On Sat, 27 Feb 2021 04:31:33 -0800 (PST), "
wrote:
Nothing to do with the frequency of the wave or its harmonics. Each
ringing event is separate and is triggered by the broadband energy of
the rising or falling edge.
** Gobbledegook.
Don't be so hasty. I know this is not instantly intuitive, but it is
exactly what happens.
The frequency of the ring is determined by
the LC circuit that is being struck by that edge.
** So a linear circuit with a resonance creates a new frequency ?
No. If you do an FFT on an edge, you will find a broad spectrum, not a
discrete one. THere will be energy at whatever frequency the resonant
circuit causing the ringing works at.
Does not do so with a sine sweep.
If you do a sine sweep you will find the resonant frequency of the
ring. That is the frequency you will see when you strike it with a
fast rising edge. Think of it like a bell. If you keep hitting it, you
will hear the individual strikes, but the frequency that comes out
will be that of the bell, and nothing to do with how quickly you hit
it.
d
--
This email has been checked for viruses by Avast antivirus software.
https://www.avast.com/antivirus