# Inverse Fourier/Laplace transform of a periodic function?

Started by August 11, 2006
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

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
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... Thanks
"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.

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
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. &#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;
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
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
"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.
________________________________________________________

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