1
$\begingroup$

enter image description here

I know these formulae and can solve problems mechanically but i never got the core concept. Please help me out with this.

$\endgroup$

marked as duplicate by MBaz, lennon310, A_A, Stanley Pawlukiewicz, Matt L. Sep 25 '18 at 8:51

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ I really doubt that this hasn't already been answered on this site. I just don't have time now to search for it. $\endgroup$ – Matt L. Sep 20 '18 at 10:58
1
$\begingroup$

A fourier series essentially breaks apart a periodic signal to represent it as an infinite sum of sine waves that are in that signal. So a fourier transform of a signal breaks apart the signal to look at all the frequencies that are in that signal (essentially looking at all the ingredients that make up that signal, ingredients being the frequencies).

A frequency response is the frequencies of an output signal compared to the input signal. Essentially, a measurement of how well the output can replicate the input.

$\endgroup$
0
$\begingroup$

If you want to truly understand the Fourier representation of a signal in layman terms, open your favourite music (WAV or MP3) in a audio player that displays the spectrum changing in realtime.

Here is an example of the spectrum of the "Lana Del Rey - Born to Die" song, at timing 01min04sec. In fact since the FFT size is 4096 (see top left) and the sampling rate is 44.1 Khz, this means that the curve here shows the frequency distribution of the song in a window 01min04,000sec to 01min04,093sec.

It shows what's going on during this 93 millisecond time-window:

  • for example, you have a peak around 220Hz at -24 dB, this is one of the loudest frequency here. That's the frequency of an A note.

  • what is the strength of the frequencies around 6000hz? Answer on the graph: -60dB

  • etc.

enter image description here

What do you do when you want to have more bass or less treble in your Hifi player? You use an audio "equalizer", and an equalizer just re-balances these frequencies. The equalizer in a audio player is often coded with Fourier transform (often on small time-window blocks one after another, this is called Short Time Fourier Transform).

Now your question: what is the Fourier series?

In this example (I use the discrete case), it means that the 93ms part of the song we were talking about, let's call it x[0], x[1], ..., x[4095], can be totally reconstructed as the sum of sinusoidal terms:

$$x[n] = \sum_{k=0}^{4095} a_k e^{i k 2 \pi n / 4096} = \sum_{k=0}^{4095} a_k (\cos(k 2 \pi n / 4096) + i \sin(k 2 \pi n / 4096)),$$

where the $a_k$ means the amplitude, the "strength" of the sinusoid $\cos(k 2 \pi n / 4096)$ in the signal (idem for $\sin(...)$).

It means that this small part of the Lana Del Rey song can be expressed as a sum of many sinusoids: a little bit of sinusoid at 20 hz + a little bit of sinuosid at 30 hz + ... + a little bit more (i.e. higher amplitude) of sinusoid at 220 hz + ...

$\endgroup$
0
$\begingroup$

Fourier had the assumption (without proof) that any continuous-time periodic signal $\tilde{x}(t)$ can be decomposed into a weighted sum of periodic sine waves $\sin(k \omega_0 t)$ and $\cos(k \omega_0 t)$ with harmonic family of fundamental frequency $\omega_0 = 2\pi f_0 = \frac{2\pi}{T_0}$. And those weights are called as the continuous-time Fourier series coefficients of the signal $\tilde{x}(t)$. Using the modern complex exponential representation mathematically stated as;

$$ \tilde{x}(t) = \sum_{k=-\infty}^{\infty} a_k e^{j k \omega_0 t } $$

which is the synthesis equation where $a_k$ is the associated complex Fourier series coefficient for the frequency $\omega_k = k \omega_0$. You can find the complex Fourier series coefficients from the analysis equation:

$$ a_k = \frac{1}{T_0} \int_{-T_0 / 2}^{T_0/2} \tilde{x}(t) e^{-j k \omega_0 t } dt $$

Frequency response $H(\omega)$ is associted with LTI systems, since they possess a unique impulse response $h(t)$ and basically is the continuous-time Fourier transform of it;

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

One can view Fourier transform as a limiting case of the Fourier series coefficients $a_k$, where as the period of the signal ($h(t)$) goes to infinity, the discrete line spectra $a_k$ merges into a continuum of frequencies whose envelope is given by the Fourier transform of $h(t)$.

$\endgroup$

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