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From: tommala on 6 Dec 2009 17:53 >On Sun, 06 Dec 2009 16:13:15 -0600, tommala wrote: > >>>On 12/6/2009 1:10 PM, tommala wrote: >>>>> On Sun, 06 Dec 2009 08:18:32 -0600, tommala wrote: >>>>> >>>>>> Hi everyone, >>>>>> I am trying to write a base band simulation of BPSK,QPSK and 16-QAM >>>>>> modulator(using only constellation points coordinates). I figured >> out >>>>>> from wikipedia.org and "Digital Modulation Technique" that >>>> constellation >>>>>> points will have following coordinates(in the complex plane): >>>>>> 1) (+-sqrt(Eb),0) for BPSK >>>>>> 2) (+-sqrt(2Eb/2),+-sqrt(2Eb/2)) for QPSK 3) >>>>>> (+-sqrt(4Eb/2);+-sqrt(4Eb/2)) and (+-3sqrt(4Eb/2);+-3sqrt(4Eb/2)) >> for >>>>>> 16-QAM >>>>>> But after simulation I get BER=f(Eb/No) for 16-QAM lower than for >> BPSK >>>>>> and QPSK. >>>>>> I think there is a mistake in constellation points representation >>>> above. >>>>>> Do You have any ideas? >>>>>> Thank You in advance for any remarks. >>>>> BPSK : . . >>>>> >>>>> . . >>>>> QPSK : >>>>> . . >>>>> >>>>> >>>>> . . . . >>>>> >>>>> . . . . >>>>> 16-QAM: >>>>> . . . . >>>>> >>>>> . . . . >>>>> >>>>> >>>>> >>>>> >>>>> >>>>> >>>>> >>>>> >>>>> -- >>>>> www.wescottdesign.com >>>>> >>>> I know how constellations look but don't know how connect its points >>>> coordinates with Eb. >>>> >>>> Dilip Warrier can You write sth more about r.How it is related to Eb? >>>> Maybe where can I find mathematical considerations of constellation >> points >>>> if such exist. >>> >>>Can you compute the SNR? Carrier to Noise ratio (C/N)? >>> >>>If so, it's a pretty straightforward transformation from S/N or C/N to >>>Eb/No. The nature of the transformation is described in many textbooks. >>> >>>Here are some hints: >>> >>>Eb is a bit energy measurement. What's the relationship between the >>>total signal power and the amount of energy proportional to a single >> bit? >>> >>>No is the noise power per unit Hz bandwidth. If you know the noise >>>power computed from the symbol constellation, that can be converted to >>>No pretty easily. >>> >>>So, to compute SNR, you need to know the signal power and the noise >>>power. Eb and No can be computed from each of those, respectively. >>> >>>-- >>>Eric Jacobsen >>>Minister of Algorithms >>>Abineau Communications >>>http://www.abineau.com >>> >>> >> I can't compute SNR nor C/N. > >Translation "I'm not qualified to do what I'm trying" > >> Even I don't have to. > >Translation "and I don't care" > >> The simulation has GUI in which user can choose (Eb/No)min; >> (Eb/No)max and (Eb/No)step (Eb/No in [dB]). >> The program compute in the loop for every i=Eb/No from (Eb/No)min to >> (Eb/No)max BER as ratio=number of wrong received bits/number of total >> transmitted bits. >> In the loop I calculate Eb as Eb=No*pow(10,i/10)where No is constant >> number e.g. 1 [W/Hz]. >> I assume Eb constant for every modulation. > >Translation "I expect to conduct my entire professional career dependent >on software that someone else writes, and I don't see a problem with that. > >> And my problem is to correct >> connect Eb with point coordinates. >> >I think your problem is bigger than that. > >-- >www.wescottdesign.com > I want to believe that You have a very bad day today and that is the only explanation for Your rudeness. Additionally Your translation skills are far from proper. I don't have to write all the time that (MY)PROGRAM THAT I'AM WRITING IS...
From: Eric Jacobsen on 6 Dec 2009 18:04 On 12/6/2009 3:13 PM, tommala wrote: >> On 12/6/2009 1:10 PM, tommala wrote: >>>> On Sun, 06 Dec 2009 08:18:32 -0600, tommala wrote: >>>> >>>>> Hi everyone, >>>>> I am trying to write a base band simulation of BPSK,QPSK and 16-QAM >>>>> modulator(using only constellation points coordinates). I figured > out >>>>> from wikipedia.org and "Digital Modulation Technique" that >>> constellation >>>>> points will have following coordinates(in the complex plane): >>>>> 1) (+-sqrt(Eb),0) for BPSK >>>>> 2) (+-sqrt(2Eb/2),+-sqrt(2Eb/2)) for QPSK 3) >>>>> (+-sqrt(4Eb/2);+-sqrt(4Eb/2)) and (+-3sqrt(4Eb/2);+-3sqrt(4Eb/2)) > for >>>>> 16-QAM >>>>> But after simulation I get BER=f(Eb/No) for 16-QAM lower than for > BPSK >>>>> and QPSK. >>>>> I think there is a mistake in constellation points representation >>> above. >>>>> Do You have any ideas? >>>>> Thank You in advance for any remarks. >>>> BPSK : . . >>>> >>>> . . >>>> QPSK : >>>> . . >>>> >>>> >>>> . . . . >>>> >>>> . . . . >>>> 16-QAM: >>>> . . . . >>>> >>>> . . . . >>>> >>>> >>>> >>>> >>>> >>>> >>>> >>>> >>>> -- >>>> www.wescottdesign.com >>>> >>> I know how constellations look but don't know how connect its points >>> coordinates with Eb. >>> >>> Dilip Warrier can You write sth more about r.How it is related to Eb? >>> Maybe where can I find mathematical considerations of constellation > points >>> if such exist. >> Can you compute the SNR? Carrier to Noise ratio (C/N)? >> >> If so, it's a pretty straightforward transformation from S/N or C/N to >> Eb/No. The nature of the transformation is described in many textbooks. >> >> Here are some hints: >> >> Eb is a bit energy measurement. What's the relationship between the >> total signal power and the amount of energy proportional to a single > bit? >> No is the noise power per unit Hz bandwidth. If you know the noise >> power computed from the symbol constellation, that can be converted to >> No pretty easily. >> >> So, to compute SNR, you need to know the signal power and the noise >> power. Eb and No can be computed from each of those, respectively. >> >> -- >> Eric Jacobsen >> Minister of Algorithms >> Abineau Communications >> http://www.abineau.com >> > > I can't compute SNR nor C/N. Even I don't have to. Are you sure? It might help you solve your problem. > The simulation has GUI in which user can choose (Eb/No)min; (Eb/No)max and > (Eb/No)step (Eb/No in [dB]). > The program compute in the loop for every i=Eb/No from (Eb/No)min to > (Eb/No)max BER as ratio=number of wrong received bits/number of total > transmitted bits. > In the loop I calculate Eb as Eb=No*pow(10,i/10)where No is constant > number e.g. 1 [W/Hz]. Since your question seems to be related to how to do this properly, you might want to check your assumptions here. > I assume Eb constant for every modulation. Is that a good idea? Again, check your assumptions. > And my problem is to correct connect Eb with point coordinates. > > P.S. > signal power=Eb*bit rate Yes. Is that constant across all modulations? > noise power=No*bandwidth Are No and/or bandwidth constant? It looks like you've made a lot of assumptions and put a lot of trust in whatever simulator you're using to perform the test. -- Eric Jacobsen Minister of Algorithms Abineau Communications http://www.abineau.com
From: steveu on 7 Dec 2009 05:21 >On Sun, 06 Dec 2009 08:18:32 -0600, tommala wrote: > >> Hi everyone, >> I am trying to write a base band simulation of BPSK,QPSK and 16-QAM >> modulator(using only constellation points coordinates). I figured out >> from wikipedia.org and "Digital Modulation Technique" that constellation >> points will have following coordinates(in the complex plane): >> 1) (+-sqrt(Eb),0) for BPSK >> 2) (+-sqrt(2Eb/2),+-sqrt(2Eb/2)) for QPSK 3) >> (+-sqrt(4Eb/2);+-sqrt(4Eb/2)) and (+-3sqrt(4Eb/2);+-3sqrt(4Eb/2)) for >> 16-QAM >> But after simulation I get BER=f(Eb/No) for 16-QAM lower than for BPSK >> and QPSK. >> I think there is a mistake in constellation points representation above. >> Do You have any ideas? >> Thank You in advance for any remarks. > >BPSK : . . > > . . >QPSK : > . . > > > . . . . > > . . . . >16-QAM: > . . . . > > . . . . > >-- >www.wescottdesign.com Not every 16-QAM constellation looks as regular as that. The V.29 constellation, for example, has a rather peculiar constellation spacing. Steve
From: Mikolaj on 7 Dec 2009 05:22 On 06-12-2009 at 23:53:11 tommala <tomasz (a)hefczyc.pl> wrote:> I want to believe that You have a very bad day today and > that is the only explanation for Your rudeness. > Additionally Your translation skills are far from proper. > I don't have to write all the time that (MY)PROGRAM THAT I'AM WRITING > IS... > I agree with Tim Wescott. His translation is very accurate and extremely polite. I've just tended to say exactly the same. You are trying to do something that you don't understand. There is no chance to understand it without hard and time consuming work that you are trying to omit. -- Mikolaj
From: tommala on 7 Dec 2009 06:28
>On 12/6/2009 3:13 PM, tommala wrote: >>> On 12/6/2009 1:10 PM, tommala wrote: >>>>> On Sun, 06 Dec 2009 08:18:32 -0600, tommala wrote: >>>>> >>>>>> Hi everyone, >>>>>> I am trying to write a base band simulation of BPSK,QPSK and 16-QAM >>>>>> modulator(using only constellation points coordinates). I figured >> out >>>>>> from wikipedia.org and "Digital Modulation Technique" that >>>> constellation >>>>>> points will have following coordinates(in the complex plane): >>>>>> 1) (+-sqrt(Eb),0) for BPSK >>>>>> 2) (+-sqrt(2Eb/2),+-sqrt(2Eb/2)) for QPSK 3) >>>>>> (+-sqrt(4Eb/2);+-sqrt(4Eb/2)) and (+-3sqrt(4Eb/2);+-3sqrt(4Eb/2)) >> for >>>>>> 16-QAM >>>>>> But after simulation I get BER=f(Eb/No) for 16-QAM lower than for >> BPSK >>>>>> and QPSK. >>>>>> I think there is a mistake in constellation points representation >>>> above. >>>>>> Do You have any ideas? >>>>>> Thank You in advance for any remarks. >>>>> BPSK : . . >>>>> >>>>> . . >>>>> QPSK : >>>>> . . >>>>> >>>>> >>>>> . . . . >>>>> >>>>> . . . . >>>>> 16-QAM: >>>>> . . . . >>>>> >>>>> . . . . >>>>> >>>>> >>>>> >>>>> >>>>> >>>>> >>>>> >>>>> >>>>> -- >>>>> www.wescottdesign.com >>>>> >>>> I know how constellations look but don't know how connect its points >>>> coordinates with Eb. >>>> >>>> Dilip Warrier can You write sth more about r.How it is related to Eb? >>>> Maybe where can I find mathematical considerations of constellation >> points >>>> if such exist. >>> Can you compute the SNR? Carrier to Noise ratio (C/N)? >>> >>> If so, it's a pretty straightforward transformation from S/N or C/N to >>> Eb/No. The nature of the transformation is described in many textbooks. >>> >>> Here are some hints: >>> >>> Eb is a bit energy measurement. What's the relationship between the >>> total signal power and the amount of energy proportional to a single >> bit? >>> No is the noise power per unit Hz bandwidth. If you know the noise >>> power computed from the symbol constellation, that can be converted to >>> No pretty easily. >>> >>> So, to compute SNR, you need to know the signal power and the noise >>> power. Eb and No can be computed from each of those, respectively. >>> >>> -- >>> Eric Jacobsen >>> Minister of Algorithms >>> Abineau Communications >>> http://www.abineau.com >>> >> >> I can't compute SNR nor C/N. Even I don't have to. > >Are you sure? It might help you solve your problem. > >> The simulation has GUI in which user can choose (Eb/No)min; (Eb/No)max and >> (Eb/No)step (Eb/No in [dB]). >> The program compute in the loop for every i=Eb/No from (Eb/No)min to >> (Eb/No)max BER as ratio=number of wrong received bits/number of total >> transmitted bits. >> In the loop I calculate Eb as Eb=No*pow(10,i/10)where No is constant >> number e.g. 1 [W/Hz]. > >Since your question seems to be related to how to do this properly, you >might want to check your assumptions here. > >> I assume Eb constant for every modulation. > >Is that a good idea? Again, check your assumptions. > >> And my problem is to correct connect Eb with point coordinates. >> >> P.S. >> signal power=Eb*bit rate > >Yes. Is that constant across all modulations? > >> noise power=No*bandwidth > >Are No and/or bandwidth constant? > >It looks like you've made a lot of assumptions and put a lot of trust in >whatever simulator you're using to perform the test. > >-- >Eric Jacobsen >Minister of Algorithms >Abineau Communications >http://www.abineau.com > I have looked on my assumptions one more time, thought about Your questions and reread notes. My conclusion after this is that constellation point's coordinates that I have mentioned earlier are correct(plus those I have missed and which Dilip Warrier has noticed) but the problem is in assumptions as You guided. Namely, to compare 16-QAM with BPSK/QPSK signal power(output from modulation)must be the same in those modulation for every generated To do this BPSK/QPSK signal amplitude has to change every symbol to be equal to signal amplitude of 16-QAM. BPSK and QPSK signal amplitude is the same:A=sqrt(Eb)but for 16-QAM it can be: A_1=2*sqrt(Eb);A_2=6*sqrt(Eb)or A_3=2*sqrt(5Eb)which depend of 4 bit sequence. Previously I assumed that Eb is constant for each modulation which conducted to much higher signal amplitude for 16-QAM than for BPSK/QPSK and consistently to lower BER for 16-QAM. |