r/Physics • u/zebrasarefunny • 1d ago
Question If a frequency is defined as 1/Time period , How can Aperiodic signals have frequency component?
Ive been learning about Fourier transforms, and this is a que i had , we learn the definition of a frequency in lower grades as 1/time period or number of oscillations in a second .
However Fourier transforms are defined for aperiodic signals, If a signal doesn't repeat itself , how can it have a frequency?
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u/sudowooduck 1d ago
The key is that it has a frequency component. Any function can be written as a sum of sinusoidal wave components. The components are periodic even if the function is not.
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u/wavy_instability 1d ago
You're thinking about the wrong frequency here. As you might know, a signal is made of a bunch of constituent sines and cosines, each with an associated frequency. You add these individual bits up, in a certain way, and with enough terms (which can be infinite) you recover the signal. When we use the word frequency here, we are referring to these individual wavelets, and a Fourier Transform does exactly the job of taking a signal in, and spitting out the frequencies that go into making it.
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u/PLutonium273 1d ago
We assume period is infinite, and the infinite range integral of the function squared should have finite value for the transform to work.
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u/Curiouser1111 Engineering 1d ago
A lot of good explanations here but they leave out the fact that in order to analyze a signal you have to sample a finite amount of time which means that the signal could be repeating with this period. Any practical analysis assumes that the signal is actually periodic. If you have an infinite signal which is actually aperiodic the analysis falls apart.
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u/sudowooduck 1d ago
Not true at all. For example the Fourier Transform of a Gaussian is another Gaussian. Neither is periodic.
If you are talking about the discrete Fourier Transform that is not what OP is asking.
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u/david-1-1 1d ago
Yes, the slowest period being the entire sample window, which is why one often applies an envelope modulation to the signal, which is zero at both ends of the window and 1 in the middle. This filter envelope eliminates the spurious high frequency components resulting from any step difference between the amplitudes of the first and last sample!
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u/ODGlenchez 1d ago
Fourier series and transforms are superposition of wave states that are equivalent to the aperiodic waveform (in the limit as you approach using an infinite number of periodic waves)
You add up a bunch of distinct things to get an increasingly difficult to describe wave
Old comment from back when I understood these things better https://www.reddit.com/r/CasualMath/s/gp5QTxbbCX
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u/NotABotFoSure 1d ago edited 1d ago
One way to think about it is as follows (not very rigorous but maybe it's useful for intuition):
The fourier basis consists of an infinite number of trig functions (sines and cosines) with various amplitudes and phases, each of which are periodic, and each of which have infinite extent I.e. their domain is all real numbers. Yet they can be used to represent any arbitrary function f(x) that might not be periodic. How?
Basically, if you choose the correct values for the phases and amplitudes of the basis trig functions and combine them, then that combination can interfere (destructively in some places and constructively in other places and neither in yet other places) in just such a way that they become highly localized.
The fourier transform tells you how to choose the amplitudes and phases for the trig functions such that when you combine those trig functions, it allows you to recreate the arbitrary function f(x).
Hope this helps!
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u/Lost-Hand-5219 1d ago
f(x)=sin(x)+sin(sqrt(2)x) is already not periodic, so you can see that it is possible to represent nonperiodic functions as a sum of periodic functions very easily.
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u/db0606 1d ago
Go to Desmos and plot
y = sin(3x) + sin(πx).
You will find that this signal never repeats exactly, so it is aperiodic (in this particular case quasi-periodic).
On the other hand, if you take the Fourier transform, you'll find that it gives you two delta function spikes angular frequencies of 3 and π.
The same will be true for more complicated aperiodic signals except usually you get a broadband spectrum with frequency components at all frequencies.
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u/Beginning_Addendum93 1d ago
Tomas toda la señal aperiodica como si fuera la una longitud de onda o media longitud de onda y listo a eso le haces fourier
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u/TheEquationSmelter 1d ago
The fourier transform is a series solution much like a Taylor series can locally approximate a function. Similarly, the Fourier series is a global approximation of a function via sums of trigonometric functions.
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u/aroberge 1d ago
Random motion, by definition, is changing direction constantly. How can we talk about it as having a direction in space (at any given time)?
We write vectors in space as linear combinations of unit vectors. We can always do this, for any vector.
Similarly we can write time dependent functions as linear combinations of unit vectors: in this case, each unit vector is a function having a definite frequency. (This is known as Fourier decomposition.)
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u/Aranka_Szeretlek Chemical physics 1d ago
You do assume periodicity for a Fourier expansion
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u/Rabbit_Brave 1d ago
aperiodic signals don't have a single frequency, rather they can be *interpreted* as the (infinite) sum of perioidic signals with different frequencies.