I totally understand the concept of fourier transform, but one thing thats bothering me is the amplitude that we plot in the frequency domain. What does that amplitude of each frequency signifies? Is it the amplitude of that particular signal component in the input signal, as it is for the frequency? Is there any other meaning or any other way we can infer the amplitude?

Thank you for your time.


5 Answers 5

  • "I totally understand the concept of Fourier transform"

Lucky you if you really do. Some of us (me, in first place) don't (in totality). The Fourier transform (and its avatars) is a prototype for duality. Duality here means that you can represent a signal on some primal domain (time) onto a dual domain (here frequency). This transformation is meant to possess useful properties, to preserve signal information, and to add insight, like a somewhat easier interpretation.

Signals are traditionally represented as "amplitudes of something for each time". These amplitudes are not always absolute. But we can hope they are linearly related to the actual physical values. And that the linearity factor does not change over time. Under the above premises, sines and cosines, or complex exponentials (cisoids), are a very good way (perhaps the best) to model the system. And in turn they can be used to represent the data, as a linear combinaison of shifted sines.

Since the sines are orthogonal, the squared magnitude of the coefficients weighting the sines is proportional to the energy of each corresponding sine frequency.

The proportional factor depends on how the coefficients are computed, often up to a scaling factor. The amplitude of a given frequency component can be directly computed with correlation. Note however that this coefficient should be taken with care, as outer conditions play a role: the finite horizon observation of a function/signal, amplitude discretization, noise, etc.

  • $\begingroup$ Thank you for your answer. So, the amplitude can be interpreted as how much sin signal of that particular frequency is there in the original signal. Am i right? $\endgroup$ Dec 31, 2019 at 7:37
  • $\begingroup$ Somehow, keeping in mind this a representation model, and the actual quantity depends on other factors (sampling, windows, quantization, etc.) $\endgroup$ Dec 31, 2019 at 15:52
  • $\begingroup$ Does this mean, in other words, the higher the amplitude of a signal in the frequency domain, the stronger is that certain frequency in the time region? $\endgroup$
    – Ben
    Mar 5, 2020 at 10:58

There are various flavors physical interpretations of the continuous Fourier Transform. Here is the one that works best for me:

The amplitude of the Fourier Transform is a metric of spectral density. If we assume that the unit's of the original time signal $x(t)$ are Volts than the units of it's Fourier Transform $X(\omega)$ will be Volts/Hertz or $V/Hz$.

Loosely speaking it's a measure of how much energy per unit of bandwidth you have. More formally speaking: the magnitude square $X(\omega)$ is a metric power spectral density and the integral/sum over all frequencies gives you the total energy of the signal.

That works both in the time and the frequency domain and the total energy can be calculated the same way (squaring and integrating) and they are the same. That's Parseval's Theorem. https://en.wikipedia.org/wiki/Parseval%27s_theorem

A somewhat different interpretation is to view the Fourier Transform as a representation through orthogonal functions. These orthogonal functions are the complex exponential and they form an ortho-normal base (more or less) . Basically that means you can represent any time domain signal as being made up of an infinite number of complex exponential. Similar to how any Lego contraption is made up of the same set of elementary Lego blocks. The Fourier Transform amplitude simply tells you how much of each Logo black are in any contraption.


The magnitude of each bin is the magnitude of that frequency component for that waveform in the time-domain, specifically when the time domain waveform is expressed as a sum of complex exponential frequencies.

In the frequency domain that includes positive and negative frequencies, each impulse for the Fourier Transform of an arbitrary waveform is the Fourier Transform of

$$Ae^{j\omega t + \phi}$$

This is a complex phasor with constant magnitude A, rotating (changing in phase) on the complex plane at rate $\omega t$ with a starting phase of $\phi$. $\omega$ is given by the position on the frequency axis and the FT value has a magnitude and phase give by $Ae^{j\phi}$.

Similarly the Fourier Series Expansion decomposes a finite duration arbitrary (single valued, analytic) waveform into a series of such spinning phasors with arbitrary magnitudes and each spinning at an integer multiple of the inverse of the duration:

$$\sum_{n=-\infty}^{\infty} A_n e^{j\omega_o n t}$$

Also note that a simple sinusoid is two of these spinning phasors rotating in opposite directions with the same magnitude as given by Euler's identity:

$$cos(\omega t) = \frac{1}{2}e^{j\omega t} + \frac{1}{2}e^{-j\omega t}$$

Hence when we have real signals, the postiive and negative frequencies are conjugage symmetric and thus can alternately be described in terms of a sum of sine waves and cosine waves. I find the exponential frequency expression more intuitive for wider application to fundamental DSP concepts when I hear the term "frequency": I think of frequency components as spinning phasors rather rotating with time than sinusoids.

This applies equally to the Discrete Time Fourier Transform (DTFT) and Discrete Fourier Transform (DFT). The difference between the two is the DTFT is the transform of a discrete time domain signal that extends from $\infty$ to $\infty$ like the Fourier Transform, while the DFT extends over a finite duration (0 to N-1) like the Fourier Series Expansion. The spectrum of the Fourier Transform and the DTFT is continuous while spectrum of the Fourier Series Expansion (where it is non-zero) and the DFT is discrete, specifically because both are time limited. So for those cases the frequency domain has "impulses" for the FSE or equivalently "unit samples" for the DFT where each impulse is the transform of the spinning phasor in time as represented above, with a constant magnitude A and starting phase $\phi$.

That said, here are some examples using the DFT given by:

$$X[k] = \frac{1}{N}\sum_{n=0}^{N-1}x[n]e^{-j (k \omega_o) n}$$

With inverse DFT given by:

$$x[n] = \sum_{n=0}^{N-1}X[k]e^{j (k \omega_o) n}$$

(Note some forms distribute the 1/N scaling to be $1/\sqrt{N}$ on each, the form I show here which is a matter of definition is consistent with my description without having to elaborate on scaling further).

Where $\omega_o$ is the fundamental frequency determined the inverse of the time duration of the signal (I describe it that way to show how it is identical to the Fourier Series Expansion, in particular the IDFT above is the reconstruction of complex scaled integer harmonics). Since the time duration of the signal in this case in N samples, then the fundamental frequency $f_o$ is 1/N and $\omega_o$ is given as:

$$\omega_o = 2\pi f_o = 2\pi/N $$

So in this case if you had a time domain waveform given as $$5e^{j\omega_o n}$$

Given this is at the fundamental frequency, it is a phasor of magnitude 5 that will rotate exactly once counter-clock-wise as n goes from 0 to N-1 in time. The DFT of this time domain waveform will be 0 in all bins except for the 2nd bin k =1, which will have $A = 5$ and and $\phi = 0$ with reference to the magnitude and phase components described earlier.

Similarly the time domain waveform given as $$5cos(\omega_o n)$$

Which is $$\frac{5}{2}e^{j\omega_o n} + \frac{5}{2}e^{-j\omega_o n}$$

Will have 0 in all but two bins: bin k = 2 and bin k = N-1 where each of those bins will have $A = 5/2$ and $\phi = 0$.

If the k = N-1 is confusing, consider from the direct equation last given above that we would expect the non-zero frequency bins to be at k = -1 and k = 1. Given the periodicity these are equivalent:

$$e^{-1 j \omega_o n} = e^{-1 j (2\pi/N) n} = e^{(N-1)j(2\pi/N)n}$$

If you divide the unit circle into N phasors ($2\pi/N$), rotating one unit clock-wise ($-2\pi/N$) is the same as rotating N-1 units counter-clockwise ($2\pi (N-1)/N)$).

Thus the magnitude of each bin is the magnitude of that frequency component for that waveform in the time-domain, specifically when the time domain waveform is expressed as a sum of complex exponential frequencies.

  • $\begingroup$ Thank you so much for the answer. The last paragraph would have been enough for me. Thanks again. $\endgroup$ Dec 31, 2019 at 7:41

The magnitude of $X(\omega)$ for a given $\omega$ signifies how much does the signal $e^{j\omega t}$ exist in the signal $x(t)$; indicated as an inner product between $x(t)$ and $e^{j\omega t}$.


In the discrete case, as when feeding a digitized signal to a DFT or FFT, each output point in the FFT spectrum is the magnitude output from a simple bandpass filter (a Goertzel filter result, or equivalent complex FIR filter).

For a rectangular windowed FFT, each of those filters has a Sinc (or Dirichlet) shaped filter response. The more input signal energy in the passband of each filter, the greater the output (amplitude or height on a frequency plot) for the FFT result bin representing that filter. The scale factor can be arbitrary (varies between FFT implementations, also graphic libraries).

  • $\begingroup$ Each or those filters approaches a Sinc shaped filter response as N goes to infinity, for lower N they can be interpreted as “aliased Sinc functions”. $\endgroup$ Dec 30, 2019 at 17:36

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.