1
$\begingroup$

I'm an EE undergrad that struggles heavily with the intuition behind the Fourier Transform (most likely due to a shoddy mathematical foundation). Specifically:

  1. From what I understand, the real part of the Fourier Transform is the Fourier Transform of the even part of the function in time domain, and the imaginary part is the transform of the odd part. But in the context of Fourier, we often say that a time domain function has n / an infinite amount of frequency components. But what exactly is a frequency component? I.e, is a function $f(t)$ represented by convention as a sum / integral of sines, cosines, or both? Or maybe it is so that we refer to the sine and cosine terms of $e^{-j\omega} = cos(\omega) - jsin(\omega)$ when we say "frequency component?"

  2. I get that the absolute value of the Fourier Transform yields the magnitude of a certain frequency component, and that the argument yields the phase shift of some frequency component. But what is that phase shift relative to? Is that phase shift relative to a sine or a cosine?

Thank you for taking your time to read this.

$\endgroup$
1
  • $\begingroup$ Welcome to SE.SP! Good question! Some perspective here. $\endgroup$
    – Peter K.
    Dec 29, 2023 at 22:28

2 Answers 2

0
$\begingroup$

Granted a precise answer to your question would (and has) fill(ed) a multitude of books, here is a stripped down answer.

  1. In the context of Fourier analysis, when we refer to a "frequency component" of a function $f(t)$, we are typically talking about the constituent sine and cosine waves that, when summed together in a specific way, form the original function $f(t)$. The Fourier Transform decomposes $f(t)$ into these sine and cosine components, each associated with a specific frequency.

    The Fourier Transform of a real-valued function, such as $f(t)$, can be complex-valued (can be real as well, depending on the symmetries of $f(t)$ - $f(t)$ even for example). This complex result encodes both the amplitude and phase information of each frequency component. The real part of this result corresponds to the coefficients of the cosine terms (even function component), and the imaginary part corresponds to the coefficients of the sine terms (odd function component).

    When we express $f(t)$ in terms of its Fourier Transform, we are essentially representing it as a sum (or integral, in the case of the continuous Fourier Transform) of these sine and cosine waves. Each sine and cosine wave is a "frequency component" of $f(t)$. The complex exponential form $e^{-j\omega t}$ (where $j$ is the imaginary unit) is often used because it compactly represents both sine and cosine terms through Euler's formula.

  2. The phase shift obtained from the Fourier Transform is the phase angle of the complex number representing a particular frequency component. This phase shift indicates how much a sine or cosine wave of that particular frequency is shifted in time relative to a reference point, typically the origin $t=0$.

    Whether the phase shift is relative to a sine or a cosine wave can depend on the convention used in the Fourier Transform definition. However, in the standard form, the cosine function is often taken as the reference. This is because the cosine wave starts at its maximum value at time zero, making it a natural reference point for measuring phase shifts.

    The argument (or angle) of the complex number from the Fourier Transform gives this phase shift. Denote the phase shift by $\phi$. A positive $\phi$ typically means the wave is shifted to the left (leading phase), and a negative $\phi$ means the wave is shifted to the right (lagging phase), relative to the reference cosine wave.

$\endgroup$
5
  • $\begingroup$ Thanks for the great answer. I now have a follow-up question: Since the phase calculated from the argument of a certain frequency component is in reference to a cosine (i.e a simple harmonic oscillator), and a frequency component consists of both sines and cosines ($e^{-j\omega}$) how does it make sense to compare it with the phase of a complex exponential? In other words, what is the intuition behind comparing the phase between a cosine which is only real with the phase of a complex exponential, which is both real and imaginary, consisting of both a sine and a cosine? $\endgroup$
    – Philip
    Dec 30, 2023 at 1:36
  • $\begingroup$ Good follow up question! does this + this answer it? To summarize, cosines can be made to look like any ratio mix of real and imaginary basis vectors (cosines and sines). To see this, consider: $$\cos(ft+\phi) = \cos(ft)\cos(\phi) - \sin(ft)\sin(\phi) = a\cos(ft) + b\sin(ft)$$ $\endgroup$
    – Jdip
    Dec 30, 2023 at 7:17
  • $\begingroup$ Yes, thank you! I got what I wanted now so you don’t have to bother with this if you don’t want to, but does this apply to complex signals too? $\endgroup$
    – Philip
    Dec 30, 2023 at 10:50
  • $\begingroup$ I’m not as familiar with the complex DFT as others on this forum, but for complex inputs, my understanding is that the DFT phase additionally conveys information about the interaction between the real and imaginary parts of the input signal. $\endgroup$
    – Jdip
    Dec 30, 2023 at 11:33
  • $\begingroup$ Got it, thanks again. Going back to your first comment, can't we then just represent all frequency components as a cosine with a phase shift, instead of splitting them up into a sine and cosine? This seems like a more intuitive convention, especially when it comes to considering the phase shift. $\endgroup$
    – Philip
    Dec 30, 2023 at 14:02
0
$\begingroup$

A frequency component of a signal in context of the Fourier transform is essentially how much of that frequency is contained in the signal. You can view the Fourier transform from the perspective of a matched filterbank. It essentially correlates the signal with a sinusoidal filter. So, a frequency component is describing how similar the signal is to a complex sinusoid at that frequency.

With respect to the phase of the Fourier transform output, this answer may be helpful. The basic idea is that the phase of the Fourier transform output at a given frequency is describing the phase offset of that frequency in the original signal. Practically, when computing the FFT, there are issues that arise from aperiodicities that affect the phase of the FFT output, and there is Matlab code demonstrating this in the other answer.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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