7
$\begingroup$

I want to ask basic question. In our mathematics classes ,while teaching the Fourier series and transform topic,the professor says that when the signal is periodic ,we should use Fourier series and Fourier transform for aperiodic signals.

My question is can't we use Fourier transform formula in case of periodic signal ? Also, Is Fourier series used always for periodic signals and Fourier transform for aperiodic signals only ?

$\endgroup$
  • 3
    $\begingroup$ you can use Fourier transform with periodic functions if you wish (provided you are comfortable with distributions), you just get $(\cal F f)(\xi)=\sum_n c(n)\delta(\xi-n)$, where $c(n)$ are the coefficients of the Fourier series. So you won't get anything new - it's simpler to stay with just Fourier series. $\endgroup$ – user8268 May 28 '15 at 19:38
4
$\begingroup$

A Fourier series is only defined for functions defined on an interval of finite length, including periodic signals, as you can see from the definition of the Fourier coefficients (in the basis $\{e^{inx}\}_{n\in\mathbb{Z}}$) $$ a_n = \frac{1}{2\pi}\int_{-\pi}^\pi f(x)e^{-inx}~dx. $$

You can't define an aperiodic signal on an interval of finite length (if you try, you'll lose information about the signal), so one must use the Fourier transform for such a signal.

In principle, one could take a Fourier transform of a periodic signal in the sense that one could extend the signal outside the interval $[-\pi,\pi]$ by zero. But the resulting Fourier transform would look like $$ \widehat{f}(p) = \frac{1}{2\pi}\int_{-\pi}^\pi f(x)e^{-ipx}~dx $$ and this isn't particularly different from the Fourier coefficients. Moreover, being able to express a periodic signal as a discrete sum of frequencies is a stronger statement than expressing it as a continuous sum via the inversion formula.

Edit: In response to the comment.

I mean "stronger" rather loosely, not in the mathematical sense of one statement implying another.

Taking the Fourier transform of a periodic signal by extending it to $0$ outside $[-\pi,\pi]$ gives no information that the usual Fourier transform would not. Furthermore, the inversion formula still holds, and therefore we see that $f$ can be recovered from its Fourier transform as a continuous sum over all frequencies.

What is remarkable about periodic functions is that one does not actually need all this information to recover the function; out of uncountably many values $\widehat{f}(p)$, one only needs the integer values $\widehat{f}(n)$ (i.e. the Fourier coefficients) to reconstruct the function.

So in this sense, for a periodic function, there is more to say than what the Fourier transform alone provides.

$\endgroup$
  • $\begingroup$ I don't get your sentence I.e.**Moreover,being able to express a periodic signal as a discrete sum of frequencies is a stronger statement than expressing it as a continuous sum via the inversion formula**. Can you explain it by taking any example? $\endgroup$ – sagar Jun 24 '15 at 17:17
  • $\begingroup$ @sagar See my edit. $\endgroup$ – Gyu Eun Lee Jun 26 '15 at 18:57
1
$\begingroup$

Fourier Series (FS) exists only for periodic signals.

Fourier Transform (FT) is derived from FS, i.e. FT is the envelope of the FS. Thus as the frequency domain became more finer, time domain enlarges making it a aperiodic signal.

So when FT is applied on a periodic signal, the result is just, T*X(k) where T is the time period of the signal and X(k) is the FS of the signal.

Fourier is just awesome!

$\endgroup$

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.