Reply by glen herrmannsfeldt March 30, 20042004-03-30
Clay S. Turner wrote:

(snip)

> The following link gives some details about how Fourier
> came up with his series.
> http://www.math.carleton.ca/~amingare/calculus/fourier-series.pdf
Somehow I always thought it was the wave equation instead of the diffusion equation. The X part is the same in both cases. Most likely not much later it would have been used for the wave equation. Vibrating strings were also in interesting problem around that time. For the wave equation it is easy to show the general solution of the form Af(x-vt)+Bg(x+vt) for arbitrary functions f and g. I still remember a physics quiz where we had to show that was the solution to the wave equation. I had never done partial derivatives before but had one hour to solve this problem. I had somehow heard that partial derivatives are like ordinary ones with everything else constant, and went on trying to do the problem. After a little while, the right answer came out! So now I never forget that problem. -- glen
Reply by Clay S. Turner March 30, 20042004-03-30



"glen herrmannsfeldt" <gah@ugcs.caltech.edu> wrote in message
news:YAaac.133061$_w.1589940@attbi_s53...

> > The modes of a vibrating string of uniform density, air column > of uniform cross section, or transmission line of uniform impedance, > for example. There are other systems with other differential > equations and corresponding eigenfunctions. > > As far as I know it, the origin of the Fourier series came > from the solutions to the differential equation for the above > systems. There were two different ways to solve it, which > came up with different answers. Consider a string, tube, > transmission line of length L. One method would allow > any periodic function with period 2L. The other method gave > an infinite sum of sines with periods integer fractions of 2L.
Hello Glen and others, The following link gives some details about how Fourier came up with his series. http://www.math.carleton.ca/~amingare/calculus/fourier-series.pdf Clay
Reply by glen herrmannsfeldt March 30, 20042004-03-30
Bob Cain wrote:
> Jerry Avins wrote: > >> Till wants to know what unique property of sine waves makes them come >> through a linear circuit unaltered (presumably in shape) while square >> waves do not. However sharp your truths, I don't think your answer to >> him was transparent at his level of understanding.
> I think it because they are Eigenvalues of a linear, time invariant > system even though I'm not sure what that means. :-)
The modes of a vibrating string of uniform density, air column of uniform cross section, or transmission line of uniform impedance, for example. There are other systems with other differential equations and corresponding eigenfunctions. As far as I know it, the origin of the Fourier series came from the solutions to the differential equation for the above systems. There were two different ways to solve it, which came up with different answers. Consider a string, tube, transmission line of length L. One method would allow any periodic function with period 2L. The other method gave an infinite sum of sines with periods integer fractions of 2L. The differential equation in those cases is Y''+A**2 Y=0, A is the eigenvalue, sin(Ax) and cos(Ax) are the eigenfunctions. Add the boundary conditions that the string is fixed at x=0 and X=L, only sin() terms are allowed, and A=n pi/L. That is how to make a violin, guitar, piano, flute, or many other musical instruments. Now, consider the vibrational modes of an oboe, with a cone shaped air column. (The vibrating reed is at the tip.) If you use spherical coordinates and consider the radial modes only, you get the n=0 spherical Bessel's equation, x**2 R'' + 2x R' + x**2 R = 0 http://mathworld.wolfram.com/SphericalBesselDifferentialEquation.html has a good explanation of the derivation, and http://mathworld.wolfram.com/SphericalBesselFunctionoftheFirstKind.html the solutions. Consider an oboe of length L. The first solution, j0(x)=sin(x)/x, and the modes again have frequencies that are integer multiples of the lowest mode. One more, to show that not all equations have harmonic solutions, consider the radial (circularly symmetric, n=0) modes of a drum, which should result if you hit it in the center. x**2 Y'' + x Y' + x**2 Y = 0 The solutions are called (cylindrical) Bessel functions, J0 for the radial modes. The second and third modes are approximately 2.295 and 3.398 times the fundamental http://www.gmi.edu/~drussell/Demos/MembraneCircle/Circle.html -- glen
Reply by glen herrmannsfeldt March 29, 20042004-03-29
Randy Yates wrote:


> I gave you the wrong reason before about why sine waves were special > in this sense. A linear system can't change the shape of a sine wave > (other than its amplitude or phase, that is) because a sine wave only > consists of one frequency. In any other waveform, multiple frequencies > are involved, and the shape of the waveform is determined by not only the > frequencies but the phases and amplitudes as well. Thus a linear system > may change the shape of such a waveform since it may change the phases > and amplitudes of each frequency component.
Well, it is a single frequency because sine and cosine are the basis functions of the Fourier transform. As derivative and integration are also linear operators, and when applied to sin() and cos() the results are still sin() and cos() of the same frequency, they are appropriate basis functions for the system. -- glen
Reply by Jerry Avins March 27, 20042004-03-27
Ronald H. Nicholson Jr. wrote:

> In article <4065044c$0$3055$61fed72c@news.rcn.com>, > Jerry Avins <jya@ieee.org> wrote: > >>Randy Yates wrote: >> >>>>Furthermore I thought about a system which only would double any frequency >>>>component present in a given Signal, and I fail to see how this system is >>>>non-linear. >>> >>>Could it be that linearity is only a necessary condition for a linear >>>system? >> >>I believe that linearity is both a necessary and a sufficient condition. > > > But linear systems are a superset of systems that are both linear and > time-shift-invariant.
That's as may be, but being linear is clearly a necessary and sufficient condition for being linear. I thought I made a joke. It evidently fell flat. Jerry -- Engineering is the art of making what you want from things you can get. &#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
Reply by Jerry Avins March 27, 20042004-03-27
Randy Yates wrote:

   ...

> Do you mean "How can we construct such a system"? Like this: > > Consider the signal x(t) that is the input to this "linear" system. > Then X(f) = F(x(t)), where F(.) denotes the Fourier transform of ".". > Then the output, y(t), is > > y(t) = F^(-1)(X(f/2)), > > where F^(-1)(.) denotes the inverse Fourier transform of ".".
...
> I'm with Till - I can't see why this system is nonlinear.
... How would you go about actually building one of these systems? When I see X(f/2), I worry about realizability and causality. Jerry -- Engineering is the art of making what you want from things you can get. &#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
Reply by Randy Yates March 27, 20042004-03-27
allnor@tele.ntnu.no (Rune Allnor) writes:

> Randy Yates <yates@ieee.org> wrote in message news:<65crwh5t.fsf@ieee.org>... >> Jerry Avins <jya@ieee.org> writes: >> >> > Randy Yates wrote: >> > >> >> Hey Till, >> >> You're right - it isn't just sine waves that a linear system will not >> >> produce new frequencies of - ANY waveform will be unaltered in >> >> frequency by a linear system. > >> > The output may contain all the component frequencies >> > of the input, but shape isn't necessarily maintained. >> >> Did I say or imply the _shape_ was maintained? In fact I did not. > > I am sure you did not, but the phrasing is somewhat unclear: It is > a matter of interpretation what you mean by "[the] waveform will be > unaltered in frequency". I suspect Jerry interpreted this as "the > magnitude function of frequency spectra can be changed by linear > systems", and he is right. I suspect you meant "the linear system does > not introduce new frequency components in the spectrum", and you are > right. I think you agree, once you find an unambiguous phrasing that > both are happy with.
Rune, Jerry, Ronald, and most of all, Till, I apologize. I had my head where the sun doesn't shine yesterday. It's definitely brighter around here today. I stated: You're right - it isn't just sine waves that a linear system will not produce new frequencies of - ANY waveform will be unaltered in frequency by a linear system. Yes, I indeed stated any "wave_FORM_", i.e., any form of wave. Aaahh! That certainly DOES imply shape. I'm an idiot. I was wrong to defend it. Till, I hope I haven't confused you. What I meant (I guess) was that any wave form will come through with the same number of frequencies present, though their amplitudes and phases may be changed (and thus the form of the wave may be changed). Yes, I know - there's also that pesky little case when the response is 0 at some frequency or set of frequencies, thus knocking them completely out. But I think ya'll get what I mean now. I think. -- % Randy Yates % "She's sweet on Wagner-I think she'd die for Beethoven. %% Fuquay-Varina, NC % She love the way Puccini lays down a tune, and %%% 919-577-9882 % Verdi's always creepin' from her room." %%%% <yates@ieee.org> % "Rockaria", *A New World Record*, ELO http://home.earthlink.net/~yatescr
Reply by Randy Yates March 27, 20042004-03-27
rhn@mauve.rahul.net (Ronald H. Nicholson Jr.) writes:

> In article <smfvgevw.fsf@ieee.org>, Randy Yates <yates@ieee.org> wrote: >>The system appears linear to me since >> >> H(a*x1(t) + b*x2(t)) = a*H(x1(t)) + b*H(x2(T)) > > Do you mean: > > H(a*x1(t) + b*x2(T)) = a*H(x1(t)) + b*H(x2(T)) > where > T = t - t0 > to combine linearity and time-shift-invariance ?
No. I simply mistyped the last t as a "T". I was just focusing on linearity for the moment. -- % Randy Yates % "Ticket to the moon, flight leaves here today %% Fuquay-Varina, NC % from Satellite 2" %%% 919-577-9882 % 'Ticket To The Moon' %%%% <yates@ieee.org> % *Time*, Electric Light Orchestra http://home.earthlink.net/~yatescr
Reply by Bob Cain March 27, 20042004-03-27
Randy Yates wrote:

> Bob Cain <arcane@arcanemethods.com> writes: > > >>Matt Timmermans wrote: >> >>>"Matt Timmermans" <mt0000@sympatico.nospam-remove.ca> wrote in message >>>news:an59c.40234$A_2.1541424@news20.bellglobal.com... >>> >>> >>>>[...] 2) Exponential functions are eigenvalues of all LTI systems [...] >>> >>>er, I mean eigenfunctions. Bob did it first -- I'm sure he meant >>>eigenfunctions too. ;-) >>> >> >>Actually no. I don't really understand eigenstuff and was just >>digging that out of associative memory. Incorrectly as it turns out >>(as usually happens when you don't understand something.) :-) > > > Hey Bob, could you show me how to design my cache (associative? :) ) like that? > That's a pretty good trick!
There's still some work to be done on the ECC code and circuitry. That's getting more and more important as the years go by. About the time I figure it out I expect I won't be able to remember it. :-) Bob -- "Things should be described as simply as possible, but no simpler." A. Einstein
Reply by Rune Allnor March 27, 20042004-03-27
Randy Yates <yates@ieee.org> wrote in message news:<65crwh5t.fsf@ieee.org>...
> Jerry Avins <jya@ieee.org> writes: > > > Randy Yates wrote: > > > >> Hey Till, > >> You're right - it isn't just sine waves that a linear system will not > >> produce new frequencies of - ANY waveform will be unaltered in > >> frequency by a linear system.
> > The output may contain all the component frequencies > > of the input, but shape isn't necessarily maintained. > > Did I say or imply the _shape_ was maintained? In fact I did not.
I am sure you did not, but the phrasing is somewhat unclear: It is a matter of interpretation what you mean by "[the] waveform will be unaltered in frequency". I suspect Jerry interpreted this as "the magnitude function of frequency spectra can be changed by linear systems", and he is right. I suspect you meant "the linear system does not introduce new frequency components in the spectrum", and you are right. I think you agree, once you find an unambiguous phrasing that both are happy with.
> But beyond the question of what I said or didn't say, your comments seem > to be aimed at how one should explain something, and THAT depends on > style and technique. This is my style. Making the truth sharp, in my > experience, usually dispells bad conclusions and sheds light on wrong > thinking.
I agree. In my experience, that requires some very meticulous attention to phrasing and terminology. But that's a "hobby horse"(?) (Norw. "kjepphest") of mine I'll keep in his stable for now. Rune