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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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.
onsdag den 6. februar 2019 kl. 20.05.05 UTC+1 skrev John Larkin:
DING DING DING DING DING DING !
You got it first. I remember now but I couldn't think of it.
That only applies if the phase difference is constant.
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
** Varying phase difference means the frequencies are not identical - f****it.
Piss off.
..... Phil
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
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.
"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
'same frequency' also works, though it's arguable whether it rules out the possibility of slight frequency difference.
NT
"Identical frequency" would be better, but "phase-locked" puts the idea more forcefully.
-- Bill Sloman, Sydney
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 .)
Am 07.02.19 um 13:51 schrieb Jeroen Belleman:
Learning by teaching (Feynman)
An unmatched impedance will give a reflected wave at the same frequency, but different amplitude and phase.
George
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
Yup. If you really want to learn some subject, write a book on it. ;)
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant
OK, I think I read too fast and missed your point.
So this is CW reflectometry?
George H.
Call it the simulated return wave.
snipped-for-privacy@gmail.com wrote in news: snipped-for-privacy@googlegroups.com:
stimulated.
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.
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