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From: illywhacker on 3 Jun 2010 06:48
On Jun 1, 5:23 am, "cfy30" <cfy30(a)n_o_s_p_a_m.yahoo.com> wrote:
> Thanks all for the inputs!
>
> It seems to me now
> 1. If both x and y are deterministic signal and periodic, orthogonality is
> the integration of x*y from 0 to T while correlation is the integration of
> x*y from 0 to n*T. The only difference between orthogonality and
> correlation is the integration length.
> 2. If both x and y are random variable, there is no orthogonality as T
> cannot be defined but correlation can still be found since T is arbitrary.
>
> Consider an ideal receiver with random data n(t) as input, and ideal I and
> Q down-converters, cos(omega*t) and sin(omega*t), respectively. The I and Q
> outputs, namely n(t)*cos(omega*t) and n(t)*sin(omega*t), they are
> uncorrelated and no orthogonal can be defined. Is this aligned with
> everyone's understanding?

As another poster has pointed out, there are two definitions of
'correlation'. I suggest that you forget the word 'correlation' when
it means 'orthogonality': use 'orthogonality' instead and you will not
get confused. Reserve 'correlation' for probabilistic statements.

Secondly, do please forget the idea of a 'deterministic signal' or a
'signal that is a random variable'. These statements have no meaning:
'random' and 'deterministic' are not properties of signals, or of the
real world external to you at all, but of your knowledge of it.
'Deterministic' means you have no uncertainty (i.e. perfect knowledge)
of the phenomennon under consideration; 'random' means you do not.

illywhacker;
From: Frank on 3 Jun 2010 09:17
On Jun 3, 12:38 pm, illywhacker <illywac...(a)gmail.com> wrote:
> On Jun 2, 1:54 pm, Frank <frank.snow...(a)gmail.com> wrote:
>
> > e.g. we form z(t) fromorthogonalfunctions x(t) and y(t), and denote
> > the inner product by <., .>
>
> > z(t) = a x(t) + b y(t)
>
> > we can uniquely determine the values of a and b by taking the
> > following
> > inner products
>
> > <z(t), x(t)> = a <x(t), x(t)> + b <y(t), x(t)> = a
> > <z(t), y(t)> = a <x(t), y(t)> + b <y(t), y(t)> = b
>
> If you know what x and y are, then taking the inner product of z with
> any two linearly independent functions will give you two simultaneous
> equations for a and b, thereby uniquely determining a nd b provided
> that x and y are also linearly independent. It has nothing to do with
> x and y beingorthogonal.
>
> illywhacker

Indeed, I was trying not to make the whole piece become too
overwhelming, but allow me to address your point and hopefully also
clarify mine.

That two functions are linearly independent means that those two
such functions may be converted to two orthogonal functions (in
the sense above) by projection allowing a conversion which gives
exactly the formulation I originally provided (and implicitly solving
the simultaneous equations you mention). In this sense, it has a
lot to do with being orthogonal since the linearly independent
functions
are in fact just formed from two orthogonal basis functions. I
apologize that my exact original choice of words was misleading and
not 100% mathematically correct, but I was hoping to make my point
without covering every nuance and making it inscrutable.

In summary, the point I wanted to make was simply that orthogonal
functions are chosen for use in communication systems because
of the (relative) simplicity of both the interpretation and
implementation.

Frank
From: illywhacker on 3 Jun 2010 11:18
On Jun 3, 3:17 pm, Frank <frank.snow...(a)gmail.com> wrote:
> On Jun 3, 12:38 pm, illywhacker <illywac...(a)gmail.com> wrote:
>
>
>
>
>
> > On Jun 2, 1:54 pm, Frank <frank.snow...(a)gmail.com> wrote:
>
> > > e.g. we form z(t) fromorthogonalfunctions x(t) and y(t), and denote
> > > the inner product by <., .>
>
> > > z(t) = a x(t) + b y(t)
>
> > > we can uniquely determine the values of a and b by taking the
> > > following
> > > inner products
>
> > > <z(t), x(t)> = a <x(t), x(t)> + b <y(t), x(t)> = a
> > > <z(t), y(t)> = a <x(t), y(t)> + b <y(t), y(t)> = b
>
> > If you know what x and y are, then taking the inner product of z with
> > any two linearly independent functions will give you two simultaneous
> > equations for a and b, thereby uniquely determining a nd b provided
> > that x and y are also linearly independent. It has nothing to do with
> > x and y beingorthogonal.
>
> > illywhacker
>
> Indeed, I was trying not to make the whole piece become too
> overwhelming, but allow me to address your point and hopefully also
> clarify mine.
>
> That two functions are linearly independent means that those two
> such functions may be converted to two orthogonal functions (in
> the sense above) by projection allowing a conversion which gives
> exactly the formulation I originally provided (and implicitly solving
> the simultaneous equations you mention). In this sense, it has a
> lot to do with being orthogonal since the linearly independent
> functions
> are in fact just formed from two orthogonal basis functions. I
> apologize that my exact original choice of words was misleading and
> not 100% mathematically correct, but I was hoping to make my point
> without covering every nuance and making it inscrutable.
>
> In summary, the point I wanted to make was simply that orthogonal
> functions are chosen for use in communication systems because
> of the (relative) simplicity of both the interpretation and
> implementation.

This last point is true, of course, but your original point was
confusing. Linear independence has no dependence on orthogonality:
this is putting the cart before the horse. Any vector space has the
notion of linear independence, but in order to have orthogonality, you
need an inner product as well. Since by definition, an inner product
is a bilinear form, it depends on linearity for its definition, but
not the other way around.

illywhacker;
From: Frank on 4 Jun 2010 11:01
On Jun 3, 5:18 pm, illywhacker <illywac...(a)gmail.com> wrote:
> On Jun 3, 3:17 pm, Frank <frank.snow...(a)gmail.com> wrote:
>
>
>
>
>
> > On Jun 3, 12:38 pm, illywhacker <illywac...(a)gmail.com> wrote:
>
> > > On Jun 2, 1:54 pm, Frank <frank.snow...(a)gmail.com> wrote:
>
> > > > e.g. we form z(t) fromorthogonalfunctions x(t) and y(t), and denote
> > > > the inner product by <., .>
>
> > > > z(t) = a x(t) + b y(t)
>
> > > > we can uniquely determine the values of a and b by taking the
> > > > following
> > > > inner products
>
> > > > <z(t), x(t)> = a <x(t), x(t)> + b <y(t), x(t)> = a
> > > > <z(t), y(t)> = a <x(t), y(t)> + b <y(t), y(t)> = b
>
> > > If you know what x and y are, then taking the inner product of z with
> > > any two linearly independent functions will give you two simultaneous
> > > equations for a and b, thereby uniquely determining a nd b provided
> > > that x and y are also linearly independent. It has nothing to do with
> > > x and y beingorthogonal.
>
> > > illywhacker
>
> > Indeed, I was trying not to make the whole piece become too
> > overwhelming, but allow me to address your point and hopefully also
> > clarify mine.
>
> > That two functions are linearly independent means that those two
> > such functions may be converted to two orthogonal functions (in
> > the sense above) by projection allowing a conversion which gives
> > exactly the formulation I originally provided (and implicitly solving
> > the simultaneous equations you mention). In this sense, it has a
> > lot to do with being orthogonal since the linearly independent
> > functions
> > are in fact just formed from two orthogonal basis functions. I
> > apologize that my exact original choice of words was misleading and
> > not 100% mathematically correct, but I was hoping to make my point
> > without covering every nuance and making it inscrutable.
>
> > In summary, the point I wanted to make was simply that orthogonal
> > functions are chosen for use in communication systems because
> > of the (relative) simplicity of both the interpretation and
> > implementation.
>
> This last point is true, of course, but your original point was
> confusing. Linear independence has no dependence onorthogonality:
> this is putting the cart before the horse. Any vector space has the
> notion of linear independence, but in order to haveorthogonality, you
> need an inner product as well. Since by definition, an inner product
> is a bilinear form, it depends on linearity for its definition, but
> not the other way around.
>
> illywhacker;


First a clarification:
I never mentioned linear independence in my original point, nor did I
say that linear independence depends on orthogonality for its
definition (although I did make a possibly invalid related assumption
- see my question further down), nor did I say that the inner-product
was the only way to reach a solution (which may have been implied, but
not intended). My original post - the related part - was a much
simpler point; simply that orthogonal functions could be summed
together and their respective weights determined by inner-product. The
intention was to show that correlation was equal to orthogonality only
where it was defined so, and that it often is defined in this way for
communication systems.

Don't get me wrong, I understand your points and appreciate the
feedback greatly. Certainly I hadn't considered the relation between
linear independence and orthogonality in that way before (or at all
really, since I'm a DSP engineer and the question doesn't arise in the
systems I typically deal with).

With that said, there is one point that I can't work out (assumption I
made really), and maybe you can answer it (and this is getting off-
topic for this group, I apologise):

If an inner-product does not exist, then is it still possible to
construct the two simultaneous equations you mentioned in your
previous post? I had assumed that this would not be possible (or not
with unique solution ?)

Frank
From: illywhacker on 4 Jun 2010 11:39
On Jun 4, 5:01 pm, Frank <frank.snow...(a)gmail.com> wrote:

> If an inner-product does not exist, then is it still possible to
> construct the two simultaneous equations you mentioned in your
> previous post? I had assumed that this would not be possible (or not
> with unique solution ?)

Sure. You have n constraints, where n is the dimensionality of the
vector space - in principle infinite in this case - and only two
unknowns. Just write down the equations for several values of t, and
you will see what I mean.

illywhacker;
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