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Propeller Head Plaza Technical and scientific discussion of amps, cables and other topics. |
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Propeller Head Plaza Technical and scientific discussion of amps, cables and other topics. |
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In Reply to: Well, actually, a fourier transform HAS to be able to detect them all... posted by real_jj on June 14, 2006 at 19:11:39:
JJ: ""
It is, after all, a lossless transform on any physically realizable signal (except perhaps things like black holes, white holes, and other things that might need renormalization).
Given that, any error in the time waveform WILL show up in the frequency analysis in some fashion.It may be hard to recognize if you don't know what to look for, but it MUST show up.""
That is what I learned back in the days, and of course, subsequently lost. sigh..
In theory, yes lossless. In practice, it's difficult to state with absolute certainty that a function is lossless and sees everything. It's certainly easy to show that something can be seen, impossible to verify blanket statements that everything can be. Until such time as a function can be shown to be transparent to FFT's, it is required to assume it can be seen..proof of burden and all..I'm not crazy, after all...:-)
jj: ""
Now, the distortion you're talking about... Could you be a bit more specific? You're talking about losses that are not linear with signal? ""I could be more specific..however, I think I'll just get all emotional and such, stomp my feet, and call you names..:-)
For non linear branch analysis of powerloss, what I haven't been able to discount is the distinction of the product power component across the wire. If you put two signals into a wire, A and B, and at the crossover (the branch), A goes into one load, B into the other, the power envelope of each load is proportional to either A2 or B2. However, the wire dissipates proportional to (A+B)2, or A2+B2 + 2AB..
2AB is the product of the independent branch currents, swings through zero to negative territory. While seemingly inconsistent with reality, this term integrates to zero and is never larger than the two squares. This does not violate conservation of energy, does not create energy, is consistent long term with reality, but is non linear nonetheless.
If one assumes that this powerloss envelope is correct (no reason has been found (by me) yet to discount it), then there is an error term at the crossover node. This is not consistent with anyone's experience, in that it has never been measured to the best of my knowledge.. Either it is there and transparent to FFT, it is there and load reactance has masked it, or it is not there and my analysis is incorrect.
This thread is too far down. If you wish to discuss more, we can either start a new thread, or it could be taken off line. It would be interesting to see if discussion via a new thread can actually be done here..no?
The four quadrant amp issue we can postpone till later..
Cheers, John
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Follow Ups
- Re: Well, actually, a fourier transform HAS to be able to detect them all... - jneutron 13:08:47 06/15/06 (6)
- Ahh, I see what you mean... there's a math error. - real_jj 13:41:14 06/15/06 (5)
- Re: Ahh, I see what you mean... there's a math error. - jneutron 14:02:41 06/15/06 (4)
- Yes, but you're missing the point.... - real_jj 14:30:30 06/15/06 (3)
- Actually, it's far more interesting than that - jneutron 06:23:52 06/16/06 (2)
- Well, I can only suggest that you look at each waveform in a simulation... - real_jj 14:22:56 06/16/06 (1)
- been there, done that. - jneutron 05:49:18 06/19/06 (0)
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