Hi there!
For a telephony application, I have to record speech with a target format
11025kHz 8bit mono. What can I do to achieve best possible quality?
I know that even the best possible quality will not compare to hi-fi,
but even with 11025khz you can produce "good" and "poor" quality.
At this moment, my equipment is a AKG C1000S microphone, a Mackie mixer
and a SoundBlaster USB to connect to my pc with SoundForge 7.
I tried to simply downsample a 44kHz recording, but its very noisy
and crackling.
I also tried a bit with noise filters and noise gates, without success.

Regards
Matthias

On Sun, 23 Jan 2005 19:40:59 +0100, "Matze N."
wrote:

Hi there!
For a telephony application, I have to record speech with a target format
11025kHz 8bit mono. What can I do to achieve best possible quality?
I know that even the best possible quality will not compare to hi-fi,
but even with 11025khz you can produce "good" and "poor" quality.
At this moment, my equipment is a AKG C1000S microphone, a Mackie mixer
and a SoundBlaster USB to connect to my pc with SoundForge 7.
I tried to simply downsample a 44kHz recording, but its very noisy
and crackling.
I also tried a bit with noise filters and noise gates, without success.

Regards
Matthias

I've recently done this for my answerphone message for my VoIP
Internet phone. I downsampled from 44.1 to 11.025 using Audition. It
worked perfectly, with just the expected top end loss, which I
mitigated with a little boost at 5kHz. I suspect a problem with your
PC setup. What happens if you record directly at 11.025 instead of
converting the sample rate?

d

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

Matze N. wrote:

Hi there!

Good to see you here. g

For a telephony application, I have to record speech with a target format
11025kHz 8bit mono. What can I do to achieve best possible quality?
I know that even the best possible quality will not compare to hi-fi,
but even with 11025khz you can produce "good" and "poor" quality.
At this moment, my equipment is a AKG C1000S microphone, a Mackie mixer
and a SoundBlaster USB to connect to my pc with SoundForge 7.

First things: the C1000 is a rather harsh sounding mic, but it's what
you have. Next, while the Mackie mixer has decent preamps, gain staging
is important, and with Mackies one must leave lots more headroom than
one might oridnarily do with other preamps/mixers. You can tap the
signal directly from the Mackie's preamp by inserting a patch cord only
to the first click of the insert point, and then routing that signal
directly to your soundcard. This bypasses the rest of ther mixer, giving
you a much more direct conncetion from source to destination.

It is possible that going all the way through the mixer and using its
high frequnecy EQ to roll-off some of what the C1000 is offering might
help, too. If you do go through the whole mixer keep the output levels
_below the yellow LED's_, nevermind the red ones. Seriously, this makes
a pretty stratling difference IME.

I tried to simply downsample a 44kHz recording, but its very noisy
and crackling.
I also tried a bit with noise filters and noise gates, without success.

Hmmmm, noisy and crackling... You're sure the mic is working properly?

--
ha

Am Sun, 23 Jan 2005 19:57:32 GMT hat hank alrich
geschrieben:

Matze N. wrote:

Hi there!

Good to see you here. g

g

I tried to simply downsample a 44kHz recording, but its very noisy
and crackling.
I also tried a bit with noise filters and noise gates, without success.

Hmmmm, noisy and crackling... You're sure the mic is working properly?

Maybe I am writing something different to what I mean... :-)
Recording at 44khz brings good quality, I have no problem here.
The problem is the conversion from 16bit to 8bit; this produces
much noise. (the whole wave is underlaid by a hissing sound)

Regards
Matthias

Matze N. wrote:
Recording at 44khz brings good quality, I have no problem here.
The problem is the conversion from 16bit to 8bit; this produces
much noise. (the whole wave is underlaid by a hissing sound)

Any 8 Bit signal has ~ 48 dB SNR at most anyway.

However, you can try to use a good dithering algorithm to shift the
noise in less-audible regions. SoundForge has a couple of choices.

Johann
--
P.S. Seine Seite ist genauso beschissen, wie sein Muell den
er ins Usenet kuebelt.
(*Tönnes über mich in )

Matze N. wrote:
Maybe I am writing something different to what I mean... :-)
Recording at 44khz brings good quality, I have no problem here.
The problem is the conversion from 16bit to 8bit; this produces
much noise. (the whole wave is underlaid by a hissing sound)

I would compress the SNOT out of the signal, then compress it again.
Then normalize.

A wave that peaks at -6 dB from absolute max in an 8-bit world is
effetively a 6-bit sample! One that peaks at -12 dB is a 4-bit
sample. The same applies to short pieces of a longer sample, in
a sense. Any time you are not near the peak, you are (for lack
of a better way to put it) running dangerously low on bits.

Because you have so few bits to work with, even cranking the signal
so high it clips a little bit would be better than leaving it too
low (in contrast to how it works in the 16-bit world).

Anyway, once you've compressed it, then downsample and convert down
to 8-bit.

Depending on the program, you may also want to apply a noise gate
as the first step. If the quiet passages are just microphone clunking
or other non-meaningful sounds, it would be much better to have
all bits zero during these quiet passages, because quiet passages
are what will sound the worst in the 8-bit world. If a noise gate
doesn't work out well, consider going through with a sound editor
and manually replacing passages that should be silent with absolute
silence.

In effect, what I'm saying is that you want to have either loud
or silent, with nothing in between. :-)

You may want to experiment with whatever different types of dithering
you have available to you. That's going to be especially critical.

- Logan

Chel van Gennip wrote:

On Sun, 23 Jan 2005 21:53:22 +0100, Matze N. wrote:


Maybe I am writing something different to what I mean... :-) Recording
at 44khz brings good quality, I have no problem here. The problem is the
conversion from 16bit to 8bit; this produces much noise. (the whole wave
is underlaid by a hissing sound)


For a telephony application start working on the 44khz signal. Add an
adjustable software bandfilter (200-3000Hz?) then do AGC tho get a high
as possible level, then do 44kHz to 11kHz conversion, and then do a
propper 16bit to 8 bit conversion. You could try some alineairity in the
16 to 8 conversion.
Play a bit with the bandfilter and agc settings to get the best result.


That ^ was exactly what you want to do.

Set your bandpass filter as steep as you can for 300-3800 Hz (or 3600 if
there's a possibility it will get sent to a modem running at 7200 Hz).
You can go higher if you want, but it won't do you much good, and there
seem to be things in the Telco cloud that can have problems if you do.

You can compress it as much as you want; on the telephone that's a big
advantage. Female voice helps, too, if you have a choice.

Anyway, the horribly low Fs and bit depths sound ridiculous, but with a
little care they sound really good, better than just talking into the
phone does.

"Matze N." wrote in message
news
Hi there!
For a telephony application, I have to record speech with a target format
11025kHz 8bit mono. What can I do to achieve best possible quality?
I know that even the best possible quality will not compare to hi-fi,
but even with 11025khz you can produce "good" and "poor" quality.
At this moment, my equipment is a AKG C1000S microphone, a Mackie mixer
and a SoundBlaster USB to connect to my pc with SoundForge 7.
I tried to simply downsample a 44kHz recording, but its very noisy
and crackling.
I also tried a bit with noise filters and noise gates, without success.

First get a good microphone, which the C1000 isn't. Next, don't normalize
the signal to 0dBFS; if you must normalize it, normalize to -2dBFS or so.
Try running a lowpass filter on the audio at about 12kHz BEFORE you
downsample to 11025. And make sure your software is set to dither the signal
when it does the downsampling/bit reduction.

Peace,
Paul

Paul Stamler wrote:
First get a good microphone, which the C1000 isn't. Next, don't normalize
the signal to 0dBFS; if you must normalize it, normalize to -2dBFS or so.

I'm curious why not. When you've only got 8 bits to work with,
you don't want to throw away 2/3 of one of your bits. With 8 bit
audio, bits are scarce, so you want to throw them away as little
as possible.

- Logan

Logan, your overall idea is largely right but your math is off by
double, so there's more of a "bit budget" available than you seem to
realize. Backing off 2 dB as Paul suggests won't even cost one-half
bit's "resolution". One bit represents ca. 6 dB dynamic range /
signal-to-noise ratio, not 3 dB as you seem to assume. When you wrote,
"A wave that peaks at -6 dB from absolute max in an 8-bit world is
effetively a 6-bit sample! One that peaks at -12 dB is a 4-bit sample"
the more nearly correct statement would have been, "A wave that peaks
at -6 dB from absolute max in an 8-bit world is effectively a 7-bit
sample! One that peaks at -12 dB is a 6-bit sample."

As a somewhat separate matter, when you wrote, "The same applies to
short pieces of a longer sample, in a sense. Any time you are not near
the peak, you are (for lack of a better way to put it) running
dangerously low on bits" that was somewhere between a truth and a
half-truth. With dithered, linear PCM the level of the noise floor
stays effectively constant regardless of the signal level. If that
noise floor is low enough, no problem; if it isn't, ... problem. It's
that simple whether we're talking about digital or analog recording.

The fallacy is in thinking too much about the behavior of individual
samples when trying to imagine the audible dynamic range of a digital
channel. The output of a DAC isn't stair-stepped, our ears don't know
when each individual sample is being presented by a DAC, and we don't
(can't) adjust to the level of each individual sample. We end up
hearing continuous sound with a certain amount of noise in it, just as
with analog recordings.

How we humans perceive signal levels and noise levels is a complex
topic, but it definitely involves a process of averaging over longer
periods of time than one sampling interval. So please consider the
donut, not just the hole.

--best regards

"Paul Stamler" wrote in message



First get a good microphone, which the C1000 isn't.

(1) Rule: Resolution reduction works best when you start out with the
cleanest possible signal. Upstream cleanliness helps.

I just got whapped in the face by this rule when I first implemented my new
DVD recorder Friday night. I was having problems with severe periodic
artifacts at the bit rate I wanted to use (6 hours). I cleaned up the
inbound cable signal by working over the wiring and putting in an amplified
splitter. Bingo, the nasty artifacts were gone!

Next, don't normalize the signal to 0dBFS; if you must normalize it,
normalize to
-2dBFS or so.

(2) Rule: Resolution reduction works best when you avoid problems
downstream. Lots of cheap low-rez gear isn't so clean over the last few dB
below FS.

Try running a lowpass filter on the audio at about 12kHz BEFORE you
downsample to 11025.

See rule (1). Only I'd pick a lower frequency - like 4.5-5.2 KHz. Or skip
worrying about this if your downsampling facility is first rate. It will
effectively do it for you.

And make sure your software is set to dither the signal when it does the
downsampling/bit reduction.

This time we're addressing the resolution reduction step itself. If you're
going to 8 bit there is no chance of zero audible noise. The only quesiton
is how relatively nice that noise will be to listen to. Carefully shaping
the dither's spectrum can pay off big time. Of course you're going to use
TPDF dither.

Good dithering used to cost big bucks, but not any longer. It is now just a
matter of making careful choices.

"Logan Shaw" wrote in message
news
Paul Stamler wrote:
First get a good microphone, which the C1000 isn't. Next, don't
normalize
the signal to 0dBFS; if you must normalize it, normalize to -2dBFS or
so.

I'm curious why not. When you've only got 8 bits to work with,
you don't want to throw away 2/3 of one of your bits. With 8 bit
audio, bits are scarce, so you want to throw them away as little
as possible.

Sorry, I quit writing too soon. The reason is to avoid generating clipping
problems during the resampling-bit reduction process. The difference between
a 48dB s/n and a 46db one isn't acute (stink is stink), but clipping is, and
one possibility for the crunchy sound the OP was getting was that the
conversion process was generating overshoots.

Peace,
Paul

David Satz wrote:

Logan, your overall idea is largely right but your math is off by
double, so there's more of a "bit budget" available than you seem to
realize. Backing off 2 dB as Paul suggests won't even cost one-half
bit's "resolution". One bit represents ca. 6 dB dynamic range /
signal-to-noise ratio, not 3 dB as you seem to assume.

I guess I'm still not clear, even though I thought I was sorta
clear, on the math. I know that 3 dB equals 10^(3/10) which is
about equal to 2. Since adding a bit doubles the range of numbers
you can represent, it *seems* like 3 dB should basically equal
one bit, but I'm willing to accept that it doesn't (especially
since it means a 16-bit signal would only have 48 dB SNR, which
I know is wrong). But I still get mixed up on that kind of thing
and don't understand why it's 6 dB per bit.

As a somewhat separate matter, when you wrote, "The same applies to
short pieces of a longer sample, in a sense. Any time you are not near
the peak, you are (for lack of a better way to put it) running
dangerously low on bits" that was somewhere between a truth and a
half-truth. With dithered, linear PCM the level of the noise floor
stays effectively constant regardless of the signal level.

Well, yes, the noise level stays constant. And if 6 dB is one bit,
it's not egregiously bad. BUT, if the signal never climbs up near
the max, then the signal-to-noise ratio suffers by virtue of
the signal being "too small". Because, of course, it's the relative
level of the two that matters.

The fallacy is in thinking too much about the behavior of individual
samples when trying to imagine the audible dynamic range of a digital
channel. The output of a DAC isn't stair-stepped, our ears don't know
when each individual sample is being presented by a DAC, and we don't
(can't) adjust to the level of each individual sample.

Basically all I was trying to say is that the ear adjusts to different
levels. Loud stuff will mask noise better than quiet stuff. If
there are 5 seconds of loud followed by 25 seconds of stuff that's
10 dB quieter, then for that 25 seconds the SNR is worse, and the
ear will adjust and be able to hear it. I wasn't talking about
down at the level of individual wavelengths, I was talking about
the larger picture where the signal varies over periods of several
seconds.

- Logan

Logan Shaw wrote:


I guess I'm still not clear, even though I thought I was sorta
clear, on the math. I know that 3 dB equals 10^(3/10) which is
about equal to 2. Since adding a bit doubles the range of numbers
you can represent, it *seems* like 3 dB should basically equal
one bit, but I'm willing to accept that it doesn't (especially
since it means a 16-bit signal would only have 48 dB SNR, which
I know is wrong). But I still get mixed up on that kind of thing
and don't understand why it's 6 dB per bit.



Bits represent the voltage of the waveform. Your dB formula is the
power formula, you need to use the voltage formula which gives 2x the
result, since power goes up with the square of the voltage (aebe - all
else being equal).


dB (V) = 20 * log10 (V2/V1)

dB (S) = 20 * log10 (S2/S1) where S is the numeric sample value.


you're using:

dB (P) = 10 * log10 (P2/P1)


It could have been done differently, but it wasn't.

Surely you don't mean 11025 kHz. You mean 11025 Hz.

--


Roger W. Norman
SirMusic Studio

"Matze N." wrote in message
news
Hi there!
For a telephony application, I have to record speech with a target format
11025kHz 8bit mono. What can I do to achieve best possible quality?
I know that even the best possible quality will not compare to hi-fi,
but even with 11025khz you can produce "good" and "poor" quality.
At this moment, my equipment is a AKG C1000S microphone, a Mackie mixer
and a SoundBlaster USB to connect to my pc with SoundForge 7.
I tried to simply downsample a 44kHz recording, but its very noisy
and crackling.
I also tried a bit with noise filters and noise gates, without success.

Regards
Matthias

Logan Shaw wrote:

Matze N. wrote:

Maybe I am writing something different to what I mean... :-)
Recording at 44khz brings good quality, I have no problem here.
The problem is the conversion from 16bit to 8bit; this produces
much noise. (the whole wave is underlaid by a hissing sound)


I would compress the SNOT out of the signal, then compress it again.
Then normalize.

A wave that peaks at -6 dB from absolute max in an 8-bit world is
effetively a 6-bit sample! One that peaks at -12 dB is a 4-bit
sample. The same applies to short pieces of a longer sample, in
a sense. Any time you are not near the peak, you are (for lack
of a better way to put it) running dangerously low on bits.

Hmmm.. since -6dBFS is half of the amplitude of 0dBFS, isn't that 7-bit
when 0dBFS is 8-bit. And -12dBFS would be 6-bit? Am I missing something?

-jp

jp wrote:

Logan Shaw wrote:

Matze N. wrote:

Maybe I am writing something different to what I mean... :-)
Recording at 44khz brings good quality, I have no problem here.
The problem is the conversion from 16bit to 8bit; this produces
much noise. (the whole wave is underlaid by a hissing sound)



I would compress the SNOT out of the signal, then compress it again.
Then normalize.

A wave that peaks at -6 dB from absolute max in an 8-bit world is
effetively a 6-bit sample! One that peaks at -12 dB is a 4-bit
sample. The same applies to short pieces of a longer sample, in
a sense. Any time you are not near the peak, you are (for lack
of a better way to put it) running dangerously low on bits.


Hmmm.. since -6dBFS is half of the amplitude of 0dBFS, isn't that 7-bit
when 0dBFS is 8-bit. And -12dBFS would be 6-bit? Am I missing something?

-jp


Ok, this was covered later in this thread ..

-jp

S O'Neill wrote:
Logan Shaw wrote:

I guess I'm still not clear, even though I thought I was sorta
clear, on the math. I know that 3 dB equals 10^(3/10) which is
about equal to 2. Since adding a bit doubles the range of numbers
you can represent, it *seems* like 3 dB should basically equal
one bit

Bits represent the voltage of the waveform. Your dB formula is the
power formula, you need to use the voltage formula which gives 2x the
result, since power goes up with the square of the voltage (aebe - all
else being equal).

dB (V) = 20 * log10 (V2/V1)

OK, it makes sense that power is the square of voltage. And it
makes sense that what you're sampling is voltage.

But in some sense, once you have the samples, you have samples,
not voltage. Voltage is just an intermediate step in the process
of creating the samples. In theory you could, for example,
create samples purely optically or something. (Attach a curved
mirror to the back of speaker cone and point a laser at it or
something.)

I suppose, though, even if you did, the samples would still need
to go into a reproduction system that produced sound whose power
was proportional to the square of the magnitude of the sample, so
maybe that's a moot point.

So, I guess where I was getting mixed up is that I was considering
decibels just to be a handy notation for expressing something on
a logarithmic scale and neglecting the context of where the sample
values came from. Which makes a strange kind of sense, since I'm
more comfortable personally with software (bits) than hardware
(voltage). :-)

- Logan

In article writes:

OK, it makes sense that power is the square of voltage. And it
makes sense that what you're sampling is voltage.

But in some sense, once you have the samples, you have samples,
not voltage. Voltage is just an intermediate step in the process
of creating the samples.

Conversely, samples are just there to create a voltage. But it doesn't
matter what voltage corresponds to a sample of 1 bit in amplitde. It
can be 1 millivolt, 1 volt, even 100 volts. But in the same system,
when the amplitude increases to two bits, the voltage will be 2
millivolts, 2 volts, or 200 volts. 200V/100V = 6 dB. 2 mV/1 mV = 6 dB.

That's my 2 cents worth, which is 6 dB more than 1 cent's worth.



--
I'm really Mike Rivers - )
However, until the spam goes away or Hell freezes over,
lots of IP addresses are blocked from this system. If
you e-mail me and it bounces, use your secret decoder ring
and reach me he double-m-eleven-double-zero at yahoo

On 25 Jan 2005 10:45:30 -0500, (Mike Rivers)
wrote:

That's my 2 cents worth, which is 6 dB more than 1 cent's worth.

I thought that money was power. Maybe things aren't so bad
after all.

Chris Hornbeck
"If that is git only stucco and Slotermeyer? Yes! Celebration dog that
or the Flipperwaldt gersput!" -the deadly WWII joke from Monty Python

I agree with you; if I was doing it, the numbers would represent power,
but using voltage they map linearly to the A/D and D/A chips.



Logan Shaw wrote:

S O'Neill wrote:

Logan Shaw wrote:


I guess I'm still not clear, even though I thought I was sorta
clear, on the math. I know that 3 dB equals 10^(3/10) which is
about equal to 2. Since adding a bit doubles the range of numbers
you can represent, it *seems* like 3 dB should basically equal
one bit


Bits represent the voltage of the waveform. Your dB formula is the
power formula, you need to use the voltage formula which gives 2x the
result, since power goes up with the square of the voltage (aebe - all
else being equal).


dB (V) = 20 * log10 (V2/V1)


OK, it makes sense that power is the square of voltage. And it
makes sense that what you're sampling is voltage.

But in some sense, once you have the samples, you have samples,
not voltage. Voltage is just an intermediate step in the process
of creating the samples. In theory you could, for example,
create samples purely optically or something. (Attach a curved
mirror to the back of speaker cone and point a laser at it or
something.)

I suppose, though, even if you did, the samples would still need
to go into a reproduction system that produced sound whose power
was proportional to the square of the magnitude of the sample, so
maybe that's a moot point.

So, I guess where I was getting mixed up is that I was considering
decibels just to be a handy notation for expressing something on
a logarithmic scale and neglecting the context of where the sample
values came from. Which makes a strange kind of sense, since I'm
more comfortable personally with software (bits) than hardware
(voltage). :-)

- Logan

On Tue, 25 Jan 2005 08:37:58 -0800, S O'Neill
wrote:

I agree with you; if I was doing it, the numbers would represent power,
but using voltage they map linearly to the A/D and D/A chips.

The numbers are the same for voltage and power; that's the
reason it's done this way. n dB change in voltage equals
n dB change in power.

Chris Hornbeck
"If that is git only stucco and Slotermeyer? Yes! Celebration dog that
or the Flipperwaldt gersput!" -the deadly WWII joke from Monty Python

Chris Hornbeck wrote:

On Tue, 25 Jan 2005 08:37:58 -0800, S O'Neill
wrote:


I agree with you; if I was doing it, the numbers would represent power,
but using voltage they map linearly to the A/D and D/A chips.


The numbers are the same for voltage and power; that's the
reason it's done this way. n dB change in voltage equals
n dB change in power.


Right, but the question is, which does a sample represent? and the
answer is voltage, so you use the voltage formula for dB.

Chris Hornbeck wrote:

On Tue, 25 Jan 2005 08:37:58 -0800, S O'Neill
wrote:


I agree with you; if I was doing it, the numbers would represent power,
but using voltage they map linearly to the A/D and D/A chips.


The numbers are the same for voltage and power; that's the
reason it's done this way. n dB change in voltage equals
n dB change in power.

OK, now I'm *REALLY* confused. I thought we just established
that n dB change in voltage equals 2*n dB change in power!

- Logan

On Tue, 25 Jan 2005 17:54:33 GMT, Logan Shaw
wrote:

The numbers are the same for voltage and power; that's the
reason it's done this way. n dB change in voltage equals
n dB change in power.

OK, now I'm *REALLY* confused. I thought we just established
that n dB change in voltage equals 2*n dB change in power!

Here's one way to think about it: the Bel is the log (base 10) of
the power ratio. Period.

All other corruptions, usages, and conventions must agree
with the above. We've found the idea of the Bel so useful
that there are a heckuva lot of extensions, often shortcutted
or otherwise abused.

Now: ten times the voltage is 100 times the power. Both are
a change of +20 dB. (at Standard Temperature and Pressure,
Greenwich Mean Time, measured in furlongs per fortnight).

Chris Hornbeck
"If that is git only stucco and Slotermeyer? Yes! Celebration dog that
or the Flipperwaldt gersput!" -the deadly WWII joke from Monty Python

Chris Hornbeck wrote:

On Tue, 25 Jan 2005 17:54:33 GMT, Logan Shaw
wrote:

The numbers are the same for voltage and power; that's the
reason it's done this way. n dB change in voltage equals
n dB change in power.

OK, now I'm *REALLY* confused. I thought we just established
that n dB change in voltage equals 2*n dB change in power!


Here's one way to think about it: the Bel is the log (base 10) of
the power ratio. Period.

All other corruptions, usages, and conventions must agree
with the above. We've found the idea of the Bel so useful
that there are a heckuva lot of extensions, often shortcutted
or otherwise abused.

OK, I gotcha. I just didn't get what you mean by "the numbers
are the same". I now think you meant something like that with
a given signal, the dB change between two levels is the same
no matter whether the quantity being discussed is power or
voltage of (sums of) bits. Whereas before, I had thought by
"n dB change in voltage equals n dB change in power" you meant
something totally different, like that log(voltage_change) equals
log(power_change). But instead you meant something like that we
must always go back to the root meaning when saying "dB".
(Unless you're trying to decrypt someone using it improperly...)

- Logan

"Matze N." wrote in newspsk2pm8dd0rk0i1
@deepblue.intranet.dcmgmbh.de:

Maybe I am writing something different to what I mean... :-)
Recording at 44khz brings good quality, I have no problem here.
The problem is the conversion from 16bit to 8bit; this produces
much noise. (the whole wave is underlaid by a hissing sound)

I use Audition for sample rate conversions all the time with no ill effect
beyond the limitations of the new sample size.

Is it recorded full volume with peaks near the maximum level of the
recording? Removing 8 bits from your sample raises the theoretical noise
floor by 48 dB.

Are you still monitoring through the Soundblaster? Perhaps its resampling
before converting to analog and creating part of your noise.

Matze N. wrote:
Hi there!
For a telephony application, I have to record speech with a target format
11025kHz 8bit mono. What can I do to achieve best possible quality?

Just out of curiosity, does it absolutely have to be 8 bit LINEAR? Most
telephony applications (and codecs used for telephony) support A-law and
ulaw compression. These compression formats map the 8 bits logarithmically,
rather than linearly. They provide a lot more precision at low levels, at
the expense of precision at the higher levels.

If it does have to be linear, try a variety of noise shaped dithers. Noise
shaping isn't as effective as it is with higher sample rates, because at
11025 Hz sampling, there is no range that is inaudible. However, the
character of the audible noise can be modified to be less annoying.

John Marvin