Hello, In case of r(t) = s(t) + i(t) + n(t) where r(t) - received signal s(t) - desired signal i(t) - interference signal n(t) - additive Gaussian noise then does the BER depend on Signal-to-{Interference + Noise} Ratio (SINR) directly For e.g. if s(t) is Binary PSK, can I use the Q-function directly by replacing the SNR by SINR to estimate the BER. Cheers, Vimal

# BER dependence on SINR

Started by ●September 15, 2004

Reply by ●September 15, 20042004-09-15

I think (but I am just a beginner) the following: if i(t) is gaussian white noise and uncorrelated with n(t), you can use the SINR directly> Hello, > > In case of > > r(t) = s(t) + i(t) + n(t) > > where r(t) - received signal > s(t) - desired signal > i(t) - interference signal > n(t) - additive Gaussian noise > > then does the BER depend on Signal-to-{Interference + Noise} Ratio > (SINR) directly > > For e.g. if s(t) is Binary PSK, can I use the Q-function directly by > replacing the SNR by SINR to estimate the BER. > > Cheers, > Vimal

Reply by ●September 15, 20042004-09-15

Hint: Consider different types of interference. Cheers, Jake vimal_bhatia2@yahoo.com (Vimal) wrote in message news:<b6fc6dda.0409151050.311b4aea@posting.google.com>...> Hello, > > In case of > > r(t) = s(t) + i(t) + n(t) > > where r(t) - received signal > s(t) - desired signal > i(t) - interference signal > n(t) - additive Gaussian noise > > then does the BER depend on Signal-to-{Interference + Noise} Ratio > (SINR) directly > > For e.g. if s(t) is Binary PSK, can I use the Q-function directly by > replacing the SNR by SINR to estimate the BER. > > Cheers, > Vimal

Reply by ●September 16, 20042004-09-16

> Hint: Consider different types of interference. > > Cheers, > Jake >If the interference, is lets say binary antipodal +/- 1 ... then can we use SINR...

Reply by ●September 18, 20042004-09-18

If the interference is another BPSK modulated signal, and if the desired user is employed with only a matched filter, then we cannot use SINR. This is as follows: After matched filtering, we have r = A1*X1 + A2*X2 + n, where A1 is the desired user's signal amplitude, X1 is the desired user's BPSK signal, A2 is the interference amplitude, X2 is the corresponding BPSK signal. It is now straightforward that, by first conditioned on X2, one can find the average BER of the desired user, which then has to be averaged over the statistics of X2. For completeness, assuming the standard deviation of n as sigma and X1 and X2 are equally likely BPSK signals, the following is the final expression for the BER: BER = 0.5*Q((A1+A2)/sigma) + 0.5*Q((A1-A2)/sigma). hope that helps Ramesh vimal_bhatia2@yahoo.com (Vimal) wrote in message news:<b6fc6dda.0409160248.6d1f88f3@posting.google.com>...> > Hint: Consider different types of interference. > > > > Cheers, > > Jake > > > > If the interference, is lets say binary antipodal +/- 1 ... then can we use SINR...

Reply by ●September 18, 20042004-09-18

> For completeness, assuming the standard deviation of n as sigma and X1 > and X2 are equally likely BPSK signals, the following is the final > expression for the BER: > > BER = 0.5*Q((A1+A2)/sigma) + 0.5*Q((A1-A2)/sigma). > > hope that helps > RameshFor more complex example, when QPSK modulation is assumed for both desired and interfered signals, can we still use a similar equation together with the following additional definitions?> BER = 0.5*( 0.5*Q((A1+A2re)/sigma) + 0.5*Q((A1-A2re)/sigma)) +0.5*( 0.5*Q((A1+A2im)/sigma) + 0.5*Q((A1-A2im)/sigma)) + where A1 = w*h1, A2re = Re{w*h2}, A2im = Im{w*h2} while w = conj{h1}/abs{h1}, in which h1 and h2 denote complex channel response and Re{}, Im{}, conj{.}, and abs{.} stand for real, image, conjugate and absolute, respectively. -- Best regards, James K. (txdiversity@hotmail.com) - Any remarks, proposal and/or indicator to text would be greatly respected. - Private opinions: These are not the opinions from my affiliation. [Home] http://home.naver.com/txdiversity

Reply by ●September 20, 20042004-09-20

Hi Vimal : you can always use the SINR directly to estimate BER but you may not get a very good estimate. If the interference is gaussian at the output of your matched filter then you'll get a good estimate , else you may well get a bad one. Best of luck - Mike "Ramesh" <ecerams@rediffmail.com> wrote in message news:b31e6a26.0409180004.5be7aca2@posting.google.com...> If the interference is another BPSK modulated signal, and if the > desired user is employed with only a matched filter, then we cannot > use SINR. This is as follows: > After matched filtering, we have > r = A1*X1 + A2*X2 + n, > where A1 is the desired user's signal amplitude, X1 is the desired > user's BPSK signal, A2 is the interference amplitude, X2 is the > corresponding BPSK signal. It is now straightforward that, by first > conditioned on X2, one can find the average BER of the desired user, > which then has to be averaged over the statistics of X2. > > For completeness, assuming the standard deviation of n as sigma and X1 > and X2 are equally likely BPSK signals, the following is the final > expression for the BER: > > BER = 0.5*Q((A1+A2)/sigma) + 0.5*Q((A1-A2)/sigma). > > hope that helps > Ramesh > > > vimal_bhatia2@yahoo.com (Vimal) wrote in messagenews:<b6fc6dda.0409160248.6d1f88f3@posting.google.com>...> > > Hint: Consider different types of interference. > > > > > > Cheers, > > > Jake > > > > > > > If the interference, is lets say binary antipodal +/- 1 ... then can weuse SINR...

Reply by ●September 28, 20042004-09-28

On Mon, 20 Sep 2004 09:12:53 +0000 (UTC), "Mike Yarwood" <mpyarwood@btopenworld.com> wrote:>Hi Vimal : you can always use the SINR directly to estimate BER but you may >not get a very good estimate. If the interference is gaussian at the output >of your matched filter then you'll get a good estimate , else you may well >get a bad one. >Best of luck - MikeMike, do you mean good performance by noting good estimation? To be more clear my question, I'd like show one example related to my question. If considered SNR is high, BER ~= 0.5*Q((A1-A2)/sigma) --> 0.5*Q(0) = 0.25 for good estimation where A1 = A2 is assumed, but for simple Gaussian approximation case which is obviously not good estimation, it is BER ~= Q(A1/(A2+sigma)) -> Q(1) = 0.16 In this case, the above result reveals that the good estimation produces worse BER. Thus, as long as my analysis is correct, the good estimation dose not (directly) mean for the good performance, right? BR, ------ James K. (txdiversity@hotmail.com) [Home] http://home.naver.com/txdiversity

Reply by ●September 29, 20042004-09-29

James K. <txdiversity@hotmail.com> wrote in message news:<3ovjl0tu6ibhlaqib5mun7sg759m3sl13l@4ax.com>...> On Mon, 20 Sep 2004 09:12:53 +0000 (UTC), "Mike Yarwood" > <mpyarwood@btopenworld.com> wrote: > > >Hi Vimal : you can always use the SINR directly to estimate BER but you may > >not get a very good estimate. If the interference is gaussian at the output > >of your matched filter then you'll get a good estimate , else you may well > >get a bad one. > >Best of luck - Mike > > Mike, do you mean good performance by noting good estimation? > > To be more clear my question, I'd like show one example related to my > question. If considered SNR is high, > > BER ~= 0.5*Q((A1-A2)/sigma) --> 0.5*Q(0) = 0.25 > > for good estimation where A1 = A2 is assumed, but for simple Gaussian > approximation case which is obviously not good estimation, it is > > BER ~= Q(A1/(A2+sigma)) -> Q(1) = 0.16 > > In this case, the above result reveals that the good estimation > produces worse BER. > > Thus, as long as my analysis is correct, the good estimation dose not > (directly) mean for the good performance, right? > > BR, > ------ > James K. (txdiversity@hotmail.com) > [Home] http://home.naver.com/txdiversityHi James, you asked:-> Mike, do you mean good performance by noting good estimation?I don't think so. I was just talking about the quality of the estimation, if it predicts low BER when it should predict high or high when it should be low then its not as good as an estimator which gives a better prediction.> Thus, as long as my analysis is correct, the good estimation dose not > (directly) mean for the good performance, right?If your analysis is correct then the better method of estimation has produced the most accurate estimate of the BER and that estimate of the BER is higher than the estimate produced by the worse estimation method , right. Best of luck -Mike