How do I determine the wavelength of a noise signal like the one below? I find it easy to understand for sine waves, but it gets tricky for me when the signal is more complex like a noise signal.

If I understand correctly a period is defined as the time it takes for the signal to cycle from one amplitude back to the same amplitude that it came from.

But many signals do not have these repeatable cycles, so how are period and wavelength defined for them? Or does it not make sense to talk about periods or wavelengths in these signals? Or do they just have many different wavelengths?

enter image description here


3 Answers 3


Let's start with some definitions (wikipedia):

  1. A signal is a periodic signal if it completes a pattern within a measurable time frame, called a period, and repeats that pattern over identical subsequent periods. The completion of a full pattern is called a cycle. A period is defined as the amount of time (expressed in seconds) required to complete one full cycle. The duration of a period, represented by $T$, may be different for each signal but it is constant for any given periodic signal.

  2. The frequency of a periodic function is the number of complete cycles that can occur per second: $$f = \cfrac{1}{T}$$

  3. The wavelength, or spatial period, is the distance a periodic wave travels during one period. For sound ($c$ being the speed of sound): $$\lambda = \frac{c}{f}$$

As you can see, if you know either of these 3 (period $T$, frequency $f$, or wavelength $\lambda$), you can compute the others.

Now to your questions:

But many signals do not have these repeatable cycles, so how are period and wavelength defined for them?

You are right, as a matter of fact most real-life signals, such as noise, are aperiodic, meaning they are NOT periodic, so they can not be characterized by a SINGLE period / wavelength / frequency.

Or does it not make sense to talk about periods or wavelength in these signals?

As stated earlier, it doesn't make sense to talk about PERIOD or WAVELENGTH (note the singular here) of an aperiodic signal. BUT:

Or do they just have many different wavelengths?

Congratulations, you just had the same exact insight French mathematician Jean-Baptiste Joseph Fourier had in 1822: that any function, whether continuous or discontinuous, can be expanded into a series of sines (to be completely accurate, into a series of complex exponentials, but don't worry about this for now). What this means is that each of these sines can be combined linearly (fancy word to say "scaled and summed") to add up to the original signal.

In that sense you could say that aperiodic signals, such as the noise you took as an example, "have many different wavelengths". In fact, they are composed of an infinite sum of sinusoidal (hence periodic) waveforms, each with a different wavelength, amplitude, and phase.

  • period / frequency / wavelength: how long it takes for that sine to complete a cycle (in seconds) / how many cycles that sine goes through in a second (in Hz) / the distance traveled by that sine in one cycle (in meters).
  • amplitude: the amount of that sine in the original signal.
  • phase: the offset of that sine relative to a sine with same frequency and 0 phase.

So, how do you get the wavelength, amplitude and phase for each of these components? Here comes the Fourier Transform.

Fourier transform

  • The Fourier transform is a framework to analyze these types of aperiodic signals. Using clever mathematics, you can transform a time-domain signal (including, but not limited to, noise) into the frequency-domain, which allows you to compute the characteristics (wavelength, amplitude and phase) for each sine wave that compose this signal.

  • The $\texttt{DFT}$ ($\texttt{D}$iscrete $\texttt{F}$ourier $\texttt{T}$ransform) is used when dealing with discrete signals, and is usually implemented using the $\texttt{FFT}$ ($\texttt{F}$ast $\texttt{F}$ourier $\texttt{T}$ransform) algorithm. Depending on your software/language (Matlab? Python? R? C?) there are libraries you can use to easily compute that.

  • Read up on $\texttt{DFT}$ theory (pay special attention to the length of the $\texttt{DFT}$ and resulting frequency resolution), experiment with the $\texttt{FFT}$ routine of your choice and come back if you have any questions!

  • 1
    $\begingroup$ A nice writeup, and nice 0-1k, it's needed. $\endgroup$ Commented Oct 3, 2022 at 19:15
  • $\begingroup$ @OverLordGoldDragon much appreciated ;) $\endgroup$
    – Jdip
    Commented Oct 3, 2022 at 19:56
  • $\begingroup$ Comment on "phase: the offset of that sine relative to a sine with same frequency and 0 phase." This is a poor representation of phase for signal processing that is often misused as it conflates time delay with phase. A time delay surely creates a phase versus frequency but the intuition from that breaks down when you get to DC (as with that explanation you would approach an infinite delay as the frequency approaches DC). We can however create complex DC signals (DC in that they are constant magnitude and phase for all time) with phase! Bin 0 of the DFT is the magnitude and phase for a signal $\endgroup$ Commented Oct 7, 2023 at 11:59
  • 1
    $\begingroup$ Phase concisely is a rotation on the complex plane. With that we can have complex samples in time or in frequency and each sample has a magnitude and phase. All samples of a real sinusoid have a phase of 0 or $\pi$. All samples of an imaginary sinusoid have a phase of $\pm \pi/2$, Combine the two and we can get all phases as we see from: $e^{j\phi} = \cos(\phi)+j\sin(\phi)$. So $\cos(\phi)$ is translating the phase of a complex sample $e^{j\phi}$ to a real number (which has zero phase). $\endgroup$ Commented Oct 7, 2023 at 12:03
  • $\begingroup$ Always appreciate your input, @DanBoschen! While your description and the “complex rotation” is the mathematically correct way to describe phase , I went for more of an intuitive answer to OP’s concerns. $\endgroup$
    – Jdip
    Commented Oct 7, 2023 at 15:32

Period and frequency are inherently defined only for periodic signals. They're easiest to understand in the context of a simple signal like a sine wave. A sine wave with period $T$ has frequency

$$ f = \frac{1}{T} $$

If this signal represents a wave traveling through space at velocity $c$, then its wavelength is

$$ \lambda = \frac{c}{f} = cT $$

A complex signal like the one you've shown doesn't have an easy-to-find period like a sine wave. To examine the behavior of such a signal we need to use a tool like the Fourier transform. The Fourier transform takes a signal like the one you've shown and finds its representation as a weighted sum of sine waves, each with a different frequency. The weight of the sine wave at a particular frequency tells us how much power the original signal has at that frequency. For example, you can't really see any patterns in the following plot of a signal's amplitude as a function of time: Plot of a signal as a function of time.

However, if we look at the Fourier coefficients of this signal as a function of frequency (also called its spectrum), you can see that the signal is just a sum of three sine waves at frequencies 5Hz, 11Hz, and 31Hz. Plot of Fourier coefficients as a function of frequency for the signal shown in the previous plot. The coefficients are zero except for spikes at 5Hz, 11Hz, and 31Hz.

In general, noise is just the component of a signal that you're not interested in, so it's hard to generalize about its spectrum. However, particular types of noise are usually defined by the power they have at different frequencies. For example, white noise is defined as having an equal amount of power at every frequency.


1.- just find the Fourier coefficients.

2.- A good explanation of how to calculate them is available from Wolfram


3.- The Fourier decomposition coefficients of a signal tell you how much of each carrier, at the different frequencies of your choice, is in the signal.

Some apparently noisy signals may hide just a couple tones, and sometimes apparently cyclic signals, judged at first sight as single carrier, they are actually thousants of different carriers overlapped.

4.- Import the signal into MATLAB and then try one of many answers available in Matheorks to find such coefficients. This one may work:


I say 'may work' because you have not supplied the actual signal, if you do so I may try it for you.

For a givem implemetation of how to find Fourier coefficient of a trace it may happen that the frequency spacing is too wide, or no enough amplitude resolution is applied, or you may need complex (as in a+1j*b) coefficients, or simply, the sample is not aignificant and you may need more samples.

5.- the image of your signal seems to come from a cheap hand held spectrum analyzer. It looks like really low resolution, or not using the right antenna, or not recording long anough.

Be aware that low resolution FFTs cause signal degradation and may even miss signal presence.

  • $\begingroup$ // just find the Fourier coefficients.// -------- you're gonna need to know the period first, and if that period is not an integer number of samples, you will need to resample an entire period so that the period of the resampled data is an integer $N$ (which is the same $N$ of the DFT). Then you have the Fourier coefficiets. $\endgroup$ Commented Oct 1, 2022 at 21:35

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.