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. Assumethat> > > f(0)=f(T)=0 for now. All the components must be sine with periodsthat> > >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 arguethat> 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 finallyhave> > 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 wayto> 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

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

Started by ●July 2, 2003

Reply by ●July 3, 20032003-07-03

Reply by ●July 3, 20032003-07-03

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

Reply by ●July 3, 20032003-07-03

"Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message news:vT%Ma.2287$Jk5.1243587@feed2.centurytel.net...> Glen, > > Whew! I'm top-posting to introduce my comments. While I've worked in > acoustics it's always been at the applied / system level. So, I've nothad> the occasion to deal with those particular differential equations in any > detail. That makes the discussion interesting! > > I'll put a DSP spin on it below. > > Regards, > > Fred > > "Glen Herrmannsfeldt" <gah@ugcs.caltech.edu> wrote in message > news:1rRMa.2466$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 beone> > > period of a periodic waveform yet. Or, you didn't say that thespectrum> > > 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 thefrequency> > spectrum at all. > > > > > > This is where we diverge. Setting the end points to zero just sets theend> 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 thatare> 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 makingit> a "period" of a periodic waveform will cause the periodic waveform to havea> serious discontinuity. > The resulting periodic waveform isn't bandlimited. It can't berepresented> 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 insin(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 theFourier> Series is limited to the number of sinusoids that you included in thefirst> 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 necessarilymeet> 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 periodicityand> 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" definitionis> 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 acceptedname> for that discrete sum but "DFT", while grammatically "correct" I don'tthink> 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

Reply by ●July 3, 20032003-07-03

"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 of2*T.> > and there are cosines. > > f''(t)+w**2*f(t)=0 is a linear differential equation. The sum ofsolutions> 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 tothe> 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

Reply by ●July 4, 20032003-07-04

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. > > FredI'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. �����������������������������������������������������������������������

Reply by ●July 4, 20032003-07-04

"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. Assumethat> > f(0)=f(T)=0 for now. All the components must be sine with periods thatare> > 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 of2*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

Reply by ●July 5, 20032003-07-05

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

Reply by ●July 6, 20032003-07-06

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. > > RuneI'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.

Reply by ●July 7, 20032003-07-07

"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 atthe> 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 havezero> 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

Reply by ●July 7, 20032003-07-07

"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 Iknow> > 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 torespectfully> > 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 becauseit> > isn't the solution to a differential equation? Do I understand? The > > function above meets the "zero at the ends" criteria even with cosinesin> > 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 - notjust> > 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 Tyou> > 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