# Why cosine basis for determining bandwidth?

Instead of Fourier transform with cosines for example, one could use a square waves basis, then a pure cosine wave would have a broad bandwidth.

The question is why the cosines seems more fundamental than let say a triangle or squares, since one could express cosines in those basis functions ?

• This is actually more of a Math SE question. But I think your example is a naive interpretation. There are more requirements than that. One of them is orthogonality I believe. There are other approaches to separating signals into orthogonal constituents other than Fourier such as "wavelets". Commented Oct 14, 2021 at 4:07
• I think they are orthogonal, but I believe sine waves come from vibration of an instrument chord or an alternator, but if we take electronic signals from computers those are square. Commented Oct 14, 2021 at 4:38

• I don't know anymore howto diagonalize that operator. I tried separation of variables $$u(x)v(t)$$ and got stuck on the ode $$u'(x)=\sqrt{cu(x)^2+C}$$, I shouls be on the wrong way. Thanks for the help. Commented Oct 14, 2021 at 5:37
• For the eigenvalue zero the eigenvectors are any functions $$f(x-ct)+g(x+ct)$$ for example Commented Oct 14, 2021 at 5:44
• Is it correct to write : $$f(x-vt)$$ leading to $$(1-\frac{v^2}{c^2})f''(u)=\lambda f(u)$$ for the eigenproblem, with $$f(u)=\sin(ku)$$ thus it is an eigenfunction for the eigenvalue $$\lambda=(1-v^2/c^2)k^2$$ ? Commented Oct 14, 2021 at 10:10