![]() ![]() |
Propeller Head Plaza Technical and scientific discussion of amps, cables and other topics. |
For Sale Ads |
Use this form to submit comments directly to the Asylum moderators for this forum. We're particularly interested in truly outstanding posts that might be added to our FAQs.You may also use this form to provide feedback or to call attention to messages that may be in violation of our content rules.
Original Message
Re: IM is the elephant...
Posted by mfc on July 29, 2003 at 18:05:20:
Most of the IM turns out to be from a slight overload problem.
I analyzed the frequency graph in terms of what each component meant.
Here is a breakdown from 13F1 to 3F2.
F1=1050Hz square wave (out to 27th harmonic)
F2=5Khz sine waveDescription Product >250uV <250uV
------------ ------- ---------------
Input Signal 13F1...13650
...............4F2-6F1................13700
...............18F1-F2...13900
................3F2-F1................13950
..............23F1-2F2...14150
...............2F2+4F1...14200
..............28F1-3F2...14400
................F2+9F1....14450
..............33F1-4F2................14650
14th Harmonic 14F1..14700
...............4F2-5F1................14750
...............19F1-F2...14950
3rd Harmonic 3F2....15000
When you zoom in on the bottom of the waveform in the time domain,
at the very tip where the Gibbs is the most pronounced. That is the
fastest transition area. That peak of the Gibbs (at the bottom) is
actually rounded off compared to the otherside (top) of the waveform.
This is only happening on the bottom of the waveform and only in the
uncompensated case.It appears that in the uncompensated case, the Gibbs area of the
waveform is boosted by the peak in frequency response. This puts the
negative peak over the edge as far as clipping goes. That seems to be
generating all the IM components.These don't appear to be influencing the harmonics of the little sine
wave. I'm going to go back and check.