HiFi vs. Computer

[Guest]

Archived from groups: rec.audio.tech (More info?)

On Fri, 03 Jun 2005 02:42:25 GMT, "Tim Martin"
<tim2718281@ntlworld.com> wrote:

>"Joe Kesselman" <keshlam-nospam@comcast.net> wrote in message
>news:brCdna2kvYJ-xAPfRVn-sA@comcast.com...
>>
>> Digital beats the accuracy of most analog media quite handily, given a
>> suprisingly small investment. The limiting factor, actually, tends to be
>> the analog hardware used to get the signal into and out of digital form.
>
>Yes. However, since digital representations of analog signals are
>compressed, there is little point agonizing over "lossy" versus "lossless"
>compresssion.

No, they're not. Please read up on digital audio before you spout this
nonsense.

> What matters is the quality delivered.

And digital far exceeds the quality of any available analogue source
signal.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

 

[Guest]

Archived from groups: rec.audio.tech (More info?)

Tim Martin wrote:
> "Richard Crowley" <rcrowley7@xprt.net> wrote in message
> news:119vfc5o62hbt00@corp.supernews.com...
>
>> How did your "analog signals" originate?
>
> A bird singing in the woods. This generates an analog
signal,
> detectable by ears.

However, the dynamic range in the woods is not all that
good, because the background noise level in the woods is
pretty high compared to bird songs, especially with the
typical distances involved. Nature can be pretty noisy.
Leaves rustle, the wind has turbulence problems, waves lap
or crash, insects buzz...

>> How did they then end up as ones and zeroes on a tape or
disc?

> It's possible to store a digital approximation of this
analog signal
> by a number of methods; I don't see that it actually
matters what
> method is used, but I suppose the most direct method is to
have some
> flexible device that vibrates with the sound of the bird
singing, and
> periodically measure the physical position of the flexible
device.

IOW a microphone.

> The more frequently we measure the position, and the more
precisely we
> measure the position, the more accurate is our digital
representation
> of the analog signal.

Wrong. While taking more measurements in the time domain
increases the highest frequency that can be reliably
discerned, it does nothing for resolution or accuracy of the
measurements. If you want accuracy, you have to improve the
quality of each measurement.

 

[Guest]

Archived from groups: rec.audio.tech (More info?)

Tim Martin wrote:
> <dpierce@cartchunk.org> wrote in message
>
news:1117671221.564785.245640@g43g2000cwa.googlegroups.com...
>
>> Beyond what may seem to be a philosphical discussion (it
isn't:
>> it's a direct and inevitable consequence of the your
basic
>> assertion and is proven rigorously in work cited above by
Nyquist
>> and Shannon), the simplem fact is that ANY system of a
finite
>> bandwidth and limited dynamic range can be EXACTLY
represented
>> by a quantized system of finite accuracy.
>
> An analog signal, such as a bird singing in the woods, has
infinite
> bandwidth.

Not really. Creating sounds at an infinitely high frequency
requires infinite amounts of energy.

> That is, there is no upper frequency f, such that any
combination of
> waves of frequency less than f, can exactly represent an
arbitrary
> analog signal, regardless of the precision of the waves.

In reality the energy in bird calls and other natural sounds
at high frequencies is limited and naturally rolling off,
and at some surprisingly low frequency, it drops below the
noise floor.

> Nyquist's Theoerem is about representation of periodic
signals; most
> sounds are not periodic signals.

All real world sounds can be sucessfully analyzed as a
collection of enveloped periodic signals. Fourier wasn't
wrong.

> Of course in practice we can specify a combinations of
waves that
> make close approximations to the bird singing; and we can
get as
> close as we like, up to the limits of whatever equipment
we use to
> detect the analog signal we are approximating.

Same basic problems and results irregardless of whether the
test equipment is analog or digital. Digital test equipment
tends to have better price performance. The lowest noise
floors in audio test equipment generally are obtained with
equipment that is mostly digital. For example, Wein bridges
have built-in dynamic range problems when you actually try
to implement one practically and economically.

 

[Guest]

Archived from groups: rec.audio.tech (More info?)

On Fri, 03 Jun 2005 02:19:12 GMT, "Tim Martin"
<tim2718281@ntlworld.com> wrote:

>
><dpierce@cartchunk.org> wrote in message
>news:1117671221.564785.245640@g43g2000cwa.googlegroups.com...
>
>> Beyond what may seem to be a philosphical discussion (it isn't:
>> it's a direct and inevitable consequence of the your basic
>> assertion and is proven rigorously in work cited above by Nyquist
>> and Shannon), the simplem fact is that ANY system of a finite
>> bandwidth and limited dynamic range can be EXACTLY represented
>> by a quantized system of finite accuracy.
>
>An analog signal, such as a bird singing in the woods, has infinite
>bandwidth.

Utter garbage! Everything generating noise in the woods has a very
well defined bandwidth,dependent on the mass/compliance resonaces of
its suspension systems. This includes the larynxes of birds. Also, the
atmosphere has a well-defined sound absorption coefficient which
increases with frequency, such that even a metre away from that bird,
you won't detect much above 100kHz. That's why bats don't have to
worry about reflections from more than a few yards away, the signal
simply doesn't get back to them.

More importantly, since you seem to understand almost nothing which
you are spouting, these are *not* analogue sugnals, they are simply
sounds. Once you have *converted* that sound with say a microphone,
you then have a signal which is an analogue of the original sound. The
live microphone feed is the analogue signal, *not* the original sound.

>That is, there is no upper frequency f, such that any combination of waves
>of frequency less than f, can exactly represent an arbitrary analog signal,
>regardless of the precision of the waves.

However, since there definitely *is* an upper frequency limit to
birdsong, that's not actually a problem.

>Nyquist's Theoerem is about representation of periodic signals; most sounds
>are not periodic signals.

Actually, it's not, except in so far as it does have a bandwidth limit
of less than half the sampling frequency.

>Of course in practice we can specify a combinations of waves that make close
>approximations to the bird singing; and we can get as close as we like, up
>to the limits of whatever equipment we use to detect the analog signal we
>are approximating.

The analogue signal from the microphone *is* the approximation of the
original sound. At least learn the *basics*, then you wouldn't make
such nonsensical statements.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

 

[Guest]

Archived from groups: rec.audio.tech (More info?)

> Nyquist's Theoerem is about representation of periodic signals; most
> sounds
> are not periodic signals.

Any arbitrary sound can be decomposed into a sum of periodic signals.
Therefore, Nyquist's theorem applies to all sounds.

 

[Guest]

Archived from groups: rec.audio.tech (More info?)

"Karl Uppiano" <karl.uppiano@verizon.net> wrote in message
news:SXTne.16106$qJ3.7554@trnddc05...
> > Nyquist's Theoerem is about representation of periodic signals; most
> > sounds
> > are not periodic signals.
>
> Any arbitrary sound can be decomposed into a sum of periodic signals.

OK; let's suppose the sound consists of silence, followed by one second of
1kHz sine wave, followed by silence.

What sum of periodic signals can this be decomposed into?

Tim

 

[Guest]

Archived from groups: rec.audio.tech (More info?)

Tim Martin wrote:
> "Karl Uppiano" <karl.uppiano@verizon.net> wrote in message
> news:SXTne.16106$qJ3.7554@trnddc05...
> > > Nyquist's Theoerem is about representation of periodic signals; most
> > > sounds
> > > are not periodic signals.
> >
> > Any arbitrary sound can be decomposed into a sum of periodic signals.
>
> OK; let's suppose the sound consists of silence, followed by one second of
> 1kHz sine wave, followed by silence.
>
> What sum of periodic signals can this be decomposed into?

It's exactly the same as a 1 kHz sine wave 100% modulated by a 0.5
Hz square wave, and such decomposes into a series of sine components
spaced 1 Hz apart spaced symmetrically about the 1 kHz component
offset from it by 0.5 Hz, (iow 1000.5, 1001.5, 1002.5, 999.5, 998.5,
997.5, ...) with amplitudes decreasing as we move away from 1 kHz by
a simple 1/n, n = 1, 3, 5, ... and so forth, all components in phase.

AND, if you insist on truning on and off the sine wave INSTANTANEOUSLY,
these sine components extend to +-infinite frequency.

Such a signal, as I am sure you will agree, could never be PRODUCED
perfectly in ANY system existing for finite time or limited to finite
energy.

 

[Guest]

Archived from groups: rec.audio.tech (More info?)

"Stewart Pinkerton" <patent3@dircon.co.uk> wrote in message
news:mhvv9118jips49mc63o6c90c1lr0qt9ng0@4ax.com...

Tim

> >Nyquist's Theoerem is about representation of periodic signals; most
sounds
> >are not periodic signals.

Stewart

> Actually, it's not, except in so far as it does have a bandwidth limit
> of less than half the sampling frequency.

Nyquist's Theorem tells us we can exactly represent the information in a
waveform by sampling it at a rate at least twice as high as the highest
frequency in the waveform

I think there's an implication here that by "highest frequency", we are
talking about the highest frequency in a Fourier transform of the original
signal. And the Fourier transform applies only to periodic signals.

Take a waveform consisting of, say a 1000Hz sine wave that is repeatedly
switched on and off at random times. This is not a periodic waveform, and
cannot be represented exactly by a Fourier transform. It has infinite
bandwidth. (Conceptually at least. As you've previously remarked, there
are physical constraints imposed by the transmission medium.)

So what happens when we try to represent that wavefom by a digital
representation (and I'm thinking here of a general-purpose digital
representation.) To keep things simple, let's suppose the amplitude of the
sine wave is small, so gives rise to only two different digital values, 0
and 1.

If we are sampling 48000 times a second, and if the sine wave is long enough
when on, our digital signal will, after some start-up sequence, consist of a
repeating pattern of 24 ones followed by 24 zeroes. Once we're in this part
of the signal, we can reproduce the original sine wave exactly (within the
resolution limits, which are not the issue here.)

But when does our reproduction start?

Our digital representation will have an initial series of values
representing the silence. Again leaving aside start-up considerations, a
series of 48000 identical values will represent one second's initial
silence, and 48001 will represent 1.000021 seconds of silence.

All silent intervals between 1 and 1.000021 seconds will be represented by
one of these values. And as there are more than 2 different analog signals
starting with between 1 and 1.000021 seconds of silence, we are losing
information in the digital representation ... that is, we are compressing
it.

All this is straightforward maths.

I think where you went wrong is failing to distinguish between the audio
frequencies contained in the signal, and the timing information contained in
the signal. In hi-fi, this doesn't usually matter - correctly reproducing
the audio frequencies gives us more than enough timing precision - but in
information theory it does.

Tim

 

[Guest]

Archived from groups: rec.audio.tech (More info?)

"Arny Krueger" <arnyk@hotpop.com> wrote in message
news:2KadnR34zaLvpD3fRVn-sw@comcast.com...

> However, the dynamic range in the woods is not all that
> good, because the background noise level in the woods is
> pretty high compared to bird songs, especially with the
> typical distances involved. Nature can be pretty noisy.

Careful - that's all part of the signal, not "noise". If we were trying to
reproduce the sound of a bird singing in the woods, we would want the
background sound, too.

> IOW a microphone.

Yes, it's pretty much what a capacitor microphone does fairly directly..
However, microphones generate electrical signals, which are affected by the
circuitry they are connected to - that is, the microphone introduces noise -
and I didn't want to confuse the issue with that, hence my suggesting we
measure the position of a moving element.

> > The more frequently we measure the position, and the more
> precisely we
> > measure the position, the more accurate is our digital
> representation
> > of the analog signal.
>
> Wrong. While taking more measurements in the time domain
> increases the highest frequency that can be reliably
> discerned, it does nothing for resolution or accuracy of the
> measurements. If you want accuracy, you have to improve the
> quality of each measurement.

Bear in mind that the context I set was an analog signal with infinite
bandwidth ... eg including sounds starting and stopping at arbitrary times.
So improving the timing resolution does improve the accuracy of the digital
representation.

If we take measurements 10,000 times a second, our resolution is 0.0001
seconds , our digital representation won't necessarily be able to
distinguish between sounds starting at 1.00001 seconds and sounds starting
at 1.00002 seconds; whereas if we take measurements 100,000 times a second,
it will be able to.

Tim

 

[Guest]

Archived from groups: rec.audio.tech (More info?)

"Arny Krueger" <arnyk@hotpop.com> wrote in message
news:sdednURrFd9Ipz3fRVn-gA@comcast.com...

> Not really. Creating sounds at an infinitely high frequency
> requires infinite amounts of energy.

The scenario does not include any sounds of infinitely high frequency. We
can stick with just two tones: one silence, and the other a single tone of
1000Hz.

What I'm saying is that it's not possible to accurately represent a sound
consisting of alternate arbitrary-length sequences of 1000Hz sine waves and
silence with any set of sine waves. regardless of the upper frequency we
use. And that it's not possible to represent them by any digital signal
with fixed sampling frequency, no matter how high that sampling frequency
is.

To put it another way: for any fixed-sampling-frequency digital
representation you specify, I can define two different non-periodic analog
signals, using only silence and 1000 Hz sine waves, that will both map to
the same digital representation.

This is trivial. .

Tim.

 

[Guest]

Archived from groups: rec.audio.tech (More info?)

Tim Martin wrote:
> "Arny Krueger" <arnyk@hotpop.com> wrote in message
> news:2KadnR34zaLvpD3fRVn-sw@comcast.com...
>
>> However, the dynamic range in the woods is not all that
>> good, because the background noise level in the woods is
>> pretty high compared to bird songs, especially with the
>> typical distances involved. Nature can be pretty noisy.
>
> Careful - that's all part of the signal, not "noise". If
we were
> trying to reproduce the sound of a bird singing in the
woods, we
> would want the background sound, too.

Agreed you want background sound, but below the incidental
sound is true and genuine noise.

>> IOW a microphone.

> Yes, it's pretty much what a capacitor microphone does
fairly
> directly.. However, microphones generate electrical
signals, which
> are affected by the circuitry they are connected to - that
is, the
> microphone introduces noise - and I didn't want to confuse
the issue
> with that, hence my suggesting we measure the position of
a moving
> element.

Condensor mics generate noise but generally have far more
output than dynamics. Dynamic mics produce far lower signals
that are far more dependent on low mic preamp noise. I
believe the most sensitive mics in actual general use are
condensers.

>>> The more frequently we measure the position, and the
more precisely we
>>> measure the position, the more accurate is our
digitalrepresentation
>>> of the analog signal.

>> Wrong. While taking more measurements in the time domain
>> increases the highest frequency that can be reliably
>> discerned, it does nothing for resolution or accuracy of
the
>> measurements. If you want accuracy, you have to improve
the
>> quality of each measurement.

> Bear in mind that the context I set was an analog signal
with infinite
> bandwidth

IOW, a figment of the imagination. Sorry, I'm stuck here in
the real world.

>... eg including sounds starting and stopping at arbitrary
> times. So improving the timing resolution does improve the
accuracy
> of the digital representation.

Back in the real world we're stuck with listeners that turn
out to have worse HF performance than plain-old common CD
audio.

> If we take measurements 10,000 times a second, our
resolution is
> 0.0001 seconds , our digital representation won't
necessarily be able
> to distinguish between sounds starting at 1.00001 seconds
and sounds
> starting at 1.00002 seconds; whereas if we take
measurements 100,000
> times a second, it will be able to.

I thought this discussion was about audio, not how many
measurements can dance on the head of an imaginary pin! ;-)

 

[Guest]

Archived from groups: rec.audio.tech (More info?)

Tim Martin wrote:

> "Arny Krueger" <arnyk@hotpop.com> wrote in message
> news:sdednURrFd9Ipz3fRVn-gA@comcast.com...

>>Tim Martin wrote:

>>> An analog signal, such as a bird singing in the woods,
has infinite
>>> bandwidth.

>> Not really. Creating sounds at an infinitely high
frequency
>> requires infinite amounts of energy.

> The scenario does not include any sounds of infinitely
high
> frequency. We can stick with just two tones: one
silence, and the
> other a single tone of 1000Hz.

I'm sorry but I was responding to your post about birds, not
some theoretical waveform.

 

[Guest]

Archived from groups: rec.audio.tech (More info?)

On Fri, 03 Jun 2005 10:47:13 GMT, "Tim Martin"
<tim2718281@ntlworld.com> wrote:

>
>"Stewart Pinkerton" <patent3@dircon.co.uk> wrote in message
>news:mhvv9118jips49mc63o6c90c1lr0qt9ng0@4ax.com...
>
>Tim
>
>> >Nyquist's Theoerem is about representation of periodic signals; most
>sounds
>> >are not periodic signals.
>
>Stewart
>
>> Actually, it's not, except in so far as it does have a bandwidth limit
>> of less than half the sampling frequency.
>
>Nyquist's Theorem tells us we can exactly represent the information in a
>waveform by sampling it at a rate at least twice as high as the highest
>frequency in the waveform
>
>I think there's an implication here that by "highest frequency", we are
>talking about the highest frequency in a Fourier transform of the original
>signal. And the Fourier transform applies only to periodic signals.

It does however apply to every frequency required for the synthesis of
an impulse, however short that impulse may be. To assume otherwise is
to misunderstand the nature of the Fourier transform.

>Take a waveform consisting of, say a 1000Hz sine wave that is repeatedly
>switched on and off at random times. This is not a periodic waveform, and
>cannot be represented exactly by a Fourier transform. It has infinite
>bandwidth. (Conceptually at least. As you've previously remarked, there
>are physical constraints imposed by the transmission medium.)

And as such, it *can* be represented by a Fourier transform containing
frequencies up to any arbitrarily high limit which you feel the medium
justifies. Infinity is never involved.

>So what happens when we try to represent that wavefom by a digital
>representation (and I'm thinking here of a general-purpose digital
>representation.) To keep things simple, let's suppose the amplitude of the
>sine wave is small, so gives rise to only two different digital values, 0
>and 1.

>If we are sampling 48000 times a second, and if the sine wave is long enough
>when on, our digital signal will, after some start-up sequence, consist of a
>repeating pattern of 24 ones followed by 24 zeroes. Once we're in this part
>of the signal, we can reproduce the original sine wave exactly (within the
>resolution limits, which are not the issue here.)

No, it won't. You really don't understand the basics of A/D
conversion, do you? The waveform will exactly replicate the original
sine wave, so there will be less ones than zeroes, in the appropriate
ratio (can't be arsed to work it out offhand, but it's around 10:14
ratio, rather than 24:24)

>But when does our reproduction start?
>
>Our digital representation will have an initial series of values
>representing the silence. Again leaving aside start-up considerations, a
>series of 48000 identical values will represent one second's initial
>silence, and 48001 will represent 1.000021 seconds of silence.
>
>All silent intervals between 1 and 1.000021 seconds will be represented by
>one of these values. And as there are more than 2 different analog signals
>starting with between 1 and 1.000021 seconds of silence, we are losing
>information in the digital representation ... that is, we are compressing
>it.

You don't understand the basics, do you? What about dither? And are
you *assuming* that the signal starts from zero? You simply have no
basis for your above statement.

>All this is straightforward maths.

Indeed it is, and you're hopelessly wrong again!

>I think where you went wrong is failing to distinguish between the audio
>frequencies contained in the signal, and the timing information contained in
>the signal. In hi-fi, this doesn't usually matter - correctly reproducing
>the audio frequencies gives us more than enough timing precision - but in
>information theory it does.

If you actually understood information theory, we wouldn't be having
this debate..............
--

Stewart Pinkerton | Music is Art - Audio is Engineering

 

[Guest]

Archived from groups: rec.audio.tech (More info?)

On Fri, 03 Jun 2005 08:38:18 GMT, "Tim Martin"
<tim2718281@ntlworld.com> wrote:

>
>"Karl Uppiano" <karl.uppiano@verizon.net> wrote in message
>news:SXTne.16106$qJ3.7554@trnddc05...
>> > Nyquist's Theoerem is about representation of periodic signals; most
>> > sounds
>> > are not periodic signals.
>>
>> Any arbitrary sound can be decomposed into a sum of periodic signals.
>
>OK; let's suppose the sound consists of silence, followed by one second of
>1kHz sine wave, followed by silence.
>
>What sum of periodic signals can this be decomposed into?

Depends on the upper frequency component of the starting corner.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

 

[Guest]

Archived from groups: rec.audio.tech (More info?)

Tim Martin wrote:

> "calmar" <calmar@calmar.ws> wrote in message
> news:slrnd9k81l.ei1.calmar@news.calmar.ws...
>
>
>>Maybe it's possible to say some general things about these ways to play
>>musik quality related?
>
>
> Well, yes. With a computer-based system, you can play the audio files
> directly, without first having to convert them to a hi-fi-friendly format
> (eg CD), thus omitting the accompany audio degradation. Whether the
> difference is detectable is debatable, of course.
>
> Also, the computer-based system can use software processing. So you can,
> for example, compensate for speaker driver impedance variations, without
> having to build it into the speaker crossover or install a hardware EQ or
> DSP system.
>
> Tim
>
>
The audio degradation going from digital-on-computer to
digital-on-audio-CD is absolutely miniscule.

From what I've seen on your posts, I think you need to forget about the
computer for a second. Start with getting an understanding of how a
square wave is a gazzilion sine waves, all slightly out of phase from
each other, summed together. Then you'll be off to a good start.

CD

 

[Guest]

Archived from groups: rec.audio.tech (More info?)

On Fri, 03 Jun 2005 13:32:36 GMT, "Tim Martin"
<tim2718281@ntlworld.com> wrote:

>
>"Arny Krueger" <arnyk@hotpop.com> wrote in message
>news:2KadnR34zaLvpD3fRVn-sw@comcast.com...
>
>> However, the dynamic range in the woods is not all that
>> good, because the background noise level in the woods is
>> pretty high compared to bird songs, especially with the
>> typical distances involved. Nature can be pretty noisy.
>
>Careful - that's all part of the signal, not "noise". If we were trying to
>reproduce the sound of a bird singing in the woods, we would want the
>background sound, too.

Fine, no argument with that in principle. Now, what *transducer* are
you going to use, which will not affect that original sound?

>> IOW a microphone.
>
>Yes, it's pretty much what a capacitor microphone does fairly directly..
>However, microphones generate electrical signals, which are affected by the
>circuitry they are connected to - that is, the microphone introduces noise -
>and I didn't want to confuse the issue with that, hence my suggesting we
>measure the position of a moving element.

Totally pointless exercise, since no such device exists - and if it
did, it would be subject to errors due to Brownian motion of that
moving element, and the positional accuracy of the detector.

>> > The more frequently we measure the position, and the more
>> precisely we
>> > measure the position, the more accurate is our digital
>> representation
>> > of the analog signal.
>>
>> Wrong. While taking more measurements in the time domain
>> increases the highest frequency that can be reliably
>> discerned, it does nothing for resolution or accuracy of the
>> measurements. If you want accuracy, you have to improve the
>> quality of each measurement.
>
>Bear in mind that the context I set was an analog signal with infinite
>bandwidth ... eg including sounds starting and stopping at arbitrary times.

Bear in mind that no such analogue signal exists.

>So improving the timing resolution does improve the accuracy of the digital
>representation.

No, it doesn't, since no such analougue signal exists.

>If we take measurements 10,000 times a second, our resolution is 0.0001
>seconds , our digital representation won't necessarily be able to
>distinguish between sounds starting at 1.00001 seconds and sounds starting
>at 1.00002 seconds; whereas if we take measurements 100,000 times a second,
>it will be able to.

So what? We already sample at 96,000 times per second in most major
recording studios, which is gross overkill for conventional studio
microphones which roll off at 18kHz or so. We're not talking schoolboy
theories here, we're talking about the *real* world.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

 

[Guest]

Archived from groups: rec.audio.tech (More info?)

On Fri, 03 Jun 2005 13:47:35 GMT, "Tim Martin"
<tim2718281@ntlworld.com> wrote:

>
>"Arny Krueger" <arnyk@hotpop.com> wrote in message
>news:sdednURrFd9Ipz3fRVn-gA@comcast.com...
>
>> Not really. Creating sounds at an infinitely high frequency
>> requires infinite amounts of energy.
>
>The scenario does not include any sounds of infinitely high frequency. We
>can stick with just two tones: one silence, and the other a single tone of
>1000Hz.

And what frequencies are involved in the *transition* from one to the
other?

>What I'm saying is that it's not possible to accurately represent a sound
>consisting of alternate arbitrary-length sequences of 1000Hz sine waves and
>silence with any set of sine waves. regardless of the upper frequency we
>use.

Yes, it is. Not *our* fault that *you* don't understand that.

> And that it's not possible to represent them by any digital signal
>with fixed sampling frequency, no matter how high that sampling frequency
>is.

Yes it is, as anyone who *understands* information theory is well
aware...............

>To put it another way: for any fixed-sampling-frequency digital
>representation you specify, I can define two different non-periodic analog
>signals, using only silence and 1000 Hz sine waves, that will both map to
>the same digital representation.
>
>This is trivial.

Indeed, so why do you keep getting it so wrong?
--

Stewart Pinkerton | Music is Art - Audio is Engineering

 

[Guest]

Archived from groups: rec.audio.tech (More info?)

"Arny Krueger" <arnyk@hotpop.com> wrote in message
news:adadnbmtJeNhOT3fRVn-rg@comcast.com...

> I thought this discussion was about audio, not how many
> measurements can dance on the head of an imaginary pin! ;-)

Well, my initial reponse was to a comment about compression. Someone said
that he felt compression was undesirable; I've simply pointed that
compression is inherent in digital representation of signals.

Having got rid of the compression bogey-man, one is free to assess any
storage scheme on its merits.

Tim

 

[Guest]

Archived from groups: rec.audio.tech (More info?)

<dpierce@cartchunk.org> wrote in message
news:1117820624.239670.28190@g47g2000cwa.googlegroups.com...
>
>
> Tim Martin wrote:
> > "Karl Uppiano" <karl.uppiano@verizon.net> wrote in message
> > news:SXTne.16106$qJ3.7554@trnddc05...
> > > > Nyquist's Theoerem is about representation of periodic signals;
most
> > > > sounds
> > > > are not periodic signals.
> > >
> > > Any arbitrary sound can be decomposed into a sum of periodic signals.
> >
> > OK; let's suppose the sound consists of silence, followed by one second
of
> > 1kHz sine wave, followed by silence.
> >
> > What sum of periodic signals can this be decomposed into?
>
> It's exactly the same as a 1 kHz sine wave 100% modulated by a 0.5
> Hz square wave ...

Wouldn't that produce a repeating signal of one second of 1000Hz alternating
with one second of silence... or have I misunderstood what you're
suggesting?

Tim

 

[Guest]

Archived from groups: rec.audio.tech (More info?)

"Stewart Pinkerton" <patent3@dircon.co.uk> wrote in message
news:dg41a15qfesbh6ib6fg84jj2nasgkh74jh@4ax.com...

Tim

> >If we are sampling 48000 times a second, and if the sine wave is long
enough
> >when on, our digital signal will, after some start-up sequence, consist
of a
> >repeating pattern of 24 ones followed by 24 zeroes. Once we're in this
part
> >of the signal, we can reproduce the original sine wave exactly (within
the
> >resolution limits, which are not the issue here.)

Srewart

> No, it won't. You really don't understand the basics of A/D
> conversion, do you? The waveform will exactly replicate the original
> sine wave, so there will be less ones than zeroes, in the appropriate
> ratio (can't be arsed to work it out offhand, but it's around 10:14
> ratio, rather than 24:24)

That's not correct. Each digital value represents a range of signal levels.
I picked the case of a sine wave; if the amplitude is x, then one of the
digital values will represent signal levels from 0 to x, and the other will
represent values from 0 to -x. (Remember, I specified a signal whose
amplitude was such as to be represented by one bit.)

The ratio of the frequencies of the two digital values represents the
offset - that is, the average value of the sine wave signal. I simply took
the case where the average is zero, so there are equal numbers of above-zero
and below-zero values.

Of course one can devise other representation systems; however, the average
of the values represented by the digital signals will be the average value
of the signal being represented.

Tim