I am very new to the area and I have a question. I came across this example, I'd appreciate it if anyone could clarify this for me. In the paper it says:

For example, we assume a multiple access scenario with 4 activated users $\mathcal{X} = \{1,2,3,4\}$. The equivalent transmission powers of 4 users are assumed as $P_1 = 20 \text{dBm}$, $P_2 = 15 \text{dBm}$, $P_3 = 10 \text{dBm}$, $P_4 = 5 \text{dBm}$ and the power of AWGN is assumed as $N_0 = 0 \text{dBm}$. Hence the theoretical throughput achieved via direct superposition scheme can be calculated as: $$ \sum\limits_{i=1}^{4} R_i \leq \log_2{\left( 1+ \dfrac{\displaystyle\sum_{i=1}^4 P_i}{N_0} \right)} = 7.19 \text{bits/s/Hz} $$

My question is how does this equation hold? In paper, additive white gaussian noise (AWGN) is given as $N(0,N_0)$.

Here is the link for the paper (page 2, Equation 3).


1 Answer 1


That seems to work out. If I do the calculation in Python:

import numpy as np

N0 = 0.001
# P1=20dBm, P2=15dBm, P3=10dBm, P4=5dBm

P1 = np.power(10.0, 20/10)*N0
P2 = np.power(10.0, 15/10)*N0
P3 = np.power(10.0, 10/10)*N0
P4 = np.power(10.0, 5/10)*N0

sum = np.log2(1 + P1/N0 + P2/N0 + P3/N0 + P4/N0)

then I get:


Which tallies with the number you get to three significant figures.


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