├── OFDM.m
└── waterfilling.m


/OFDM.m:
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 1 | %% Characteristics & Transmitter
 2 | c =[1 1];
 3 | N = length(c);
 4 | L=512; % Subcarriers
 5 | P0 = 1000; % Transmit power
 6 | Lx = 100; % simulate Lx blocks
 7 | G = abs(fft(c,L)); % L point FFT of channel
 8 | SNR = (G.^2); % SNR for a channel, with noise variance 1.
 9 | D = waterfilling(SNR,P0);
10 | Diag = diag(D);
11 | T = dftmtx(L)/sqrt(L); % L-point DFT matrix, scaled to be unitary
12 | x = randn(1,L*Lx); XX = zeros(L,Lx); % signal
13 | XX(:) = x; % demultiplex
14 | XX2 = Diag*XX;
15 | X = T'*XX2; % IDFT
16 | S = [X(end-N+2:end,:); X]; % cyclic prefix
17 | s = S(:); % multiplex
18 | %% Noisy and Dispersive Channel
19 | v = 0.0 * (randn(Lx*(L+N-1),1) + 1i*randn(Lx*(L+N-1),1))/sqrt(2);
20 | r = filter(c,1,s) + v; % transmission over channel
21 | %% OFDM Receiver and Equaliser
22 | R= zeros(L+N-1,Lx);
23 | R(:) = r; % demultiplex
24 | R2 = T*R(N:end,:); % prefix removal and DFT
25 | S_hat = pinv(Diag)*pinv(diag(fft(c,L)))*R2; % ZF equalisation
26 | %% MSE
27 | for i= 1:Lx,
28 | MSE(:,i) = sum((S_hat(:,i) - XX(:,i)).^2)/L;
29 | end
30 | 


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/waterfilling.m:
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 1 | function D = waterfilling(SNR,P0);
 2 | % D = waterfilling(SNR,P0);
 3 | %
 4 | % Perform waterfilling to maximise the capacity of a system with
 5 | % subchannels with given SNRs under a constraint P0 for the transmit
 6 | % power.
 7 | %
 8 | % Input parameters:
 9 | %   SNR     column vector of subchannel SNRs  
10 | %   P0      total transmit power budget
11 | %
12 | % Output parameter:
13 | %   D       column vector containing optimised subchannel gains
14 | %           applicable at the transmitter's precoder.
15 | %  
16 | % S. Weiss, UoS, 2/5/2007
17 |   
18 | L = length(SNR);      % number of subchannels  
19 | epsilon = 10^(-14);   % avoid division by zero
20 |   
21 | % order the inverse subchannel SNRs from smallest to largest 
22 | [SNRinv,Index] = sort(1./(SNR+epsilon)); 
23 | 
24 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
25 | % part 1:
26 | % determine how many subchannels K<=L can be supported 
27 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
28 | K=L;   % first consider taking all (K=L) subchannels
29 | mu = SNRinv(K);
30 | P = sum(mu-SNRinv);
31 | while (P>P0),
32 |   K=K-1;
33 |   mu = SNRinv(K);
34 |   P = sum((mu-SNRinv).*((mu-SNRinv)>0));
35 | end;    
36 | % K now contains the number of subchannels that can be "covered by
37 | % water" (i.e. utilised for transmission)
38 | 
39 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
40 | % part 2:
41 | % P is the power required to service the K strongest subchannels,
42 | % i.e. water just reaches up to the Kth subchannel.
43 | % The remaining (P0-P) is uniformly poored across these K subchannels.
44 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
45 | mu = mu + (P0-P)/K;
46 | SNRinv = SNRinv(:);   % enforce column vector 
47 | D2 = max(zeros(L,1), mu-SNRinv);
48 | 
49 | % turn power into gain and re-order subchannels
50 | D(Index) = sqrt(D2);
51 | 


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