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Hi,
I have few questions :
After going through the .mdl file, it seems that
1) Preambles are being sent based on orthogonality in time i.e. sending one preamble per transmitter in round robin fashion. Hence, channel estimation is done using the above concept.
2) The MIMO Detection has been done using ZF detector. Could you please elaborate what is
YA*[HBB,-HAB] AND YB*[-HBA, HAA] ?
Any help is greatly appreciated.
Thanks,
Mauvais Garcon.
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You're right on both counts.
The YA*[HBB,-HAB] AND YB*[-HBA, HAA] blocks are time-shared multipliers that realize part of the matrix multiplication for the ZF equalization. The Y terms are the received symbols from antennas A and B; the H terms are the channel coefficients (where HAB is the estimated channel coefficient from Tx antenna A to Rx antenna B).
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hi,
I thought about your answer. But is matrix rows [Hbb -Hab] [-Hba Haa] some kind of reduction of moore-penrose pseudoinverse ? Because ideally,
Azf = /H\ * r; where /H\ = moore-penrose pseudoinverse.
Also, from the literature, /H\ = (Ht * H)^-1 * Ht where Ht is the hermitian transpose. Now, I am not able to figure out how did you guyz reduce this matrix.
Thanks,
Vijayant Bhatnagar.
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We're doing a true (non-pseudo) matrix inverse.
The transmitter sends vector X (X = [x_A;x_B], where each x_ is a complex sample from the OFDM transmitter).
The receiver receives Y = [y_A; y_B], where each y_ is a complex sample received at one antenna.
Y = HX, where H is the 2x2 channel matrix (H = [h_AA h_BA; h_AB h_BB]).
So the equalized symbols are X_eq = inv(H) * Y. Because H is just 2x2, we do the inverse directly (inv(H) = [h_BB -h_AB; -h_BA h_AA]./det(H)), then multiply the resulting matrix by the received vector Y to recover the transmitted vector X.
There are, of course, many ways to improve this approach, especially in ways that would let it work over wider ranges of channel coefficients.
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I actually figured out that the above method is valid because we have a matrix with full rank (assuming linear independence and Tx & Rx = 2). Hence calculation of inverse is justified. Anyways, I appreciate your answer.
I would also like to quote that there are few "expensive" methods like : greville's method, frobenius method etc.
Best,
Vijayant Bhatnagar
Last edited by veebee (2008-Apr-28 15:18:02)
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