Low freq "analog" of Nyquist? ( possibly naive question )

Glen,

Whew!  I'm top-posting to introduce my comments.  While I've worked in
acoustics it's always been at the applied / system level.  So, I've not had
the occasion to deal with those particular differential equations in any
detail.  That makes the discussion interesting!

I'll put a DSP spin on it below.

Regards,

Fred

"Glen Herrmannsfeldt" <gah@ugcs.caltech.edu> wrote in message
news:1rRMa.2466$NW6.2669@rwcrnsc51.ops.asp.att.net...
>
> "Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message
> news:%cQMa.2283$Jk5.1212136@feed2.centurytel.net...
> >
> > "Glen Herrmannsfeldt" <gah@ugcs.caltech.edu> wrote in message
> > news:i0LMa.1321$Ey6.785@rwcrnsc52.ops.asp.att.net...
> > >
>
> (big snip)
>
>
> > > Consider a signal, f, sampled over time T, from t=0 to t=T.  Assume
that
> > > f(0)=f(T)=0 for now.  All the components must be sine with periods
that
> > >re multiples of 2T.
> >
> > Not yet.  You didn't say that this was going to be considered to be one
> > period of a periodic waveform yet.  Or, you didn't say that the spectrum
> > would be discrete.
>
> With f(0) = f(T) = 0 they pretty much have to be.   Now, you can argue
that
> it is an arbitrary restriction, but I just said to assume it.  f(0)=0
> removes all cos() terms.  f(T) = sin( w T) = 0
> allows only w that are integer multiples of pi/T, a discrete spectrum.
>

 and.......

>
> The f(0) = f(T) = 0 makes it periodic.
>

and ...

>
> Having fixed the end conditions makes the solutions periodic and the
> spectrum discrete.  Very important to musicians.
>
> > Having accepted the spectral aliasing by sampling in time, we finally
have
> > discrete, periodic time and discrete, periodic frequency.
>
> > When you said: "assume that f(0)=f(T)=0" then you denied the later
> > assertion: "it is known to have a maximum frequency component < Fn".
> > Perhaps you didn't mean they were tied together but it seemed so.
>
> I only fixed the function at two points.  That doesn't limit the frequency
> spectrum at all.
>
>

This is where we diverge.  Setting the end points to zero just sets the end
points to zero.  The DSP math wouldn't leap to assume a discrete spectrum /
periodic temporal function.  That takes another step in the math.  AT this
stage there is a continuous spectrum that is determined by the Fourier
Transform of a time-limited function - whether the end points at the time
limits are zero or not.

Consider the continuous Fourier Transform of such a function - as you have
done.
You said:
"Consider a signal, f, sampled over time T, from t=0 to t=T.  Assume that
f(0)=f(T)=0 for now.  All the components must be sine with periods that are
multiples of 2T."

Hmmmm.... I'm not so sure about this.  Here's a counter example:
This is a waveform, that has zeros at 0 and T.
cos(pi*T) - cos(3*pi*T) + sin(3*pi*T)
The sine has period 3*pi*T/2*pi = (3/2)*T which is not a multiple of 2*T.
and there are cosines.

So, taking a segment T of this waveform made up of "sinuoids" and making it
a "period" of a periodic waveform will cause the periodic waveform to have a
serious discontinuity.
The resulting periodic waveform isn't bandlimited.  It can't be represented
by a finite Fourier Series.  The Fourier Series of this waveform goes on
infinitely.
Now, of course this one does have a period so we might have thrown in sin(T)
just for fun and looked for a solution where there are common zeros if one
exists - but it is guaranteed to not be a periodic function and there will
be a discontinuity if we assume one based on zero end points.

*However* if you limit the formation of waves to those that are sums of
sinusoids whose periods are multiples of 2*T to begin with, then the Fourier
Series is limited to the number of sinusoids that you included in the first
place.  *Then* you can pick zero points of the waveform that are spaced at
T.  But this is a special waveform to begin with.  /simply creating a
waveform that has zeros spaced at some arbitrary T doesn't necessarily meet
this more restrictive criterion.

I think the bottom line is this:
If you start with a definition of a waveform that has a finite Fourier
Series to begin with, then you may time-limit it to one period with no
affect.  Once you've done this time limiting, you may assume periodicity and
may compute the Fourier Series, which will be finite.  A circular
discussion.

****************************

> Now, FFT has some uniform requirements, and it is certainly a better way
to
> do it, but I don't see such a restriction on DFT.

Yes, I guess it depends on how you define "DFT".  The "normal" definition is
a sum with regularly spaced terms.  But a continuous Fourier Transform of
non-equally spaced "samples" / diracs can be reduced to a discrete sum of
irregularly spaced exponential terms.  I don't know there's an accepted name
for that discrete sum but "DFT", while grammatically "correct" I don't think
is it.

You made me think ...... I may have messed up somewhere.

Fred

Richard Owlett <rowlett@atlascomm.net> wrote in message news:<vg67rsivcmn27a@corp.supernews.com>...
> I understand Nyquist specifying a minimum sampling rate to determine 
> the high frequency component of a signal.
> 
> What happens at at the other end of the spectrum?
> 
> I.E. Is there a minimum time window required?
> 
> E.G. If the signal has a significant 1 Hz component and sample window 
> was .1 sec with a sample rate of 10 kHz, would an FFT portray the 1 Hz 
> component?

Interesting question... There is, of course, the Fourier limit that has 
been discussed in another thread. A 0.1 s window at 10 kHz would be 
some 1000 samples in the time series. That's also how many bins you would 
get for the spectrum, so each bin would be 10 Hz wide. From a 
non-parametric point of view, the 1 Hz component would appear as a 
DC component. 

Intuitively, that makes sense. Assume your sampling starts at t=0 an that 
the zero-mean signal is on the form cos(wt). During the short observation 
window the signal becomes almost constant and would be indistinguishable 
from the DC...

On the other hand, if you do the same sort of argument but with sin(wt) 
instead of cos(wt), the waveform should appear as a ramp function
inside your observation window...  the first-order naive conclusion 
would be that you get some sort of overharmonics, due to the "almost 
linear" ramp. 

All of a sudden there is a phase aspect involved. Perhaps one should 
sit down with pencil and paper and work out the maths... in any respect, 
your "naive" question may very well have touched upon one of those 
Diabolic Details that are all over DSP.

Rune

"Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message
news:vT%Ma.2287$Jk5.1243587@feed2.centurytel.net...
> Glen,
>
> Whew!  I'm top-posting to introduce my comments.  While I've worked in
> acoustics it's always been at the applied / system level.  So, I've not
had
> the occasion to deal with those particular differential equations in any
> detail.  That makes the discussion interesting!
>
> I'll put a DSP spin on it below.
>
> Regards,
>
> Fred
>
> "Glen Herrmannsfeldt" <gah@ugcs.caltech.edu> wrote in message
> news:1rRMa.2466$NW6.2669@rwcrnsc51.ops.asp.att.net...
> >
> > "Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message
> > news:%cQMa.2283$Jk5.1212136@feed2.centurytel.net...
> > >
> > > "Glen Herrmannsfeldt" <gah@ugcs.caltech.edu> wrote in message
> > > news:i0LMa.1321$Ey6.785@rwcrnsc52.ops.asp.att.net...
> > > >
> >
> > (big snip)
> >
> >
> > > > Consider a signal, f, sampled over time T, from t=0 to t=T.
> > > >  Assume that f(0)=f(T)=0 for now.
> > > > All the components must be sine with period that
> > > > are multiples of 2T.

I am not sure if this is what you are commenting on or not.  It is supposed
to be frequencies that are multiples of a sine with period 2 T, so the
periods are integer fractions of 2 T.

> > > Not yet.  You didn't say that this was going to be considered to be
one
> > > period of a periodic waveform yet.  Or, you didn't say that the
spectrum
> > > would be discrete.
> >
> > With f(0) = f(T) = 0 they pretty much have to be.   Now, you can argue
> that
> > it is an arbitrary restriction, but I just said to assume it.  f(0)=0
> > removes all cos() terms.  f(T) = sin( w T) = 0
> > allows only w that are integer multiples of pi/T, a discrete spectrum.
> >
>
>  and.......
>
> >
> > The f(0) = f(T) = 0 makes it periodic.
> >
>
> and ...
>
> >
> > Having fixed the end conditions makes the solutions periodic and the
> > spectrum discrete.  Very important to musicians.
> >
> > > Having accepted the spectral aliasing by sampling in time, we finally
> have
> > > discrete, periodic time and discrete, periodic frequency.
> >
> > > When you said: "assume that f(0)=f(T)=0" then you denied the later
> > > assertion: "it is known to have a maximum frequency component < Fn".
> > > Perhaps you didn't mean they were tied together but it seemed so.
> >
> > I only fixed the function at two points.  That doesn't limit the
frequency
> > spectrum at all.
> >
> >
>
> This is where we diverge.  Setting the end points to zero just sets the
end
> points to zero.  The DSP math wouldn't leap to assume a discrete spectrum
/
> periodic temporal function.  That takes another step in the math.  AT this
> stage there is a continuous spectrum that is determined by the Fourier
> Transform of a time-limited function - whether the end points at the time
> limits are zero or not.
>
> Consider the continuous Fourier Transform of such a function -
> as you have done.

> You said:
> "Consider a signal, f, sampled over time T, from t=0 to t=T.  Assume that
> f(0)=f(T)=0 for now.  All the components must be sine with periods that
are
> multiples of 2T."
>
> Hmmmm.... I'm not so sure about this.  Here's a counter example:
> This is a waveform, that has zeros at 0 and T.
> cos(pi*T) - cos(3*pi*T) + sin(3*pi*T)
> The sine has period 3*pi*T/2*pi = (3/2)*T which is not a multiple of 2*T.
> and there are cosines.

f''(t)+w**2*f(t)=0 is a linear differential equation.  The sum of solutions
to it is also a solution.

So the series solution f(t)=Sum(n=0 to infinity)  A(n) cos(n w t) + B(n)
sin(n w t) is substituted and each term must separately be a solution to the
equation.

f(0)=0  means that A(n)=0 for all A, as cos(0)=1

f(T)=0, so B(n) sin(n w T)=0  This is true when w=pi/T for all integer n,
and any B(n).

It is these equation that limits it to periodic solutions.  You can use
different boundary conditions and will get slightly less convenient answers,
but still periodic.

> So, taking a segment T of this waveform made up of "sinuoids" and making
it
> a "period" of a periodic waveform will cause the periodic waveform to have
a
> serious discontinuity.
> The resulting periodic waveform isn't bandlimited.  It can't be
represented
> by a finite Fourier Series.  The Fourier Series of this waveform goes on
> infinitely.
> Now, of course this one does have a period so we might have thrown in
sin(T)
> just for fun and looked for a solution where there are common zeros if one
> exists - but it is guaranteed to not be a periodic function and there will
> be a discontinuity if we assume one based on zero end points.
>
> *However* if you limit the formation of waves to those that are sums of
> sinusoids whose periods are multiples of 2*T to begin with, then the
Fourier
> Series is limited to the number of sinusoids that you included in the
first
> place.  *Then* you can pick zero points of the waveform that are spaced at
> T.  But this is a special waveform to begin with.  /simply creating a
> waveform that has zeros spaced at some arbitrary T doesn't necessarily
meet
> this more restrictive criterion.

They aren't limited until the boundary conditions are applied.   With those
boundary conditions they will be periodic, but with others they won't be.

> I think the bottom line is this:
> If you start with a definition of a waveform that has a finite Fourier
> Series to begin with, then you may time-limit it to one period with no
> affect.  Once you've done this time limiting, you may assume periodicity
and
> may compute the Fourier Series, which will be finite.  A circular
> discussion.
>
> ****************************
>
> > Now, FFT has some uniform requirements, and it is certainly a
> > better way to
> > do it, but I don't see such a restriction on DFT.
>
> Yes, I guess it depends on how you define "DFT".  The "normal" definition
is
> a sum with regularly spaced terms.  But a continuous Fourier Transform of
> non-equally spaced "samples" / diracs can be reduced to a discrete sum of
> irregularly spaced exponential terms.  I don't know there's an accepted
name
> for that discrete sum but "DFT", while grammatically "correct" I don't
think
> is it.
>
> You made me think ...... I may have messed up somewhere.

That was the first time I tried to do that.  I think it is just because it
is so rare to do a non-uniform sampling.  The subject comes up once in a
while, but in certain other places the math gets much harder.  It seems to
me that it isn't so much harder here, though.  Certainly in the majority of
cases DFT is used with uniform samples.  Whether it is implied or not, I
don't know.

-- glen

"Glen Herrmannsfeldt" <gah@ugcs.caltech.edu> wrote in message
news:gq1Na.5327$NW6.3810@rwcrnsc51.ops.asp.att.net...
>
> > This is a waveform, that has zeros at 0 and T.
> > cos(pi*T) - cos(3*pi*T) + sin(3*pi*T)
> > The sine has period 3*pi*T/2*pi = (3/2)*T which is not a multiple of
2*T.
> > and there are cosines.
>
> f''(t)+w**2*f(t)=0 is a linear differential equation.  The sum of
solutions
> to it is also a solution.
>
> So the series solution f(t)=Sum(n=0 to infinity)  A(n) cos(n w t) + B(n)
> sin(n w t) is substituted and each term must separately be a solution to
the
> equation.
>

Glen,

I didn't ever say anything about differential equations - although I know
that you had earlier which I didn't understand but didn't worry about at the
time.  I figured perhaps there was a nice analogy...

I tried to define f(t) but made an error.  The objective was to define a
function that would be zero at t=0 and at t=T and in order to respectfully
create a counter example to your assertion about things zero at these
points.  I should have said:

f(t)=cos(pi*t/T) - cos(3*pi*t/T) + sin(3*pi*t/T)
Using your notation with more definition:
w=pi/T
f(t)=sum of A(n)cos(nwt) + B(n)sin(nwt); n=0 to inf
A(0) is 0; A(1)=1.0, A(2)=0, A(3)=-1.0 all other A(n)=0
B(3)=1.0

OK, I think I have it right this time....
The sum is zero at t=0 and t=T.

But, it seems that you proscribe a function with this structure because it
isn't the solution to a differential equation?  Do I understand?  The
function above meets the "zero at the ends" criteria even with cosines in
the sum because the two cosines sum to zero at the ends.  The added
requirement about being able to remove one of the terms and still have zero
is a new requirement....and is restrictive for what we're talking about.

Being the solution to a differential equation isn't necessary in this
discussion or I'm really going to learn something interesting!!  I think
this because *any* sinusoid is fair game as a signal component - not just
ones that happen to fit nicely in T.  For example sin(t) and sin(pi*t) are
both allowed and their sum has no definable period - so there is no T you
can pick where their sum will "behave".

Otherwise, there's a circular argument.....

I hope this helps....

Fred

Fred Marshall wrote:
> 
  ...
 
> You said:
> "Consider a signal, f, sampled over time T, from t=0 to t=T.  Assume that
> f(0)=f(T)=0 for now.  All the components must be sine with periods that are
> multiples of 2T."
> 
> Hmmmm.... I'm not so sure about this.  Here's a counter example:
> This is a waveform, that has zeros at 0 and T.
> cos(pi*T) - cos(3*pi*T) + sin(3*pi*T)
> The sine has period 3*pi*T/2*pi = (3/2)*T which is not a multiple of 2*T.
> and there are cosines.

How about a single frequency? Sin(pi*T) will have the spectrum of a
full-wave rectifier.
> 
> So, taking a segment T of this waveform made up of "sinuoids" and making it
> a "period" of a periodic waveform will cause the periodic waveform to have a
> serious discontinuity.
> The resulting periodic waveform isn't bandlimited.  It can't be represented
> by a finite Fourier Series.  The Fourier Series of this waveform goes on
> infinitely.
> Now, of course this one does have a period so we might have thrown in sin(T)
> just for fun and looked for a solution where there are common zeros if one
> exists - but it is guaranteed to not be a periodic function and there will
> be a discontinuity if we assume one based on zero end points.
> 
> *However* if you limit the formation of waves to those that are sums of
> sinusoids whose periods are multiples of 2*T to begin with, then the Fourier
> Series is limited to the number of sinusoids that you included in the first
> place.  *Then* you can pick zero points of the waveform that are spaced at
> T.  But this is a special waveform to begin with.  Simply creating a
> waveform that has zeros spaced at some arbitrary T doesn't necessarily meet
> this more restrictive criterion.
> 
> I think the bottom line is this:
> If you start with a definition of a waveform that has a finite Fourier
> Series to begin with, then you may time-limit it to one period with no
> affect.  Once you've done this time limiting, you may assume periodicity and
> may compute the Fourier Series, which will be finite.  A circular
> discussion.
> 
> ****************************
> 
> > Now, FFT has some uniform requirements, and it is certainly a better way
> to
> > do it, but I don't see such a restriction on DFT.
> 
> Yes, I guess it depends on how you define "DFT".  The "normal" definition is
> a sum with regularly spaced terms.  But a continuous Fourier Transform of
> non-equally spaced "samples" / diracs can be reduced to a discrete sum of
> irregularly spaced exponential terms.  I don't know there's an accepted name
> for that discrete sum but "DFT", while grammatically "correct" I don't think
> is it.
> 
> You made me think ...... I may have messed up somewhere.
> 
> Fred

I'm sure there must be restrictions somewhere, if only in the necessary
SNR. Suppose I have a signal bandlimited to 5 KHz, sampled for a second
with 10,001 samples. Surely, there is a limit to how close those samples
can be. Suppose one sample is taken at t = 0, another at t = 1 sec, and
the rest concentrated in a millisecond period that includes t = .5 sec.
Is reasonable reconstruction possible? Has my inspection equipment
failed again?

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

"Jerry Avins" <jya@ieee.org> wrote in message
news:3F04F015.71E8BC70@ieee.org...
> Fred Marshall wrote:
> >
>   ...
>
> > You said:
> > "Consider a signal, f, sampled over time T, from t=0 to t=T.  Assume
that
> > f(0)=f(T)=0 for now.  All the components must be sine with periods that
are
> > multiples of 2T."
> >
> > Hmmmm.... I'm not so sure about this.  Here's a counter example:
> > This is a waveform, that has zeros at 0 and T.
> > cos(pi*T) - cos(3*pi*T) + sin(3*pi*T)
> > The sine has period 3*pi*T/2*pi = (3/2)*T which is not a multiple of
2*T.
> > and there are cosines.
>
> How about a single frequency? Sin(pi*T) will have the spectrum of a
> full-wave rectifier.

Exactly.. that was the point.  It can't be fully or accurately represented
by a finite series, a finite discrete spectrum, even though it is zero at
the end points.

> >
.........................
> I'm sure there must be restrictions somewhere, if only in the necessary
> SNR. Suppose I have a signal bandlimited to 5 KHz, sampled for a second
> with 10,001 samples. Surely, there is a limit to how close those samples
> can be. Suppose one sample is taken at t = 0, another at t = 1 sec, and
> the rest concentrated in a millisecond period that includes t = .5 sec.
> Is reasonable reconstruction possible? Has my inspection equipment
> failed again?

No Jerry.  But you *knew* that this was about theory and not practice ...
:-)
I wouldn't want to try with even one derivative being used.  OK in theory,
messy in practice.  Nice to know about on occasion.

Here's a simple queston:
Q: Differential equations are theoretical models that attempt to approximate
reality.
How does one practically solve a differential equation for continuous time
without using analytical methods (math with pencil and paper or an
analytical program)?  No digital arithmetic please - thus the "continuous
time" requirement.

Fred

James Horn <jimhorn@svn.net> wrote:

>Hi Richard - Actually, Nyquist doesn't have anything to say on this.  The 
>sampling theorem refers to essentially infinite sequences.  For your case 
>(a short sequence compared to the desired frequency), the limits are 
>determined by the analysis technique you use.
>
>For FFTs, the resolution is fixed and, with 0.1 seconds of sample, limited 
>to 10 Hz (i.e. will return levels of 10, 20, 30, etc. Hz *only*).  Other, 
>more involved techniques may be able to discern your 1Hz signal but you 
>have to realize that the signal to noise requirements will be fierce.

Jim,

I have an idea ... but could you explain a bit more about the fierce
signal to noise requirements for discerning the 1 Hz signal?

Regards,

Robert

www.gldsp.com

( modify address for return email )

www.numbersusa.com
www.americanpatrol.com

Rune Allnor wrote:

> Richard Owlett <rowlett@atlascomm.net> wrote in message news:<vg67rsivcmn27a@corp.supernews.com>...
> 
>>I understand Nyquist specifying a minimum sampling rate to determine 
>>the high frequency component of a signal.
>>
>>What happens at at the other end of the spectrum?
>>
>>I.E. Is there a minimum time window required?
>>
>>E.G. If the signal has a significant 1 Hz component and sample window 
>>was .1 sec with a sample rate of 10 kHz, would an FFT portray the 1 Hz 
>>component?
> 
> 
> Interesting question... There is, of course, the Fourier limit that has 
> been discussed in another thread. A 0.1 s window at 10 kHz would be 
> some 1000 samples in the time series. That's also how many bins you would 
> get for the spectrum, so each bin would be 10 Hz wide. From a 
> non-parametric point of view, the 1 Hz component would appear as a 
> DC component. 
> 
> Intuitively, that makes sense. Assume your sampling starts at t=0 an that 
> the zero-mean signal is on the form cos(wt). During the short observation 
> window the signal becomes almost constant and would be indistinguishable 
> from the DC...
> 
> On the other hand, if you do the same sort of argument but with sin(wt) 
> instead of cos(wt), the waveform should appear as a ramp function
> inside your observation window...  the first-order naive conclusion 
> would be that you get some sort of overharmonics, due to the "almost 
> linear" ramp. 
> 
> All of a sudden there is a phase aspect involved. Perhaps one should 
> sit down with pencil and paper and work out the maths... in any respect, 
> your "naive" question may very well have touched upon one of those 
> Diabolic Details that are all over DSP.
> 
> Rune

I've paused to digest the responses received.
Previously Mr. Avins stated I was not asking A "wrong question".
However, I'm not sure I'm asking THE "right question".

I've some 'off-the-wall' ideas about improving speech recognition.
They require recognizing artifacts of tenths of seconds duration of a 
fundamental of multiple kHz.

I think what I'm looking for is an explanation of how to combine the 
output of transforms of m  n  sec duration windows into something that 
resembles a transform of a m*n second window without losing resolution.

"Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message
news:tj4Na.2297$Jk5.1256042@feed2.centurytel.net...

> I didn't ever say anything about differential equations - although I know
> that you had earlier which I didn't understand but didn't worry about at
the
> time.  I figured perhaps there was a nice analogy...

Well, different people have different ideas about why things work the way
they do.

The reason sin() and cos() are popular basis functions is because they are
solutions for the simplest second order differential equation, and that
equation comes up in many places.

Different differential equations have different solutions and different
basis functions will be used.  Problems with cylindrical symmetry have
Bessel function solutions, for example.  Sometimes it is easier to build
something cylindrical, and whether one likes Bessel functions or not (I
don't especially like them), one is stuck with them.

> I tried to define f(t) but made an error.  The objective was to define a
> function that would be zero at t=0 and at t=T and in order to respectfully
> create a counter example to your assertion about things zero at these
> points.  I should have said:
>
> f(t)=cos(pi*t/T) - cos(3*pi*t/T) + sin(3*pi*t/T)
> Using your notation with more definition:
> w=pi/T
> f(t)=sum of A(n)cos(nwt) + B(n)sin(nwt); n=0 to inf
> A(0) is 0; A(1)=1.0, A(2)=0, A(3)=-1.0 all other A(n)=0
> B(3)=1.0
>
> OK, I think I have it right this time....
> The sum is zero at t=0 and t=T.
>
> But, it seems that you proscribe a function with this structure because it
> isn't the solution to a differential equation?  Do I understand?  The
> function above meets the "zero at the ends" criteria even with cosines in
> the sum because the two cosines sum to zero at the ends.  The added
> requirement about being able to remove one of the terms and still have
zero
> is a new requirement....and is restrictive for what we're talking about.
>
> Being the solution to a differential equation isn't necessary in this
> discussion or I'm really going to learn something interesting!!  I think
> this because *any* sinusoid is fair game as a signal component - not just
> ones that happen to fit nicely in T.  For example sin(t) and sin(pi*t) are
> both allowed and their sum has no definable period - so there is no T you
> can pick where their sum will "behave".
>
> Otherwise, there's a circular argument.....

A complete set of basis functions is sufficient to solve the equations.
That is why things like Fourier transform work.  It might be that your
solution can is equal to the expansion that I gave, using the appropriate
trig. identities.  I didn't try to show it.

-- glen

"Glen Herrmannsfeldt" <gah@ugcs.caltech.edu> wrote in message
news:qSaOa.58648$fG.41271@sccrnsc01...
>
> "Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message
> news:tj4Na.2297$Jk5.1256042@feed2.centurytel.net...
>
> > I didn't ever say anything about differential equations - although I
know
> > that you had earlier which I didn't understand but didn't worry about at
> the
> > time.  I figured perhaps there was a nice analogy...
>
> Well, different people have different ideas about why things work the way
> they do.
>
> The reason sin() and cos() are popular basis functions is because they are
> solutions for the simplest second order differential equation, and that
> equation comes up in many places.
>
> Different differential equations have different solutions and different
> basis functions will be used.  Problems with cylindrical symmetry have
> Bessel function solutions, for example.  Sometimes it is easier to build
> something cylindrical, and whether one likes Bessel functions or not (I
> don't especially like them), one is stuck with them.
>
> > I tried to define f(t) but made an error.  The objective was to define a
> > function that would be zero at t=0 and at t=T and in order to
respectfully
> > create a counter example to your assertion about things zero at these
> > points.  I should have said:
> >
> > f(t)=cos(pi*t/T) - cos(3*pi*t/T) + sin(3*pi*t/T)
> > Using your notation with more definition:
> > w=pi/T
> > f(t)=sum of A(n)cos(nwt) + B(n)sin(nwt); n=0 to inf
> > A(0) is 0; A(1)=1.0, A(2)=0, A(3)=-1.0 all other A(n)=0
> > B(3)=1.0
> >
> > OK, I think I have it right this time....
> > The sum is zero at t=0 and t=T.
> >
> > But, it seems that you proscribe a function with this structure because
it
> > isn't the solution to a differential equation?  Do I understand?  The
> > function above meets the "zero at the ends" criteria even with cosines
in
> > the sum because the two cosines sum to zero at the ends.  The added
> > requirement about being able to remove one of the terms and still have
> zero
> > is a new requirement....and is restrictive for what we're talking about.
> >
> > Being the solution to a differential equation isn't necessary in this
> > discussion or I'm really going to learn something interesting!!  I think
> > this because *any* sinusoid is fair game as a signal component - not
just
> > ones that happen to fit nicely in T.  For example sin(t) and sin(pi*t)
are
> > both allowed and their sum has no definable period - so there is no T
you
> > can pick where their sum will "behave".
> >
> > Otherwise, there's a circular argument.....
>
> A complete set of basis functions is sufficient to solve the equations.
> That is why things like Fourier transform work.  It might be that your
> solution can is equal to the expansion that I gave, using the appropriate
> trig. identities.  I didn't try to show it.


Glen,

I'm trying to discover if there's something I'm missing here - because then
I'd learn something.  So, your persistence in sticking with the conversation
is appreciated!

At the moment, I'm stuck on the idea that the conditions you've asserted are
more restrictive than we generally apply in signal analysis and processing.

When you say:
>A complete set of basis functions is sufficient to solve the equations.
> That is why things like Fourier transform work.

I'm not really sure which "equations" you're referring to here.
Now, I know that the Fourier Transform is a nice tool to use for solving
differential equations - as is the Laplace Transform.
At the same time, I believe that the Fourier Transform is useful in
analyzing and manipulating waveforms of a very general variety that go
beyond situations of differential equations with nice boundary conditions.

I don't think you'll find trig identities that will work because the
functions I provided have infinite derivatives at the edges - well, I think
that's the correct way to put it.  There is 1/2 cycle of a sine in one term
and 3/2 cycle of a sine in another term.
Now, if I relate these terms to a vibrating string with nodes at the ends,
then there must be an "image" vibrating string that moves in "opposition"
(?) that takes up another string length in order for the entire span of the
real string and the image string to make up an entire "period".  And, these
are standing waves.  If there's an excitation that is not at a resonant
frequency then there will be a travelling wave, right?  Not my field of
expertise to express or manipulate these equations but the analogy seems OK.

So, one might double the frequency span by doubling the temporal epoch and
encompass the sin(t) and sin(3*t/2) types of terms and have continuous
derivatives at the edges of the newly defined "period".  But there's still
the possibility of sin(pi*t) which will surely not fit in any period you
might define if it includes the other two terms.....

So, I'm still stuck on the assertion that time-limiting a function or,
equivalently, having a time-limited function has nothing to do with
frequency domain sampling - whether the time function is zero at the ends or
not.
The frequency domain sampling can be viewed as a result of considering the
time-limited function to be a single period of a periodic waveform.  That
should cause no problems but the Fourier Series *can* be of infinite extent.

If the Fourier Series is indeed infinite, then subsequent temporal sampling
will cause frequency domain aliasing and the character of the function is
(usually irrevocably) changed.  If this change is accepted, then there is
generally a periodic time / periodic frequency - discrete frequency /
discrete time representation that's accepted.  However, accepting the
aliasing is necessary  to be able to assert that an arbitrary time-limited
function can be expressed as a finite discrete sequence or periodic sequence
in frequency.

Fred