Fair comparison between time domain equalizer and OFDM
in [DSP]
|
Prev: MAKE UPTO $5000 MONTHLY! $2000 INYOUR FIRST 30 DAYS!
Next: Last Call for Papers Reminder (extended): World Congress on Engineering and Computer Science WCECS 2010
From: cpshah99 on 16 Jul 2010 10:39 >cpshah99 wrote: >> Hi All >> >> So far I have worked on time domain equalizers such as LE and DFE. Recently >> I have started to work on OFDM systems. And I am trying to compare the >> performance for Proakis channel B which is three tap channel. >> >> First system: map the info bits to BPSK, pass the symbols through channel, >> add noise. The channel output I write it as y=conv(h,x)+noise, where >> h=[0.407 0.815 0.407] and x= +/- 1. At receiver perform LE using MMSE under >> the assumption that the channel is knonwn at the receiver. I get the exact >> BER plot given in Proakis comms edition 4. >> >> Second system: map the info bits to BPSK, perform IFFT, add cyclic prefix, >> pass it through channel and add noise. The channel output >> y=conv(h,x)+noise, where x=IFFT(X,N), X=+/- 1 and N=2048. At receiver, >> remove the CP, truncate the signal and performe one tap MMSE equalization >> i.e. \hat{X}=fft(H,N)/fft(y,N) and then take hard decision on real part. >> > >shoudln't this be \hat{X}=fft(y,N)/fft(h,N) ? > >Laurent > Hi everybody... Thanks for your reply. sorry for the typo. it is indeed \hat{X}=fft(y,N)/fft(h,N). When i say the BER is way off, i mean there is a huge gap in performance. For example, system 1 gives BER of 10^-4 at 30 dB, where as system 2 gives BER of 10^-2 at 30 dB. When I remove the noise and send only +1s, after the equalization, I do not see phase rotation. And I also get 0 BER without noise on any channel. My matlab code is acurate, as it passes thru all the test cases. I can post it if anybody want to have a look. I do not think, even coding wil any good job since I can introducing coding to single carrier systems as well and it wil also perform well. I feel something has to be changed. Any hint where I could be making a mistake? Chintan
From: Mark on 16 Jul 2010 12:15 On Jul 16, 10:39 am, "cpshah99" <cpshah99(a)n_o_s_p_a_m.rediffmail.com> wrote: > >cpshah99 wrote: > >> Hi All > > >> So far I have worked on time domain equalizers such as LE and DFE. > Recently > >> I have started to work on OFDM systems. And I am trying to compare the > >> performance for Proakis channel B which is three tap channel. > > >> First system: map the info bits to BPSK, pass the symbols through > channel, > >> add noise. The channel output I write it as y=conv(h,x)+noise, where > >> h=[0.407 0.815 0.407] and x= +/- 1. At receiver perform LE using MMSE > under > >> the assumption that the channel is knonwn at the receiver. I get the > exact > >> BER plot given in Proakis comms edition 4. > > >> Second system: map the info bits to BPSK, perform IFFT, add cyclic > prefix, > >> pass it through channel and add noise. The channel output > >> y=conv(h,x)+noise, where x=IFFT(X,N), X=+/- 1 and N=2048. At receiver, > >> remove the CP, truncate the signal and performe one tap MMSE > equalization > >> i.e. \hat{X}=fft(H,N)/fft(y,N) and then take hard decision on real > part. > > >shoudln't this be \hat{X}=fft(y,N)/fft(h,N) ? > > >Laurent > > Hi everybody... > > Thanks for your reply. > > sorry for the typo. it is indeed \hat{X}=fft(y,N)/fft(h,N). > > When i say the BER is way off, i mean there is a huge gap in performance. > For example, system 1 gives BER of 10^-4 at 30 dB, where as system 2 gives > BER of 10^-2 at 30 dB. > > When I remove the noise and send only +1s, after the equalization, I do not > see phase rotation. And I also get 0 BER without noise on any channel. > > My matlab code is acurate, as it passes thru all the test cases. I can post > it if anybody want to have a look. > > I do not think, even coding wil any good job since I can introducing coding > to single carrier systems as well and it wil also perform well. > > I feel something has to be changed. > > Any hint where I could be making a mistake? > > Chintan- Hide quoted text - > > - Show quoted text - you are saying you are testing in simulation a BPSK system and OFDM system with 2 impairments, 1) noise and 2)channel you are saying when tested with both impairments, the BPSK BER is much better. you are saying when tested with only 1 impairment, i.e without the noise impairment, both are good... so do the reverse case, test with only one imparment, the noise impairment but without the channel impairment... Mark
From: cpshah99 on 16 Jul 2010 12:50 >you are saying you are testing in simulation a BPSK system and OFDM >system with 2 impairments, 1) noise and 2)channel > >you are saying when tested with both impairments, the BPSK BER is much >better. > >you are saying when tested with only 1 impairment, i.e without the >noise impairment, both are good... > >so do the reverse case, test with only one imparment, the noise >impairment but without the channel impairment... > >Mark > Hi Mark I am testing two systems, one uses time domain LE and other uses OFDM. In both the cases I have used BPSK and a three tap proakis channel. The BER plot of system one i.e. time domain equalization, is absolutely correct as it matches with the BER plot given in the book. However, my problem starts with this OFDM system, where if I remove the channel impairment and just add noise, which is basically AWGN, I get the exact theoretical BER plot. Also, when I remove the noise and add channel impairment, I get 0 BER. So guys, I am not sure where to look. I am sure some of you must have done similar stuff. Thanks. Chintan
From: cpshah99 on 16 Jul 2010 14:08 OK boys Here is an update. As we all know that each subcarrier in OFDM undergoes flat fading. So what I did that I changed the channel in my simulation i.e. the channel is changing per OFDM symbol. Now, when I measure the BER, it is exactly same to the theoretical BER plot of BPSK on flat-Rayleigh fading channel. The length of channel is 4 taps and the length of CP is 5. Does this result make sense? Your opinion and help is always appreciated. Thanks Regards Chintan
From: Sebastian Doht on 16 Jul 2010 16:34
On 16 Jul., 16:39, "cpshah99" <cpshah99(a)n_o_s_p_a_m.rediffmail.com> wrote: > >cpshah99 wrote: > >> Hi All > > >> So far I have worked on time domain equalizers such as LE and DFE. > Recently > >> I have started to work on OFDM systems. And I am trying to compare the > >> performance for Proakis channel B which is three tap channel. > > >> First system: map the info bits to BPSK, pass the symbols through > channel, > >> add noise. The channel output I write it as y=conv(h,x)+noise, where > >> h=[0.407 0.815 0.407] and x= +/- 1. At receiver perform LE using MMSE > under > >> the assumption that the channel is knonwn at the receiver. I get the > exact > >> BER plot given in Proakis comms edition 4. > > >> Second system: map the info bits to BPSK, perform IFFT, add cyclic > prefix, > >> pass it through channel and add noise. The channel output > >> y=conv(h,x)+noise, where x=IFFT(X,N), X=+/- 1 and N=2048. At receiver, > >> remove the CP, truncate the signal and performe one tap MMSE > equalization > >> i.e. \hat{X}=fft(H,N)/fft(y,N) and then take hard decision on real > part. > > >shoudln't this be \hat{X}=fft(y,N)/fft(h,N) ? > > >Laurent > > Hi everybody... > > Thanks for your reply. > > sorry for the typo. it is indeed \hat{X}=fft(y,N)/fft(h,N). > > When i say the BER is way off, i mean there is a huge gap in performance. > For example, system 1 gives BER of 10^-4 at 30 dB, where as system 2 gives > BER of 10^-2 at 30 dB. > > When I remove the noise and send only +1s, after the equalization, I do not > see phase rotation. And I also get 0 BER without noise on any channel. > > My matlab code is acurate, as it passes thru all the test cases. I can post > it if anybody want to have a look. > > I do not think, even coding wil any good job since I can introducing coding > to single carrier systems as well and it wil also perform well. > > I feel something has to be changed. > > Any hint where I could be making a mistake? > > Chintan It is probably not the source of your errors, but you wrote that you are performing an MMSE equalization while the formula you post is essentially a zero-forcing (ZF) equalizer. It has been a few years since I dealt with this topic but shouldn't the MMSE equalizer (which in my understanding should be a Winer-Filter) be something like Y(f) = (conj(H(f)))/((abs(H(f))).^2+(1/SNR)) There should be difference in performance for low SNR. Greetz, Sebastian |