Reference no: EM131313444
Download the MATLAB program "ofdm_alloc.m", which simulates power and bit allocation for a 64 channel OFDM system, assuming that all channel symbol streams have equal power allocation Pk = E [|sk[m]|2] = 1, and that the total input power P = ΣPk = 64. This program has step-by-step comments telling you what to do, but you need to fill in the missing MATLAB code by replacing the question marks with one or more lines of code.
There is more than one way to fill in the missing code, and anything that works will earn full credit. You should refer to the OFDM lecture notes (especially pages 11-13 and 15-19) and/or the OFDM reading assignment (especially pages 15-18) when doing this problem.
In addition to your MATLAB code, for this problem you need to turn in:
(i) the five Eb/N0 values that give Pb = 10-6 for each of the five modulations (this is the EbNOth array):
(ii) the graph of |H[n]|2 versus n;
(iii) the bit allocations for the 64 subbands for the first two runs (Ki array in the code);
(iv) the total mean square channel frequency response power EHtot_mean (which should be close to 6.4), and
(v) the mean value Ktot mean of the total number of bits allocated to each 64-subchannel OFDM frame. Note that it is normal for some subbands in the Ki array to have zero allocated bits; this happens when that subband's Et To is not higher than the Eh/No threshold for QPSK to achieve the target Pb of 10-6.
d. Modify the program to adaptively allocate the symbol powers Pk according to text equation 13.74, and then turn in items (iii)-(v) listed in part
a. Has tot _mean increased with this non-equal power allocation? Can you explain why or why not by comparing the bit allocations for the first two runs (item (iii)) between parts c and d?
e. Now set the sigma 2 variable (which specifies the AWGN noise variance) to 0.01, instead of its original value of 0.04, and repeat parts c and d, again turning in items (iii)-(v) for each case. Which power allocation scheme (equal power or adaptive power) gives higher Ktot_mean under this higher SNR scenario? Can you explain why the results in this case are different than when sigma 2 equals 0.04?
f. Go back to the equal power allocation case of part c, and modify the program so that only QPSK is used (if the Eb/N0 is above the QPSK threshold). Again turn in items (iii)-(iv). How has Ktot_mean changed, and why?
g. Now fill in the part of the code where the average bit error rate Pbav over all runs is computed, and turn in the Pbav for the conditions in parts c, d, e, and f. Is Pbav less than the target Pb of 10-6 in all cases? (Hint: It should be!) Do the results for Pb help to explain the differences in the total number of bits Ktot mean for these cases?
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