Trigonometry terminology

A sin wave and cosine wave are orthogonal to each other which means their correlation over a cycle is 0. All of fourier theory is based on this.

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bulegoge
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onsdag den 6. februar 2019 kl. 20.05.05 UTC+1 skrev John Larkin:

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Reply to
Lasse Langwadt Christensen

DING DING DING DING DING DING !

You got it first. I remember now but I couldn't think of it.

Reply to
jurb6006

That only applies if the phase difference is constant.

Reply to
jurb6006

onsdag den 6. februar 2019 kl. 23.17.39 UTC+1 skrev snipped-for-privacy@gmail.com:

if it is not then the frequencies must be different

Reply to
Lasse Langwadt Christensen

** Varying phase difference means the frequencies are not identical - f****it.

Piss off.

..... Phil

Reply to
Phil Allison

How else could they differ?

They aren't necessarily correlated--the cross correlation of sin and cos is zero.

I'd just say "of different amplitudes and phases".

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
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Reply to
Phil Hobbs

Grin, that's my opinion, say what you mean. Fancy words are just going to confuse me.

To me synchronous, implies phase shift = 0, sure I can set my synchronous detector otherwise, (say in quadrature to find the zero.) but if I'm using it right it should be close to zero phase shift. (or 180 :^)

George H.

Reply to
George Herold

"Phase-locked" strikes me as better. As has been pointed out, "correlated" has a strict mathematical interpretation, and if the two waveforms were 90 or 270 degrees out of phase the correlation would be exactly zero.

--
Bill Sloman, Sydney
Reply to
bill.sloman

'same frequency' also works, though it's arguable whether it rules out the possibility of slight frequency difference.

NT

Reply to
tabbypurr

"Identical frequency" would be better, but "phase-locked" puts the idea more forcefully.

--
Bill Sloman, Sydney
Reply to
bill.sloman

I'd be using the term in a write-up about frequency-domain reflectometry. I'm trying to put together a clear argument that a terminating impedance different from Z0 is equivalent to a signal of equal frequency, but different in amplitude and phase being injected in the backward direction from a Z0 impedance source.

In other words, that one can make a terminating Z0 impedance look like any impedance by putting another source of the correct phase and amplitude in series with it.

As you can see from the above, I have trouble wording this clearly and concisely.

Thanks for your collective input!

Jeroen Belleman (You don't understand a theory if you can't explain it to your grandmother .)

Reply to
Jeroen Belleman

Am 07.02.19 um 13:51 schrieb Jeroen Belleman:

Learning by teaching (Feynman)

Reply to
Gerhard Hoffmann

An unmatched impedance will give a reflected wave at the same frequency, but different amplitude and phase.

George

Reply to
George Herold

Thanks George, but I clearly failed to get my point across. My point was that this is indistinguishable from injecting a well-chosen backwards travelling signal through a matched impedance.

Jeroen Belleman

Reply to
Jeroen Belleman

Yup. If you really want to learn some subject, write a book on it. ;)

Cheers

Phil Hobbs

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Dr Philip C D Hobbs 
Principal Consultant 
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Reply to
Phil Hobbs

OK, I think I read too fast and missed your point.

So this is CW reflectometry?

George H.

Reply to
George Herold

Call it the simulated return wave.

Reply to
bloggs.fredbloggs.fred

snipped-for-privacy@gmail.com wrote in news: snipped-for-privacy@googlegroups.com:

stimulated.

Reply to
DecadentLinuxUserNumeroUno

Look you illiterate sot, (am I doing well here ?) that is almost what I said but what if both of them are locked to WWV or some shit ? One would not be derived from the other.

Reply to
jurb6006

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