Shannon ad fc/2 tricks

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Pergo, why do you continue to feed it??

Because i'm curious to see his math demonstration about modulation. 😉

Modulation has a strong math definition, that produces a typical spectrum.
If we simply apply the definition and see the spectrum, we can clearly see that's not a modulation.

But gumo73 won't listen. 🙂
So...i'm curious about his mathematical demonstration.
 
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Because i'm curious to see his math demonstration about modulation. 😉

Modulation has a strong math definition, that produces a typical spectrum.
If we simply apply the definition and see the spectrum, we can clearly see that's not a modulation.

But gumo73 won't listen. 🙂
So...i'm curious about his mathematical demonstration.


If there is 2 sine with opposite phase and the same Vmax have digital median null .
What you think about seeing in the fft graph
 
Mostly Clear exsample

If you select the samples multiple of 3 and putting others samples to 0
see below sine0.jpeg original is only a sine at 15khz sampled at 44100
You can see one of 2 sine result of modulation
Now I'm attempt to simulate errors of settling time and amplitude .
If is right the median of the two sines become higest than 0 of ideal case ( ideal DAC)
 

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Errata corrige

I have made a mistake when i have hypnotized there is 2 sine modulation inside the
samples .
fo=15000
fc=44100
ris=2^16/2
maxi=200
maxsine=0.9
k=1:maxi
xt=floor(maxsine*sin(2*pi*(k/fc)*fo)*ris)
In the real there are 3 main sine modulation
xt1 made from samples multiple of 3
xt2 made from smaples multiple of 3 plus 1
xt3 made from samples multiple of 3 plus 2
Their median under time domain xm=(xt1+xt2+xt3)/3=0 (on ideal case)
See below the graphs where is on evidence xt1 or xt2 or xt3 by green pen
 

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Last edited:
Near fc/6

on matlab
printf("Samplerate test \n")
fo=7550
fc=44100
ris=2^16
maxsamples=200
maxi=floor(maxsamples/6)*6
maxsine=1
k=1:maxi
xt=floor(maxsine*sin(2*pi*(k/fc)*fo)*ris)
In this case the modulation is not 3-phase but 6-phase
original is
xt8
first phase
xt8-1 is composed only from samples multiple of 6
2nd phase
xt8-2 shifted by one sample from xrt8-1
3nd phase
xt8-3 shifted by one sample from xrt8-2
4nd phase
xt8-4 shifted by one sample from xrt8-3
5nd phase
xt8-5 shifted by one sample from xrt8-4
and evidenced one by one by greeen pen
is lost one graph for 5 phase 🙁
 

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