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. If I want to find its inverse Fourier transform, I know I should try either DTFT/IDTFT, or Fourier series, for such a periodic function, and the target function f(t) in the time-domain should be a discrete comb-like function. However, instead of using DTFT/IDTFT, Fourier series, if I still want to use Inverse Fourier Transform, and it is f(t)=Integrate(F(u)*exp(i*u*t), u from -inf to +inf), However, the integration does not exist for such a periodic function F(u) when integrating on R. Is there a nice and elegant way to unify the Foureir transform and Fourier Series and starting with the above integral which seemingly does not exist and via sampling function or the dirac-delta function and finally still reach the target function f(t) which is discrete signal? Same question for the inverse Laplace transform, if the integrand is periodic with 2*pi, e.g. the F(s) function is a function of exp(s), where s is the variable in Laplace transform, it will be periodic with 2*pi on the line integral from c-i*inf to c+i*inf when doing inverse Laplace transform. Is there a way to reconcile and start with this periodic function integration and obtain a discrete f(t) in time-domain? I have seen this unification of Fourier series and Fourier transform somewhere but it just elapsed my memory. Could you give some pointers? Thanks

# Inverse Fourier/Laplace transform of a periodic function?

Started by ●August 11, 2006

Reply by ●August 11, 20062006-08-11

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

Reply by ●August 11, 20062006-08-11

robert bristow-johnson wrote:> 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-jHi Robert, 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. I am looking for a rigorous derivation of the above steps... Thanks

Reply by ●August 11, 20062006-08-11

"Lucy" <comtech.usa@gmail.com> wrote in message news:1155315242.518235.60420@h48g2000cwc.googlegroups.com... | 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. | | If I want to find its inverse Fourier transform, I know I should try | either DTFT/IDTFT, or Fourier series, for such a periodic function, and | the target function f(t) in the time-domain should be a discrete | comb-like function. | | However, instead of using DTFT/IDTFT, Fourier series, if I still want | to use Inverse Fourier Transform, and it is | | f(t)=Integrate(F(u)*exp(i*u*t), u from -inf to +inf), | | However, the integration does not exist for such a periodic function | F(u) when integrating on R. | | Is there a nice and elegant way to unify the Foureir transform and | Fourier Series and starting with the above integral which seemingly | does not exist and via sampling function or the dirac-delta function | and finally still reach the target function f(t) which is discrete | signal? | | Same question for the inverse Laplace transform, if the integrand is | periodic with 2*pi, e.g. the F(s) function is a function of exp(s), | where s is the variable in Laplace transform, it will be periodic with | 2*pi on the line integral from c-i*inf to c+i*inf when doing inverse | Laplace transform. | | Is there a way to reconcile and start with this periodic function | integration and obtain a discrete f(t) in time-domain? | | I have seen this unification of Fourier series and Fourier transform | somewhere but it just elapsed my memory. | | Could you give some pointers? | | Thanks Integrating from -inf to +inf will be a bit of problem, but remember you are dealing with a periodic function. So what happens if you integrate from -50 to +50 instead? The end points are going to be nasty, but the middle, on either side of zero, is going to give you something very close to what you seek. After all is said and done a square wave is just a set of sinewaves. For example, sin(x) + sin (3x)/3 + ...+sin( 2K-1)/(2K-1)+sin((2N-1)x)/(2N-1) is periodically square. In this spreadsheet example, N = 32: http://www.androcles01.pwp.blueyonder.co.uk/FT.JPG You can see the signal is "ringing" because insufficient terms were used, but is clearly square as expected. One often sees this on an oscilloscope from a square wave generator feeding into a reactive circuit. While it is very nice to be mathematically correct, sin( inf*x) / inf = 0 and practically speaking, small values become insignificant, particularly when computing and you cannot get to inf anyway. It's kinda like writing out pi to a million places when 22/7 is good enough. Good luck. Androcles.

Reply by ●August 11, 20062006-08-11

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. > > If I want to find its inverse Fourier transform, I know I should try > either DTFT/IDTFT, or Fourier series, for such a periodic function, and > the target function f(t) in the time-domain should be a discrete > comb-like function. > > However, instead of using DTFT/IDTFT, Fourier series, if I still want > to use Inverse Fourier Transform, and it is > > f(t)=Integrate(F(u)*exp(i*u*t), u from -inf to +inf), > > However, the integration does not exist for such a periodic function > F(u) when integrating on R. > > Is there a nice and elegant way to unify the Foureir transform and > Fourier Series and starting with the above integral which seemingly > does not exist and via sampling function or the dirac-delta function > and finally still reach the target function f(t) which is discrete > signal?The delta-function is the standard way I have seen, starting with the identity integral(exp(i*u*t) exp(i*v*t) dt) = 2*pi*delta(u-v) (I'm not 100% confident of the 2*pi, this is by memory). So if you have a periodic function f(t) = sum(k=0,inf) a_k * exp(i*k*w0*t) then its Fourier transform is proportional to sum(k=0,inf) a_k*delta(u-k*w0) I'm not sure how rigorous that is. To do this rigorously, you need to suitably generalize integration so the result above is meaningful. You probably have to start with the rigorous theory of distributions. - Randy

Reply by ●August 11, 20062006-08-11

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. �����������������������������������������������������������������������

Reply by ●August 11, 20062006-08-11

On 11 Aug 2006 09:54:02 -0700, "Lucy" <comtech.usa@gmail.com> 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.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. Otoh, if you didn't actually mean that f was a _function_ when you called it a function, there certainly are _distributions_ that have periodic Fourier transforms. In this case f will be a sum of "delta functions" supported on the integers. You should think of f as a sequence instead of a function, and F is then just the sum of the Fourier _series_ of f - you then revover the _coefficients_ f(n) by the usual formula for Fourier coefficients.>If I want to find its inverse Fourier transform, I know I should try >either DTFT/IDTFT, or Fourier series, for such a periodic function, and >the target function f(t) in the time-domain should be a discrete >comb-like function. > >However, instead of using DTFT/IDTFT, Fourier series, if I still want >to use Inverse Fourier Transform, and it is > >f(t)=Integrate(F(u)*exp(i*u*t), u from -inf to +inf), > >However, the integration does not exist for such a periodic function >F(u) when integrating on R. > >Is there a nice and elegant way to unify the Foureir transform and >Fourier Series and starting with the above integral which seemingly >does not exist and via sampling function or the dirac-delta function >and finally still reach the target function f(t) which is discrete >signal? > >Same question for the inverse Laplace transform, if the integrand is >periodic with 2*pi, e.g. the F(s) function is a function of exp(s), >where s is the variable in Laplace transform, it will be periodic with >2*pi on the line integral from c-i*inf to c+i*inf when doing inverse >Laplace transform. > >Is there a way to reconcile and start with this periodic function >integration and obtain a discrete f(t) in time-domain? > >I have seen this unification of Fourier series and Fourier transform >somewhere but it just elapsed my memory. > >Could you give some pointers? > >Thanks************************ David C. Ullrich

Reply by ●August 11, 20062006-08-11

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. > > I am looking for a rigorous derivation of the above steps...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. Lucy, can you accept the following identity?: +inf +inf SUM{ delta(t - n) } = SUM{ exp(i*2*pi*k*t) } n=-inf k=-inf if you can accept that, there is not a problem. if you cannot i dunno where to go with it. the rigorous math guys don't like how us electrical engineers deal with the dirac delta function (we call it the "unit impulse function"). this is because the dirac delta "function" is not a function in the regular sense of the word, but we EE guys treat it pretty much like one (and i, for one, am pretty unapologetic about it). r b-j

Reply by ●August 11, 20062006-08-11

"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? One more term is necessary and sufficient, of course. What an odd question to ask... Androcles Engineering is the art of making things others want operate safely and at low cost. ________________________________________________________

Reply by ●August 11, 20062006-08-11

Lucy wrote:> robert bristow-johnson wrote: > > 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 > > Hi Robert, > > 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. > > I am looking for a rigorous derivation of the above steps...This rests on the fact that FT takes convolution to multiplication and vice-versa. Then one notices that any periodic function can be written as convolution of its single period with an infinite train of delta functions spaced by the distance equal to one period (T). Finally, you need to know that FT of such train of delta functions is another train of delta functions of slightly different weight and reciprocally spaced spikes (1/T). Then you have the connection between the FT and the Fourier series. Suppose f is your periodic function and g is this function restricted to a single period (i.e., set to zero outside that period), then (letting * denote convolution and x multiplication): FT(f) = FT(g * [train of delta functions spaced by T]) = = FT(g) x FT([train of delta functions spaced by T]) = = integral_(single period) (g) x [train of delta functions spaced by 1/T] = = FT of one period of f sampled every 1/T seconds. -- Jan Bielawski