1
$\begingroup$

Fourier transform for aperiodic signal is given by

$$ X(\omega) = \int\limits_{t=-\infty}^{+\infty} x(t) e^{-j \omega t} dt. \quad (1) $$

Fourier series for periodic signal is given by

$$ y(t) = \sum\limits_{m=0}^{+\infty} a_m \cos(w_m t) + \sum\limits_{m=0}^{+\infty}b_m \sin(w_m t). \quad (2)$$

I want to ask that

Why Fourier series has summation and Fourier transform has integration symbol in their respective formulae although both $x(t)$ and $y(t)$ are continuous signals only? Is it just because one is used for aperiodic and one is for periodic signal?

$\endgroup$
  • $\begingroup$ The Fourier transform and Fourier series are different things. There are connections (in functional analysis you deal with something called Pontryagin duality where you learn exactly how they're connected), but what you're asking is like asking "why is there a summation in Taylor series but an integral to find areas?". They're two different things. $\endgroup$ – user223391 Jun 4 '15 at 17:19
  • 1
    $\begingroup$ your question was asked before , i think you can find an explanation here : math.stackexchange.com/questions/221137/… $\endgroup$ – mounir ben salem Jun 4 '15 at 17:24
  • $\begingroup$ @mvw sir,1.in case of Laplace transform ->we multiplied $exp(-p.t)$ term by Fourier transform formula to obtain Laplace transform formula which is useful when we can't find Fourier transform of function which is continuously increasing with time. 2.for z transform -> it is first of all used only in case of discrete signals which is derived from Fourier transform multiplying it with radius $r$ 3.I don't know about wavelets $\endgroup$ – pandu Jun 4 '15 at 17:56
4
$\begingroup$

There are a number of different ways to pose this connection. One way is to view the Fourier transform of a periodic function as a distribution, in which case it is a sum of Dirac deltas with weights corresponding to the coefficients in the Fourier series. In this sense the Fourier transform is really the fundamental object.

Another way is to realize the Fourier transform as a limit of Fourier series. Specifically, you can take a function and cut it off outside some large interval. Then you can take the Fourier series of the new function on the interval. Note that there are some questions about how this Fourier series converges if the new function is not periodic. But we are at least guaranteed convergence in $L^2$ if the original function was $L^2$.

Now when you take the size of the interval to infinity, these Fourier series converge in a sense to the Fourier transform. What is going on here is that as the space scale you are considering gets larger, the frequency gap between the relevant Fourier modes is getting smaller. (This is a variant of the uncertainty principle.) As the space scale gets very large, the frequency gap becomes very small, until in the limit there is a continuum of frequencies.

$\endgroup$
  • $\begingroup$ Can you recommend a math book that discusses the Fourier transform of a periodic function as a distribution? If the answer is "any good book on Fourier analysis", please name your favorite one. $\endgroup$ – littleO Dec 27 '16 at 23:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.