"robert bristow-johnson" <rbj@audioimagination.com> wrote in message news:1155316423.969659.269300@75g2000cwc.googlegroups.com...> > Lucy wrote: > > HI, > > > > If a function F(u) is known to be a Fourier transform of some function > > f(t), and F(u) is periodic in u, which can be deemed as frequency > > variable, with a period 2*pi. > > sounds like u is in angular frequency and the result is a string of > dirac-delta functions with weights or coefficients equal to the sampled > continuous-time function that is the inverse Fourier transform of the > one period of F(u). this is just the basic sampling theorem, sorta in > reverse. > > first do > > g(t) = Integrate(F(u)*exp(i*u*t), u from -pi to +pi) > > then sample g(t) at every unit value of t and attach to a dirac > impulse. > > f(t) = g(t)*SUM{ delta(t - n)) ( -inf < n < +inf) > > or > > f(t) = SUM{ g(n) * delta(t - n)) ( -inf < n < +inf) > > r b-j >Easier way is to take the Fourier transform of the complex Fourier series (since it is periodic). The exponentials jwt terms give rise to an impulse train weighted by the complex Fourier coefficients. M.P -- Posted via a free Usenet account from http://www.teranews.com

# Inverse Fourier/Laplace transform of a periodic function?

Started by ●August 11, 2006

Reply by ●August 12, 20062006-08-12

Reply by ●August 12, 20062006-08-12

David C. Ullrich wrote:> > I'm not sure whether this is a math post or one of those posts > about the Fourier transform where we don't want to worry about > actually getting the details straight. > > If you do want to get the mathematical details straight and if > you mean exactly what you say, you should note that there's no > such function (except for the zero function). The Fourier transform > of a function is never periodic.them's fightin' words. :-) but i think that's because we disagree over the nature of the dirac delta "function". i s'pose that's at the root of this, David? r b-j

Reply by ●August 12, 20062006-08-12

Sorcerer wrote:> "Jerry Avins" <jya@ieee.org> wrote in message > news:-_qdnYd3q7Z5QkHZnZ2dnUVZ_rCdnZ2d@rcn.net... > | Sorcerer wrote: > | ... > | > | > You can see the signal is "ringing" because insufficient terms were > used, > | > | Oh? How many terms are needed to make the ringing amplitude diminish? > | > | Jerry > | -- > | Engineering is the art of making what you want from things you can get. > > To diminish? As in make less, such as 999 < 1000?No; to make the peak amplitude of the Gibbs oscillation diminish, as I wrote.> One more term is necessary and sufficient, of course. What an > odd question to ask...Yes, that would have been an odd question, but I didn't ask it. I asked my question tongue in cheek. The overshoot due to Gibbs's phenomenon has a fixed value of about 11% for a square wave, regardless on the number of terms. More terms put the peaks closer together and bunches them closer to the step, but the peaks adjacent to the step remain the same size. Finally, when the number of terms --> infinity, the whole set of peaks merges into a delta, and the deltas of opposite sign (pre-ring and post-ring), both being *at* the step, engage in mutual annihilation. http://www.sosmath.com/fourier/fourier3/gibbs.html But for n = infinity - 1, the Gibbs amplitude is about the same as it is for n = 3. So you can see the signal ringing no matter how many terms you use. There is no "enough". (But there is a Blackman window.) Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������

Reply by ●August 12, 20062006-08-12

Jerry Avins wrote:> ... The overshoot due to Gibbs's > phenomenon has a fixed value of about 11% for a square wave, regardless > on the number of terms. ... ���Klong! 18% Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������

Reply by ●August 12, 20062006-08-12

"Jerry Avins" <jya@ieee.org> wrote in message news:scydnds9iaNlmEPZnZ2dnUVZ_uWdnZ2d@rcn.net... | Sorcerer wrote: | > "Jerry Avins" <jya@ieee.org> wrote in message | > news:-_qdnYd3q7Z5QkHZnZ2dnUVZ_rCdnZ2d@rcn.net... | > | Sorcerer wrote: | > | ... | > | | > | > You can see the signal is "ringing" because insufficient terms were | > used, | > | | > | Oh? How many terms are needed to make the ringing amplitude diminish? | > | | > | Jerry | > | -- | > | Engineering is the art of making what you want from things you can get. | > | > To diminish? As in make less, such as 999 < 1000? | | No; to make the peak amplitude of the Gibbs oscillation diminish, as I | wrote. | | > One more term is necessary and sufficient, of course. What an | > odd question to ask... | | Yes, that would have been an odd question, but I didn't ask it. | | I asked my question tongue in cheek. The overshoot due to Gibbs's | phenomenon has a fixed value of about 11% for a square wave, regardless | on the number of terms. More terms put the peaks closer together and | bunches them closer to the step, but the peaks adjacent to the step | remain the same size. Finally, when the number of terms --> infinity, | the whole set of peaks merges into a delta, and the deltas of opposite | sign (pre-ring and post-ring), both being *at* the step, engage in | mutual annihilation. http://www.sosmath.com/fourier/fourier3/gibbs.html | But for n = infinity - 1, the Gibbs amplitude is about the same as it is | for n = 3. So you can see the signal ringing no matter how many terms | you use. There is no "enough". (But there is a Blackman window.) Yeah, ok... There's a fat farad (capacitor) as well :-) You may find a few on your motherboard, square signals are popular in computing, edge triggered flip-flops don't care much about overshoot. Some even like it. In reality closing a switch is generating the infinite set of sine waves and can cause radio interference no matter which station you are tuned to. Androcles Engineering is the art of adding fat farads when needed. ____________________________________

Reply by ●August 12, 20062006-08-12

On 12 Aug 2006 07:14:16 -0700, "robert bristow-johnson" <rbj@audioimagination.com> wrote:> >David C. Ullrich wrote: >> >> I'm not sure whether this is a math post or one of those posts >> about the Fourier transform where we don't want to worry about >> actually getting the details straight. >> >> If you do want to get the mathematical details straight and if >> you mean exactly what you say, you should note that there's no >> such function (except for the zero function). The Fourier transform >> of a function is never periodic. > >them's fightin' words. :-)If you don't bother to read the rest of what I wrote it would seem so.>but i think that's because we disagree over the nature of the dirac >delta "function". i s'pose that's at the root of this, David?No, we don't disagree on that. Evidently what we disagree on is what the word "function" means. So what?>r b-j************************ David C. Ullrich

Reply by ●August 12, 20062006-08-12

Sorcerer wrote: ...> Engineering is the art of adding fat farads when needed. > ____________________________________That too. Just don't tell a kid who doesn't know better that Mr. Gibbs will calm down if he gets enough terms. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������

Reply by ●August 12, 20062006-08-12

"Jerry Avins" <jya@ieee.org> wrote in message news:xNOdnVtssZHg0EPZnZ2dnUVZ_vydnZ2d@rcn.net... ... The overshoot happens, Fourier was correct, Gibbs was correct. Incidentally, overshoot is more voltage than was supplied. http://tinyurl.com/n67uy Androcles Engineering is the art of flying Concorde using analogue computers when digital technology was still making its debut, kid who doesn't know better :-) _____________________________________________________

Reply by ●August 12, 20062006-08-12

Jerry Avins wrote:> > That too. Just don't tell a kid who doesn't know better that Mr. Gibbs > will calm down if he gets enough terms.I don't get it why wouldn't you tell someone that. There are an infinite number of ways you can weight the sinusoids to approximate a square wave some will have more ripple some have less. The more terms you have the closer you can get to an ideal square wave. That's essentially what fir filter design is all about. You're not going to tell that kid if he designs a half-band filter he is going to have to live with 18% (or whatever the amount you have claimed) ripple are you? -jim ----== Posted via Newsfeeds.Com - Unlimited-Unrestricted-Secure Usenet News==---- http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups ----= East and West-Coast Server Farms - Total Privacy via Encryption =----

Reply by ●August 13, 20062006-08-13

robert bristow-johnson wrote:> Lucy wrote:>>Yes, you can do that. But I am looking for a derivation which naturally >>goes from the integral from -inf to +inf, and then to the integral from >>-pi to +pi, and then to the sampling for the inversed function f(t), as >>you've said.> well, it depends on how rigorous you want to be. the folks on sci.math > might blanch at the way we comp.dspers treat the Dirac impulse > function.Not to mention quantum mechanics, where it came from, as far as I know. There is a story about someone rewriting a QM book to remove all delta functions, because they didn't believe in them. Believe them or not, they do make many explanations much simpler.> Lucy, can you accept the following identity?:> +inf +inf > SUM{ delta(t - n) } = SUM{ exp(i*2*pi*k*t) } > n=-inf k=-infIt looks strange with k, which should be wave number or wave vector, but otherwise I agree. -- glen