I'm struggling with calculating the power on Pattern Division Multiple Access (PDMA) when transmitting on Four subcarriers

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samson zitha
samson zitha 2018-1-26
评论： Monika Singh 2023-2-24
I'm simulation the Pattern Division Multiple Access (PDMA)
where K: is number of subcarriers/channels, well after the function PDMA_signal thats were the problem starts when
i'm transmitting more than one QPSK symbol for each user k.
For example1: a working example
[signals,HS] = PDMA_signal(modulatedSymbols,PDMA_patterns,channel_gain);
signals =
-2.4742 + 2.4742i
0.3640 + 0.8079i
1.7584 - 1.7584i
2.5434 + 1.6466i
signal_power = sum(abs(signals).^2)/K;
signal_power =
7.0981
noise_power = signal_power/snr;
noise_power =
7.0981
sigma = sqrt(noise_power);
sigma =
2.6642
noise = sigma *(randn(K,size(signals,2))+1i*randn(K,size(signals,2)))/sqrt(2);
noise =
1.3682 + 0.0104i
0.6090 - 0.7717i
4.2299 - 1.5291i
-0.0663 - 0.3606i
y = signals + noise;
y =
-1.1060 + 2.4845i
0.9730 + 0.0363i
5.9883 - 3.2875i
2.4771 + 1.2860i
For example2: example where i need help
[signals,HS] = PDMA_signal(modulatedSymbols,PDMA_patterns,channel_gain);
signals =
0.1947 - 0.1947i -2.2074 + 3.2085i
0.9024 - 0.9024i 2.3989 - 0.2145i
0.7649 - 0.7649i -1.2776 + 2.9950i
-2.5269 + 2.5269i -1.6450 + 1.6450i
signal_power = sum(abs(signals).^2)/K;
signal_power =
3.9113 9.2456
noise_power = signal_power/snr;
noise_power =
3.9113 9.2456
sigma = sqrt(noise_power);
sigma =
1.9777 3.0407
noise = sigma *(randn(K,size(signals,2))+1i*randn(K,size(signals,2)))/sqrt(2);
noise =
?
y = signals + noise;
y =
?
My problem is calculating the signal of more than two symbols together not separate?

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