EbNo and SNR
in [DSP]
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From: TZ on 8 Oct 2009 01:18 Hi Jason, Thanks for your reply. The reason I was asking that was because I was struggling with a different but related issue: Apparently Eb/No I have found out, is NOT equal to SNR. Upon further investigation, I found this: Eb/No = SNR / rho , where rho = 'spectral efficiency', and its units are 'bits/Hz'. For a BPSK system apparently, rho = 2 bits/Hz. Why is this the case? And is this related to what we just discussed? Furthermore, what would the 'spectral efficiency' be for QPSK? Or 4-QAM?, etc etc?... Thanks a million, Jason. -TZ
From: maury001 on 8 Oct 2009 10:51 On Oct 8, 12:18 am, "TZ" <tarin.ziy...(a)gmail.com> wrote: > Hi Jason, > > Thanks for your reply. > > The reason I was asking that was because I was struggling with a different > but related issue: > > Apparently Eb/No I have found out, is NOT equal to SNR. Upon further > investigation, I found this: > > Eb/No = SNR / rho > > , where rho = 'spectral efficiency', and its units are 'bits/Hz'. > > For a BPSK system apparently, rho = 2 bits/Hz. > > Why is this the case? And is this related to what we just discussed? > Furthermore, what would the 'spectral efficiency' be for QPSK? Or 4-QAM?, > etc etc?... > > Thanks a million, Jason. > -TZ If you google "eb/no explained" you will see a link to an article written by Jim Pearce that not only explaines this, but gives an example. Maurice Givens
From: Eric Jacobsen on 8 Oct 2009 11:54 On 10/8/2009 7:51 AM, maury001(a)core.com wrote: > On Oct 8, 12:18 am, "TZ"<tarin.ziy...(a)gmail.com> wrote: >> Hi Jason, >> >> Thanks for your reply. >> >> The reason I was asking that was because I was struggling with a different >> but related issue: >> >> Apparently Eb/No I have found out, is NOT equal to SNR. Upon further >> investigation, I found this: >> >> Eb/No = SNR / rho >> >> , where rho = 'spectral efficiency', and its units are 'bits/Hz'. >> >> For a BPSK system apparently, rho = 2 bits/Hz. >> >> Why is this the case? And is this related to what we just discussed? >> Furthermore, what would the 'spectral efficiency' be for QPSK? Or 4-QAM?, >> etc etc?... >> >> Thanks a million, Jason. >> -TZ > > If you google "eb/no explained" you will see a link to an article > written by Jim Pearce that not only explaines this, but gives an > example. > > Maurice Givens Nice find. That's a really good article with a good discussion following. BTW, Phil Karn's point about channel BW is important. -- Eric Jacobsen Minister of Algorithms Abineau Communications http://www.abineau.com
From: Raymond Toy on 8 Oct 2009 12:05 >>>>> "TZ" == TZ <tarin.ziyaee(a)gmail.com> writes: TZ> Hi Jason, TZ> Thanks for your reply. TZ> The reason I was asking that was because I was struggling with a different TZ> but related issue: TZ> Apparently Eb/No I have found out, is NOT equal to SNR. Upon further TZ> investigation, I found this: People define SNR in many different ways. Typically SNR is the received signal power divided by the received noise power. But Eb/No is pretty well defined. Eb is the energy per bit and No is the power spectral density. No is not the received noise power, because the receive filter removes a fair amount of noise. TZ> Eb/No = SNR / rho TZ> , where rho = 'spectral efficiency', and its units are 'bits/Hz'. TZ> For a BPSK system apparently, rho = 2 bits/Hz. Why wouldn't rho closer to 3? 8 = 2^3 after all. But the conversion from SNR to Eb/No depends on the number of bits and the filter bandwidth. The filter BW will give the relationship between the received noise to the noise spectral density. Ray
From: Eric Jacobsen on 8 Oct 2009 13:22 On 10/8/2009 9:05 AM, Raymond Toy wrote: >>>>>> "TZ" == TZ<tarin.ziyaee(a)gmail.com> writes: > > TZ> Hi Jason, > > TZ> Thanks for your reply. > > TZ> The reason I was asking that was because I was struggling with a different > TZ> but related issue: > > TZ> Apparently Eb/No I have found out, is NOT equal to SNR. Upon further > TZ> investigation, I found this: > > People define SNR in many different ways. Typically SNR is the > received signal power divided by the received noise power. But Eb/No > is pretty well defined. Eb is the energy per bit and No is the power > spectral density. No is not the received noise power, because the > receive filter removes a fair amount of noise. > > TZ> Eb/No = SNR / rho > > TZ> , where rho = 'spectral efficiency', and its units are 'bits/Hz'. > > TZ> For a BPSK system apparently, rho = 2 bits/Hz. > > Why wouldn't rho closer to 3? 8 = 2^3 after all. But the conversion > from SNR to Eb/No depends on the number of bits and the filter > bandwidth. The filter BW will give the relationship between the > received noise to the noise spectral density. > > Ray Actually, for uncoded BPSK rho is 1 bits/Hz, even less if coding is applied. I suspect you read the B as an 8. Time to go up a notch on the reading glasses. ;) Also, for single-carrier systems with Nyquist filtering in the receiver the "channel" bandwidth is generally equal to the symbol rate. In other words, the 3dB points of the filter match the symbol rate due to the Nyquist (or matched) pulse filter. So, generally, the noise BW matches the symbol rate, which makes the straightforward computation above practical as long as one knows how many bits/symbol the system transports. -- Eric Jacobsen Minister of Algorithms Abineau Communications http://www.abineau.com
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