
> I must apologize for not meaning literally what I said. When I wrote
> "correlation tells you the sum of the squares of the difference
> between 2 signals", I mean:
>
> Given 2 signals, both with energy E
>
> (E - their inner product) = half the energy of (their difference)
Ah, so. Thanks. Your math did say that but the words were
at odds.
Bob
--
"Things should be described as simply as possible, but no
simpler."
A. Einstein
Reply by Skint●October 25, 20042004-10-25
On Mon, 25 Oct 2004 17:16:24 -0400, Jerry Avins <jya@ieee.org> wrote:
>Skint wrote:
>> On Mon, 25 Oct 2004 20:42:32 +0200, Stephan M. Bernsee
>> <spam@dspdimension.com> wrote:
>>
>>
>>>On 2004-10-25 03:53:38 +0200, Skint
>>><wayneREMOVECAPSmarsh@dsl.pipexREMOVE.cXXXXom> said:
>>>
>>>
>>>>For example, with a signal bounded between -1 and 1, and values of
>>>>around 0.3 in the target signal, won't the resulting cross-correlation
>>>>signal always be highest in the areas where the original signal
>>>>averages around its peak? i.e. 0.3*1 > 0.3*0.3, by far. Surely in that
>>>>case the correlation signal is going to show higher correlation where
>>>>there is none?
>>>
>>>I'm sorry but I don't really understand your question. Can you be a bit
>>>more specific as to how your two data sets look like that you're trying
>>>to cross correlate?
>>
>>
>> It's okay, I've worked it out in my head now. I was looking form the
>> wrong philosophy as it were.
>>
>> Another, unrelated thought on the subject - could a correlation signal
>> be generated by the inverse of the standard deviation of the shifted
>> signals with respect to each other? This seems more instantly
>> intuitive.
>
>It ain't intuitive to me. I have no idea what you mean. I am confused on
>several points:
>
>I thought that a correlation is a number, not a signal.
>
>Standard deviation is a statistical measure related to the width of a
>probability density. What does the standard deviation of two signals
>mean? How is it calculated?
Okay, yes. I don't have any formal mathematical background, as you can
tell. I know that what I'm suggesting isn't exactly standard
deviation, but it seemed a fair enough description at the time.
What I mean was that instead of multiplying and summing/integrating a
window against a signal to produce the output, you could instead take
the difference of each sample in comparison, and take the average of
that. The inverse of this result is then used as the output signal
(as when the windows are perfectly aligned the result would be
smaller, when we actually want it to be bigger).
Does that make sense at all to anybody? Anywhere? Somebody?
Reply by Jerry Avins●October 25, 20042004-10-25
Skint wrote:
> On Mon, 25 Oct 2004 20:42:32 +0200, Stephan M. Bernsee
> <spam@dspdimension.com> wrote:
>
>
>>On 2004-10-25 03:53:38 +0200, Skint
>><wayneREMOVECAPSmarsh@dsl.pipexREMOVE.cXXXXom> said:
>>
>>
>>>For example, with a signal bounded between -1 and 1, and values of
>>>around 0.3 in the target signal, won't the resulting cross-correlation
>>>signal always be highest in the areas where the original signal
>>>averages around its peak? i.e. 0.3*1 > 0.3*0.3, by far. Surely in that
>>>case the correlation signal is going to show higher correlation where
>>>there is none?
>>
>>I'm sorry but I don't really understand your question. Can you be a bit
>>more specific as to how your two data sets look like that you're trying
>>to cross correlate?
>
>
> It's okay, I've worked it out in my head now. I was looking form the
> wrong philosophy as it were.
>
> Another, unrelated thought on the subject - could a correlation signal
> be generated by the inverse of the standard deviation of the shifted
> signals with respect to each other? This seems more instantly
> intuitive.
It ain't intuitive to me. I have no idea what you mean. I am confused on
several points:
I thought that a correlation is a number, not a signal.
Standard deviation is a statistical measure related to the width of a
probability density. What does the standard deviation of two signals
mean? How is it calculated?
Your inverse, is it multiplicative? Additive?
Standard deviation is always positive. Correlation can be positive or
negative.
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by Skint●October 25, 20042004-10-25
On Mon, 25 Oct 2004 20:42:32 +0200, Stephan M. Bernsee
<spam@dspdimension.com> wrote:
>On 2004-10-25 03:53:38 +0200, Skint
><wayneREMOVECAPSmarsh@dsl.pipexREMOVE.cXXXXom> said:
>
>> For example, with a signal bounded between -1 and 1, and values of
>> around 0.3 in the target signal, won't the resulting cross-correlation
>> signal always be highest in the areas where the original signal
>> averages around its peak? i.e. 0.3*1 > 0.3*0.3, by far. Surely in that
>> case the correlation signal is going to show higher correlation where
>> there is none?
>
>I'm sorry but I don't really understand your question. Can you be a bit
>more specific as to how your two data sets look like that you're trying
>to cross correlate?
It's okay, I've worked it out in my head now. I was looking form the
wrong philosophy as it were.
Another, unrelated thought on the subject - could a correlation signal
be generated by the inverse of the standard deviation of the shifted
signals with respect to each other? This seems more instantly
intuitive.
Reply by Kedi●October 25, 20042004-10-25
Hi Bob,
I must apologize for not meaning literally what I said. When I wrote
"correlation tells you the sum of the squares of the difference
between 2 signals", I mean:
Given 2 signals, both with energy E
(E - their inner product) = half the energy of (their difference)
Regards.
KD
Reply by Stephan M. Bernsee●October 25, 20042004-10-25
On 2004-10-25 03:53:38 +0200, Skint
<wayneREMOVECAPSmarsh@dsl.pipexREMOVE.cXXXXom> said:
> For example, with a signal bounded between -1 and 1, and values of
> around 0.3 in the target signal, won't the resulting cross-correlation
> signal always be highest in the areas where the original signal
> averages around its peak? i.e. 0.3*1 > 0.3*0.3, by far. Surely in that
> case the correlation signal is going to show higher correlation where
> there is none?
I'm sorry but I don't really understand your question. Can you be a bit
more specific as to how your two data sets look like that you're trying
to cross correlate?
--
Stephan M. Bernsee
http://www.dspdimension.com
Reply by Bob Cain●October 25, 20042004-10-25
Rick Lyons wrote:
> Good Hell! I wonder what caused "The Ghost" to make
> that wisecrack.
He's obsessed with me and now knows most of the places where
I hang out so that he can stalk me as he's been doing for
years on alt.sci.physics.acoustics but on a broader front.
It's rather sad to be so filled with hatred.
Bob
--
"Things should be described as simply as possible, but no
simpler."
A. Einstein
Reply by Rick Lyons●October 25, 20042004-10-25
On Sun, 24 Oct 2004 16:12:18 -0700, Bob Cain
<arcane@arcanemethods.com> wrote:
>
>
>The Ghost wrote:
>
>> Bob Cain <arcane@arcanemethods.com> wrote in message news:<clf0jt01vlm@enews2.newsguy.com>...
>>
>>
>>>As I said, just a nit. The real content was illuminating
>>>for me. One of the problems with being self taught is that
>>>the lack of a disciplined course can result in a hit or miss
>>>base of knowledge.
>>
>>
>> Based on your track record, it is appears to be mostly misses and few
>> if any sutstantive hits.
>
>Print this out and take it to your next therapy session.
>
> http://www.apa.org/journals/psp/psp7761121.html
>
>I'm sure your mental health professional will appreciate the
>insight it gives relative to your pathology.
>
>It'll be alright, Gary. New drugs are emerging for about
>any mental ailment so relief shouldn't be that far away for you.
>
>
>Bob
Good Hell! I wonder what caused "The Ghost" to make
that wisecrack.
[-Rick-]
Reply by Rick Lyons●October 25, 20042004-10-25
On Sat, 23 Oct 2004 18:27:05 -0700, Bob Cain
<arcane@arcanemethods.com> wrote:
>
>
>Kedi wrote:
>> On 2nd thought, Bob, you are not "quite right". You are "absolutely
>> right" in saying that I should have used the word "inner product" in
>> all my previous posts.
>
>As I said, just a nit. The real content was illuminating
>for me. One of the problems with being self taught is that
>the lack of a disciplined course can result in a hit or miss
>base of knowledge.
(snipped)
Whew! You're not kiddin' !!
But that's one of the great values of this news
group. The guys here sure help us improve
our aim.
[-Rick-]