[Don Pearce[_3_]]

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

[[email protected]]

Don Pearce 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.

** No it ain't.

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.

** You are here claiming it does.

If you do an FFT on an edge,

** Not the situation.

The *square wave* can be band limited to just a few times the ringing frequency.
Only needs ONE harmonic close enough to that frequency to get a damped sine result.

Does not do so with a sine sweep.

If you do a sine sweep you will find the resonant frequency of the ring.

** But not *create* it - the maximum is at the exact same frequency as the input sine.

That is the frequency you will see when you strike it with a
fast rising edge.

** Not the case with a normal, square wave test.



...... Phil

[Don Pearce[_3_]]

On Sat, 27 Feb 2021 15:20:57 -0800 (PST), "
wrote:

Don Pearce 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.

** No it ain't.

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.

** You are here claiming it does.

If you do an FFT on an edge,

** Not the situation.

The *square wave* can be band limited to just a few times the ringing frequency.
Only needs ONE harmonic close enough to that frequency to get a damped sine result.

Does not do so with a sine sweep.

If you do a sine sweep you will find the resonant frequency of the ring.

** But not *create* it - the maximum is at the exact same frequency as the input sine.

That is the frequency you will see when you strike it with a
fast rising edge.

** Not the case with a normal, square wave test.



..... Phil


Phil, you need to think. Forget square waves and harmonics for a
moment. You can trigger ringing with a single transient edge. No
repetition or harmonics needed.

d

--
This email has been checked for viruses by Avast antivirus software.
https://www.avast.com/antivirus

[[email protected]]

Don Pearce wrote:
=================


Phil, you need to think.

** No way, YOU do - ****head.


Forget square waves and harmonics for a moment.

** No way - cos THAT IS ACTUALLY the topic here.

FYI:

I suspect you know you are wrong, by dint of being in the wrong context.
You favorite hiding hole, for the last 20 plus years.

Along with all the other smelly, know nothing, usenet rodents.

**** off.


....... Phil

[John Williamson]

On 28/02/2021 09:46, wrote:

Along with all the other smelly, know nothing, usenet rodents.

**** off.

Genuine question. If your opinion of us is so low, why in heaven do you
keep posting here?

Your posts are always framed in unhelpful ways and insult the
intelligence of the poster you are replying to. This does not impress
anyone, least of all the person you are insulting.

This is why you are in many people's killfiles, so we only see your
potty mouthed outpourings when someone replies to them.


--
Tciao for Now!

John.

[[email protected]]

John Williamson wrote:
==================

Genuine question.

** Like hell it is.

If your opinion of us is so low,

**Who is "us" and when did you get elected to represented them ??

why in heaven do you keep posting here?

** Why do you ? You don't seem to know anything.

Your posts are always framed in unhelpful ways

** On the contrary - my posts are VERY helpful.

and insult the intelligence of the poster you are replying to.

** I never insult intelligence - only stupidity and arrogance.


This is why you are in many people's killfiles,

** Using the killfile is like burying your head in the sand.
Have to be an ostrich to do that.

FYI Pearce is a bull****ting idiot and massive fake.


....... Phil