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

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