0
$\begingroup$

Any one know of work on non-periodic but deterministic power signals? Now one member in this class would be the quasi periodic signals. I wonder if there is a generalized Fourier analysis of non-periodic deterministic power signals.

$\endgroup$
  • $\begingroup$ Fourier analysis does apply to non-periodic signals... The spectrum is just continuous rather than discrete. The conditions of existence of the Fourier transform of a signal is that it has to be deterministic and can be integrated on $\mathbb{R}$ or $\mathbb{C}$ $\endgroup$ – Florent Jul 20 '17 at 1:54
  • $\begingroup$ Also a generalisation of the Fourier transform is the Laplace transform (cf wikipedia) $\endgroup$ – Florent Jul 20 '17 at 1:57
  • $\begingroup$ I talk about a power signal here. Fourier transforms are only defined for some energy signals. $\endgroup$ – John Woods Jul 21 '17 at 15:06
2
$\begingroup$

Consider the signal $$x(t) = \sum_{n=-\infty}^\infty \operatorname{rect}(t-n)\sin(2\pi nt)$$ which consists of $|n|$ periods of the signal $\sin(2\pi nt)$ in the interval $\left(n-\frac 12,n+\frac 12\right)$ for each integer $n$. Clearly, $x(t)$ is not periodic. Equally clearly, $x(t)$ is a power signal with average power $\frac 12$, and thus it is a nonperiodic but deterministic power signal. I leave it to the OP to determine whether the signal in question has a Fourier transform in the generalized sense (meaning it involves impulses) or not.

| improve this answer | |
$\endgroup$
  • $\begingroup$ $x(t)$ is even continuous, but the derivatives are not at the $\operatorname{rect}(\cdot)$ boundaries. how 'bout this one? : $$x(t) = \sum_{n=-\infty}^\infty e^{-\pi(t-n)^2} \sin(2\pi nt)$$ $\endgroup$ – robert bristow-johnson Jul 17 '18 at 2:27
  • $\begingroup$ These last two proposals seem to qualify for non-periodic power signals. Now is there a generalized Fourier analysis to represent either or both of these? $\endgroup$ – John Woods Jul 18 '18 at 14:41
  • $\begingroup$ Short-Time Fourier Transform (STFT) ? $\endgroup$ – robert bristow-johnson Jul 18 '18 at 19:19
0
$\begingroup$

Chaotic Time Series are deterministic but not information preserving. The term orbit is often used because they have cycles but are not periodic.

http://sprott.physics.wisc.edu/chaostsa/

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.