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Old 18th February 2015, 06:14 PM   #1
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Default Shannon ad fc/2 tricks

Hi guys , reading the Shannon theorem i have found a trick
Is quiet that theorem tell the maximum frequencies for your signal from analog to digital becomes a maximum value of fc/2 where fc is a sample rate .
But if you have a pure sine tone at frequencies 15khz and
using acquisition asynchronous Samples describing
a 2 other tones at low frequencies generated from data output of dac
the calculus is sin(2*3.14159*15000*kt) and for example
t = 1/44100 if iu put this under a spreadsheet and print result of ideal smple rate you can see the envelope
the problem start about at fc/9 and have minimums at
fc/3 and fc/9 . Nearest this values the acquisition create a sine frequncies modulation .
I think the problem is the sinc on atempt of recostruction
of signal next post the immages of results
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Old 18th February 2015, 06:42 PM   #2
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Default Results of smples

this is an image of samples
the 2 tones of modulation have no the same phase
Attached Images
File Type: jpeg Shannon.jpeg (67.8 KB, 164 views)
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Old 18th February 2015, 06:54 PM   #3
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Default About fc/6

Nearest Fc/6 another trick sine have 7500hz .
If the sinc on the conversion DA is not assailable to an impulse the result is another modulation in the end
the samples lost the phase information
Attached Images
File Type: jpeg Shannon-8000.jpeg (58.2 KB, 161 views)

Last edited by gumo73; 18th February 2015 at 06:56 PM.
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Old 18th February 2015, 07:24 PM   #4
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Default nearest fc/9

In this case the values have a small modulation
fo = 5600hz .
This is my examples .
I think the same problem becomes with digital filters wit high
fault rates
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File Type: jpeg Shannon-4900.jpeg (53.7 KB, 149 views)
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Old 18th February 2015, 07:31 PM   #5
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Default The vinyl

I think in the end we need the vinyl for Hi-end audio chains
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Old 18th February 2015, 08:20 PM   #6
DF96 is offline DF96  England
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You need to find out what a reconstruction filter does. Then you will understand why hi-end audio chains can be built without vinyl.
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Old 18th February 2015, 08:37 PM   #7
gmarsh is offline gmarsh  Canada
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This isn't a digital problem, if you combine two tones and cut them into vinyl, you'll see the same "envelope" effect on playback.
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Old 19th February 2015, 10:13 AM   #8
DF96 is offline DF96  England
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The OP is seeing ultrasonic images, which the reconstruction filter will remove. These images, if left unfiltered, will cause beats with the wanted sound as he shows in his plots. It is quite common for people who don't understand digital audio to frighten themselves in this way.
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Old 19th February 2015, 12:13 PM   #9
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Quote:
Originally Posted by DF96 View Post
The OP is seeing ultrasonic images, which the reconstruction filter will remove. These images, if left unfiltered, will cause beats with the wanted sound as he shows in his plots. It is quite common for people who don't understand digital audio to frighten themselves in this way.
The beats are visual only, a 1000Hz and a 1001Hz signal combined in any context will have an apparent 1Hz envelope but there is no 1Hz content. A non-linear transfer function is needed to create actual signal content, vinyl has lots of that.
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Last edited by scott wurcer; 19th February 2015 at 12:18 PM.
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Old 19th February 2015, 01:20 PM   #10
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Default simulation

There is no combination about any 2 tones .
the result is ideal ADC by a sheet or on matlab or on octave
or others mathematical programs
The print is only samples without no interpolation the
calculus is sin( 2*3.14159*fo*kt) where kt is the time of samples k= 1,2,3,,.....,n or
k=,number-start,+1,+2,+3,+4,+n,number-end , t=1/44100Hz , fo=15000Hz
Printed are the result .
On sheet by example k=12 t=1/44100Hz
| k | Amplitude |
| 12 | =sin( 2*3.14159*15000*12* 2.2676e-05)|
| 13 | =sin( 2*3.14159*15000*13* 2.2676e-05)|
| 14 | =sin( 2*3.14159*15000*14* 2.2676e-05)|
.........
| n | =sin( 2*3.14159*15000*n* 2.2676e-05)|

a variation you can put a delay

Last edited by gumo73; 19th February 2015 at 01:31 PM.
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