audio recording on IC -help wanted

[Green Xenon [Radium]]

On Sep 29, 8:59 pm, dpl...@radagast.org (Dave Platt) wrote in
http://groups.google.com/group/uk.tech.digital-tv/msg/872fdf9b0c765bd0 :

In article <46fefb70$0$7493$4c368...@roadrunner.com>,
Green Xenon [Radium] <gluceg...@excite.com> wrote:

Sure it is possible. It won't be of any assistance to you (or to
anyone else since you can't put 8 pounds in a 1-pound sack.
Bandwidth is bandwidth, no matter how you slice it. Do you
really think that nobody has thought of various methods of
transmitting video over low-bandwidth paths? If so, you need
to study not only science and engineering but the HISTORY
of science and engineering.

There are 8-bits-per-symbol, only 1 baud. The symbol is split into 8
parts. Each part is 1-bit.

How is this “putting 8 pounds in a 1-pound sack?”

It’s more like “putting 1 pound in a 1-pound sack”

You are using the word "split" without defining what you are meaning,
and I think you may be using the word "bit" in two different ways (as
in "a single binary digit's worth of information" and "a part")...
maybe I'm wrong about the latter.

Sure, you can carry 8 bits of information per symbol, at a rate of one
baud. That's an information rate of 8 bits per second, which isn't a
terribly useful amount of information.

What I *think* you are saying is something along the lines of

"If there's no way to carry N bits of information per symbol (where
N is a large number) at a rate of 1 symbol per second, in a
specified bandwidth, over a channel having a certain amount of
dynamic range / signal-to-noise ratio available, then I'd
like to somehow divide each symbol into eight sub-symbols which
each carry N/8 of the information, but which still make up only one
symbol per second. My hope is that these eight sub-symbols could
be easier to transmit, somehow, than a single symbol carrying N
bits, and that I could thus transmit them more easily but without
using any more bandwidth. Can I do that?"

If that's what you're asking, the answer is "no".

Once you hit the theoretical information limit of the communication
channel (which is set by the bandwidth you use and by the amount of
noise on the channel) you can't do any better than that. No matter
how you modulate (changing the baud rate, the amount of information
per symbol, etc.) you can't do better than this.

And, as we've been trying to make clear, your goal is quite far beyond
the theoretical limit. You just can't get there. You're trying to
put all of Lake Erie in a water-glass.

Let it go, Radium.

Within physical-possibility, what is the largest amount of
bits-per-symbol [assuming a baud-rate of only 1-bit-per-symbol] that can
be reached without the highest-voltage causing any clipping, generating
any temperatures above 70 Fahrenheit, resulting in any harm to
anyone/anything [including the equipment itself], or shortening the life
of the equipment and without the lowest-voltage being lost in the noise?
What is the maximum-possible amount of discrete levels between the
highest and lowest voltage in such a signal?

An 8-bit signal can have a maximum of 256 different voltage levels
between the highest and lowest voltage. Right? Go too high and the
signal clips, go too low and the signal will not be recognized.

 

[Roderick Stewart]

In article <46ff32e0$0$32538$4c368faf@roadrunner.com>, Green Xenon [Radium]
wrote:

And, as we've been trying to make clear, your goal is quite far beyond
 > the theoretical limit.  You just can't get there.  You're trying to
 > put all of Lake Erie in a water-glass.
 
 > Let it go, Radium.

Within physical-possibility, what is the largest amount of 
bits-per-symbol [assuming a baud-rate of only 1-bit-per-symbol] that can 
be reached without the highest-voltage causing any clipping, generating 
any temperatures above 70 Fahrenheit, resulting in any harm to 
anyone/anything [including the equipment itself], or shortening the life 
of the equipment and without the lowest-voltage being lost in the noise? 
What is the maximum-possible amount of discrete levels between the 
highest and lowest voltage in such a signal?

An 8-bit signal can have a maximum of 256 different voltage levels 
between the highest and lowest voltage. Right? Go too high and the 
signal clips, go too low and the signal will not be recognized.

I tell you what. Why don't you build one of these magic modulators, show us
that it works, and then we'll believe you.

If it does, I might be encouraged to have a go at making that machine I
invented at the age of about seven, even though my dad told me it wouldn't
work. My plan was to use an electric motor to turn a dynamo, which would
generate electricity, which would then be used to power the electric motor.
After it had been patiently explained to me that the amount of electricity
generated would not be enough to power the motor, my next idea was to
interpose some gears or pulleys so that the motor would turn the generator
a little faster, thereby generating more electricity. My understanding of
the physical world had some way to go, but we all have to start somewhere.

Rod.

 

[Dave Platt]

In article <46ff32e0$0$32538$4c368faf@roadrunner.com>,
Green Xenon [Radium] <glucegen1@excite.com> wrote:

Within physical-possibility, what is the largest amount of
bits-per-symbol [assuming a baud-rate of only 1-bit-per-symbol]

You just made a meaningless statement. Baud rate is measured in
symbols per second, not bits per symbol. I assume that you meant to
say "assuming a baud-rate of one symbol per second."

that can
be reached without the highest-voltage causing any clipping, generating
any temperatures above 70 Fahrenheit, resulting in any harm to
anyone/anything [including the equipment itself], or shortening the life
of the equipment and without the lowest-voltage being lost in the noise?
What is the maximum-possible amount of discrete levels between the
highest and lowest voltage in such a signal?

And, another important constraint for actual usability is this: you
have to make sure that the minimum-detectable difference between two
different symbol levels can be detected reliably and accurately at
_all_ levels. In the language of analog-to-digital converts this is
referred to as "linearity" and "no missing codes".

[As a counterexample: the human hearing system can hear down to 0 dBa
or so... a sound level which is just barely above the noise created by
the random collision of air molecules with the eardrum. And, we can
hear sounds up to around 120 decibels above that, before damage starts
to result. That's a pretty wide dynamic range. However, it's not
linear... if we're listening to a loud sound (say, at 110 dB or so),
small sounds are completely lost... you can't hear somebody whispering
10 feet away when you're listening to a rock concert.]

An 8-bit signal can have a maximum of 256 different voltage levels
between the highest and lowest voltage. Right? Go too high and the
signal clips, go too low and the signal will not be recognized.

Right.

At audio-quality sampling rates (say, 50,000 baud) you can buy
converters that are linear down to around 22 bits, I think...
marketers call them "24-bit" converters but they aren't actually
linear down to those levels. At instrumentation rates like 1 baud (1
sample per second), with filtering and averaging being applied to
eliminate the noise, you can do rather better.

I don't know quite what the state-of-the-art is for measuring signals
at such low baud rates as you are referring to. I'd guess that it's
somewhere in the range of 28-30 bits.

30 bits is roughly a billion-to-one ratio between the smallest signal
and the largest. Crudely put, it would mean that you might have a
circuit which has to handle voltages of up to 1000 volts, and has to
be able to generate, and then measure voltage differences of a
*millionth* of a volt, at all of these levels. That's going to be
technologically difficult, to say the least.

The *theoretical* limit is somewhat higher than this, but not enough
to help you achieve what you wish. It'll be limited at the low end by
the thermal noise level (a 50-ohm resistance at room temperature
generates -174 dBm of noise over a 1 Hz bandwidth) and at the high end
by whatever voltage you think your equipment can handle without
damage.

Even being extremely generous, and saying 32 bits of linear resolution
(and thus reliable data) per symbol, you aren't going to get video
across it. 32 bits per second is somewhere between "fast Morse code"
and "old Teletype teleprinter or ticker-tape" bandwidth.

Note that this is the generation and measurement limit and assumes an
interference-free communication link (e.g. a well-shielded cable), and
is *not* what you can get away with in a real-world radio
transmission! The background noise level on LF radio frequencies is
higher than this, due to both manmade and atmospheric electrical
noise.

As another poster has pointed out, there's a damned good reason why
nobody uses very-low-baud-rate modulations to send large amounts of
high-speed data over a narrow-bandwidth channel, despite a century or
more of research and study and competition in the fields of radio and
electronic communication. It just doesn't work, and the reasons why
it doesn't are well understood by those who practice in the field.

--
Dave Platt <dplatt@radagast.org> AE6EO
Friends of Jade Warrior home page: http://www.radagast.org/jade-warrior
I do _not_ wish to receive unsolicited commercial email, and I will
boycott any company which has the gall to send me such ads!

 

[Green Xenon [Radium]]

On Sep 30, 10:42 am, dpl...@radagast.org (Dave Platt) wrote in
http://groups.google.com/group/uk.tech.digital-tv/msg/82ff55b36ba97dbf :

In article <46ff32e0$0$32538$4c368...@roadrunner.com>,
Green Xenon [Radium] <gluceg...@excite.com> wrote:

Within physical-possibility, what is the largest amount of
bits-per-symbol [assuming a baud-rate of only 1-bit-per-symbol]

You just made a meaningless statement. Baud rate is measured in
symbols per second, not bits per symbol. I assume that you meant to
say "assuming a baud-rate of one symbol per second."

Yes. I meant "assuming a baud-rate of one symbol per second."

Sorry.

F-------king typos!!!!!!!!!!!!!!!!!!!!!!!!!!

that can
be reached without the highest-voltage causing any clipping, generating
any temperatures above 70 Fahrenheit, resulting in any harm to
anyone/anything [including the equipment itself], or shortening the life
of the equipment and without the lowest-voltage being lost in the noise?
What is the maximum-possible amount of discrete levels between the
highest and lowest voltage in such a signal?

And, another important constraint for actual usability is this: you
have to make sure that the minimum-detectable difference between two
different symbol levels can be detected reliably and accurately at
_all_ levels. In the language of analog-to-digital converts this is
referred to as "linearity" and "no missing codes".

[As a counterexample: the human hearing system can hear down to 0 dBa
or so... a sound level which is just barely above the noise created by
the random collision of air molecules with the eardrum. And, we can
hear sounds up to around 120 decibels above that, before damage starts
to result. That's a pretty wide dynamic range. However, it's not
linear... if we're listening to a loud sound (say, at 110 dB or so),
small sounds are completely lost... you can't hear somebody whispering
10 feet away when you're listening to a rock concert.]

An 8-bit signal can have a maximum of 256 different voltage levels
between the highest and lowest voltage. Right? Go too high and the
signal clips, go too low and the signal will not be recognized.

Right.

At audio-quality sampling rates (say, 50,000 baud) you can buy
converters that are linear down to around 22 bits, I think...
marketers call them "24-bit" converters but they aren't actually
linear down to those levels. At instrumentation rates like 1 baud (1
sample per second), with filtering and averaging being applied to
eliminate the noise, you can do rather better.

I don't know quite what the state-of-the-art is for measuring signals
at such low baud rates as you are referring to. I'd guess that it's
somewhere in the range of 28-30 bits.

30 bits is roughly a billion-to-one ratio between the smallest signal
and the largest. Crudely put, it would mean that you might have a
circuit which has to handle voltages of up to 1000 volts, and has to
be able to generate, and then measure voltage differences of a
*millionth* of a volt, at all of these levels. That's going to be
technologically difficult, to say the least.

The *theoretical* limit is somewhat higher than this, but not enough
to help you achieve what you wish. It'll be limited at the low end by
the thermal noise level (a 50-ohm resistance at room temperature
generates -174 dBm of noise over a 1 Hz bandwidth) and at the high end
by whatever voltage you think your equipment can handle without
damage.

Even being extremely generous, and saying 32 bits of linear resolution
(and thus reliable data) per symbol, you aren't going to get video
across it. 32 bits per second is somewhere between "fast Morse code"
and "old Teletype teleprinter or ticker-tape" bandwidth.

Note that this is the generation and measurement limit and assumes an
interference-free communication link (e.g. a well-shielded cable), and
is *not* what you can get away with in a real-world radio
transmission! The background noise level on LF radio frequencies is
higher than this, due to both manmade and atmospheric electrical
noise.

As another poster has pointed out, there's a damned good reason why
nobody uses very-low-baud-rate modulations to send large amounts of
high-speed data over a narrow-bandwidth channel, despite a century or
more of research and study and competition in the fields of radio and
electronic communication. It just doesn't work, and the reasons why
it doesn't are well understood by those who practice in the field.

If 400-nanometer-wavelength coherent light was used in place of electric
signals, would QM in this system be able to pack in more
bits-per-symbol. AFAIK, optical signals can have a greater dynamic range
than electric signals. That is one reason that optic fibers are
replacing coaxial cables. Less noise.

 

[Dave Platt]

In article <46ffea44$0$24264$4c368faf@roadrunner.com>,
Green Xenon [Radium] <glucegen1@excite.com> wrote:

If 400-nanometer-wavelength coherent light was used in place of electric
signals, would QM in this system be able to pack in more
bits-per-symbol. AFAIK, optical signals can have a greater dynamic range
than electric signals. That is one reason that optic fibers are
replacing coaxial cables. Less noise.

*sigh*

I give up.

--
Dave Platt <dplatt@radagast.org> AE6EO
Friends of Jade Warrior home page: http://www.radagast.org/jade-warrior
I do _not_ wish to receive unsolicited commercial email, and I will
boycott any company which has the gall to send me such ads!

 

[Richard Crowley]

"Green Xenon [Radium]" wrote ...

"Richard Crowley" wrote :
"Green Xenon [Radium]" wrote ...
Anyways, my next question. Is it possible to split the symbol of a
1
baud, 8-bit-per-symbol signal
into 8 bits? I.e. in this case, a single
symbol would be split into 8 different
parts each carrying one of the
bits. Is this possible? If so, would
this be of any assistance to me?

Sure it is possible. It won't be of any assistance to you (or to
anyone else since you can't put 8 pounds in a 1-pound sack.
Bandwidth is bandwidth, no matter how you slice it. Do you
really think that nobody has thought of various methods of
transmitting video over low-bandwidth paths? If so, you need
to study not only science and engineering but the HISTORY
of science and engineering.

There are 8-bits-per-symbol, only 1 baud. The symbol is split into 8
parts. Each part is 1-bit.

How is this “putting 8 pounds in a 1-pound sack?”

It’s more like “putting 1 pound in a 1-pound sack”

Which weighs more: one 8-pound sack, or eight 1-pound sacks?
Yes, it is a trick question, but you are the master of trick questions.

 

[Bob Myers]

"Richard Crowley" <rcrowley@xp7rt.net> wrote in message
news:13g11vn3e4ksc08@corp.supernews.com...

Which weighs more: one 8-pound sack, or eight 1-pound sacks?
Yes, it is a trick question, but you are the master of trick questions.

Or we might ask our boy Radium this one:

Which weighs more, a pound of feathers or a pound of
gold? ;-)


Bob M.

 

[John Williamson]

Bob Myers wrote:

"Richard Crowley" <rcrowley@xp7rt.net> wrote in message
news:13g11vn3e4ksc08@corp.supernews.com...
Which weighs more: one 8-pound sack, or eight 1-pound sacks?
Yes, it is a trick question, but you are the master of trick questions.

Or we might ask our boy Radium this one:

Which weighs more, a pound of feathers or a pound of
gold? ;-)


Errm...

1 troy ounce (Gold) = 1.09714 imperial ounces (Feathers)
12 Troy ounces = 1 Troy pound (Gold)
16 ounces = 1 pound (Feathers)

1 (troy) pound Gold = 0.822857 pounds (Imperial) feathers

Sorry to be picky...;-)

--
Tciao for Now!

John.

 

[Don Bowey]

On 10/1/07 2:18 PM, in article fdrnc7$mi8$1@stable.tornevall.net, "John
Williamson" <johnwilliamson@btinternet.com> wrote:

Bob Myers wrote:
"Richard Crowley" <rcrowley@xp7rt.net> wrote in message
news:13g11vn3e4ksc08@corp.supernews.com...
Which weighs more: one 8-pound sack, or eight 1-pound sacks?
Yes, it is a trick question, but you are the master of trick questions.

Or we might ask our boy Radium this one:

Which weighs more, a pound of feathers or a pound of
gold? ;-)


Errm...

Errm? Clearing your throat?

1 troy ounce (Gold) = 1.09714 imperial ounces (Feathers)
12 Troy ounces = 1 Troy pound (Gold)
16 ounces = 1 pound (Feathers)

1 (troy) pound Gold = 0.822857 pounds (Imperial) feathers

Sorry to be picky...;-)

To be picky......... I'm sure it was assumed that both the feathers and gold
would be measured with the same scale.

 

[Bob Myers]

"John Williamson" <johnwilliamson@btinternet.com> wrote in message
news:fdrnc7$mi8$1@stable.tornevall.net...

Bob Myers wrote:
"Richard Crowley" <rcrowley@xp7rt.net> wrote in message
news:13g11vn3e4ksc08@corp.supernews.com...
Which weighs more: one 8-pound sack, or eight 1-pound sacks?
Yes, it is a trick question, but you are the master of trick questions.

Or we might ask our boy Radium this one:

Which weighs more, a pound of feathers or a pound of
gold? ;-)


Errm...
1 troy ounce (Gold) = 1.09714 imperial ounces (Feathers)
12 Troy ounces = 1 Troy pound (Gold)
16 ounces = 1 pound (Feathers)

1 (troy) pound Gold = 0.822857 pounds (Imperial) feathers

Sorry to be picky...;-)

Oh, fine - just GIVE the answer away!

Why, pray tell, did you THINK I was suggesting this
question be asked of our dear friend Radium? Didn't
occur to you that there might actually be a reason I
chose those two particular materials?

Bob M.

 

[Bob Myers]

"Don Bowey" <dbowey@comcast.net> wrote in message
news:C326B53C.791C0%dbowey@comcast.net...

To be picky......... I'm sure it was assumed that both the feathers and
gold
would be measured with the same scale.

Sorry, Don - bad assumption, which was the whole point of
the "trick question." But John W. just HAD to go and spoil
the fun before we could possibly see Radium jump in....sigh...

Bob M.

 

[Don Bowey]

On 10/1/07 6:05 PM, in article fds5gg$lah$1@usenet01.boi.hp.com, "Bob Myers"
<nospamplease@address.invalid> wrote:

"Don Bowey" <dbowey@comcast.net> wrote in message
news:C326B53C.791C0%dbowey@comcast.net...
To be picky......... I'm sure it was assumed that both the feathers and
gold
would be measured with the same scale.

Sorry, Don - bad assumption, which was the whole point of
the "trick question." But John W. just HAD to go and spoil
the fun before we could possibly see Radium jump in....sigh...

Bob M.

Well, Radium sets hooks far better that he takes one. I suspect he has no
interest in other's questions.

 

[Michael A. Terrell]

Bob Myers wrote:

"John Williamson" <johnwilliamson@btinternet.com> wrote in message
news:fdrnc7$mi8$1@stable.tornevall.net...
Bob Myers wrote:
"Richard Crowley" <rcrowley@xp7rt.net> wrote in message
news:13g11vn3e4ksc08@corp.supernews.com...
Which weighs more: one 8-pound sack, or eight 1-pound sacks?
Yes, it is a trick question, but you are the master of trick questions.

Or we might ask our boy Radium this one:

Which weighs more, a pound of feathers or a pound of
gold? ;-)


Errm...
1 troy ounce (Gold) = 1.09714 imperial ounces (Feathers)
12 Troy ounces = 1 Troy pound (Gold)
16 ounces = 1 pound (Feathers)

1 (troy) pound Gold = 0.822857 pounds (Imperial) feathers

Sorry to be picky...;-)


Oh, fine - just GIVE the answer away!

Why, pray tell, did you THINK I was suggesting this
question be asked of our dear friend Radium? Didn't
occur to you that there might actually be a reason I
chose those two particular materials?

Bob M.

No, other than he's a gold bricking bird brain.


--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida

 

[Halmyre]

In article <fdrnc7$mi8$1@stable.tornevall.net>,
johnwilliamson@btinternet.com says...

Bob Myers wrote:
"Richard Crowley" <rcrowley@xp7rt.net> wrote in message
news:13g11vn3e4ksc08@corp.supernews.com...
Which weighs more: one 8-pound sack, or eight 1-pound sacks?
Yes, it is a trick question, but you are the master of trick questions.

Or we might ask our boy Radium this one:

Which weighs more, a pound of feathers or a pound of
gold? ;-)


Errm...
1 troy ounce (Gold) = 1.09714 imperial ounces (Feathers)
12 Troy ounces = 1 Troy pound (Gold)
16 ounces = 1 pound (Feathers)

1 (troy) pound Gold = 0.822857 pounds (Imperial) feathers

Sorry to be picky...;-)

There's a bit in one of Richard Feynmann's books where he's talking
about borrowing the US's silver reserves to make generator windings for
the Manhattan Project during WW2. The Treasury agreed to the request,
but took offence for being asked for several tons of silver, pointing
out that "we actually measure silver in ounces"...

--
Halmyre

 

[legg]

forgery

 

[legg]

forgery

 

[legg]

forgery

 

[Joerg]

David DiGiacomo wrote:

In article <p3TIj.4014$p24.3690@nlpi061.nbdc.sbc.com>,
Joerg <notthisjoergsch@removethispacbell.net> wrote:
I usually need more extreme tap ratios (or impedance ratios). But even
these 3.58MHz will likely go lalaland soon because analog TV is gone for
good in the US next year. It's going to be a 100% cut, there won't be
any analog stations left by the end of February. Meaning no chroma
carriers either.

Not true, there will still be LPTV and analog cable... not to mention
Canada, Mexico, etc. Also, parts that are useful in ATSC to NTSC
converter boxes will be popular for a while longer.

Right, but the sets will most likely do all this in a baseband DSP
setting. The ATSC to NTSC modulators won't have LC, just a 3.58MHz
crystal. Maybe not even that if they contain a modulator and generate it
from another reference. Last time at CostCo it seemed the regular sets
have already gone extinct, all flat screens.

The converter coupons arrived yesterday BTW. However, so far it looks
like OTA DTV isn't going to very reliable out here, pretty much as I had
expected :-(

--
Regards, Joerg

http://www.analogconsultants.com/

"gmail" domain blocked because of excessive spam.
Use another domain or send PM.

 

[gearhead]

On Apr 12, 8:36 am, Hammy <spa...@hotmail.com> wrote:

Am I reading the datasheet to the TLV3702? It says that the input
voltages can exceed the rail voltage? Here is the datasheet.

http://focus.ti.com/lit/ds/symlink/tlv3702-q1.pdf

If that is the case does anyone know of any other comparators that are
capable of this and can operate off a single supply upto 12v?
Preferably cheaper.I've been wading through Digikey and that's the
only one I can find.

I know it's slow but I don't need nanosecond transitions uS are good
enough.

Linear makes the LT1716, inputs can go to 44 v regardless of supply,
according to the datasheet.
Looks like it costs a little more than the TI comparator.
I cross-posted this to some other newsgroups in hopes of hearing of
other high-side comparators. I didn't know about the TLV370x until I
saw your post. That's a real micropower chip. Take a look at the
graphs for output. You really don't want to put much of a load on it.

You also have the option of using a high-side op amp as a comparator,
especially since you don't need nanosecond response times. FET op
amps like the TL08x are very common and fairly cheap. You can even
get them with nulling pins.

 

[gearhead]

On Apr 12, 10:26 am, gearhead <nos...@billburg.com> wrote:

On Apr 12, 8:36 am, Hammy <spa...@hotmail.com> wrote:

Am I reading the datasheet to the TLV3702? It says that the input
voltages can exceed the rail voltage? Here is the datasheet.

http://focus.ti.com/lit/ds/symlink/tlv3702-q1.pdf

If that is the case does anyone know of any other comparators that are
capable of this and can operate off a single supply upto 12v?
Preferably cheaper.I've been wading through Digikey and that's the
only one I can find.

I know it's slow but I don't need nanosecond transitions uS are good
enough.

Linear makes the LT1716, inputs can go to 44 v regardless of supply,
according to the datasheet.
Looks like it costs a little more than the TI comparator.
I cross-posted this to some other newsgroups in hopes of hearing of
other high-side comparators.  I didn't know about the TLV370x until I
saw your post.  That's a real micropower chip.  Take a look at the
graphs for output.  You really don't want to put much of a load on it.

You also have the option of using a high-side op amp as a comparator,
especially since you don't need nanosecond response times.  FET op
amps like the TL08x are very common and fairly cheap.  You can even
get them with nulling pins.

And LT6700 common mode goes to 18v regardless of supply.