Capacity for QAM
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From: hramcha on 19 Mar 2010 08:01 Guys, I need a clarification on shannon capacity equation for QAM systems with coding. Lets assume M-QAM with code rate R Let m = log2(M) My goal is to find at what SNR (Eb/No) we achieve capacity with the shannon equation. C= Wlog2(1+SNR). I am a little confused. can someone please explain with an example value of M and R Thanks H
From: Mark on 19 Mar 2010 12:13 On Mar 19, 8:01 am, "hramcha" <hramcha(a)n_o_s_p_a_m.gmail.com> wrote: > Guys, > > I need a clarification on shannon capacity equation for QAM systems with > coding. > > Lets assume M-QAM with code rate R > > Let m = log2(M) > > My goal is to find at what SNR (Eb/No) we achieve capacity with the shannon > equation. > > C= Wlog2(1+SNR). > > I am a little confused. can someone please explain with an example value of > M and R > > Thanks > > H The Shannon equation is also known as the Shannon "BOUND". As such it is the theoretical best capactiy that you can ever get for a given SNR. Any actual combination of modualtion and coding will be worse then the upper bound. So you can make a plot of the Shannon bound as a line and then your specific modulation and coding as a point and see how close your system has approached the upper bound. The Shanoon equation does not tell you the actual capacity of any particualr system, it tells you the theoretical upper limit. Mark
From: Tim Wescott on 19 Mar 2010 12:38 Mark wrote: > On Mar 19, 8:01 am, "hramcha" <hramcha(a)n_o_s_p_a_m.gmail.com> wrote: >> Guys, >> >> I need a clarification on shannon capacity equation for QAM systems with >> coding. >> >> Lets assume M-QAM with code rate R >> >> Let m = log2(M) >> >> My goal is to find at what SNR (Eb/No) we achieve capacity with the shannon >> equation. >> >> C= Wlog2(1+SNR). >> >> I am a little confused. can someone please explain with an example value of >> M and R >> >> Thanks >> >> H > > The Shannon equation is also known as the Shannon "BOUND". As such it > is the theoretical best capactiy that you can ever get for a given > SNR. Any actual combination of modualtion and coding will be worse > then the upper bound. So you can make a plot of the Shannon bound as > a line and then your specific modulation and coding as a point and see > how close your system has approached the upper bound. > > The Shanoon equation does not tell you the actual capacity of any > particualr system, it tells you the theoretical upper limit. > > Mark Moreover, the Shannon bound isn't a bound like the wall of a room that you might lean against, or a speed limit on a freeway (that of course none of us ever exceed :-) ). It's more like the speed of light -- one can, with great effort and under special circumstances, reach this bound in this fallen and imperfect world, but it takes a lot of effort and it is fragile. In reality most systems don't approach it all that closely, and the ones that do use more elaborate modulation schemes than simple QAM. -- Tim Wescott Control system and signal processing consulting www.wescottdesign.com
From: Eric Jacobsen on 19 Mar 2010 18:20 On 3/19/2010 9:38 AM, Tim Wescott wrote: > Mark wrote: >> On Mar 19, 8:01 am, "hramcha" <hramcha(a)n_o_s_p_a_m.gmail.com> wrote: >>> Guys, >>> >>> I need a clarification on shannon capacity equation for QAM systems with >>> coding. >>> >>> Lets assume M-QAM with code rate R >>> >>> Let m = log2(M) >>> >>> My goal is to find at what SNR (Eb/No) we achieve capacity with the >>> shannon >>> equation. >>> >>> C= Wlog2(1+SNR). >>> >>> I am a little confused. can someone please explain with an example >>> value of >>> M and R >>> >>> Thanks >>> >>> H >> >> The Shannon equation is also known as the Shannon "BOUND". As such it >> is the theoretical best capactiy that you can ever get for a given >> SNR. Any actual combination of modualtion and coding will be worse >> then the upper bound. So you can make a plot of the Shannon bound as >> a line and then your specific modulation and coding as a point and see >> how close your system has approached the upper bound. >> >> The Shanoon equation does not tell you the actual capacity of any >> particualr system, it tells you the theoretical upper limit. >> >> Mark > > Moreover, the Shannon bound isn't a bound like the wall of a room that > you might lean against, or a speed limit on a freeway (that of course > none of us ever exceed :-) ). It's more like the speed of light -- one > can, with great effort and under special circumstances, reach this bound > in this fallen and imperfect world, but it takes a lot of effort and it > is fragile. In reality most systems don't approach it all that closely, > and the ones that do use more elaborate modulation schemes than simple QAM. > There've been some BPSK and QPSK systems that have gotten within a dB or so (or even less) of capacity. It's not trivial to maintain synchronization at that low of an SNR, but it has been done. -- Eric Jacobsen Minister of Algorithms Abineau Communications http://www.abineau.com
From: dvsarwate on 19 Mar 2010 21:24
On Mar 19, 11:38 am, Tim Wescott <t...(a)seemywebsite.now> wrote: > > Moreover, the Shannon bound isn't a bound like the wall of a room that > you might lean against, or a speed limit on a freeway (that of course > none of us ever exceed :-) ). It's more like the speed of light -- one > can, with great effort and under special circumstances, reach this bound > in this fallen and imperfect world, but it takes a lot of effort and it > is fragile. In reality most systems don't approach it all that closely, > and the ones that do use more elaborate modulation schemes than simple QAM. > Actually, the Shannon bound *is* like a speed limit on a freeway. It is quite possible for a communication system to transmit at rates above the rate given by Shannon's capacity formula just as it is possible to exceed the speed limit on the freeway. It is just that it is impossible to transmit **reliably** at rates above the capacity. If one wants to continue the analogy, it is not possible to exceed the speed limit and simultaneously achieve good gas mileage. For any given transmit power and bandwidth (and noise spectral density), if 16-QAM gives good error rates at (say) 50% of the Shannon capacity, then using 1024-QAM with the same transmit power, symbol rate (and hence bandwidth) will give a system that will operate at 125% of the Shannon capacity. Of course, the BER will be so close to 50% that nobody will want to use such a system, but hey, it *is* transmitting 10 bits per symbol rather than 4, and *is* operating above capacity as was wanted. Who cares about the BER? Hope this muddies the waters further... --Dilip Sarwate |