I am working on MIMO capacity. The model is
Y_{M*1}=G_{M*N}S_{N*1}+N_{M*1}
The entries of G are generated by rand(0,1).
C = log_2( det(I_{N*N} + SNR * S*S' * G'*G) )
In waterpouring,
C_1 = log_2( det(I_{N*N} + SNR * Gamma * lamda(G'*G)) )
In equal power allocation,
C_2 = log_2( det(I_{N*N} + SNR * I_{N*N} * G'*G) )
The simulations show that C_2 > C_1 in low SNR region.
So strange !
That would seem strange. But ...
Did you run across this paper?
http://jwcn.eurasipjournals.com/content/2014/1/228
Very well written, I think. The main intent of the paper is to show how a modified waterpouring algorithm, in which you select some but not necessarily all of the MIMO antennas, can help (for cost reasons), compared with the normal waterpouring approach. But to get to this, they also cover the equal power MIMO case, compared with the waterpouring algorithm.
Their results show a dependency on (1) knowledge of channel conditions by the transmitter (or lack of knowledge) and (2) on the correlation matrix.
It's logical that if channel conditions are unknown, best go with equal power MIMO, at all SNR values. And conversely, waterpouring gives the best results when channel conditions are completely known. Again, this holds true for all SNR values.
More to the point, though, if you assume an identity matrix for the correlation matrix, all paths uncorrelated, they show that for all SNR values, the best results are those which use the same power to all MIMO antennas. And the opposite is also true. For a completely correlated case, you're better off with just one transmit antenna, also true for all SNR values. (Figures 5 and 6.)
So, as far as this paper is saying, it would appear that the advantages of the waterpouring algorithm occur if you know the channel conditions well and if the paths are somewhere between completely correlated and completely uncorrelated. And that this holds true for all SNR values.
Excuse me, Mr. Moradnoory, but all you have done is copy what I wrote? And you even got a whopping 5 upvotes for the answer! Congratulations are in order!
Or am I not seeing some more text that you contributed in your answer?
can anyone plz tell me the cdf formula without using H matrix..???
Because i m trying to plot CDF vs Capacity graph for my MIMO-OFDM simulink model..
and in simulink H matrix is directly computed from the model .
also how to plot trhouhput vs no. of users or capacity vs no. of users graph for mimo??
any help or suggestion will be heartedly appretiated ..
i m in great trouble for my project .
The water-pouring algorithm can NOT be poorer than equal-power allocation at ANY SNR. If implemented correctly it should provide the optimum solution which is for low SNR: putting all the power in one dimension (being the eigenmode for MIMO). In fact at low SNR the relative capacity gap of water-filling (WF) from equal power allocation is high.
May be the problem is in implementing your WF capacity or in the WF algo itself. In the WF capacity expression, Gamma should be a diagonal matrix and so it lamda which is the diag of the svd of H'H. You are also missing to divide the SNR by Nt in both equations.
The det is on the # of allowable eigenmodes by the WF algo. You can also express it as a sum of log terms running through the virtual MIMO channel, it should yield the same answer.
I tried to go through the WF algo code but it is not commented at all. You need to clearly define all your parameters. You can always do some sanity check by getting the sum of the WF algo output and check if it meets the total Power budget. And also check that at low SNR, it yields one power value.
I would suggest reading the relevant sections in the following neat book chapters: http://www.eecs.berkeley.edu/~dtse/Chapters_PDF/Fundamentals_Wireless_Communication_chapter7.pdf and http://www.ifp.illinois.edu/~pramodv/Chapters_PDF/Fundamentals_Wireless_Communication_chapter8.pdf
@Ramy M. Abdallah
Thanks very much. Through debugging for several days, I found that the algorithm has no problem. But the problem lies in the eigenvalue decomposition. If I_{N*N} >> SNR * Gamma * lamda(G'*G)) , then wf is better than epa.
On the other hand, in the high SNR region, epa achieves the similar capacity as the wf.
Thanks again.
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