Eric Jacobsen wrote:


> On Sat, 17 May 2008 15:55:02 -0700, "Green Xenon [Radium]"
> <glucegen1@excite.com> wrote:


>> Let's say I am using QAM and want a baud-rate of only 
>> 1-symbol-per-second but I want there to be 1-billion-bits-per-symbol. 
>> How many phases do I need?


> 
> You can do it with one if you have log2(1B) amplitudes.   Two if you
> have log2(1B)/2, four with log2(1B)/4, amplitudes, etc., etc.   With
> QAM you can encode information in both the phase and the amplitude.


If I have log2(1B) phases, I need only one amplitude. Right?

Green Xenon [Radium] wrote:
> Eric Jacobsen wrote:
> 
> 
>> You can do it with *one* if you have log2(1B) amplitudes. 
> 
> 'One' of what?
> 
>> *Two* if you
>> have log2(1B)/2, *four* with log2(1B)/4, amplitudes, etc., etc.
> 
> "Two" and "four" of what?

Once again, I apologize for responding without reading properly. The 
'one', 'two', and 'four' are the amounts of phases.

Sorry for the annoyance cause by the above response.

Green Xenon [Radium] wrote:
> Eric Jacobsen wrote:
> 
> 
>> Duh, I mean 2^1B amplitudes, (2^1B)/2, etc....reversed the
>> relationship there...
> 
> 
> In the '2^1B', does the 'B' stand for the bit-resolution?

Sorry, I should've realized the that B stands for billion.

My bad.

Eric Jacobsen wrote:


> Duh, I mean 2^1B amplitudes, (2^1B)/2, etc....reversed the
> relationship there...


In the '2^1B', does the 'B' stand for the bit-resolution?

Eric Jacobsen wrote:


> You can do it with *one* if you have log2(1B) amplitudes. 

'One' of what?

> *Two* if you
> have log2(1B)/2, *four* with log2(1B)/4, amplitudes, etc., etc.

"Two" and "four" of what?

>  With
> QAM you can encode information in both the phase and the amplitude.


Really, 1-symbol-per-second with a billion-bits-per-symbol is possible 
-- even if there is only changes in phase states but with the 
peak-to-peak amplitude being constant?

On Sat, 17 May 2008 16:57:38 -0700, Eric Jacobsen
<eric.jacobsen@ieee.org> wrote:

>On Sat, 17 May 2008 15:55:02 -0700, "Green Xenon [Radium]"
><glucegen1@excite.com> wrote:
>
>>glen herrmannsfeldt wrote:
>>> Green Xenon [Radium] wrote:
>>> (snip)
>>> 
>>>> Ok. What is the maximum amount of phases used in QAM?
>>> 
>>> http://en.wikipedia.org/wiki/Constellation_diagram
>>> 
>>>> If more phases are used, does this mean there are more 
>>>> bits-per-symbol? Or is the amount of bits-per-symbol determined by the 
>>>> amounts of amplitudes?
>>> 
>>> It depends on the number of constellation points.  As the
>>> constellation doesn't need to have a power of two points,
>>> the bits/symbol might not be an integer.  That allows some
>>> symbols for other uses, or, if it is about a half integer
>>> then two symbols could hold some number of bits.
>>> 
>>> Twelve points hold 3.5 bits, so two symbols would be 7 bits,
>>> with a few symbols left over.
>>> 
>>> Also, you count the phases differently as phase/magnitude
>>> than as two quadrature signals, but the result is the same.
>>> 
>>> -- glen
>>> 
>>
>>
>>Let's say I am using QAM and want a baud-rate of only 
>>1-symbol-per-second but I want there to be 1-billion-bits-per-symbol. 
>>How many phases do I need?
>
>You can do it with one if you have log2(1B) amplitudes.   Two if you
>have log2(1B)/2, four with log2(1B)/4, amplitudes, etc., etc.   With
>QAM you can encode information in both the phase and the amplitude.

Duh, I mean 2^1B amplitudes, (2^1B)/2, etc....reversed the
relationship there...

>
>>I am guessing I would need 2^1,000,000,000 different phases to achieve 
>>this. Right?
>>
>>A bit-resolution results in 2^ state of that bit-resolution. E.G. an 
>>8-bit resolution results in 2^8 different states -- or 256 states.
>Eric Jacobsen
>Minister of Algorithms
>Abineau Communications
>http://www.ericjacobsen.org
>
>Blog: http://www.dsprelated.com/blogs-1/hf/Eric_Jacobsen.php
Eric Jacobsen
Minister of Algorithms
Abineau Communications
http://www.ericjacobsen.org

Blog: http://www.dsprelated.com/blogs-1/hf/Eric_Jacobsen.php

On Sat, 17 May 2008 15:55:02 -0700, "Green Xenon [Radium]"
<glucegen1@excite.com> wrote:

>glen herrmannsfeldt wrote:
>> Green Xenon [Radium] wrote:
>> (snip)
>> 
>>> Ok. What is the maximum amount of phases used in QAM?
>> 
>> http://en.wikipedia.org/wiki/Constellation_diagram
>> 
>>> If more phases are used, does this mean there are more 
>>> bits-per-symbol? Or is the amount of bits-per-symbol determined by the 
>>> amounts of amplitudes?
>> 
>> It depends on the number of constellation points.  As the
>> constellation doesn't need to have a power of two points,
>> the bits/symbol might not be an integer.  That allows some
>> symbols for other uses, or, if it is about a half integer
>> then two symbols could hold some number of bits.
>> 
>> Twelve points hold 3.5 bits, so two symbols would be 7 bits,
>> with a few symbols left over.
>> 
>> Also, you count the phases differently as phase/magnitude
>> than as two quadrature signals, but the result is the same.
>> 
>> -- glen
>> 
>
>
>Let's say I am using QAM and want a baud-rate of only 
>1-symbol-per-second but I want there to be 1-billion-bits-per-symbol. 
>How many phases do I need?

You can do it with one if you have log2(1B) amplitudes.   Two if you
have log2(1B)/2, four with log2(1B)/4, amplitudes, etc., etc.   With
QAM you can encode information in both the phase and the amplitude.

>I am guessing I would need 2^1,000,000,000 different phases to achieve 
>this. Right?
>
>A bit-resolution results in 2^ state of that bit-resolution. E.G. an 
>8-bit resolution results in 2^8 different states -- or 256 states.
Eric Jacobsen
Minister of Algorithms
Abineau Communications
http://www.ericjacobsen.org

Blog: http://www.dsprelated.com/blogs-1/hf/Eric_Jacobsen.php

glen herrmannsfeldt wrote:
> Green Xenon [Radium] wrote:
> (snip)
> 
>> Ok. What is the maximum amount of phases used in QAM?
> 
> http://en.wikipedia.org/wiki/Constellation_diagram
> 
>> If more phases are used, does this mean there are more 
>> bits-per-symbol? Or is the amount of bits-per-symbol determined by the 
>> amounts of amplitudes?
> 
> It depends on the number of constellation points.  As the
> constellation doesn't need to have a power of two points,
> the bits/symbol might not be an integer.  That allows some
> symbols for other uses, or, if it is about a half integer
> then two symbols could hold some number of bits.
> 
> Twelve points hold 3.5 bits, so two symbols would be 7 bits,
> with a few symbols left over.
> 
> Also, you count the phases differently as phase/magnitude
> than as two quadrature signals, but the result is the same.
> 
> -- glen
> 


Let's say I am using QAM and want a baud-rate of only 
1-symbol-per-second but I want there to be 1-billion-bits-per-symbol. 
How many phases do I need?

I am guessing I would need 2^1,000,000,000 different phases to achieve 
this. Right?

A bit-resolution results in 2^ state of that bit-resolution. E.G. an 
8-bit resolution results in 2^8 different states -- or 256 states.

Green Xenon [Radium] wrote:
(snip)

> Ok. What is the maximum amount of phases used in QAM?

http://en.wikipedia.org/wiki/Constellation_diagram

> If more phases are used, does this mean there are more bits-per-symbol? 
> Or is the amount of bits-per-symbol determined by the amounts of 
> amplitudes?

It depends on the number of constellation points.  As the
constellation doesn't need to have a power of two points,
the bits/symbol might not be an integer.  That allows some
symbols for other uses, or, if it is about a half integer
then two symbols could hold some number of bits.

Twelve points hold 3.5 bits, so two symbols would be 7 bits,
with a few symbols left over.

Also, you count the phases differently as phase/magnitude
than as two quadrature signals, but the result is the same.

-- glen