Ferkans — Interactive Telecom Tutor

How Fast Does Capacity Grow with SNR?

The exact capacity formulas of Sections 15.3 and 15.4 are powerful but complex. A simpler and often more insightful characterisation asks: at high SNR, how does capacity scale?

The answer is captured by the multiplexing gain (or degrees of freedom), which counts the number of independent "pipes" that the channel provides. This single number determines the high-SNR slope of the capacity curve and is the key quantity for comparing different antenna configurations.

Definition:
Degrees of Freedom (DoF)

The degrees of freedom (DoF) of a MIMO channel is the pre-log factor in the high-SNR capacity expansion:

$\mathrm{DoF} = \lim_{\text{SNR} \to \infty} \frac{C(\text{SNR})}{\log_2(\text{SNR})}$

The DoF represents the number of independent data streams that can be communicated at rates growing logarithmically with SNR. For a SISO channel, $\mathrm{DoF} = 1$ .

Definition:
Multiplexing Gain

The multiplexing gain $r$ of a MIMO communication scheme is defined as

$r = \lim_{\text{SNR} \to \infty} \frac{R(\text{SNR})}{\log_2(\text{SNR})}$

where $R(\text{SNR})$ is the data rate achieved by the scheme at a given SNR. A scheme that achieves the maximum multiplexing gain $r = \mathrm{DoF} = \min(n_t, n_r)$ is called a full-multiplexing scheme (e.g., V-BLAST with ML detection).

The multiplexing gain is a property of a specific scheme (coding + detection), while DoF is a property of the channel. The channel DoF is the maximum achievable multiplexing gain.

Definition:
High-SNR MIMO Capacity Approximation

At high SNR, the ergodic capacity of an i.i.d. Rayleigh $n_t \times n_r$ MIMO channel with no CSIT has the asymptotic expansion

$C(\text{SNR}) = \min(n_t, n_r)\log_2(\text{SNR}) + \mathcal{O}(1)$

More precisely, with $m = \min(n_t, n_r)$ and $n = \max(n_t, n_r)$ :

$C(\text{SNR}) \approx m\log_2\!\left(\frac{\text{SNR}}{n_t}\right) + \mathbb{E}\!\left[\sum_{i=1}^{m}\log_2 \lambda_i\right]$

The first term grows without bound; the second is a constant (the "offset") determined by the eigenvalue distribution.

Theorem: DoF of a MIMO Channel

For an $n_t \times n_r$ MIMO channel with i.i.d. Rayleigh fading and no CSIT, the degrees of freedom are

$\mathrm{DoF} = \min(n_t, n_r)$

That is, the capacity at high SNR grows as $C \sim \min(n_t, n_r) \log_2(\text{SNR})$ .

This means that a MIMO system provides a linear capacity increase with the number of antennas (on the smaller side), at no cost in bandwidth or total transmit power.

The i.i.d. Rayleigh channel matrix is full-rank with probability 1 (its $\min(n_t, n_r)$ singular values are all strictly positive). Each nonzero singular value contributes one spatial stream, and at high SNR, even the weakest stream contributes $\sim \log_2(\text{SNR})$ bits.

Proof

Upper bound

From Telatar's formula:

$C = \mathbb{E}\!\left[\sum_{i=1}^{m}\log_2\!\left(1 + \frac{\text{SNR}}{n_t}\lambda_i\right)\right] \leq m\, \mathbb{E}\!\left[\log_2\!\left(1 + \frac{\text{SNR}}{n_t}\lambda_{\max}\right)\right]$

Since $\lambda_{\max}$ is bounded in distribution, $\log_2(1 + \frac{\text{SNR}}{n_t}\lambda_{\max}) = \log_2(\text{SNR}) + \mathcal{O}(1)$ . Thus $C \leq m \log_2(\text{SNR}) + \mathcal{O}(1)$ .

Lower bound

For any $i$ , $\log_2(1 + \frac{\text{SNR}}{n_t}\lambda_i) \geq \log_2(\frac{\text{SNR}}{n_t}\lambda_i) = \log_2(\text{SNR}) + \log_2(\lambda_i/n_t)$ when $\text{SNR}$ is large enough that $\frac{\text{SNR}}{n_t}\lambda_i \gg 1$ .

Since $\Pr[\lambda_{\min} > 0] = 1$ for i.i.d. Rayleigh, all $m$ terms contribute, giving

$C \geq m\log_2(\text{SNR}) + \sum_i \mathbb{E}[\log_2(\lambda_i/n_t)]$

Combining: $\mathrm{DoF} = \lim \frac{C}{\log_2(\text{SNR})} = m = \min(n_t, n_r)$ . $\blacksquare$

,

Example: Comparing DoF for Different Antenna Configurations

Compare the degrees of freedom and approximate capacity at $\text{SNR} = 20$ dB for the following configurations: (a) $1 \times 1$ (SISO), (b) $4 \times 1$ (MISO), (c) $1 \times 4$ (SIMO), (d) $2 \times 2$ , (e) $4 \times 4$ .

Solution

Compute DoF for each configuration

Configuration	$n_t$	$n_r$	$\mathrm{DoF} = \min(n_t, n_r)$
SISO	1	1	1
MISO	4	1	1
SIMO	1	4	1
$2 \times 2$	2	2	2
$4 \times 4$	4	4	4

Approximate capacities at 20 dB

Using $C \approx \mathrm{DoF} \times \log_2(\text{SNR}/n_t) + \text{offset}$ :

(a) SISO: $C \approx 1 \times \log_2(100) + \text{offset} \approx 6.6 + \text{offset}$

(b) MISO: DoF = 1 but array gain adds $\log_2(4) = 2$ bits: $C \approx \log_2(100) + 2 \approx 8.6$ bits/s/Hz

(c) SIMO: Similar to MISO (array gain from combining): $C \approx \log_2(100 \times 4) \approx 8.6$ bits/s/Hz

(d) $2 \times 2$ : $C \approx 2\log_2(50) + \text{offset} \approx 11.3 + \text{offset}$

(e) $4 \times 4$ : $C \approx 4\log_2(25) + \text{offset} \approx 18.6 + \text{offset}$

The key insight: MISO and SIMO have the same DoF as SISO (they only provide array gain, not multiplexing gain). True capacity scaling requires antennas on both sides. $\blacksquare$

Quick Check

A MIMO system is upgraded from $2 \times 2$ to $4 \times 4$ while keeping the total transmit power and bandwidth constant. At high SNR, by approximately how much does the capacity increase?

The capacity approximately doubles (high-SNR slope doubles from 2 to 4)

The capacity increases by 3 dB (factor of 2 in linear scale)

The capacity increases by a constant offset but the slope stays the same

The capacity quadruples because the channel matrix has $4\times$ more entries

Correction:

The capacity approximately doubles (high-SNR slope doubles from 2 to 4)

The DoF doubles from $\min(2,2) = 2$ to $\min(4,4) = 4$ . At high SNR, $C \sim \mathrm{DoF} \times \log_2(\text{SNR})$ , so the capacity roughly doubles.

Why This Matters: MIMO Configurations in Wireless Standards

The DoF formula $\min(n_t, n_r)$ directly shapes antenna configurations in wireless standards:

Wi-Fi 6 (802.11ax): Up to $8 \times 8$ MIMO for 8 spatial streams, delivering up to 9.6 Gbps peak throughput.
5G NR Release 15: Downlink supports up to 8-layer MIMO ( $n_t$ up to 32 at gNB, $n_r$ up to 8 at UE), with the bottleneck being UE antenna count.
5G NR Massive MIMO: 64T64R (64 transmit, 64 receive at the base station) serving multiple users. The DoF is shared among users via MU-MIMO: each user sees a few streams, but the total system DoF approaches 64.

The practical limitation is always the device with fewer antennas --- typically the user equipment. This is why 5G focuses on massive arrays at the base station combined with MU-MIMO to serve many users simultaneously.

See full treatment in Chapter 16

Key Takeaway

The degrees of freedom $\mathrm{DoF} = \min(n_t, n_r)$ is the single most important number in MIMO theory. It means capacity grows linearly with the number of antennas (on the smaller side) at no cost in bandwidth or total power. This linear scaling is the fundamental reason MIMO is deployed in every modern wireless standard from Wi-Fi to 5G. However, realising these DoF requires a full-rank channel (rich scattering) and sufficient CSI.

Why This Matters: MIMO Capacity and Information Theory

The MIMO capacity formulas in this chapter are direct extensions of the parallel Gaussian channel capacity from Chapter 11. The ITA book develops the information-theoretic foundations more deeply: the converse proofs via Fano's inequality, the role of the maximum entropy property of Gaussian distributions, and the connection to rate-distortion theory for MIMO source coding. The MIMO channel capacity with CSIT is a special case of the general water-filling solution for parallel channels.

See full treatment in Capacity with Diversity

Degrees of freedom (DoF)

The pre-log factor in the high-SNR capacity expansion: $\mathrm{DoF} = \lim_{\text{SNR}\to\infty} C(\text{SNR})/\log_2(\text{SNR})$ . For MIMO, $\mathrm{DoF} = \min(n_t, n_r)$ .

Multiplexing gain

The rate of growth of a MIMO scheme's data rate relative to $\log_2(\text{SNR})$ at high SNR. The maximum achievable multiplexing gain equals the channel DoF.

Degrees of Freedom and Multiplexing Gain