Ok, this is the way I see it: you have bit rate and sample rate.

The sample rate is the number of times per second that the signal is cut up.
The Nyquist Theorem states that the highest frequency reproduced is half of
the sample rate. This is too long and convoluted for me to understand or
explain. But so long as the sample rate is CD-quality (44.1 kHz), we'll have
a possible frequency of 22.05 kHz, which is out of normal human hearing and
definitely out of the range of electric guitar.

The bit rate is the dynamic range of a unit. Let's say the input is 1 volt
and the converter has 1 bit. It can be either 0 or 1. So if the signal is
closer to 1 volt, it will be 1, and if it's closer to 0 volts, it will be 0.
Obviously, you'd have full signal then nothing. With 2 bits, you double that
dynamic range. Each additional bit doubles that range so you end up with 2^N
different volume levels for each bit rate N.

Let's also say you have a guitar signal that comes in at only half the
volume the input buffer wants to see. You are wasting half the bits right
there, squashing your dynamic range. Likewise, if you have buzzing or
humming noise, it masks the lower bits squashing your dynamic range again.

A compressor will take a volume level over a certain threshold and reduce
it. You can make up for this reduction by increasing gain, making the louder
parts just as loud as before, and the softer parts louder. If done subtly,
you will just notice a smoother input.

A compressed signal has the POTENTIAL to increase dynamic range by ensuring
that your maximum peak levels are not too much higher than the rest of your
playing. Let's say that you hit the strings on a guitar as hard as you can
and it peaks at twice the normal playing volume. Again, you're wasting bits
and dynamic range right there. So a compressor will get your guitar
operating at a higher average level, increasing the number of bits used and
thus dynamic range.

Here I will BS a bit because I don't know exactly what happens (I'm an audio
enthusiast, not an EE major). A buffer will take a high impedance signal and
output a low impedance signal. This signal is less susceptible to
high-frequency and volume loss, perhaps?

Furthermore, impedance is not a flat number; it changes based on frequency.
I'm imagining that by converting this impedance, you are also flattening the
impedance of a guitars pickup, for instance. By doing so, certain
frequencies are not picked up louder because of different impedance, but
because of different output characteristics. In this way, you will not have
spikes or dips in the impedance of a pickup, which I propose, decreases the
dynamic range.

In my opinion, a well-designed, hum- and noise-free, buffer with a hint of
compression (i.e. a Valvulator), can actually increase dynamic range.

Am I understanding this correctly? Feel free to rip my comments to shreds.

jaric wrote:

Ok, this is the way I see it: you have bit rate and sample rate.

You have bit _depth_ and sample _rate_

The sample rate is the number of times per second that the signal is cut up.
The Nyquist Theorem states that the highest frequency reproduced is half of
the sample rate.

The highest frequency that can be reporduced is _less than half_ the
smaple rate.

This is too long and convoluted for me to understand or
explain. But so long as the sample rate is CD-quality (44.1 kHz), we'll have
a possible frequency of 22.05 kHz, which is out of normal human hearing and
definitely out of the range of electric guitar.
The bit rate is the dynamic range of a unit. Let's say the input is 1 volt
and the converter has 1 bit. It can be either 0 or 1. So if the signal is
closer to 1 volt, it will be 1, and if it's closer to 0 volts, it will be 0.
Obviously, you'd have full signal then nothing. With 2 bits, you double that
dynamic range. Each additional bit doubles that range so you end up with 2^N
different volume levels for each bit rate N.

Let's also say you have a guitar signal that comes in at only half the
volume the input buffer wants to see. You are wasting half the bits right
there, squashing your dynamic range. Likewise, if you have buzzing or
humming noise, it masks the lower bits squashing your dynamic range again.

A compressor will take a volume level over a certain threshold and reduce
it. You can make up for this reduction by increasing gain, making the louder
parts just as loud as before, and the softer parts louder. If done subtly,
you will just notice a smoother input.

A compressed signal has the POTENTIAL to increase dynamic range by ensuring
that your maximum peak levels are not too much higher than the rest of your
playing.

You have just described _decreasing_ the dynamic range using a
compressor.

Let's say that you hit the strings on a guitar as hard as you can
and it peaks at twice the normal playing volume. Again, you're wasting bits
and dynamic range right there. So a compressor will get your guitar
operating at a higher average level, increasing the number of bits used and
thus dynamic range.

Here I will BS a bit because I don't know exactly what happens (I'm an audio
enthusiast, not an EE major). A buffer will take a high impedance signal and
output a low impedance signal. This signal is less susceptible to
high-frequency and volume loss, perhaps?

Furthermore, impedance is not a flat number; it changes based on frequency.
I'm imagining that by converting this impedance, you are also flattening the
impedance of a guitars pickup, for instance. By doing so, certain
frequencies are not picked up louder because of different impedance, but
because of different output characteristics. In this way, you will not have
spikes or dips in the impedance of a pickup, which I propose, decreases the
dynamic range.

In my opinion, a well-designed, hum- and noise-free, buffer with a hint of
compression (i.e. a Valvulator), can actually increase dynamic range.

That's a questionable opinion.

Am I understanding this correctly? Feel free to rip my comments to shreds.

I think folks will have some inpout here.

--
ha

On Wed, 24 May 2006 06:51:13 GMT, "jaric" wrote:

Ok, this is the way I see it: you have bit rate and sample rate.

The sample rate is the number of times per second that the signal is cut up.
The Nyquist Theorem states that the highest frequency reproduced is half of
the sample rate. This is too long and convoluted for me to understand or
explain. But so long as the sample rate is CD-quality (44.1 kHz), we'll have
a possible frequency of 22.05 kHz, which is out of normal human hearing and
definitely out of the range of electric guitar.


Almost right. 22.05kHz is the first impossible frequency. The wanted
stuff must be below the Nyquist frequency.

The bit rate is the dynamic range of a unit. Let's say the input is 1 volt
and the converter has 1 bit. It can be either 0 or 1. So if the signal is
closer to 1 volt, it will be 1, and if it's closer to 0 volts, it will be 0.
Obviously, you'd have full signal then nothing. With 2 bits, you double that
dynamic range. Each additional bit doubles that range so you end up with 2^N
different volume levels for each bit rate N.

No. Bit rate is a term used in transmitting audio. It is the product
of the sampling rate and the bit depth. Bit depth is the term you are
looking for. For the second part, you are sort of right but you are
ignoring dithering, which negates the effect you describe. A 1 bit
converter will correctly distinguish levels all the way down to zero -
there is no step. Like analogue audio, of course, it will have noise
associated with it. But if you really are measuring DC voltages, you
can average the dithered signal to get as accurate a measure as you
need.

Let's also say you have a guitar signal that comes in at only half the
volume the input buffer wants to see. You are wasting half the bits right
there, squashing your dynamic range. Likewise, if you have buzzing or
humming noise, it masks the lower bits squashing your dynamic range again.

No. Dynamic range is the difference between the smallest and biggest
*possible* signals a system will reproduce. How much of that you make
use of is up to you. Current pop music may use 2dB, while the best
classical may have perhaps 55dB.

A compressor will take a volume level over a certain threshold and reduce
it. You can make up for this reduction by increasing gain, making the louder
parts just as loud as before, and the softer parts louder. If done subtly,
you will just notice a smoother input.

You got what happens right, but maybe the conclusion is rather
questionable. These days it is just louder.

A compressed signal has the POTENTIAL to increase dynamic range by ensuring
that your maximum peak levels are not too much higher than the rest of your
playing. Let's say that you hit the strings on a guitar as hard as you can
and it peaks at twice the normal playing volume. Again, you're wasting bits
and dynamic range right there. So a compressor will get your guitar
operating at a higher average level, increasing the number of bits used and
thus dynamic range.

No. Compression always (and indeed must) decreases dynamic range. The
number of bits you are currently using is not dynamic range. The
dynamic range of a piece of music is the difference between the
loudest bits and the softest bits. WHat you are describing is just
loudness.

Here I will BS a bit because I don't know exactly what happens (I'm an audio
enthusiast, not an EE major). A buffer will take a high impedance signal and
output a low impedance signal. This signal is less susceptible to
high-frequency and volume loss, perhaps?

Pretty much right. Look at it as making the signal much more robust,
so it can be fed to assorted bits of kit without fear of changing it.

Furthermore, impedance is not a flat number; it changes based on frequency.
I'm imagining that by converting this impedance, you are also flattening the
impedance of a guitars pickup, for instance. By doing so, certain
frequencies are not picked up louder because of different impedance, but
because of different output characteristics. In this way, you will not have
spikes or dips in the impedance of a pickup, which I propose, decreases the
dynamic range.

Yes. Guitar pickups are slightly different to most audio kit in that
they actually use their varying impedance with frequency to produce
their characteristic sound in conjunction with guitar amps and guitar
leads. Forget dynamic range in connection with this - it has no
relevance.

In my opinion, a well-designed, hum- and noise-free, buffer with a hint of
compression (i.e. a Valvulator), can actually increase dynamic range.

No. Nothing can increase dynamic range (apart from expansion, the
exact opposite of compression). When you add stages, each will have
its own noise level and the dynamic range will, however slightly,
reduce.

Am I understanding this correctly? Feel free to rip my comments to shreds.

You've got bits of it spot on, but you need to understand what dynamic
range is and not apply it wrongly. Let me sum up:

For a piece of equipment, the dynamic range is the difference between
the biggest level it can reproduce without clipping and the level of
the noise.

For a piece of music, the dynamic range is the difference between the
loudest part and the quietest part.

The key word here is difference. It is a range you are describing.

d

--
Pearce Consulting
http://www.pearce.uk.com

On Wed, 24 May 2006 07:33:02 GMT, (Don Pearce)
wrote:

you are sort of right but you are
ignoring dithering

Which I beleive is the source of essentially *all* of these
misunderstandings. Certainly was true of mine.

The surprising and counter-intuitive fact about digital
storage is that it can (ideally, of course) make a
*perfect* copy, plus the small noise added to dither
and after bandpass filtering. The interesting bit lies
in the dither: information storage is actually increased
by adding noise; like that. Great stuff and enormously
fun to learn.

The PERFECT part is the hardest to appreciate, because
of, ... well, just because.

Thanks, as always,

Chris Hornbeck

Chris Hornbeck wrote:

The interesting bit lies
in the dither: information storage is actually increased
by adding noise

Prompted by the universal rule that you can't get something for nothing,
I have to doubt this. I think the answer is that dither gets you more
resolution at low frequencies but less at high frequencies.

It just so happens that's acoustically desirable because it changes the
quantization noise from something signal-related to a constant white
noise that is aurally less distracting at the same level.

Anahata

On Thu, 25 May 2006 09:18:10 +0100, Anahata
wrote:

Chris Hornbeck wrote:

The interesting bit lies
in the dither: information storage is actually increased
by adding noise

Prompted by the universal rule that you can't get something for nothing,
I have to doubt this. I think the answer is that dither gets you more
resolution at low frequencies but less at high frequencies.

No, resolution is not frequency dependent. It is constant all the way
up to the Nyquist frequency.

It just so happens that's acoustically desirable because it changes the
quantization noise from something signal-related to a constant white
noise that is aurally less distracting at the same level.

I would disagree with your analysis about level. The level of the
quantization products is much higher than the level of noise caused by
dithering. Think of it as all the energy being piled into tall heaps
rather than being spread out.

d

--
Pearce Consulting
http://www.pearce.uk.com

Don Pearce wrote:

No, resolution is not frequency dependent. It is constant all the way
up to the Nyquist frequency.

Perhaps resolution isn't the right word, but it's obvious that the error
on any individual sample is higher if noise has been added than if it
hasn't, and averaging that error over several samples will reduce it.
Perhaps I should have said signal to noise ratio, compared in different
frequency bands.

Think of it as all the energy being piled into tall heaps
rather than being spread out.

Yes, but it's the same total energy, which is what I meant by level.
I don't think we're in fundamental disagreement.

Anahata

Anahata wrote in news:4475683b$0$18226
:

Chris Hornbeck wrote:

The interesting bit lies
in the dither: information storage is actually increased
by adding noise

Prompted by the universal rule that you can't get something for nothing,
I have to doubt this. I think the answer is that dither gets you more
resolution at low frequencies but less at high frequencies.

Dither lets you hear signal down farther in the noise.

It just so happens that's acoustically desirable because it changes the
quantization noise from something signal-related to a constant white
noise that is aurally less distracting at the same level.

Exactly. Removing the correlation allows your brain to "hear through it"
to the signal content at lower levels.

"Anahata" wrote in message

Chris Hornbeck wrote:

The interesting bit lies
in the dither: information storage is actually increased
by adding noise

Prompted by the universal rule that you can't get
something for nothing, I have to doubt this.

Your doubts are well-founded. In fact, application of dither slightly
reduces information storage capacity. But, you don't give up something to
get nothing. Instead, you obtain a more listenable product and a more
generally useful product because quantization error is randomized (which you
say near the end of your post).

I think the
answer is that dither gets you more resolution at low
frequencies but less at high frequencies.

Not true in general, but certainly true in useful special cases.

Dither is often weighted or shaped to concentrate its energy at frequencies
that are outside the range where the ear is most sensitive. Then, your
statement is almost perfectly correct - resolution is increased at lower
frequencies at the cost of reduced resolution at the highest frequencies.
Since the ear is not so sensitive at the highest frequencies, this loss of
high frequency resolution is a more than acceptable trade-off.

It just so happens that's acoustically desirable because
it changes the quantization noise from something
signal-related to a constant white noise that is aurally
less distracting at the same level.

100% right, except that this is no happenstance - it is exactly per design.
OK, so "It just so happens" is a figure of speech. ;-)

On Thu, 25 May 2006 09:18:10 +0100, Anahata
wrote:

The interesting bit lies
in the dither: information storage is actually increased
by adding noise

Prompted by the universal rule that you can't get something for nothing,
I have to doubt this. I think the answer is that dither gets you more
resolution at low frequencies but less at high frequencies.

It just so happens that's acoustically desirable because it changes the
quantization noise from something signal-related to a constant white
noise that is aurally less distracting at the same level.

You and all subsequent posters understand the technique
perfectly, but until corrected I'll have to stand by my
assertion that dither allows us to hear (because of
critical bands) below the noise floor, just like in
analog recording.

Perhaps Bob Cain will weigh in here and, as so often,
open my eyes.

Much thanks, as always,

Chris Hornbeck

"Chris Hornbeck" wrote in
message
On Thu, 25 May 2006 09:18:10 +0100, Anahata
wrote:

The interesting bit lies
in the dither: information storage is actually increased
by adding noise

Prompted by the universal rule that you can't get
something for nothing, I have to doubt this. I think
the answer is that dither gets you more resolution at
low frequencies but less at high frequencies.

It just so happens that's acoustically desirable because
it changes the quantization noise from something
signal-related to a constant white noise that is aurally
less distracting at the same level.

You and all subsequent posters understand the technique
perfectly, but until corrected I'll have to stand by my
assertion that dither allows us to hear (because of
critical bands) below the noise floor, just like in
analog recording.

Below noise-floor hearing also happens in quantized systems without dither.
It just sounds nastier.

Chris Hornbeck wrote:

You and all subsequent posters understand the technique
perfectly, but until corrected I'll have to stand by my
assertion that dither allows us to hear (because of
critical bands) below the noise floor, just like in
analog recording.

Perhaps Bob Cain will weigh in here and, as so often,
open my eyes.

Everybody's pretty well covered it. The dither doesn't create the
ability to hear signal below noise, it just makes it undistorted by
decorrelating the error and the signal thus making the error spectrum
independant of that of the signal. We hear the true spectrum of the
signal with hiss on it rather than hearing the signal with all kinds
of its harmonics on top of it.

For very low level signals, at the level of the low order bit, the
undithered quantization distortion will be a large percentage of the
signal. Our ears are less tolerant of that much distortion than they
are of the noise that dither replaces it with.

The power in the dither noise is actually greater than that of the
quantization noise but the signal is more "realistic" in the presence
of the noise (uncorrelated error) than in the presence of the
distortion (correlated error) allowing it to remain intelligible at
levels lower than are possible without dither.

There are those who claim to hear differences in high level signals
that depend not just on the presence of dither but on the dither
algorithm used. Of this I remain skeptical.


Bob
--

"Things should be described as simply as possible, but no simpler."

A. Einstein

The sample rate is the number of times per second that the signal is cut
up.
The Nyquist Theorem states that the highest frequency reproduced is half
of
the sample rate. This is too long and convoluted for me to understand or
explain. But so long as the sample rate is CD-quality (44.1 kHz), we'll
have
a possible frequency of 22.05 kHz, which is out of normal human hearing
and
definitely out of the range of electric guitar.

Actually you'll have a possible frequency of just under 22.05kHz, but close
enough.

The bit rate is the dynamic range of a unit. Let's say the input is 1 volt
and the converter has 1 bit. It can be either 0 or 1. So if the signal is
closer to 1 volt, it will be 1, and if it's closer to 0 volts, it will be
0.
Obviously, you'd have full signal then nothing. With 2 bits, you double
that
dynamic range. Each additional bit doubles that range so you end up with
2^N
different volume levels for each bit rate N.

Yes.

Let's also say you have a guitar signal that comes in at only half the
volume the input buffer wants to see. You are wasting half the bits right
there, squashing your dynamic range. Likewise, if you have buzzing or
humming noise, it masks the lower bits squashing your dynamic range again.

No; if your signal is only half the volume your input buffer wants to see,
you're down 6dB, and you're wasting one bit (the one of the far left). So if
you're working in a 16-bit system, you're effectively using 15 bits, for a
theoretical dynamic range of 90dB, rather than 96dB.

A compressor will take a volume level over a certain threshold and reduce
it. You can make up for this reduction by increasing gain, making the
louder
parts just as loud as before, and the softer parts louder. If done subtly,
you will just notice a smoother input.

Right, more or less.

A compressed signal has the POTENTIAL to increase dynamic range by
ensuring
that your maximum peak levels are not too much higher than the rest of
your
playing. Let's say that you hit the strings on a guitar as hard as you can
and it peaks at twice the normal playing volume. Again, you're wasting
bits
and dynamic range right there. So a compressor will get your guitar
operating at a higher average level, increasing the number of bits used
and
thus dynamic range.

Not really, because practically speaking the dynamic range of any electric
guitar signal I've come across is less than the dynamic range of a 16-bit
recording system, never mind the 24-bit systems that are the norm for
recording these days. (The limiting factor is usually the noise of the
amplifier plus the noise of the pickups.) So unless you're drastically over-
or under-recording the signal, you'll get the same effective dynamic
range -- and if you compress it, you'll get the compressed dynamic range,
which is of course less than the uncompressed.

In effect, your limiting factor, assuming reasonable recording practices, is
the source material, not the digital medium.

Here I will BS a bit because I don't know exactly what happens (I'm an
audio
enthusiast, not an EE major). A buffer will take a high impedance signal
and
output a low impedance signal. This signal is less susceptible to
high-frequency and volume loss, perhaps?

Less susceptible to high-frequency loss from load capacitance, at least, and
less susceptible to volume loss from loading.

Furthermore, impedance is not a flat number; it changes based on
frequency.
I'm imagining that by converting this impedance, you are also flattening
the
impedance of a guitars pickup, for instance. By doing so, certain
frequencies are not picked up louder because of different impedance, but
because of different output characteristics. In this way, you will not
have
spikes or dips in the impedance of a pickup, which I propose, decreases
the
dynamic range.

You're not flattening the impedance of the guitar's pickup; if you put a
buffer, say, inside the guitar, you're isolating the guitar's impedance from
the outside world (cables, amplifier input impedance, etc.). Most of the
time that will change the tone of the pickup -- for better or worse,
depending on your preferances. It won't change the dynamic range, though,
other than making it a tiny bit lower due to the additional noise of the
buffer.

In my opinion, a well-designed, hum- and noise-free, buffer with a hint of
compression (i.e. a Valvulator), can actually increase dynamic range.

Nope.

Am I understanding this correctly? Feel free to rip my comments to shreds.

No; you're making the initial assumption that the limit on dynamic range
when digitally recording or processing a guitar signal is the digital
recording process, whereas it's actually the guitar/amp system itself.

Peace,
Paul

The sample rate is the number of times per second that the signal is cut up.
The Nyquist Theorem states that the highest frequency reproduced is half of
the sample rate. This is too long and convoluted for me to understand or
explain. But so long as the sample rate is CD-quality (44.1 kHz), we'll have
a possible frequency of 22.05 kHz, which is out of normal human hearing and
definitely out of the range of electric guitar.

Close enough. Practically, the cut-off will be a bit under half the
sample rate.



The bit rate is the dynamic range of a unit. Let's say the input is 1 volt
and the converter has 1 bit. It can be either 0 or 1. So if the signal is
closer to 1 volt, it will be 1, and if it's closer to 0 volts, it will be 0.
Obviously, you'd have full signal then nothing. With 2 bits, you double that
dynamic range. Each additional bit doubles that range so you end up with 2^N
different volume levels for each bit rate N.

Again, close enough. Though the term is bit DEPTH. Bit rate means
something else



A compressed signal has the POTENTIAL to increase dynamic range by ensuring
that your maximum peak levels are not too much higher than the rest of your
playing. Let's say that you hit the strings on a guitar as hard as you can
and it peaks at twice the normal playing volume. Again, you're wasting bits
and dynamic range right there. So a compressor will get your guitar
operating at a higher average level, increasing the number of bits used and
thus dynamic range.

No. This is where you're going wrong. A compressor REDUCES dynamic
range. That's why it's called a compressor. Having squashed the
peaks out of the dynamics you there is then room to move the whole
signal up. This can make the music sound louder and punchier. But
you are actually using LESS dynamic range.

A compressed signal has the POTENTIAL to increase dynamic range by
ensuring that your maximum peak levels are not too much higher than the
rest of your playing. Let's say that you hit the strings on a guitar as hard
as you can and it peaks at twice the normal playing volume. Again, you're
wasting bits and dynamic range right there. So a compressor will get you
guitar operating at a higher average level, increasing the number of bits
used and thus dynamic range.

All other replies so far have been spot on. I would merely like to say
that the idea which you expressed in this paragraph would make a kind
of sense if the universe were very different from the way it actually
is--if the main source of noise in a 16-bit recording were the limit of
those 16 bits, and if the signal coming from your guitar were
relatively free of noise, hum, RFI, etc., and if all circuitry in the
path between your guitar and the compressor were similarly noise-free.

In that hypothetical case you would be lifting the guitar signal higher
(on average) above the noise floor of the A/D converter--which is what
I think you probably mean by "increas[ing] your dynamic range" here.

But in reality, as soon as you set your recording levels so that your
loudest sound approaches 0 dB (full scale), the noise floor of a 16-bit
recording is actually far below the noise floor of your guitar signal.
So using a limiter would actually worsen the situation, as the others
have pointed out. You would also be squashing the dynamic range of your
musical signal, which robs it of a great deal of its liveness and
expressive character.

--best regards