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Does more bits-per-character but less characters-per-second makebetter use of bandwidth?
What do you mean by "better use"?
G
Green Xenon [Radium]
Hi:
Sorry for my persistence on this question but it really does interest me
and I’ve not been able to find answers doing my own research. I
apologize profusely if anyone is annoyed.
If you have more bits-per-character but less characters-per-second,
would this be more efficient in terms of bandwidth usage?
I ask because I am aware that more bits-per-symbol but fewer
symbols-per-second makes better use of bandwidth. So I am guessing that
this is analogous.
Thanks for your understanding, cooperation, and assistance,
Radium
Sorry for my persistence on this question but it really does interest me
and I’ve not been able to find answers doing my own research. I
apologize profusely if anyone is annoyed.
If you have more bits-per-character but less characters-per-second,
would this be more efficient in terms of bandwidth usage?
I ask because I am aware that more bits-per-symbol but fewer
symbols-per-second makes better use of bandwidth. So I am guessing that
this is analogous.
Thanks for your understanding, cooperation, and assistance,
Radium
G
Green Xenon [Radium]
in
http://groups.google.com/group/sci.electronics.design/msg/5400f61a68bf9e84?hl=en&
:
As explained below, using more bits-per-symbol with fewer
symbols-per-second makes better use of bandwidth than using more
symbols-per-second with fewer bits-per-symbol.
http://en.wikipedia.org/wiki/Baud
“Conveying more than one bit per symbol has advantages. It reduces the
time required to send a given quantity of data, and allows modern
modems, FDDI and 100/1000 Mbit/s Ethernet LANs, and others, to achieve
high data rates. An optimal symbol set design takes into account channel
bandwidth, desired information rate, noise characteristics of the
channel and the receiver, and receiver and decoder complexity. A typical
2400-bit/s modem transmits at 600 baud (600 symbol/s), where each
quadrature amplitude modulation symbol carries four bits of information.
1000 Mbit/s Ethernet LAN cables use many wire pairs and many bits per
symbol to encode their data payloads. 1000BASE-T uses four wire pairs
and two data bits per symbol to get a symbol rate of 125MBd.”
http://groups.google.com/group/sci.electronics.design/msg/5400f61a68bf9e84?hl=en&
:
As explained below, using more bits-per-symbol with fewer
symbols-per-second makes better use of bandwidth than using more
symbols-per-second with fewer bits-per-symbol.
http://en.wikipedia.org/wiki/Baud
“Conveying more than one bit per symbol has advantages. It reduces the
time required to send a given quantity of data, and allows modern
modems, FDDI and 100/1000 Mbit/s Ethernet LANs, and others, to achieve
high data rates. An optimal symbol set design takes into account channel
bandwidth, desired information rate, noise characteristics of the
channel and the receiver, and receiver and decoder complexity. A typical
2400-bit/s modem transmits at 600 baud (600 symbol/s), where each
quadrature amplitude modulation symbol carries four bits of information.
1000 Mbit/s Ethernet LAN cables use many wire pairs and many bits per
symbol to encode their data payloads. 1000BASE-T uses four wire pairs
and two data bits per symbol to get a symbol rate of 125MBd.”
P
Phil Hobbs
Green said:
Shannon's theorem says that the channel capacity (i.e. absolute maximum
number of bits per second transmitted) in a noisy channel is
C = BW*log_2(1+SNR).
The low SNR limit of this is
C ~ BW*SNR/ln(2) (SNR << 1)
which is what we expect from signal averaging, and the high SNR limit is
C ~ BW*log_2(SNR) (SNR >> 1)
which basically says that by using N-level signalling, you can get
log_2(N) times as much data if you have N times more SNR. If you have 1
nW of noise, you can fit 1024 levels in with the same error rate as if
you have two levels and a 1.024 uW of noise. Unfortunately since that's
only 10 bits' worth, you get only 10 times more data for 30 dB more
transmitted power (assuming the noise is additive).
Cheers,
Phil Hobbs
G
Green Xenon [Radium]
On Sep 14, 7:38 am, Phil Hobbs <[email protected]>
wrote in
http://groups.google.com/group/sci.electronics.design/msg/82e7987327f3532b?hl=en&
:
Thanks but I still don't understand, would “more bits-per-character but
less characters-per-second” use less bandwidth than “less
bits-per-character but more characters-per-second”?
wrote in
http://groups.google.com/group/sci.electronics.design/msg/82e7987327f3532b?hl=en&
:
Thanks but I still don't understand, would “more bits-per-character but
less characters-per-second” use less bandwidth than “less
bits-per-character but more characters-per-second”?
M
MooseFET
(A)
It really needs to be worked out in terms of how much information is
really being sent per second.
(B)
It really needs to be worked out in terms of how much information is
really being sent per second.
(C)
IItt rreeaallllyy nneeeeddss ttoo bbee wwoorrkkeedd oouutt iinn
tteerrmmss oogf hhooww mmuucchh iinnffoorrmmaattiioonn iiss
rreeaallllyy bbeeiinngg sseenntt ppeerr sseeccoonndd.
(A) carried information. (B) was a repeat of (A) so no information
was carried. (C) took 16 bits per character.
When you add bits to a character, there is no added information in the
data stream unless those bits mean something.
P
Paul E. Schoen
Green Xenon said:
It seems that the baud rate differs from bit rate on devices like modems,
which use combinations of frequencies and amplitudes to pack more bits into
each symbol. But at the level of a USART, or RS-232, there is only one bit
per symbol, consisting of a logic high/low, or mark/space. Asynchronous
communication also requires at least a start bit, and one or two stop bits,
with the actual data contained in a 5 to 8 bit code. So, for a given value
of bits/sec, you can transfer more data per second with an 8 bit code than
a 5 bit code. The sequence of bits is called a frame, which might be
considered a symbol. 8 bits are needed to transmit 6 bits of information,
and 10 bits for 8 bits of information. So, at 4800 bps, you can transmit
600 or 480 frames per second, corresponding to 600*6=3600 or 480*8=3840
bits/second of data.
There may also be more overhead if a parity bit is used, and even more with
CRC error correction. But without error correction, there is a chance of
bad data being received, which could be disastrous for critical binary
data, but more tolerable for text, which has some built-in redundancy which
makes it readable even if some characters are missing or incorrect.
I found a fairly simple explanation of the RS232 spec at:
<http://www.lammertbies.nl/comm/info/RS-232_specs.html>
I'm not a communications expert, but I have had to deal with transmitting
data on a serial port. For my Ortmaster product, I need to transmit 10 or
12 bit A/D readings at 2400 per second. I use two 8 bit frames, with 1
start and 1 stop bit each, so I have two extra bits per frame. I use these
for error detection, by incrementing a two bit counter for each reading. If
the counters do not match, I know that a frame has been missed, so I
resynchronize and set the erroneous sample to zero. I use 57600 bps for
communication. The absolute minimum would be 48k.
Paul
P
Phil Hobbs
Green said:
How are you sending 'characters' if not as sequences of bits, then?
Cheers,
Phil Hobbs
W
whit3rd
If the signal is binary (has only two recognizable states), one can
only send a
sequence of (at most) one bit symbols. If the signal can take on
values
-2, -1, 0,+1, +2, then there are five symbols, each having a little
over two bits
of information. Gigabit ethernet uses such a multi-level signal
scheme.
The 'SNR' part of Shannon's theorem represents the limit of how many
symbols
can be represented, and acts as the number of bits per symbol (it's
not the real
number of bits, just a MAXIMUM value, a limit). For gigabit
ethernet, there
are more symbols, but less noise tolerance, than for 100baseT. But
the two
standards use the same bandwidth and baud rate.
S
Skybuck Flying
Bit means Binary Digit.
Shannon says the binary digit is the most efficient way of encoding
information.
However Skybuck says it might not always be practical
Your hands have ten fingers.
The way people count is:
.. = finger down
| = finger up
- = hand seperator
......-..... = 0
|....-..... = 1
||...-..... = 2
|||..-..... = 3
||||.-..... = 4
|||||-..... = 5
|||||-|.... = 6
|||||-||... = 7
|||||-|||.. = 8
|||||-||||. = 9
|||||-||||| = 10
This numbering is called a decimal digit.
The way computers count is to flip/flop each finger (using as many
combinations as possible):
The number 10 is a special number, which indicates a transfer has taken
place in the positional numbering system.
The way computers count is:
......-..... = 0
|....-..... = 1
..|...-..... = 2
||...-..... = 3
...|..-..... = 4
|.|..-..... = 5
..||..-..... = 6
|||..-..... = 7
....|.-..... = 8
|..|.-..... = 9
..|.|.-..... = 10
||.|.-..... = 11
...||.-..... = 12
|.||.-..... = 13
..|||.-..... = 14
||||.-..... = 15
.....|-..... = 16
|...|-..... = 17
..|..|-..... = 18
||..|-..... = 19
...|.|-..... = 20
|.|.|-..... = 21
..||.|-..... = 22
|||.|-..... = 23
....||-..... = 24
|..||-..... = 25
..|.||-..... = 26
||.||-..... = 27
...|||-..... = 28
|.|||-..... = 29
..||||-..... = 30
|||||-..... = 31
......-|.... = 32
With just one hand a computer can encode 32 different values (including
zero).
For each extra finger the number of values doubles.
Let's see if it's possible to improve on the binary digital numbering
system.
Suppose a finger can have 3 positions instead of two:
.. = finger down
+ = finger half way
| = finger up
......-..... = 0
+....-..... = 1
|....-..... = 2
..+...-..... = 3
++...-..... = 4
|+...-..... = 5
..|...-..... = 6
+|...-..... = 7
||...-..... = 8
...+..-..... = 9
+.+..-..... = 10
|.+..-..... = 11
..++..-..... = 12
+++..-..... = 13
|++..-..... = 14
..|+..-..... = 15
+|+..-..... = 16
||+..-..... = 17
...|..-..... = 18
+.|..-..... = 19
|.|..-..... = 20
..+|..-..... = 21
++|..-..... = 22
|+|..-..... = 23
..||..-..... = 24
+||..-..... = 25
|||..-..... = 26
....+.-..... = 27
+..+.-..... = 28
|..+.-..... = 29
..+.+.-..... = 30
++.+.-..... = 31
|+.+.-..... = 32
..|.+.-..... = 33
+|.+.-..... = 34
||.+.-..... = 35
...++.-..... = 36
+.++.-..... = 37
|.++.-..... = 38
..+++.-..... = 39
++++.-..... = 40
|+++.-..... = 41
..|++.-..... = 42
+|++.-..... = 43
||++.-..... = 44
...|+.-..... = 45
+.|+.-..... = 46
|.|+.-..... = 47
..+|+.-..... = 48
++|+.-..... = 49
|+|+.-..... = 50
..||+.-..... = 51
+||+.-..... = 52
|||+.-..... = 53
Well at this point I am gonna stop since it looks like the tertiary
numbering system is more efficient.
However shannon says no, the binary digit is the most efficient encoding so
there must be a way to encode more efficient.
This would require "repacking" the system above.
You would have to cut off the top of your finger, and keep the bottom and
middle together to be able to alternate.
.. = bottom half
+ = middle half
| = top cut off
Then they can be used as a binary system:
Suppose there were only 3 fingers:
The possibilities become:
.... = 0
+.. = 1
..+. = 2
++. = 3
...+ = 4
+.+ = 5
..++ = 6
+++ = 7
....| = 8
+..| = 9
..+.| = 10
++.| = 11
...+| = 12
+.+| = 13
..++| = 14
+++| = 15
....|| = 16
+..|| = 17
..+.|| = 18
++.|| = 19
...+|| = 20
+.+|| = 21
..++|| = 22
+++|| = 23
....||| = 24
+..||| = 25
..+.||| = 26
++.||| = 27
...+||| = 28
+.+||| = 29
..++||| = 30
+++||| = 31
Let's compare:
The tertiary system had 26 + zero = 27 values for 3 fingers with 3 positions
The binary system had 31 + zero = 32 values for 3 fingers with 2 positions,
and 3 seperated tops.
Now pretend it's possible to make a new finger from the 3 seperated tops.
This would give one new finger.
So there would be four fingers with one extra top.
Now suppose we had 4 fingers to start with, we would have had 4 tops and
that would have given two perfect new fingers for a total of 6 fingers.
So then it's easy to see that with only 4 fingers it's possible to encode
2^6 = 64 different values.
Following this example:
If we cut each finger in three pieces and reattached the top two.
We still have 10 alternating fingers plus 20 spare parts.
If each spare part can be reconnected to the hand we have 30 alternating
fingers.
So we can encode 2^30 = 4294967296 values with a binary system for our
hands/finger bones.
The question now is:
How many times can you divide a finger ?
Or how many different positions can you make with your fingers ?
(I would call this: How many values can you encode per symbol ?)
For example you could bend them slightly...
But then it becomes increasingly difficult for other people to see which
value you are expressing.
For example: it's to difficult to see the rotation in degrees.
For example: the finger has become too small... like a few micrometers.
This introduces errors in the "symbol" encoding.
Now once you have figured out a comfortable encoding per symbol.
A different question is:
How many hands can you transmit per second ?
Or in other words:
How quickly can you move all your fingers at once to make a new number and
how fast can somebody else read this and determine the number, before you
make the next number with your hands.
This is the ammount of "numbers" or "digits" per second.
So I will give you a formula which is maybe more clear to you, instead of
the baud formula:
Skybuck's Formula for Bandwidth: Digits per second * Values per Digit.
I would find the word "symbol" confusing because a digit is visually
represented by a symbol.
Thus for a digit or number to be visualized each value needs it's own
symbol.
For our decimal system:
0 = symbol 0
1 = symbol 1
2 = symbol 2
3 = symbol 3
4 = symbol 4
5 = symbol 5
6 = symbol 6
7 = symbol 7
8 = symbol 8
9 = symbol 9
No more symbols.
Every other decimal number consists out of these digits or symbols
For example:
345345345
So I hope this answers your question a little bit.
To increase bandwidth you can do two things:
Increase digits per second.
or
Increase values per digit.
But because increasing the number of values is kinda the same as increasing
the number of digits per second if values are re-encoded to be binary it's
more or less the same.
So if you decrease one and increase the other you are just shifting bits and
possibly error rates, the total bandwidth stays more or less the same.
It depends how much you increase the one and decrease the other
Bye,
Skybuck.
Shannon says the binary digit is the most efficient way of encoding
information.
However Skybuck says it might not always be practical
Your hands have ten fingers.
The way people count is:
.. = finger down
| = finger up
- = hand seperator
......-..... = 0
|....-..... = 1
||...-..... = 2
|||..-..... = 3
||||.-..... = 4
|||||-..... = 5
|||||-|.... = 6
|||||-||... = 7
|||||-|||.. = 8
|||||-||||. = 9
|||||-||||| = 10
This numbering is called a decimal digit.
The way computers count is to flip/flop each finger (using as many
combinations as possible):
The number 10 is a special number, which indicates a transfer has taken
place in the positional numbering system.
The way computers count is:
......-..... = 0
|....-..... = 1
..|...-..... = 2
||...-..... = 3
...|..-..... = 4
|.|..-..... = 5
..||..-..... = 6
|||..-..... = 7
....|.-..... = 8
|..|.-..... = 9
..|.|.-..... = 10
||.|.-..... = 11
...||.-..... = 12
|.||.-..... = 13
..|||.-..... = 14
||||.-..... = 15
.....|-..... = 16
|...|-..... = 17
..|..|-..... = 18
||..|-..... = 19
...|.|-..... = 20
|.|.|-..... = 21
..||.|-..... = 22
|||.|-..... = 23
....||-..... = 24
|..||-..... = 25
..|.||-..... = 26
||.||-..... = 27
...|||-..... = 28
|.|||-..... = 29
..||||-..... = 30
|||||-..... = 31
......-|.... = 32
With just one hand a computer can encode 32 different values (including
zero).
For each extra finger the number of values doubles.
Let's see if it's possible to improve on the binary digital numbering
system.
Suppose a finger can have 3 positions instead of two:
.. = finger down
+ = finger half way
| = finger up
......-..... = 0
+....-..... = 1
|....-..... = 2
..+...-..... = 3
++...-..... = 4
|+...-..... = 5
..|...-..... = 6
+|...-..... = 7
||...-..... = 8
...+..-..... = 9
+.+..-..... = 10
|.+..-..... = 11
..++..-..... = 12
+++..-..... = 13
|++..-..... = 14
..|+..-..... = 15
+|+..-..... = 16
||+..-..... = 17
...|..-..... = 18
+.|..-..... = 19
|.|..-..... = 20
..+|..-..... = 21
++|..-..... = 22
|+|..-..... = 23
..||..-..... = 24
+||..-..... = 25
|||..-..... = 26
....+.-..... = 27
+..+.-..... = 28
|..+.-..... = 29
..+.+.-..... = 30
++.+.-..... = 31
|+.+.-..... = 32
..|.+.-..... = 33
+|.+.-..... = 34
||.+.-..... = 35
...++.-..... = 36
+.++.-..... = 37
|.++.-..... = 38
..+++.-..... = 39
++++.-..... = 40
|+++.-..... = 41
..|++.-..... = 42
+|++.-..... = 43
||++.-..... = 44
...|+.-..... = 45
+.|+.-..... = 46
|.|+.-..... = 47
..+|+.-..... = 48
++|+.-..... = 49
|+|+.-..... = 50
..||+.-..... = 51
+||+.-..... = 52
|||+.-..... = 53
Well at this point I am gonna stop since it looks like the tertiary
numbering system is more efficient.
However shannon says no, the binary digit is the most efficient encoding so
there must be a way to encode more efficient.
This would require "repacking" the system above.
You would have to cut off the top of your finger, and keep the bottom and
middle together to be able to alternate.
.. = bottom half
+ = middle half
| = top cut off
Then they can be used as a binary system:
Suppose there were only 3 fingers:
The possibilities become:
.... = 0
+.. = 1
..+. = 2
++. = 3
...+ = 4
+.+ = 5
..++ = 6
+++ = 7
....| = 8
+..| = 9
..+.| = 10
++.| = 11
...+| = 12
+.+| = 13
..++| = 14
+++| = 15
....|| = 16
+..|| = 17
..+.|| = 18
++.|| = 19
...+|| = 20
+.+|| = 21
..++|| = 22
+++|| = 23
....||| = 24
+..||| = 25
..+.||| = 26
++.||| = 27
...+||| = 28
+.+||| = 29
..++||| = 30
+++||| = 31
Let's compare:
The tertiary system had 26 + zero = 27 values for 3 fingers with 3 positions
The binary system had 31 + zero = 32 values for 3 fingers with 2 positions,
and 3 seperated tops.
Now pretend it's possible to make a new finger from the 3 seperated tops.
This would give one new finger.
So there would be four fingers with one extra top.
Now suppose we had 4 fingers to start with, we would have had 4 tops and
that would have given two perfect new fingers for a total of 6 fingers.
So then it's easy to see that with only 4 fingers it's possible to encode
2^6 = 64 different values.
Following this example:
If we cut each finger in three pieces and reattached the top two.
We still have 10 alternating fingers plus 20 spare parts.
If each spare part can be reconnected to the hand we have 30 alternating
fingers.
So we can encode 2^30 = 4294967296 values with a binary system for our
hands/finger bones.
The question now is:
How many times can you divide a finger ?
Or how many different positions can you make with your fingers ?
(I would call this: How many values can you encode per symbol ?)
For example you could bend them slightly...
But then it becomes increasingly difficult for other people to see which
value you are expressing.
For example: it's to difficult to see the rotation in degrees.
For example: the finger has become too small... like a few micrometers.
This introduces errors in the "symbol" encoding.
Now once you have figured out a comfortable encoding per symbol.
A different question is:
How many hands can you transmit per second ?
Or in other words:
How quickly can you move all your fingers at once to make a new number and
how fast can somebody else read this and determine the number, before you
make the next number with your hands.
This is the ammount of "numbers" or "digits" per second.
So I will give you a formula which is maybe more clear to you, instead of
the baud formula:
Skybuck's Formula for Bandwidth: Digits per second * Values per Digit.
I would find the word "symbol" confusing because a digit is visually
represented by a symbol.
Thus for a digit or number to be visualized each value needs it's own
symbol.
For our decimal system:
0 = symbol 0
1 = symbol 1
2 = symbol 2
3 = symbol 3
4 = symbol 4
5 = symbol 5
6 = symbol 6
7 = symbol 7
8 = symbol 8
9 = symbol 9
No more symbols.
Every other decimal number consists out of these digits or symbols
For example:
345345345
So I hope this answers your question a little bit.
To increase bandwidth you can do two things:
Increase digits per second.
or
Increase values per digit.
But because increasing the number of values is kinda the same as increasing
the number of digits per second if values are re-encoded to be binary it's
more or less the same.
So if you decrease one and increase the other you are just shifting bits and
possibly error rates, the total bandwidth stays more or less the same.
It depends how much you increase the one and decrease the other
Bye,
Skybuck.
S
Skybuck Flying
Skybuck Flying said:Bit means Binary Digit.
Shannon says the binary digit is the most efficient way of encoding
information.
However Skybuck says it might not always be practical
Your hands have ten fingers.
The way people count is:
. = finger down
| = finger up
- = hand seperator
.....-..... = 0
|....-..... = 1
||...-..... = 2
|||..-..... = 3
||||.-..... = 4
|||||-..... = 5
|||||-|.... = 6
|||||-||... = 7
|||||-|||.. = 8
|||||-||||. = 9
|||||-||||| = 10
This numbering is called a decimal digit.
The way computers count is to flip/flop each finger (using as many
combinations as possible):
The number 10 is a special number, which indicates a transfer has taken
place in the positional numbering system.
The way computers count is:
.....-..... = 0
|....-..... = 1
.|...-..... = 2
||...-..... = 3
..|..-..... = 4
|.|..-..... = 5
.||..-..... = 6
|||..-..... = 7
...|.-..... = 8
|..|.-..... = 9
.|.|.-..... = 10
||.|.-..... = 11
..||.-..... = 12
|.||.-..... = 13
.|||.-..... = 14
||||.-..... = 15
....|-..... = 16
|...|-..... = 17
.|..|-..... = 18
||..|-..... = 19
..|.|-..... = 20
|.|.|-..... = 21
.||.|-..... = 22
|||.|-..... = 23
...||-..... = 24
|..||-..... = 25
.|.||-..... = 26
||.||-..... = 27
..|||-..... = 28
|.|||-..... = 29
.||||-..... = 30
|||||-..... = 31
.....-|.... = 32
With just one hand a computer can encode 32 different values (including
zero).
For each extra finger the number of values doubles.
Let's see if it's possible to improve on the binary digital numbering
system.
Suppose a finger can have 3 positions instead of two:
. = finger down
+ = finger half way
| = finger up
.....-..... = 0
+....-..... = 1
|....-..... = 2
.+...-..... = 3
++...-..... = 4
|+...-..... = 5
.|...-..... = 6
+|...-..... = 7
||...-..... = 8
..+..-..... = 9
+.+..-..... = 10
|.+..-..... = 11
.++..-..... = 12
+++..-..... = 13
|++..-..... = 14
.|+..-..... = 15
+|+..-..... = 16
||+..-..... = 17
..|..-..... = 18
+.|..-..... = 19
|.|..-..... = 20
.+|..-..... = 21
++|..-..... = 22
|+|..-..... = 23
.||..-..... = 24
+||..-..... = 25
|||..-..... = 26
...+.-..... = 27
+..+.-..... = 28
|..+.-..... = 29
.+.+.-..... = 30
++.+.-..... = 31
|+.+.-..... = 32
.|.+.-..... = 33
+|.+.-..... = 34
||.+.-..... = 35
..++.-..... = 36
+.++.-..... = 37
|.++.-..... = 38
.+++.-..... = 39
++++.-..... = 40
|+++.-..... = 41
.|++.-..... = 42
+|++.-..... = 43
||++.-..... = 44
..|+.-..... = 45
+.|+.-..... = 46
|.|+.-..... = 47
.+|+.-..... = 48
++|+.-..... = 49
|+|+.-..... = 50
.||+.-..... = 51
+||+.-..... = 52
|||+.-..... = 53
Well at this point I am gonna stop since it looks like the tertiary
numbering system is more efficient.
However shannon says no, the binary digit is the most efficient encoding
so there must be a way to encode more efficient.
This would require "repacking" the system above.
You would have to cut off the top of your finger, and keep the bottom and
middle together to be able to alternate.
. = bottom half
+ = middle half
| = top cut off
Then they can be used as a binary system:
Suppose there were only 3 fingers:
The possibilities become:
... = 0
+.. = 1
.+. = 2
++. = 3
..+ = 4
+.+ = 5
.++ = 6
+++ = 7
...| = 8
+..| = 9
.+.| = 10
++.| = 11
..+| = 12
+.+| = 13
.++| = 14
+++| = 15
...|| = 16
+..|| = 17
.+.|| = 18
++.|| = 19
..+|| = 20
+.+|| = 21
.++|| = 22
+++|| = 23
...||| = 24
+..||| = 25
.+.||| = 26
++.||| = 27
..+||| = 28
+.+||| = 29
.++||| = 30
+++||| = 31
Let's compare:
The tertiary system had 26 + zero = 27 values for 3 fingers with 3
positions
The binary system had 31 + zero = 32 values for 3 fingers with 2
positions, and 3 seperated tops.
Now pretend it's possible to make a new finger from the 3 seperated tops.
This would give one new finger.
So there would be four fingers with one extra top.
Now suppose we had 4 fingers to start with, we would have had 4 tops and
that would have given two perfect new fingers for a total of 6 fingers.
So then it's easy to see that with only 4 fingers it's possible to encode
2^6 = 64 different values.
Following this example:
If we cut each finger in three pieces and reattached the top two.
We still have 10 alternating fingers plus 20 spare parts.
If each spare part can be reconnected to the hand we have 30 alternating
fingers.
So we can encode 2^30 = 4294967296 values with a binary system for our
hands/finger bones.
The question now is:
How many times can you divide a finger ?
Or how many different positions can you make with your fingers ?
(I would call this: How many values can you encode per symbol ?)
Little mistake I made here because I tried to adept to somebody's confusing
term of symbol.
Correction:
I like to think of a symbol as being one unique value.
So I ment:
(I would call this: How many different/unique values (or symbols) can you
encode/make per digit ?)
For example you could bend them slightly...
But then it becomes increasingly difficult for other people to see which
value you are expressing.
For example: it's to difficult to see the rotation in degrees.
For example: the finger has become too small... like a few micrometers.
This introduces errors in the "symbol" encoding.
Now once you have figured out a comfortable encoding per symbol.
A different question is:
How many hands can you transmit per second ?
Or in other words:
How quickly can you move all your fingers at once to make a new number and
how fast can somebody else read this and determine the number, before you
make the next number with your hands.
This is the ammount of "numbers" or "digits" per second.
So I will give you a formula which is maybe more clear to you, instead of
the baud formula:
Skybuck's Formula for Bandwidth: Digits per second * Values per Digit.
I would find the word "symbol" confusing because a digit is visually
represented by a symbol.
Thus for a digit or number to be visualized each value needs it's own
symbol.
For our decimal system:
0 = symbol 0
1 = symbol 1
2 = symbol 2
3 = symbol 3
4 = symbol 4
5 = symbol 5
6 = symbol 6
7 = symbol 7
8 = symbol 8
9 = symbol 9
No more symbols.
Every other decimal number consists out of these digits or symbols
For example:
345345345
So I hope this answers your question a little bit.
To increase bandwidth you can do two things:
Increase digits per second.
or
Increase values per digit.
But because increasing the number of values is kinda the same as
increasing the number of digits per second if values are re-encoded to be
binary it's more or less the same.
So if you decrease one and increase the other you are just shifting bits
and possibly error rates, the total bandwidth stays more or less the same.
It depends how much you increase the one and decrease the other
Bye,
Skybuck.
Bye,
Skybuck
K
krw
Shannon says no such thing.
However Skybuck says it might not always be practical
Skybuck *never* has anything practical.
No, they have four.
How would you know what people do?
Exactly!
<More SkyDuck nonsense snipped>
P
Phil Hobbs
Sure. That's why I didn't understand the OP's confusion. Multilevelwhit3rd said:
signalling depends on high SNR, but you win logarithmically with
transmitted power, rather than linearly with bandwidth or number of
parallel channels.
Cheers,
Phil Hobbs
S
Skybuck Flying
Misquoting idiot.
K
krw
Yes, perhaps I did misquote idiot. Shannon would be happy though.
No information was lost.
S
Skybuck Flying
krw said:
Shannon would be sad, great information replaced with the lunacy of a
lunatic.
Bye,
Skybuck.
K
krw
Yes, Shannon is turning in his grave at your mere mention of his
name. Please go away and learn something (then don't come back).
M
MooseFET
Not always.
Actually Shannon is no longer with us and he never said that. In
Shannons and many other people's writings, the bit is the unit of
information.
However Skybuck says it might not always be practical
... and a guy named James says "Help! Help! The aliens are sucking my
brain out".
Only if you count your thumbs as fingers and you never worked in a saw
mill.
You've never been to China and have flexible hands.
...!!!-..... = 3
No, it is called weird hand signals. A decimal digit would be only
one finger.
Most computers do this in most cases. It isn't a hard and fast rule
though. Some also have an extra bit for every several that is only
true for one clock cycle.
You need to define "improve"
We've known about base 3 for years and years.
[....]
No, it just looks like base 3.
You have two transistors, design a circuit that remembers 3 states.
Stop claiming he said things he didn't.
[...]
No lets not. Lets take the case where the fingers can be in 1,000,023
positions, with only one finger you can do much better than you silly
base and limited 3 method. For that matter, you have over looked the
problem of irrational numbers. If the finger can have exp(sqrt(2) *
pi), you can encode numbers that would normally need an infinite
number of fingers using just two.
N
nospam
Skybuck Flying (who I kill filed ages ago) is a good example of a person
who is not smart enough to know he isn't smart.
Probably the product of a modern education where pupils must never be
criticised or allowed to fail because it might damage their self esteem.
http://www.bbc.co.uk/apps/ifl/schoo...iz=decimalsfoundationtest&templateStyle=maths
Try putting some wrong answers in the above test. If you don't know what
1.1 + 2.4 is you will not be told you stupid or lazy or even wrong, you
will be told "Hard luck. Try again!." The dim pupils of today don't know
they are dim they must just think they are really unlucky.
--
who is not smart enough to know he isn't smart.
Probably the product of a modern education where pupils must never be
criticised or allowed to fail because it might damage their self esteem.
http://www.bbc.co.uk/apps/ifl/schoo...iz=decimalsfoundationtest&templateStyle=maths
Try putting some wrong answers in the above test. If you don't know what
1.1 + 2.4 is you will not be told you stupid or lazy or even wrong, you
will be told "Hard luck. Try again!." The dim pupils of today don't know
they are dim they must just think they are really unlucky.
--
