HiFi vs. Computer

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Tim Martin wrote:
> "Arny Krueger" <arnyk@hotpop.com> wrote in message
> news:B5-dnaJC2MZQ4jjfRVn-rg@comcast.com...
>
>> What you don't seem to realize Tim is the fact that all
of
>> the limiations that you've been obsessing over relate to
>> both analog and digital signals.
>
> That's incorrect: analog signals are continuous. Between
any two
> points in time where the analog signal is changing, one
can find
> another point in time where the value of the analog signal
differs
> from its values at either of the first two points in time.

That harkens back to the myth that all analog signals have
infinite resolution and infinite bandwidth.

> This is not true for digital representations of analog
signals.

It's just that in the digital domain, the limitations on
resolution and bandwidth are very obvious and explicit.
However, in the digital domain the resolution and bandwidth
are not limited by physical considerations other than
computational bandwidth and precision which is more readily
extenstible or managable.

> Of course, there are limits in how accurately one can
represent analog
> signals, whether the representation is digital or analog.

That's my point.

> But the limits are in communication channels and storage
schemes, not in the
> signals themselves.

You can always find more communcations bandwidth and storage
media. You can't extend analog dynamic range much beyond
about 130 dB, usually not more than 110 or so.

 

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"Arny Krueger" <arnyk@hotpop.com> wrote in message
news:mvGdnY6A15feVjjfRVn-rg@comcast.com...

> That's my point.

It's also my point.

And because digital and analog representations cannot represent original
analog signals exactly, it doesn't in itself matter if we use compression
when storing those representations; what matters is whether the reproduction
of the stored data is close enough to the original analog signal to suit our
purposes.

As you say, in the digital domain, the limitations are explicit, which is
why it seemed easier to explain the point referring only to digital
representations (which are what the person mentioning compression wanted to
store anyway.) But of course exactly the same applies to analog
representations.

Tim

 

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On Tue, 07 Jun 2005 10:10:09 GMT, "Tim Martin"
<tim2718281@ntlworld.com> wrote:

>
>"Stewart Pinkerton" <patent3@dircon.co.uk> wrote in message
>news:itbaa1tt16evbb2mi5ue5so8pvnv9ubsbo@4ax.com...
>
>> You don't even understand your own arguments, do you? You talked about
>> signal, and about noise.
>
>I did say that what Arny referred to as the "background noise" of the woods
>was part of the signal.

Except that, below a certain level, it just isn't. Shame that you
don't realise this.

> And I did say that I avoided assuming the use of a
>microphones as the measurement device, because microphones are connected to
>electrical circuits and so introduce noise.

That's because you are ignoring reality, as in most of your posts. How
else would you produce an analogue of the original sonic event? BTW,
the fact that microphones have self-noise has *nothing* to do with the
circuits to which they are connected. You need to learn some *very*
basic physics.

>I don't see how either of those comments can be interpreted as a remark
>about dynamic range.

That simply spotlights your ignorance.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

 

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On Tue, 07 Jun 2005 15:00:28 GMT, "Tim Martin"
<tim2718281@ntlworld.com> wrote:

>
>"Arny Krueger" <arnyk@hotpop.com> wrote in message
>news:B5-dnaJC2MZQ4jjfRVn-rg@comcast.com...
>
>> What you don't seem to realize Tim is the fact that all of
>> the limiations that you've been obsessing over relate to
>> both analog and digital signals.
>
>That's incorrect: analog signals are continuous. Between any two points in
>time where the analog signal is changing, one can find another point in time
>where the value of the analog signal differs from its values at either of
>the first two points in time.

Clearly, you have absolutely *no* understanding of the real physical
world. Go learn about uncertainty, before you spout such nonsense
here.

>This is not true for digital representations of analog signals.

Bullshit. Clearly, you have absolutely *no* understanding of the real
physical world. Go learn about uncertainty, before you spout such
nonsense here.

>Of course, there are limits in how accurately one can represent analog
>signals, whether the representation is digital or analog. But the limits
>are in communication channels and storage schemes, not in the signals
>themselves.

No, there are very well defined limits in the signals themselves.
Clearly, you have absolutely *no* understanding of the real physical
world. Go learn about uncertainty, before you spout such nonsense
here.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

 

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On Tue, 07 Jun 2005 17:32:46 GMT, "Tim Martin"
<tim2718281@ntlworld.com> wrote:

>And because digital and analog representations cannot represent original
>analog signals exactly,

Sure they can, provided that the resolution and bandwidth limits of
the system exceed those of the original event. In the case of the
ubiquitous 24/96, that is a given for anything short of a military jet
doing a touch and go over the microphone at midnight in the desert!

> it doesn't in itself matter if we use compression
>when storing those representations;

It does if we use *lossy* compression!

> what matters is whether the reproduction
>of the stored data is close enough to the original analog signal to suit our
>purposes.

Indeed, and even with 16/44, that will be the case for any music
master tape.

>As you say, in the digital domain, the limitations are explicit, which is
>why it seemed easier to explain the point referring only to digital
>representations (which are what the person mentioning compression wanted to
>store anyway.) But of course exactly the same applies to analog
>representations.

Indeed, except that all known analogue storage systems are inferior to
24/96 digital, and the majority are inferior to 16/44.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

 

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Tim Martin wrote:
> "Richard Crowley" <rcrowley7@xprt.net> wrote in message
> news:11ads6hp0chg949@corp.supernews.com...
>
> > Congratulations, you have re-invented the microphone.
>
> I think by "microphone" people normally mean a transducer that converts
> changes in air pressure into electrical signals.
> It's possible to measure movement without that, and in fact there are
> commercial devices using optical measurement to measure such movements.

And then the optical signal is converted into an electrical signal.
So what's the difference (subtle hint: NONE!)

> Such devices are not generally termed "microphones".

They ARE called microphones if their bandwidth is over the
audio range. If lower, they're often called "barometers." If
higher, they're called "ultrasonic detectors."

The point being, and one which, I suspect, your quite aware of
despite your obvious troll-like behavior to date, is that the
physical operating prionciple behind what you assign different
names to is utterly irrelevant semantic obfuscation on your part.

There are bascially two classes of transducers that perform these
tasks. One class includes true pressure transducers, and include
barometers and audio microphones such as some condenser microphones
and piezo microphones whose output signal is a direct function of
diaphragm position. The second class is velocity transducers,
which include anemometers and other types of audio microphones,
whose output is a function of instantaneous disphragm velocity.
The difference, for the purpose of discussion, is irrelevant,
especially since you can take the output of a pressure transducer
and differentiate it WRT time and get velocity, or take the output
of a velocity transducer and intergaret it WRT time and get position.

Your ability to drown in your own swirl of semantic two-step
continues to be a source of amusement for us all.

 

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Arny Krueger schreef:
> Laurence Payne wrote:
> > On Fri, 03 Jun 2005 02:19:12 GMT, "Tim Martin"
> > <tim2718281@ntlworld.com> wrote:
> >
> >> An analog signal, such as a bird singing in the woods,
> has infinite
> >> bandwidth.
> >
> > No it isn't. It is constrained by all sorts of physical
> limitations,
> > both in the bird and in the environment.
>
> Agreed. The spectral content of most natural sounds will
> reveal the inevitable high end roll-off if measurered with a
> mic capable of going out to about 50 KHz. In a few cases one
> must go out to a few 100 KHz.

Simpler yet, infinity is even ruled out by the very vehicle of sound
propagation itself which can never be infinite. If anything is acoustic
in nature, it can't be of infinite bandwidth otherwise it would violate
Newton's basic law of conservation of energy. To say that any "sound"
has infinite bandwidth is to say that inertia doesn't exist.

 

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Tim Martin wrote:
> <dpierce@cartchunk.org> wrote in message
> news:1118245149.952932.31770@g43g2000cwa.googlegroups.com...
>
> Just to remind you, the point is:
>
> "Any digital storage of an analog signal compresses it.
>
> "That is, for any method of storing an analog signal in x bits, it is
> possible to devise a digital storage mechanism using >x bits which can be
> used to reproduce a more accurate rendition of the original analog signal."
>
> Notice that the statement has nothing to do with electrical signals,
> transducers and so on.

And just to remind you, not that it did any good the last time,
that your point is just plain wrong, despite the numerous
irrelevancies and flat out nonsense you've brought to the table.

 

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"Stewart Pinkerton" <patent3@dircon.co.uk> wrote in message
news:2osba118e8g7l2hbi8daj9v41qo37niast@4ax.com...

> That's because you are ignoring reality, as in most of your posts. How
> else would you produce an analogue of the original sonic event?

I explained that. One can measure the position of a vibrating diaphragm,

Tim

 

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"Tim Martin" <tim2718281@ntlworld.com> wrote in message
news:fGApe.1829$m4.93@newsfe3-gui.ntli.net...
>
> "Stewart Pinkerton" <patent3@dircon.co.uk> wrote in message
> news:2osba118e8g7l2hbi8daj9v41qo37niast@4ax.com...
>
>> That's because you are ignoring reality, as in most of your posts.
>> How
>> else would you produce an analogue of the original sonic event?
>
> I explained that. One can measure the position of a vibrating
> diaphragm,

Congratulations, you have re-invented the microphone.
Unfortunately for you, about 100 years too late. Some
study of history and theory really would go a long ways
toward helping you understand what is happening here
in the real world.

 

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"Stewart Pinkerton" <patent3@dircon.co.uk> wrote in message
news:10tba1h8lpuhj674hu1a1v1n7na630d1f7@4ax.com...

> >That's incorrect: analog signals are continuous. Between any two points
in
> >time where the analog signal is changing, one can find another point in
time
> >where the value of the analog signal differs from its values at either of
> >the first two points in time.
>
> Clearly, you have absolutely *no* understanding of the real physical
> world. Go learn about uncertainty, before you spout such nonsense
> here.

As usual, instead of commenting on the point made, you put up a smokescreen.

However, if you think what I said is incorrect, please show it. In the case
of the bird singing in the woods, demonstrate there are two points in
time,where the analog signal in the interval between those two ponts in time
is unchanging.

Tim

 

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Tim Martin wrote:
> "Arny Krueger" <arnyk@hotpop.com> wrote in message
> news:B5-dnaJC2MZQ4jjfRVn-rg@comcast.com...
>
> > What you don't seem to realize Tim is the fact that all of
> > the limiations that you've been obsessing over relate to
> > both analog and digital signals.
>
> That's incorrect: analog signals are continuous.

And you keep making the same fundamental error over and over again,
when you equate continuous with infinite resolution. It just ain't
so, matter how many times you repeat it.

> Between any two points in time where the analog signal is
> changing, one can find another point in time where the value
> of the analog signal differs from its values at either of
> the first two points in time.

And what you utterly fail or refuse to realize is that ALL
the values in between those two points in a continuous
representation are ENTIRELY defined by the bandwidth of the
system. If I know the bandwidth, and I know the data on either
side of of your two points, and those two points are separated
by 1/2*bandwidth, then I can exactly predict the value at ANY
arbitrary point in between.

> This is not true for digital representations of analog signals.

Once again, other than stating it over and over and over agin, you
have failed completely to deomstrate your assertion.

> Of course, there are limits in how accurately one can represent analog
> signals, whether the representation is digital or analog.

Yup, and those limits are imposed first by the fact that NO signal
can have infinite resolution to begin with, your irrational yet
exuberant insistance on infinite energy and time not withstanding
at all. Secondly, the limits are imposed by the bandwidth of the
system and its dynamic range.

> But the limits
> are in communication channels and storage schemes, not in the signals
> themselves.

Not if the signals themselves don't exist for infinite time, have
infinite bandwidth and have inifite dynamic range.

No matter how many times you repeat you incorrect assertion that
"digital compresses", no matter how many times you misapply,
deliberately or otherwise, the principles of Fourier decomposition,
no matter how many times you invoke physically unrealizable
conditions or theoretically ridiculous signals, no matter how many
times you attempt to sidestep the issue with irrelevancies in
an attempt to support your assertion, the assertion still remains
incorrect.

 

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Tim Martin wrote:
> However, if you think what I said is incorrect, please show it. In the case
> of the bird singing in the woods, demonstrate there are two points in
> time,where the analog signal in the interval between those two ponts in time
> is unchanging.

No one ever said that the signal is unchanging. That's a strawman
that you alone erected.

What one CAN say, most definitely, is that the number of possible paths
the signal CAN take between the two points is limited by the interval
between those two points and the bandwidth of the signal. Make the
bandwidth narrow enough, or make the two points close enough, and the
signal is uniquely constrained. At that point, then, if you "sample"
the signal at those intervals, you can have complete knowledge of the
signal wihout having INFINITE knowledge of the signal.

In other words, once you have sampled the signal at a rate fast
enough as defined by the bandwidth of the signal, smapling any
faster WILL NOT give you ANY more information about that signal.
Any extra data is simply redundant and provides no new information
that we could not have already derived from the data we have.

Your repeated assertion is based on the fallacy that any signal,
such as a bird singing in the woods, has infinite information
content. It does not. Once you finally come to that realization,
then you'll see the nonsense of your assertion. You do not need
infinite data to represent finite information.

Now, go ponder your position in the light that no signal has
infinite information content. If you still hold to your position,
please don't bother us until you are able to support that position
with equal the rigor the likes of Shannon applied to the sampling
theorem.

 

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"Richard Crowley" <rcrowley7@xprt.net> wrote in message
news:11ads6hp0chg949@corp.supernews.com...

> Congratulations, you have re-invented the microphone.

I think by "microphone" people normally mean a transducer that converts
changes in air pressure into electrical signals.

It's possible to measure movement without that, and in fact there are
commercial devices using optical measurement to measure such movements.

Such devices are not generally termed "microphones".

Tim

 

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<dpierce@cartchunk.org> wrote in message
news:1118245149.952932.31770@g43g2000cwa.googlegroups.com...

> > I think by "microphone" people normally mean a transducer that converts
> > changes in air pressure into electrical signals.
> > It's possible to measure movement without that, and in fact there are
> > commercial devices using optical measurement to measure such movements.
>
> And then the optical signal is converted into an electrical signal.
> So what's the difference (subtle hint: NONE!)

The lack of need for generating an electrical signal from the movement of
the air may mean that one can use a detection system with physical
characteristics different from microphones.

Tim

 

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"Wessel Dirksen" wrote in...
> Simpler yet, infinity is even ruled out by the very vehicle of sound
> propagation itself which can never be infinite. If anything is acoustic
> in nature, it can't be of infinite bandwidth otherwise it would violate
> Newton's basic law of conservation of energy. To say that any "sound"
> has infinite bandwidth is to say that inertia doesn't exist.

Mr. Martin never directly addressed the question of whether he is
posting his messages from the same universe the rest of us are in.
He may not know what "inertia" is in his Martin Universe. 🙂 His
responses and questions suggest that he may not operating from
the same frame of reference as the rest of us.

 

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<dpierce@cartchunk.org> wrote in message
news:1118245149.952932.31770@g43g2000cwa.googlegroups.com...

Just to remind you, the point is:

"Any digital storage of an analog signal compresses it.

"That is, for any method of storing an analog signal in x bits, it is
possible to devise a digital storage mechanism using >x bits which can be
used to reproduce a more accurate rendition of the original analog signal."

Notice that the statement has nothing to do with electrical signals,
transducers and so on.

Tim

 

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On Wed, 08 Jun 2005 16:39:54 GMT, "Tim Martin"
<tim2718281@ntlworld.com> wrote:

>
><dpierce@cartchunk.org> wrote in message
>news:1118245149.952932.31770@g43g2000cwa.googlegroups.com...
>
>Just to remind you, the point is:
>
>"Any digital storage of an analog signal compresses it.
>
>"That is, for any method of storing an analog signal in x bits, it is
>possible to devise a digital storage mechanism using >x bits which can be
>used to reproduce a more accurate rendition of the original analog signal."
>
>Notice that the statement has nothing to do with electrical signals,
>transducers and so on.

Yes, and it's still bollocks, just as it was the first time you
spouted it.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

 

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On Wed, 08 Jun 2005 11:31:23 GMT, "Tim Martin"
<tim2718281@ntlworld.com> wrote:

>
>"Stewart Pinkerton" <patent3@dircon.co.uk> wrote in message
>news:2osba118e8g7l2hbi8daj9v41qo37niast@4ax.com...
>
>> That's because you are ignoring reality, as in most of your posts. How
>> else would you produce an analogue of the original sonic event?
>
>I explained that. One can measure the position of a vibrating diaphragm,

Not with real-world equipment. These aren't mind games, we're talking
about reality here. Besides, you'd have the same self-noise problems
with your proposed measurement system, and *really* serious frequency
response issues.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

 

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On Wed, 08 Jun 2005 11:38:14 GMT, "Tim Martin"
<tim2718281@ntlworld.com> wrote:

>
>"Stewart Pinkerton" <patent3@dircon.co.uk> wrote in message
>news:10tba1h8lpuhj674hu1a1v1n7na630d1f7@4ax.com...
>
>> >That's incorrect: analog signals are continuous. Between any two points in
>> >time where the analog signal is changing, one can find another point in time
>> >where the value of the analog signal differs from its values at either of
>> >the first two points in time.
>>
>> Clearly, you have absolutely *no* understanding of the real physical
>> world. Go learn about uncertainty, before you spout such nonsense
>> here.
>
>As usual, instead of commenting on the point made, you put up a smokescreen.

No, you simply fail to understand the basic physics, as has been all
too obvious all along.

>However, if you think what I said is incorrect, please show it.

About a dozen people already have, but you appear to be too ignorant
and/or stupid to understand what is being explained to you.

> In the case
>of the bird singing in the woods, demonstrate there are two points in
>time,where the analog signal in the interval between those two ponts in time
>is unchanging.

Simple, if you have *any* grasp of basic physics. There is a noise
floor in any acoustic event, whereby there is uncertainty about the
sound pressure at any moment in time. Hence, where you have a sound
such as birdsong, where a sinusoid may be assumed to be increasing as
it passes through the zero crossing point in a positive direction, the
noise floor may be moving in the opposite direction with the same
fundamental frequency at that time interval - in an entirely
unpredictable way, since it is *random* in nature. In such a case, the
sound pressure will remain constant for that time interval. Any
transducer will record an analogue signal representative of that
event, as a flat line.

BTW, will you *please* understand that the original sound is *not* an
analogue signal, it is a physical event. For God's sake at least buy a
*dictionary*, if physics texts are too difficult for you.
--

Stewart Pinkerton | Music is Art - Audio is Engineering