Think a solutions
My simulation come from studying the Calculus to understand the dac nos and why they have many realizations,
Beside them is the trick i found on Shannon theory .
My simulation come from studying the Calculus to understand the dac nos and why they have many realizations,
Beside them is the trick i found on Shannon theory .
NOS DAC usually omit the reconstruction filter, hence they cannot replicate the signal which emerged from the anti-aliasing filter, hence they cannot be considered as hi-fi. However, some people prefer the sound and some people like to confuse themselves by plotting what they claim is the square-wave response.
The sound preference for NOS DAC can perhaps be explained by assuming that the ultrasonic images (which conventional hi-fi DACs filter away) can somehow mimic the effects of ultrasonic transients (which the anti-aliasing filter has filtered away). This is because we have no pitch sensitivity at such high frequencies (and most of us can't hear them at all!).
The confusion about square-wave response just shows that there are a lot of people who don't understand digital audio or, perhaps in some cases, merely hope that their customers don't understand digital audio.
So, a NOS DAC could produce the type of output you showed but that is not a flaw in Shannon but a flaw in NOS DACs.
The sound preference for NOS DAC can perhaps be explained by assuming that the ultrasonic images (which conventional hi-fi DACs filter away) can somehow mimic the effects of ultrasonic transients (which the anti-aliasing filter has filtered away). This is because we have no pitch sensitivity at such high frequencies (and most of us can't hear them at all!).
The confusion about square-wave response just shows that there are a lot of people who don't understand digital audio or, perhaps in some cases, merely hope that their customers don't understand digital audio.
So, a NOS DAC could produce the type of output you showed but that is not a flaw in Shannon but a flaw in NOS DACs.
Images
A nos dont have the same flat band than the oversampling dac have . But don't have a sinc on reconstruction filter and the sinc is not unitary at 15khz
then not have any modulations result on this components
See that the ideal filter on fourier domain as a rect but under time domain as a sinc ,
on nos techniques no sinc is multiplied by samples .
This means you don't find any noise addition to the original signal
A nos dont have the same flat band than the oversampling dac have . But don't have a sinc on reconstruction filter and the sinc is not unitary at 15khz
then not have any modulations result on this components
See that the ideal filter on fourier domain as a rect but under time domain as a sinc ,
on nos techniques no sinc is multiplied by samples .
This means you don't find any noise addition to the original signal
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Sinc on recostruction filter
I the example isn't present any reconstruction filter and no beats is possible because is a ADC ideal from a continuous
signal .
But the ADC is too close theory to the DAC theory
if the sinc on reconstruction filter as the same as a ideal impulse the discrete integral (summation) is nearest
x(t)*1 where 1 is the integration of unity impluse
I the example isn't present any reconstruction filter and no beats is possible because is a ADC ideal from a continuous
signal .
But the ADC is too close theory to the DAC theory
if the sinc on reconstruction filter as the same as a ideal impulse the discrete integral (summation) is nearest
x(t)*1 where 1 is the integration of unity impluse
Anyone who has ever tuned a guitar is aware that the beats are still there in the original (acoustic) domain.
You're feeding it sine waves. Resonance is irrelevant. Two sine waves added together show beats.
You are talking about beats arising from unfiltered ultrasonic images. These are unlikely to be audible (to humans) as they are at too high a frequency, and in a hi-fi DAC will be filtered away. Hence your original conclusion that "I think in the end we need the vinyl for Hi-end audio chains" (post 5) is unfounded.
Note that a time domain plot can be misleading.
Note that a time domain plot can be misleading.
ultrasonic images
On Shannon theory you have a aliasing only if signal that you converting from analogue to digital have frequencies or in oder hand components up of fc/2 where fc=Sample rate frequency . On may example the component or all inside this limit
On Shannon theory you have a aliasing only if signal that you converting from analogue to digital have frequencies or in oder hand components up of fc/2 where fc=Sample rate frequency . On may example the component or all inside this limit
Simulation
width' out any type of interpolation
how you calculus this triangles ?
On your simulation you don't create a pure samples from a pure sine and print it by a vertical lines or columnsgumo, which of the green or blue traces in this plot has more signal power?
width' out any type of interpolation
how you calculus this triangles ?
Why would I redo an incorrect calculation? Band-limit, sample, reconstruct, that's the correct way.
Simulation
Signals x(t) are pure tone under limit band of Shannon
Fc is 44100Hz more of 2 X 15000khz
Sine as a one component of Fourier transfer !
ADC calculation take the samples on equidistant times and is not affected by jitter or others things
k is an integer
t=1/44100hz
tau= is additional time but unneeded for Calculation
Yu can crate a smples by calculus x(kt)=sin(2*pi*fo*(kt+tau)
where you see an error
Signals x(t) are pure tone under limit band of Shannon
Fc is 44100Hz more of 2 X 15000khz
Sine as a one component of Fourier transfer !
ADC calculation take the samples on equidistant times and is not affected by jitter or others things
k is an integer
t=1/44100hz
tau= is additional time but unneeded for Calculation
Yu can crate a smples by calculus x(kt)=sin(2*pi*fo*(kt+tau)
where you see an error
Obviously, true, but uninteresting.
Sin(wn/Fs) is not a pure sine wave when n is a sequence of integers, rather it is a sine wave multiplied by the dirac delta, which can hanwavingly be thought of as 1/n * {1,sin (fs), sin (2fs), sin (3fs), sin(4fs)...... to infinity}. We will not get into negative frequencies here they are not really needed to comprehend this.
The tricky part is when you convert back from the samples, which actually do NOT uniquely represent sin(2*pi*fo*(kt+tau) but rather have components at N*(Fs +- k) for all integer N.
This is inherent to the sampling process, thus for example the sample sequence represented by a sampled 15KHz sine at 44.1Fs, is IDENTICAL to and indistinguishable from the one representing a 29.1Khz sine sampled at 44.1K (Actually sometimes a valid thing to do), or the sample sequence representing the sine at 59.1K sampled at 44.1K......
It is only by knowing that we band limited before sampling and therefore that anything at or above Fs/2 must perforce be an alias that we can filter to reconstruct the correct sine wave from the infinite number of possibilities.
Your mistake is that you are seeing the sample values as being the signal, when in fact they represent points on the signal (but also points on all the possible subsampled signals).
You really need to read up on some more discreet time maths before crying that the sky is falling on some really well understood basic signal processing theory.
Regards, Dan.
Sin(wn/Fs) is not a pure sine wave when n is a sequence of integers, rather it is a sine wave multiplied by the dirac delta, which can hanwavingly be thought of as 1/n * {1,sin (fs), sin (2fs), sin (3fs), sin(4fs)...... to infinity}. We will not get into negative frequencies here they are not really needed to comprehend this.
The tricky part is when you convert back from the samples, which actually do NOT uniquely represent sin(2*pi*fo*(kt+tau) but rather have components at N*(Fs +- k) for all integer N.
This is inherent to the sampling process, thus for example the sample sequence represented by a sampled 15KHz sine at 44.1Fs, is IDENTICAL to and indistinguishable from the one representing a 29.1Khz sine sampled at 44.1K (Actually sometimes a valid thing to do), or the sample sequence representing the sine at 59.1K sampled at 44.1K......
It is only by knowing that we band limited before sampling and therefore that anything at or above Fs/2 must perforce be an alias that we can filter to reconstruct the correct sine wave from the infinite number of possibilities.
Your mistake is that you are seeing the sample values as being the signal, when in fact they represent points on the signal (but also points on all the possible subsampled signals).
You really need to read up on some more discreet time maths before crying that the sky is falling on some really well understood basic signal processing theory.
Regards, Dan.
I did, Octave just draws lines from one point to the next. Here's the exact same data, but using stem() instead of plot().
What signal has the higher level, blue or green?
infinite images
Ok when you going on on discrete domain you have infinite images positioned on k*fs+fo k*fs-fo
But on my simulation i obtain the samples describing
another sine at low frequencies no aliasing is possible
because fo is inside fc/2 .
And I can't have others sine described from samples.
On the reconstruction filter spline pass trough
the original samples , and if the samples in my simulation described a frequencies at low values inside fc/2 .
The calculus to made samples is too simple because delta Dirac as the times of sampling
Ok when you going on on discrete domain you have infinite images positioned on k*fs+fo k*fs-fo
But on my simulation i obtain the samples describing
another sine at low frequencies no aliasing is possible
because fo is inside fc/2 .
And I can't have others sine described from samples.
On the reconstruction filter spline pass trough
the original samples , and if the samples in my simulation described a frequencies at low values inside fc/2 .
The calculus to made samples is too simple because delta Dirac as the times of sampling
simulation
k= integer
2*3.14159*15000 = angular velocity of pulsation in rad/sec
On octave
k=1:341
tc=1/44100
xt=sine(2*3.14159*15000*k*tc)
plot (k,xt)
if you wont't to see a lo frequency of modulation going nearest fc/3
xt=sine(2*3.14159*14800*k*tc)
plot (k,xt)
tc=time of samplerateI did, Octave just draws lines from one point to the next. Here's the exact same data, but using stem() instead of plot().
What signal has the higher level, blue or green?
k= integer
2*3.14159*15000 = angular velocity of pulsation in rad/sec
On octave
k=1:341
tc=1/44100
xt=sine(2*3.14159*15000*k*tc)
plot (k,xt)
if you wont't to see a lo frequency of modulation going nearest fc/3
xt=sine(2*3.14159*14800*k*tc)
plot (k,xt)
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