Wireless

What are the Degrees-of-Freedom of a MIMO point-to-point channel?

The intuition behind degrees-of-freedom in a wireless channel is to describe the number of interference-free transmissions that the channel can support simultaneously. This is also sometimes called the multiplexing gain.

Jake

Oct 26, 2021 • 3 min read

The intuition behind degrees-of-freedom (DoF) in a wireless channel would be the number of interference-free transmissions that the channel can support simultaneously. This is also sometimes called the multiplexing gain. Here, we will show you how to get from the classical definition of DoF in multi-antenna (MIMO) point-to-point (PTP) channel to the multiplexing gain.

Let's consider a realization of point-to-point PTP MIMO channel matrix \(\mathbf{H}\) with \(N = \text{rank}(\mathbf{H})\). Now, using SVD and optimal power loading over the \(N\) channels (\(p_1, \ldots, p_N\)) given the transmit power limit \(P\), we get the PTP capacity in form

\[
C(P) = \sum_{n=1}^N \log_2 (1 + {\lambda_n p_n \over N_0})
\]

In the following, we always consider \(p_n\) to represent the optimal power loading with respect to the corresponding channel \(H\) and given power budget \(P\).

Next, we follow the classic definition of DoF. That is the ratio between the channel capacity and PTP capacity of a single-link Gaussian channel. In the PTP MIMO case, it becomes

\[
\text{DoF} = \lim_{P \to \infty} {C(P) \over \log_2(1 + {P / (N N_0)})}
\]

Notice, how we define the SNR in the Gaussian channel as \(P /(N N_0)\), where the transmit power is \(P \over N\). This gives us a reasonable point of comparison between the two.

Let's see what happens to the capacity if we let the power budget approach infinity.

\[
\lim_{P \to \infty} C(P) = \lim_{P \to \infty} \sum_{n=1}^N \log_2 (1 + {\lambda_n p_n \over N_0})
\]

Here, \({\lambda_n p_n \over N_0}\) dominates over 1 as \(P \to \infty\), and we can approximate the capacity at the limit by

\[
\lim_{P \to \infty} \sum_{n=1}^N \log_2 ({\lambda_n p_n \over N_0}) =
\lim_{P \to \infty} \sum_{n=1}^N (\log_2 ({p_n \over N_0}) + \log_2(\lambda_n ))
\]

Again, \(\lambda_n\) are fixed coefficients and \(\log_2 ({p_n \over N_0})\) dominates as \(P \to \infty\). We can focus our high capacity approximation to

\[ \lim_{P \to \infty} \sum_{n=1}^N \log_2({p_n \over N_0}) \]

We can see how the optimal power allocation behaves for our approximated capacity limit by considering the following optimization problem.

\[
\underset{p_n}{\text{max.}} \sum_{n=1}^N \log_2({p_n \over N_0})\ \text{s.t.}\ \sum_{n=1}^N p_n = P
\]

The optimal power loading can be found by having a look at the stationary points of the Lagrangian with respect to our variables \(p_n, n = 1, \ldots, N\)

\[
\Delta_{p_n} (\sum_{n=1}^N \log_2({p_n \over N_0}) - \beta(\sum_{n=1}^N p_n - P)) = 0.
\]

The dual variable \(\beta\) is by definition a non-negative value. This leads us to the solution for \(p_n\)
\[
{1 \over {p_n}} - \beta = 0 \Leftrightarrow p_n = {1 \over \beta}, n = 1,\ldots,N
\]

Thus, at high SNR, power is equally divided among the channels and optimal power allocation becomes

\[
p_n = {P \over N}
\]

This leads to the following limit for the PTP MIMO channel, at high SNR,

\[ \lim_{P \to \infty} \sum_{n=1}^N \log_2({p_n \over N_0}) = \lim_{P \to \infty} \sum_{n=1}^N \log_2({P \over {N N_0}}) = \lim_{P \to \infty} N \log_2({P \over {N N_0}}) \]

Finally, we can plug it into the definition of DoF and get

\[
\text{DoF} = \lim_{P \to \infty} {C(P) \over \log_2(1 + {P / (N N_0)})} = \lim_{P \to \infty} {C(P) \over \log_2({P/N_0})} = \lim_{P \to \infty} {N \log_2({P / (N N_0)}) \over \log_2({P / (N N_0)})} = N,
\]
where \(N\) is essentially the spatial multiplexing gain of the corresponding MIMO channel.

Further reading