Ferkans — Interactive Telecom Tutor

Sharing the Radio Resource

Wireless communication is inherently a shared medium: all transmitters radiate into the same physical space. The fundamental challenge of multiple access is to allow $K$ users to share a common channel of bandwidth $W$ and duration $T$ with controlled interference. The simplest approach is orthogonal multiple access, which partitions the resource space — frequency, time, or code — so that each user occupies a non-overlapping slice. Orthogonal schemes eliminate multi-user interference entirely, at the cost of giving each user only a fraction of the total resource. Whether this trade-off is favourable depends on the channel conditions and the number of users.

Definition:
FDMA, TDMA, and OFDMA

Let the total resource consist of bandwidth $W$ Hz and frame duration $T$ seconds, serving $K$ users.

FDMA (Frequency Division Multiple Access): The bandwidth is divided into $K$ non-overlapping sub-bands of width $W_k$ , with $\sum_{k=1}^{K} W_k = W$ . User $k$ transmits continuously over sub-band $k$ for the entire frame. The achievable rate is:

$R_k^{\text{FDMA}} = W_k \log_2\!\left(1 + \frac{P_k |h_k|^2}{W_k \sigma^2}\right)$

where $P_k$ is the power allocated to user $k$ and $\sigma^2$ is the noise power spectral density.

TDMA (Time Division Multiple Access): The frame is divided into $K$ non-overlapping time slots of duration $T_k$ , with $\sum_{k=1}^{K} T_k = T$ . User $k$ transmits over the full bandwidth $W$ during slot $k$ . The achievable rate is:

$R_k^{\text{TDMA}} = \frac{T_k}{T}\, W \log_2\!\left(1 + \frac{P_k |h_k|^2}{W \sigma^2} \cdot \frac{T}{T_k}\right)$

where $P_k T/T_k$ is the instantaneous power (burst power) during the active slot.

OFDMA (Orthogonal Frequency Division Multiple Access): The bandwidth is divided into $N_{\text{sc}}$ orthogonal subcarriers, and each subcarrier-symbol resource element is assigned to one user. Let $\mathcal{S}_k$ denote the set of subcarriers assigned to user $k$ , with $\bigcup_k \mathcal{S}_k = \{1, \ldots, N_{\text{sc}}\}$ and $\mathcal{S}_i \cap \mathcal{S}_j = \emptyset$ for $i \neq j$ . The rate for user $k$ is:

$R_k^{\text{OFDMA}} = \sum_{n \in \mathcal{S}_k} \Delta f \, \log_2\!\left(1 + \frac{p_{k,n} |h_{k,n}|^2}{\Delta f \, \sigma^2}\right)$

where $\Delta f = W/N_{\text{sc}}$ is the subcarrier spacing, $p_{k,n}$ is the power on subcarrier $n$ , and $h_{k,n}$ is user $k$ 's channel on subcarrier $n$ .

All three schemes are special cases of orthogonal resource partitioning. FDMA partitions in frequency, TDMA in time, and OFDMA provides a two-dimensional partition in the time-frequency grid, offering the greatest scheduling flexibility.

,

Theorem: Sum-Rate Penalty of Orthogonal Multiple Access

Consider a $K$ -user Gaussian multiple-access channel with total bandwidth $W$ , noise PSD $\sigma^2$ , and user channel gains $\{|h_k|^2\}$ . Let each user have power $P_k$ and define $\text{SNR}_{k} = P_k |h_k|^2 / (W\sigma^2)$ .

The sum capacity of the MAC (achievable with joint decoding) is:

$C_{\text{sum}} = W \log_2\!\left(1 + \sum_{k=1}^{K} \text{SNR}_{k}\right)$

Any orthogonal scheme that allocates a fraction $\alpha_k$ of the resource to user $k$ ( $\sum_k \alpha_k = 1$ ) achieves at most:

$R_{\text{sum}}^{\text{orth}} = W \sum_{k=1}^{K} \alpha_k \log_2\!\left(1 + \frac{\text{SNR}_{k}}{\alpha_k}\right)$

The gap $C_{\text{sum}} - R_{\text{sum}}^{\text{orth}} \geq 0$ , with equality if and only if $K = 1$ .

Orthogonal access forces each user to occupy only a fraction $\alpha_k$ of the resource. While each user enjoys a higher "per-slice" SNR of $\text{SNR}_{k}/\alpha_k$ (because it concentrates its power into a smaller resource), the pre-log factor $\alpha_k$ reduces the rate. The concavity of the logarithm implies:

$\sum_k \alpha_k \log_2\!\left(1 + \frac{\text{SNR}_{k}}{\alpha_k}\right) \leq \log_2\!\left(1 + \sum_k \text{SNR}_{k}\right)$

by Jensen's inequality applied in the appropriate direction. Non-orthogonal schemes (superposition coding with SIC) avoid this loss by allowing all users to share the entire resource simultaneously.

Proof

Jensen's inequality argument

Define $f(x) = \log_2(1 + x)$ , which is concave. We need the reverse direction. Instead, consider the sum directly.

By the log-sum inequality, for $\alpha_k > 0$ with $\sum_k \alpha_k = 1$ and $x_k = \text{SNR}_{k}/\alpha_k$ :

$\sum_k \alpha_k \log_2(1 + x_k) \leq \log_2\!\left(\sum_k \alpha_k(1 + x_k)\right) = \log_2\!\left(1 + \sum_k \text{SNR}_{k}\right)$

The first inequality follows from the concavity of $\log_2$ and Jensen's inequality:

$\sum_k \alpha_k f(x_k) \leq f\!\left(\sum_k \alpha_k x_k\right)$

Note: $\sum_k \alpha_k x_k = \sum_k \text{SNR}_{k}$ and $\sum_k \alpha_k \cdot 1 = 1$ , so $\sum_k \alpha_k(1+x_k) = 1 + \sum_k \text{SNR}_{k}$ .

Equality condition

Equality in Jensen's inequality holds if and only if all $x_k$ are equal, i.e., $\text{SNR}_{k}/\alpha_k = c$ for all $k$ . But this also requires $\sum_k \alpha_k = 1$ , which gives $\alpha_k = \text{SNR}_{k} / \sum_j \text{SNR}_{j}$ . Even with this optimal resource allocation, the orthogonal sum rate equals $C_{\text{sum}}$ only if $K = 1$ .

For $K \geq 2$ with unequal SNRs, the gap is strictly positive. The loss is most significant at high SNR, where the MAC sum capacity grows as $\log_2(\sum_k \text{SNR}_{k})$ while orthogonal schemes lose the multiplexing gain. $\blacksquare$

,

Orthogonal MA Resource Grid Comparison

Visualise how FDMA, TDMA, and OFDMA partition the time-frequency resource grid among $K$ users. Each user is assigned a distinct colour. Observe how OFDMA provides finer granularity than FDMA or TDMA, enabling better adaptation to frequency-selective fading. Adjust the number of users $K$ to see how the grid becomes more fragmented.

Parameters

K

(number of users)4

Example: OFDMA Capacity for Four Users

An OFDMA system has total bandwidth $W = 10$ MHz divided into $N_{\text{sc}} = 64$ subcarriers. Four users ( $K = 4$ ) each receive 16 subcarriers. All users have equal power $P_k = 100$ mW, noise PSD $\sigma^2 = 10^{-9}$ W/Hz, and flat-fading channel gains $|h_1|^2 = 4$ , $|h_2|^2 = 2$ , $|h_3|^2 = 1$ , $|h_4|^2 = 0.5$ .

(a) Compute each user's rate with equal subcarrier allocation. (b) Compute the MAC sum capacity (non-orthogonal bound). (c) What is the sum-rate loss due to orthogonal access?

Solution

Per-user SNR

Subcarrier spacing: $\Delta f = W/64 = 156.25$ kHz.

Per-subcarrier noise power: $\Delta f \cdot \sigma^2 = 156250 \times 10^{-9} = 1.5625 \times 10^{-4}$ W.

Per-subcarrier power: $p_{k,n} = P_k / 16 = 6.25$ mW.

Per-subcarrier SNR for user $k$ : $\text{SNR}_{k,n} = p_{k,n} |h_k|^2 / (\Delta f \cdot \sigma^2)$

$\text{SNR}_{1} = \frac{6.25 \times 10^{-3} \times 4}{1.5625 \times 10^{-4}} = 160$ $\text{SNR}_{2} = 80, \quad \text{SNR}_{3} = 40, \quad \text{SNR}_{4} = 20$

Per-user rate (equal allocation)

Each user gets 16 subcarriers:

$R_k = 16 \times \Delta f \times \log_2(1 + \text{SNR}_{k})$

$R_1 = 16 \times 156.25 \times 10^3 \times \log_2(161) = 2.5 \times 10^6 \times 7.33 = 18.33 \;\text{Mbps}$ $R_2 = 2.5 \times 10^6 \times 6.34 = 15.85 \;\text{Mbps}$ $R_3 = 2.5 \times 10^6 \times 5.36 = 13.40 \;\text{Mbps}$ $R_4 = 2.5 \times 10^6 \times 4.39 = 10.98 \;\text{Mbps}$

Sum rate: $R_{\text{sum}}^{\text{OFDMA}} = 58.56$ Mbps.

MAC sum capacity

Total SNR: $\sum_k \text{SNR}_{k} = P_k |h_k|^2 / (W\sigma^2)$ per user over the full bandwidth.

$\text{SNR}_{k}^{\text{full}} = P_k |h_k|^2 / (W \sigma^2) = 0.1 \times |h_k|^2 / (10^7 \times 10^{-9}) = 10 |h_k|^2$

$\sum_k \text{SNR}_{k}^{\text{full}} = 10(4 + 2 + 1 + 0.5) = 75$

$C_{\text{sum}} = W \log_2(1 + 75) = 10^7 \times 6.25 = 62.5 \;\text{Mbps}$

Sum-rate loss: $62.5 - 58.56 = 3.94$ Mbps (6.3% loss). $\blacksquare$

Quick Check

In OFDMA, if user 1 has a deep fade on subcarriers 1--16 but strong gain on subcarriers 49--64, what should the scheduler do compared to a fixed FDMA assignment?

Assign subcarriers 49--64 to user 1 (exploit frequency selectivity)

Keep the fixed assignment since orthogonality is preserved either way

Give all 64 subcarriers to user 1 since it has the best peak channel

Reduce user 1 power to compensate for the fading

Correction:

Assign subcarriers 49--64 to user 1 (exploit frequency selectivity)

OFDMA's key advantage over FDMA is the ability to assign subcarriers dynamically based on channel conditions. By giving each user the subcarriers where its channel is strongest, OFDMA exploits multi-user diversity in the frequency domain, potentially approaching the MAC sum capacity.

Common Mistake: Orthogonal Access Is Not Capacity-Achieving

Mistake:

Assuming that FDMA, TDMA, or OFDMA achieves the multi-user capacity because they eliminate interference.

Correction:

Orthogonal schemes eliminate multi-user interference but at the cost of giving each user only a fraction of the total resource. The MAC capacity region is strictly larger than the orthogonal rate region for $K \geq 2$ . Specifically, the sum-rate corner point of the MAC region requires superposition coding with successive interference cancellation (SIC), not orthogonal partitioning.

At low SNR, the loss is small because $\log(1 + x) \approx x$ is nearly linear and Jensen's gap vanishes. At high SNR, the loss can be significant: each orthogonal user loses a factor of $\alpha_k$ in the pre-log, reducing the multiplexing gain from $\log(\text{SNR})$ to $\alpha_k \log(\text{SNR}/\alpha_k)$ .

OFDMA partially recovers the loss through frequency-domain scheduling (multi-user diversity), but the fundamental gap relative to non-orthogonal schemes remains.

Why This Matters: OFDMA in LTE and 5G NR

LTE (4G) adopted OFDMA for the downlink, with SC-FDMA (a DFT-precoded variant) for the uplink. The resource grid is partitioned into resource blocks of 12 subcarriers $\times$ 7 OFDM symbols (one slot), and the scheduler assigns resource blocks to users every 1 ms TTI.

5G NR generalises this with flexible numerology: subcarrier spacings of 15, 30, 60, 120, or 240 kHz, and slot durations from 1 ms down to 62.5 $\mu$ s. Both downlink and uplink use CP-OFDM (OFDMA), with DFT-spread OFDM as an optional uplink waveform. The scheduler operates on a 2D time-frequency grid, exploiting both frequency-selective fading and temporal variations for multi-user diversity gain.

FDMA

Frequency Division Multiple Access: a multiple access scheme in which each user is assigned a non-overlapping frequency sub-band for the entire frame duration. Used in 1G cellular (AMPS) and as a component of hybrid access schemes.

Related: TDMA, OFDMA

TDMA

Time Division Multiple Access: a multiple access scheme in which each user is assigned a non-overlapping time slot during which it transmits over the full bandwidth. Used in 2G cellular (GSM) and satellite systems.

Related: FDMA, OFDMA

OFDMA

Orthogonal Frequency Division Multiple Access: a multiple access scheme based on OFDM in which individual subcarriers or groups of subcarriers are assigned to different users, enabling fine-grained frequency-domain scheduling. The downlink access scheme for LTE and the primary access scheme for 5G NR.

Related: FDMA, TDMA

Orthogonal Multiple Access (FDMA/TDMA/OFDMA)