1
$\begingroup$

The motivation behind the fourier transform is to somehow represent a non-periodic signal as a sum of sinusoids just as we do with the fourier series for periodic signals, correct?

With the Fourier series we have, $$x(t) = x(t+T) = \sum_{k = -\infty}^{\infty} X[k]e^{j\frac{2\pi}{T} t}$$

where $T$ is the period of our signal and $X[K]$ is the weight or coefficient of one specific sinusoid frequency, is that correct?

$$$$

On the other hand, we can't represent non-periodic the same way because each part of the signal is not (necessarily) the same. We attempt to remedy this issue by using Fourier transforms. Let's take the continuous time Fourier transform (the only one I've learned):

Let $x(t)$ be some non-periodic signal, let's say it looks like a signal that has a square wave first, then a triangle wave, then something random.

We take the Fourier transform:

$$X(\omega) = \int_{-\infty} ^{\infty} x(t)e^{-j\omega t}dt$$

where I believe $\omega = \frac{2\pi k}{T}$. In this case, $T$ here is some period of time we choose to repeat the signal to attempt to make a periodic signal... if that makes any sense to anyone out there

Now here's where my confusion really gets complicated.

Transferring the same logic as we did with the Fourier series $X(\omega)$ should be the the weight or coefficient of one particular sinusoid frequency that makes up the signal $x(t)$, right? Or is that just completely wrong? What's the interpretation of $X(\omega)$ here?

Also, we said our signal $x(t)$ was a square wave first, then a triangle wave, then something random. Shouldn't the $X(\omega)$ for a particular $\omega$ be different for the first part (square wave) than it is for the second part(triangle wave), and so on? How come we only get one value of $X(\omega)$?

$\endgroup$
7
  • $\begingroup$ Related, also animated $\endgroup$ Sep 6 at 17:24
  • $\begingroup$ "We take the Fourier transform: $$X(\omega) = \int_{-\infty} ^{\infty} x(t)e^{-j\omega t}dt$$ where I believe $\omega = \frac{2\pi k}{T}$." STOP right there. Your belief is incorrect and everything else falls to the ground right there. $\endgroup$ Sep 6 at 18:48
  • $\begingroup$ @Dilip Sarwate Okay, then should we take $\omega $to be any non-negative, real number? My confusion is still the same. The signal is still aperiodic, so if we sum up sinusoids(which have some periodicity) we won't get an aperiodic signal, right? $\endgroup$
    – BigBear
    Sep 6 at 19:27
  • $\begingroup$ @DilipSarwate Okay, I've had a moment to think about it some more. Let me see if I can get this right. With the Fourier Series(FS), we had $\omega$ be some multiple of $\frac{2\pi}{T}$ so we didn't have really "fine grained" frequencies. Here with DTFT, we have $\omega$ as any real non-negative number. This change let's us describe arbitrary signals because we have so many frequencies to choose from? Is that right? If so, that's INCREDIBLE! What a powerful tool! $\endgroup$
    – BigBear
    Sep 6 at 21:06
  • 1
    $\begingroup$ I mentioned it because you just commented that FS doesn't have fine grained frequencies but DTFT has any real non-negative frequencies. That's all becuase that FS is periodic in time domain so it has discrete frequencies, and DTFT is non-periodic in time domain so it has continuous frequencies. $\endgroup$
    – ZR Han
    Sep 7 at 3:03
1
$\begingroup$

If the Fourier transform

$$X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\tag{1}$$

of a function $x(t)$ exists, then the function $x(t)$ can be expressed in terms of its Fourier transform:

$$x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}X(\omega)e^{j\omega t}d\omega\tag{2}$$

Eq. $(2)$ is also called inversion formula. Note that $\omega$ is a real continuous variable. I.e., in general all frequencies are necessary to represent a non-periodic function $x(t)$, not just a discrete set of frequencies, as is the case for periodic functions.

The Fourier transform $X(\omega)$ is generally a complex-valued function. Its magnitude is often called the spectrum of $x(t)$. It represents the frequency content of $x(t)$. However, the phase of $X(\omega)$ is equally important for representing $x(t)$. The squared magnitude $|X(\omega)|^2$ is called the energy density of $x(t)$.

Even though mathematically unsatisfactory, it may help intuitively to think of $x(t)$ as a function whose "period" approaches infinity (which makes it non-periodic), and, consequently, the distance between the frequencies needed for its representation becomes smaller and smaller, until, in the limit, $\omega$ becomes a continuous variable, and the sum of the original Fourier series becomes and integral.

$\endgroup$
2
  • $\begingroup$ Thanks for the follow-up answer, really appreciate it :-) Could you elaborate on the importance of the "spectrum" and its phase please? $\endgroup$
    – BigBear
    Sep 7 at 15:51
  • 1
    $\begingroup$ @BigBear: The spectrum describes the frequency content of a signal, e.g., "low pass" if there are only/mainly low frequencies present, etc. The phase describes the temporal details of the signal. There are infinitely many different low pass signals with the same spectrum, but their phases are all different. $\endgroup$
    – Matt L.
    Sep 7 at 16:04

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.