# How are simultaneous transmissions decoded at the receiver?

I am finding it hard to understand how does a receiver differentiate between the interference and the signal intended for the corresponding receiver. For example in a case where two user pairs are communicating on the same channel simultaneously in a device-to-device (D2D) communication fashion (Scenario 1 in the attached figure), there are many papers that show that the achievable rate of the receiver-1 is given by: \begin{align} \log_2(1+ P_1 h_1/(P_2 f_2 +\sigma^2)) \end{align} where $$\P_1\$$ is the transmission power of transmitter-1, $$\h_1\$$ is the channel gain from transmitter-1 to receiver-1, similarly $$\f_2\$$ is the gain of interference channel from transmitter-2 to receiver-1, $$\P_2\$$ denotes the transmission power of the transmitter-2 and $$\\sigma^2\$$ is the variance of additive white Gaussian noise. Then the achievable rate of the receiver-2 is given by:

\begin{align} \log_2(1+ P_2 h_2/(P_1f_1+\sigma^2)) \end{align} The signal received at the receiver-1 will contain both symbols: \begin{align} y=\sqrt{P_1 h_1}x_1+\sqrt{P_2 f_2}x_2 \end{align} where $$\x_1\$$ is the symbol intended for the receiver-1 and $$\x_2\$$ is the symbol for receiver-2. I am unable to understand how the receivers separate these symbols, after separation how do they know which symbol was sent by their transmitter and not by the other one?

The best solution I could find in some papers is through a technique called successive interference cancellation where we first decode the signal received with more power, then we subtract it from the total received signal and after that, we decode the signal that was transmitted with less power. Is there some way we can directly decode the signal transmitted with less power?

What happens if somehow we have $$\P_1 h_1\$$=$$\P_2 f_2\$$? Especially, in the uplink scenario where two transmitters send data to the same receiver simultaneously (Scenario 2 of the attached figure). If the SINR of both the transmitters are the same, can the receiver decode both signals? How is it possible? I do understand that when the interference is very low the receiver can use direct decoding while treating the interference as noise, but why do so many published papers consider that communication is still possible in high interference scenarios?

I have read papers considering similar scenarios [1],[2], [3], but I failed to find any paper that could help me in answering these questions. I do know that there are practical systems that operate in similar scenarios (e.g. GPS receivers, receive signals from multiple satellites simultaneously on the same frequency), but I could not find how the signals are separated in this case.

I do understand that there are so many questions I have asked and that I am very confused, therefore, please know that I will also really appreciate someone pointing me towards a paper, article, or a book that addresses these questions.

[1] Andreotti, R., Marchetti, L., Sanguinetti, L., & Debbah, M. (2014, December). Distributed power control over interference channels using ACK/NACK feedback. In 2014 IEEE Global Communications Conference (pp. 4186-4190). IEEE.

[2] I. AlQerm and B. Shihada, ”Enhanced Online Q-Learning Scheme for Energy Efficient Power Allocation in Cognitive Radio Networks,” IEEE Wireless Communication and Networking Conference (WCNC), 2019, pp. 1-6.

[3] A. Attar, O. Holland, M. R. Nakhai and A. H. Aghvami, ”Interference-limited resource allocation for cognitive radio in orthogonal frequency division multiplexing networks,” in IET Communications, vol. 2, no. 6, pp. 806-814, July 2008.

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• Are you leaving out some basic context about what kind of systems you're asking about? For example, is this related to CDMA systems or some other class of systems? Certainly this isn't how it works when FM radio stations interfere with each other, for example. Giving links to some of the papers you referred to might help us understand what you're asking about. Commented Oct 28, 2021 at 2:27
• @ThePhoton thank you for the suggestion, I have added references to some of the papers.
– ZZ1
Commented Oct 28, 2021 at 6:49

You will always get interference if two TX-RX pairs utilise some transmission channel (i.e. use the same frequency bands to transmit) at the same time. The extent to which the interference will hamper operation will depend on the manner in which transmission takes place. Usually, you would want to multiplex the signals to minimise interference. This can be done in time (TX-RX pairs operate at different times), in frequency (TX-RX pairs utilise different frequencies) and using codes. The first two methods are pretty intuitive to understand, code based set-ups (CDMA) need a little bit of thinking about. I will illustrate a small worked out example of the core idea below.

Assume you wanted to transmit some signal with amplitude $$\A\$$ from sender $$\1\$$ to receiver $$\1\$$ and you also wanted to transmit some signal $$\B\$$ from sender $$\2\$$ to receiver $$\2\$$ (both signals using frequency $$\f_1\$$). First looking at the first TX-RX pair - instead of sending the signal at $$\f_1\$$, you would instead send $$\n\$$ smaller amplitude signals at frequencies $$\f_1\$$ to $$\f_n\$$. The first TX-RX pair would do this by first agreeing to utilise a common code which could just be randomly generated $$\n\$$ element array of $$\-1\$$'s and $$\1\$$'s. Let the code used by the first pair be denoted as $$\r_A^n\$$ and the $$\i^{th}\$$ element of the code be denoted as $$\r_A^n(i)\$$.

We can then send a signal with amplitude $$\\frac{Ar_A^n(1)}{n}\$$ at frequency $$\f_1\$$, also $$\\frac{Ar_A^n(2)}{n}\$$ at frequency $$\f_2\$$ and so on. At the receiver the signal received over frequency $$\i\$$ is multiplied by $$\r_A^n(i) \$$ to recover $$\A/n\$$. Because: $$\ \left(\frac{Ar_A^n(i)}{n} \times r_A^n(i) = \frac{A}{n}\right)\$$. Summing all these $$\A/n\$$ values up will allow us to recover $$\A\$$. Now if another TX-RX pair is trying to do the same thing then instead of receiving $$\\frac{Ar_A^n(i)}{n}\$$ you would receive $$\\frac{Ar_A^n(i)}{n} + \frac{Br_B^n(i)}{n}\$$.

$$\left( \frac{Ar_A^n(i)}{n} + \frac{Br_B^n(i)}{n} \right) \times r_A^n(i) = \frac{A}{n} + \frac{Br_B^n(i)r_A^n(i)}{n}$$

Summing these up like we did before

$$A + \frac{1}{n}\sum _{i = 1} ^{n} Br_B^n(i)r_A^n(i)$$

Taking the expected value

$$\mathbb{E} \left \{ A + \frac{B}{n}\sum _{i = 1} ^{n} r_B^n(i)r_A^n(i) \right \} = A + 0$$

The above is due to the fact that if $$\r_B^n(i)r_A^n(i)\$$ is equally likely to evaluate to $$\-1\$$ as $$\1\$$. Then values in the second term will cancel themselves out when you sum them up. If you also had additive noise you would receive and process the quantity below at frequency $$\i\$$

$$\left( \frac{Ar_A^n(i)}{n} + \frac{Br_B^n(i)}{n} + N(i) \right) \times r_A^n(i) = \frac{A}{n} + \frac{Br_B^n(i)r_A^n(i)}{n} + N(i)r_A^n(i)$$

Which still will sum up to an expected value of $$\A\$$. So the above illustrates the general idea of spreading a signal over a wider bandwidth using a shared code (CDMA). In practice, you be mapping a smaller bandwidth to a much larger frequency band (and not just mapping a single frequency) and the codes would be chosen more carefully to ensure orthogonality but the core idea stays the same.