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> Lagrange had insisted that such an approach could not be used to represent signals with corners, i.e., discontinuous slopes, such as in square waves. Isn't t
by comboy 4y ago
> Lagrange had insisted that such an approach could not be used to represent signals with corners, i.e., discontinuous slopes, such as in square waves.
Isn't that true though? You can only approximate it, right?
- AceJohnny2 4y agoIn practice, yes you can only approximate it, because in theory it requires an infinite series of sinusoidal waves. But since maths is all about the theory, Laplace's objections were silly. (though I admit I don't know what the state of acceptance of infinite series was at that time) Also, it's fine to cut off the infinite series of sinusoidal waves because all physical systems cut off around a given frequency, for example human hearing around ~20kHz.
- armadsen 4y agoAlso, in practice, there's no such thing as a truly sharp corner, which makes sense intuitively based on the math, I think. But real systems have inertia, friction, stray resistance/capacitance/inductance, etc. so you can't have a truly perfect square wave in a physical or electrical system in the real world.
- tuatoru 4y ago* Lagrange's objections
- AceJohnny2 4y agowhoops! thanks. Too late to edit.
- andrepd 4y agoNot quite, that's not the problem. We're in the realm of maths here and therefore of pure abstraction, so that's not the point. Adding terms to the Fourier series of a square wave you have an overshoot that does not converge to zero as the number of terms goes to infinity. This is called the Gibbs phenomenon https://en.wikipedia.org/wiki/Gibbs_phenomenon https://en.wikipedia.org/wiki/Gibbs_phenomenon
- retrac 4y agoIn the same sense that a circle does not exist, because you can only approximate it with an arbitrary number of points/lines. But an infinite number of points the same distance from the origin is a good definition of a circle. In a similar way, an infinite set of sine waves approximating a square wave is a square wave, mathematically speaking.
- cornel_io 4y agoSort of. The error at a square wave's "cliff" does shrink in width as you add higher and higher frequencies to the approximation, but the height of the error does not diminish, so you have to be explicit about what you mean by "converge". If you considered MAX(ABS(approximation(x) - square(x))) as your error metric it would not go to zero even in the limit; for most intents and purposes we don't care about that, though, and in any case the limiting behavior is much nicer for "normal" functions that don't have discontinuities like a square wave.
- anigbrowl 4y agoCorners in signals are like limits in calculus or irrational numbers in arithmetic. Once you accept them as an abstraction that you can't touch them directly all the anxiety dissipates.
- hoseja 4y agoPerfectly square waves also can't physically exist, as long as there is a speed limit on the universe.