I have a signal $x(t)$ with bandwidth $B_x$, and I am taking its cosine to create $y(t) = cos(x(t))$. After checking the spectrum with FFT, it seems that $y(t)$ is also bandlimited. But, is there a way to prove it? If so, what is the bandwidth?
I have tried applying the integrals from the Fourier series definition, but these integrals have to be evaluated numerically, so they don´t provide a rigorous, analytical proof.
Thanks in advance.
It's not strictly bandlimited.
The Taylor series of a cosine is
$$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} ... $$
If we look at the function $ g(x) = cos(cos(x))$ we get
$$g(x) = 1 - \frac{\cos(x)^2}{2!} + \frac{\cos(x)^4}{4!} - \frac{\cos(x)^6}{6!} + \frac{\cos(x)^8}{8!} ... $$
That means for an input of a single frequency cosine (which is clearly bandlimited), we get an infinite sum of even harmonics which is not bandlimited.
It may look this way, since the harmonics decay really fast. The denominator is $(2n)!$ which grows really really quickly so only the first few harmonics are numerically relevant. However there is no hard band limited.
In practice you will find often that the bulk of the energy will be contained within twice the original bandwidth since the second harmonic dominates.
