On Mon, 12 May 2008 11:50:48 -0700 (PDT), "sarwate@YouEyeYouSee.edu" <dvsarwate@gmail.com> wrote:>On May 12, 12:12 am, Eric Jacobsen <eric.jacob...@ieee.org> wrote: > >> >Jerry >> >> Yes, and if the constellation were strictly PSK, i.e., traced a >> circle, that would be all you'd need. With a QAM grid, though, one >> constellation point has potentially eight neighbors (not just two, >> like on a line or in a circle), and, ideally, you'd like only one bit >> to differ with each of them. With low-order QAM (e.g., 16-QAM), >> that's not possible, so one tries to manage the decision regions to >> maximize the Euclidean distance while minimizing the Hamming distance. >> >> Eric Jacobsen >> Minister of Algorithms >> Abineau Communicationshttp://www.ericjacobsen.org > >In all the messages in this thread with splendid examples >of MATLAB code and snippets of C operators and so on, >there has been little attention paid to the specific application >to QAM, and I think that Eric's comment about low-order QAM >is misleading in some respects. My response discusses only >M-QAM where M = 2^{2n} as opposed to the general M-QAM >desired by the OP, with 16-QAM as an example. > >An 2^{2n}-QAM signal is the sum of two 2^n-ASK signals transmitted >on the I and Q channels, with each channel carrying n bits. The >mapping of data bits to signal constellation point on each channel >must correspond to Gray coding, but the Gray coding is not >explicit. For example, in 16-QAM with 4 data bits (b_3,b_2,b_1,b_0), >one could send b_3 and b_2 using 4-ASK on the I channel and >b_1 and b_0 using 4-ASK on the Q channel. (Some might prefer >to send the odd-numbered bits on the I channel and the even- >numbered bits on the Q channel instead). The mapping of >data bits to signal point is *effectively* Gray coding, with (for >example) 00, 01, 11, and 10 corresponding to signal levels >-3, -1, +1, and +3 respectively. The DATA bits themselves >don't need to be transformed to Gray code explicitly because if >I give you bits 11, you cannot say whether I mean 3 (standard >binary coding) or 2 (Gray coding) UNLESS I tell you what >coding I was using when I produced the 11. All the modulator >needs to know is that if the modulator input is 11, it needs to >produce output +1 and if the input is 10, it needs to produce >+3 etc. Now, those who want to implement the modulator as >a LINEAR modulator can write code which says that my input >can take on values 00, 01, 11, and 10 which is Gray-coded >(actually it isn't Gray-coded or anything: it is raw data) which >thus represents 0, 1, 2, and 3 respectively, The output I have >to produce is -3 + 2A where A is the integer value of my >Gray-coded data input. But remember that the data source >just handed the modulator two bits without saying anything >about whether the bits are Gray-coded or not. It is the LINEAR >modulator which is treating these bits as Gray-coded and >converting them to standard binary representation (that is, >applying the inverse Gray-code mapping!) so as to be able to >use its magic formula -3+2A to produce the output. A >modulator that is not quite as anal-retentive might just use >a LUT to translate data bits 00, 01, 11, and 10 into signal levels >-3, -1, +1, and +3 which uses Gray coding implicitly but not >explicitly. > >Returning to 16-QAM, the 4-bit nybble (0111) is split into 01 and >11, and mapped to the signal constellation point (-1, +1) via >two modulators. The 16-QAM constellation thus looks like > > -3 -1 +1 +3 > >+3 0010 0110 1110 1010 > >+1 0011 0111 1111 1011 > >-1 0001 0101 1101 1001 > >-3 0000 0100 1100 1000 > >where each constellation point differs from its 4 NEAREST >neighbors (distance 2) in one bit and from its 4 NEXT >NEAREST neighbors by 2 bits. The latter have less effect >on the overall BER as compared to the 4 NEAREST neighbors >and so it matters less that the difference is two bits. > >If the signal point (-1, +1) is transmitted and (X, Y) received, >then X and Y are each quantized into 4 levels >(-3, -1, +1, +3) and if the quantized X and Y are -1 and +1 >respectively, then the demodulator output is 0111. There >is no need for the demodulator to do do anything else, or >to translate the bits from Gray code back to standard binary >representation etc. All that is built in to the demodulator >LUTs. The harder way of doing this is to remember that >the modulator used the formula -3+2A to get +1, and so >A is given by (+1 + 3)/2 = 2, but since Gray coding is >being used, the output should be the Gray-coded version >of 2, viz. 11. > >--Dilip SarwateThanks, Dilip. That says pretty much what I was trying to say (not sure where I became misleading, but it's all good). I think you've hit the points of potential ambiguity pretty well. Eric Jacobsen Minister of Algorithms Abineau Communications http://www.ericjacobsen.org
Gray code for QAM
Started by ●May 11, 2008
Reply by ●May 12, 20082008-05-12
Reply by ●May 12, 20082008-05-12
Eric Jacobsen wrote: ...> Yes, and if the constellation were strictly PSK, i.e., traced a > circle, that would be all you'd need. With a QAM grid, though, one > constellation point has potentially eight neighbors (not just two, > like on a line or in a circle), and, ideally, you'd like only one bit > to differ with each of them. With low-order QAM (e.g., 16-QAM), > that's not possible, so one tries to manage the decision regions to > maximize the Euclidean distance while minimizing the Hamming distance.Oh. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������
Reply by ●May 14, 20082008-05-14
On May 12, 11:50 pm, "sarw...@YouEyeYouSee.edu" <dvsarw...@gmail.com> wrote:> On May 12, 12:12 am, Eric Jacobsen <eric.jacob...@ieee.org> wrote: > > > >Jerry > > > Yes, and if the constellation were strictly PSK, i.e., traced a > > circle, that would be all you'd need. With a QAM grid, though, one > > constellation point has potentially eight neighbors (not just two, > > like on a line or in a circle), and, ideally, you'd like only one bit > > to differ with each of them. With low-order QAM (e.g., 16-QAM), > > that's not possible, so one tries to manage the decision regions to > > maximize the Euclidean distance while minimizing the Hamming distance. > > > Eric Jacobsen > > Minister of Algorithms > > Abineau Communicationshttp://www.ericjacobsen.org > > In all the messages in this thread with splendid examples > of MATLAB code and snippets of C operators and so on, > there has been little attention paid to the specific application > to QAM, and I think that Eric's comment about low-order QAM > is misleading in some respects. My response discusses only > M-QAM where M = 2^{2n} as opposed to the general M-QAM > desired by the OP, with 16-QAM as an example. > > An 2^{2n}-QAM signal is the sum of two 2^n-ASK signals transmitted > on the I and Q channels, with each channel carrying n bits. The > mapping of data bits to signal constellation point on each channel > must correspond to Gray coding, but the Gray coding is not > explicit. For example, in 16-QAM with 4 data bits (b_3,b_2,b_1,b_0), > one could send b_3 and b_2 using 4-ASK on the I channel and > b_1 and b_0 using 4-ASK on the Q channel. (Some might prefer > to send the odd-numbered bits on the I channel and the even- > numbered bits on the Q channel instead). The mapping of > data bits to signal point is *effectively* Gray coding, with (for > example) 00, 01, 11, and 10 corresponding to signal levels > -3, -1, +1, and +3 respectively. The DATA bits themselves > don't need to be transformed to Gray code explicitly because if > I give you bits 11, you cannot say whether I mean 3 (standard > binary coding) or 2 (Gray coding) UNLESS I tell you what > coding I was using when I produced the 11. All the modulator > needs to know is that if the modulator input is 11, it needs to > produce output +1 and if the input is 10, it needs to produce > +3 etc. Now, those who want to implement the modulator as > a LINEAR modulator can write code which says that my input > can take on values 00, 01, 11, and 10 which is Gray-coded > (actually it isn't Gray-coded or anything: it is raw data) which > thus represents 0, 1, 2, and 3 respectively, The output I have > to produce is -3 + 2A where A is the integer value of my > Gray-coded data input. But remember that the data source > just handed the modulator two bits without saying anything > about whether the bits are Gray-coded or not. It is the LINEAR > modulator which is treating these bits as Gray-coded and > converting them to standard binary representation (that is, > applying the inverse Gray-code mapping!) so as to be able to > use its magic formula -3+2A to produce the output. A > modulator that is not quite as anal-retentive might just use > a LUT to translate data bits 00, 01, 11, and 10 into signal levels > -3, -1, +1, and +3 which uses Gray coding implicitly but not > explicitly. > > Returning to 16-QAM, the 4-bit nybble (0111) is split into 01 and > 11, and mapped to the signal constellation point (-1, +1) via > two modulators. The 16-QAM constellation thus looks like > > -3 -1 +1 +3 > > +3 0010 0110 1110 1010 > > +1 0011 0111 1111 1011 > > -1 0001 0101 1101 1001 > > -3 0000 0100 1100 1000 > > where each constellation point differs from its 4 NEAREST > neighbors (distance 2) in one bit and from its 4 NEXT > NEAREST neighbors by 2 bits. The latter have less effect > on the overall BER as compared to the 4 NEAREST neighbors > and so it matters less that the difference is two bits. > > If the signal point (-1, +1) is transmitted and (X, Y) received, > then X and Y are each quantized into 4 levels > (-3, -1, +1, +3) and if the quantized X and Y are -1 and +1 > respectively, then the demodulator output is 0111. There > is no need for the demodulator to do do anything else, or > to translate the bits from Gray code back to standard binary > representation etc. All that is built in to the demodulator > LUTs. The harder way of doing this is to remember that > the modulator used the formula -3+2A to get +1, and so > A is given by (+1 + 3)/2 = 2, but since Gray coding is > being used, the output should be the Gray-coded version > of 2, viz. 11. > > --Dilip SarwateThanks Dilip. Krishna ~blogs @ http://www.dsplog.com
