Green Xenon [Radium] wrote:> QAM uses two carrier waves that are 90-degrees out of phases with each > each and amplitude-modulates them. QAM only has two phases but can have > more than two amplitude levels. Is there any modulation scheme that does > the opposite -- i.e. the two carrier waves have only two amplitudes but > with more than two phases?Usually each has a diagram with phase and amplitude indicated by a point for each allowed combination. Consider it as a phase magnitude plot in complex space. http://www.google.com/imgres?imgurl=http://www.blondertongue.com/QAM-Transmodulator/QAM_Anal4.gif&imgrefurl=http://www.blondertongue.com/QAM-Transmodulator/QAM_defined.php&h=367&w=371&sz=40&tbnid=yPNp9zooSk4J:&tbnh=121&tbnw=122&prev=/images%3Fq%3Dconstellation%2Bdiagram&hl=en&sa=X&oi=image_result&resnum=1&ct=image&cd=2 http://en.wikipedia.org/wiki/Constellation_diagram For efficient coding the points should have approximately equal space between them, which would not be true for a large number of phases and only two amplitudes. It isn't too far off for up to about six or eight, though. Note that it isn't required that the allowed values for I and Q be independent. If they are, the result is a square array with a square outline, but many other combinations are possible. Limiting the amplitude removes the corner points for a square array with a circular outline. -- glen
Modulation scheme with multiple phases and a two amplitudes?
Started by ●May 16, 2008
Reply by ●May 16, 20082008-05-16
Reply by ●May 16, 20082008-05-16
emeb wrote:> Let's do a bit of careful deconstruction on the OP:>>QAM uses two carrier waves that are 90-degrees out of phases with each >>each and amplitude-modulates them.> I assume from this that you're referring to I/Q complex baseband > modulation. In this you're correct - the phases are fixed at 90deg and > by varying amplitudes we can create any 2-D vector. Different types/ > orders of QAM will have different sets of discrete amplitudes.The circular 8QAM can be described as either one amplitude and eight phases, or in I/Q space as quadrature signals each with five amplitudes (-1, -sqrt(1/2), 0, sqrt(1/2), 1) but only eight combinations allowed. I could also see 12 points with two amplitudes and six phases each.>>QAM only has two phases but can have >>more than two amplitude levels. Is there any modulation scheme that does >>the opposite -- i.e. the two carrier waves have only two amplitudes but >>with more than two phases?> Interesting idea. It seems that mathematically there are an infinite > number of ways to generate 2-D vectors with various combinations of > amplitude & phase. The end results are all the same though, and using > quadrature carriers (that are orthogonal and hence give maximum > coverage of a space with minimum required amplitude variation of the I/ > Q carriers) is simplest.> I suppose if you were trying to avoid a patent that this might be one > way to do it though.For generating and decoding, yes. But the allowed combinations of constellation points can be described either as two orthogonal signals, possibly with restrictions on the allowed points, or as phase and amplitude. Done carefully that might get around a patent. -- glen
Reply by ●May 16, 20082008-05-16
Randy Yates wrote: (someone wrote)>>>>QAM uses two carrier waves that are 90-degrees out of phases with each >>>>each and amplitude-modulates them. QAM only has two phases but can have >>>>more than two amplitude levels. Is there any modulation scheme that does >>>>the opposite -- i.e. the two carrier waves have only two amplitudes but >>>>with more than two phases?(snip)> The answer is that, for simple amplitude modulation, the dimension of > the vector space for this type of modulation is two - that is > essentially what you've already stated. The reason that this is the case > is that basis vectors in this space are sinusoids, and there can only be > a maximum of two linearly independent vectors (at the same frequency) of > this type.Using that argument it should be possible to show that there is no advantage to three-phase power. Note that the modulation method and the resulting constellation are independent. One could do 3QAM (three phase) using quadrature carriers or three different phases. http://en.wikipedia.org/wiki/Constellation_diagram There are many ways to arrange N points in a two dimensional space. -- glen
Reply by ●May 16, 20082008-05-16
Eric Jacobsen wrote:> QAM > constellations have way more than two phases in the possible symbols.Ok. What is the maximum amount of phases used in QAM? 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?
Reply by ●May 16, 20082008-05-16
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
Reply by ●May 17, 20082008-05-17
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.
Reply by ●May 17, 20082008-05-17
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
Reply by ●May 17, 20082008-05-17
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.phpEric Jacobsen Minister of Algorithms Abineau Communications http://www.ericjacobsen.org Blog: http://www.dsprelated.com/blogs-1/hf/Eric_Jacobsen.php
Reply by ●May 18, 20082008-05-18
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?
Reply by ●May 18, 20082008-05-18
