Ferkans — Interactive Telecom Tutor

From Time to Frequency

Section 11.3 treated flat-fading channels where the gain varies in time but is constant across the signal bandwidth. Wideband wireless channels are frequency-selective: different frequency sub-bands experience different gains. The key insight is that a frequency-selective channel can be decomposed into parallel independent sub-channels, and the capacity is found by water-filling across frequency — allocating more power to sub-bands with higher gain.

Theorem: Capacity of Parallel Independent Channels

Consider $K$ parallel independent AWGN sub-channels with gains $\{g_k\}_{k=1}^K$ and a total power constraint $\sum_k P_k \leq P$ . The sub-channel model is

$y_k = g_k x_k + w_k, \quad w_k \sim \mathcal{N}(0, N_0), \quad k = 1,\ldots,K$

The capacity is

$C = \sum_{k=1}^K \log_2\!\left(1 + \frac{|g_k|^2 P_k}{N_0}\right)$

maximised by the water-filling power allocation:

$P_k = \left(\mu - \frac{N_0}{|g_k|^2}\right)^+$

where $\mu$ is chosen so that $\sum_k P_k = P$ .

Sub-channels with $|g_k|^2 < N_0/\mu$ receive no power.

Each sub-channel is an independent AWGN channel. The total capacity is the sum of individual capacities. The only coupling is through the total power constraint, which is resolved by water-filling: equalise the total (power + noise) across active sub-channels.

Proof

Independence and additivity

Since the sub-channels are independent, the joint capacity decomposes as

$C = \max_{\{P_k\}} \sum_k \log_2\!\left(1 + \frac{|g_k|^2 P_k}{N_0}\right)$

subject to $\sum_k P_k \leq P$ and $P_k \geq 0$ .

KKT conditions

The Lagrangian is

$\mathcal{L} = \sum_k \log_2\!\left(1 + \frac{|g_k|^2 P_k}{N_0}\right) - \lambda\!\left(\sum_k P_k - P\right)$

Setting $\partial\mathcal{L}/\partial P_k = 0$ :

$\frac{|g_k|^2/N_0}{(1 + |g_k|^2 P_k/N_0)\ln 2} = \lambda$

Solving for $P_k$ and enforcing non-negativity gives $P_k = (\mu - N_0/|g_k|^2)^+$ with $\mu = 1/(\lambda\ln 2)$ . $\blacksquare$

, ,

Definition:
Water-Filling Power Allocation Across Frequency

For a frequency-selective channel with transfer function $H(f)$ and bandwidth $B$ , discretise the band into $K$ sub-channels of width $\Delta f = B/K$ , each with gain $|H(f_k)|^2$ . The water-filling solution allocates power density

$S(f_k) = \left(\mu - \frac{N_0}{|H(f_k)|^2}\right)^+$

The resulting capacity is

$C = \sum_{k=1}^K \Delta f \cdot \log_2\!\left(1 + \frac{|H(f_k)|^2 S(f_k)}{N_0}\right)$

In the continuous limit ( $K \to \infty$ ):

$C = \int_B \log_2\!\left(1 + \frac{|H(f)|^2 S(f)}{N_0}\right) df$

,

Water-Filling Power Allocation Animation

Cinematic animation of the water-filling algorithm. The noise floor

N_0/|H_k|^2

forms the uneven bottom of a vessel, and power (water) is poured in until reaching the constant level

\mu

. Sub-channels in deep fades remain dry.

Water fills up to a constant level

\mu

. Sub-channels below

1/\mu

receive no power — it is more efficient to redirect that power to stronger sub-channels.

Water-Filling in Frequency

Visualise the optimal water-filling power allocation across frequency for a frequency-selective channel. The channel gain $|H(f)|^2$ is shown inverted (as the "bottom of the vessel"), and the allocated power fills up to the water level $\mu$ . Sub-carriers in deep fades receive no power.

Parameters

Number of sub-carriers32

Average SNR (dB)15

Channel model

Example: Water-Filling with 4 Sub-Channels

A frequency-selective channel is decomposed into $K = 4$ sub-channels with gains $|g_k|^2 = \{4, 1, 0.25, 2\}$ and noise power $N_0 = 1$ . The total power budget is $P = 4$ . Find the water-filling power allocation and the total capacity.

Solution

Compute noise floors

The effective noise floor for each sub-channel is $N_0/|g_k|^2$ :

Sub-channel 1: $1/4 = 0.25$ Sub-channel 2: $1/1 = 1.0$ Sub-channel 3: $1/0.25 = 4.0$ Sub-channel 4: $1/2 = 0.5$

Find the water level $\mu$

Try all 4 sub-channels active: $\sum_k (\mu - N_0/|g_k|^2) = P$

$4\mu - (0.25 + 1.0 + 4.0 + 0.5) = 4$

$4\mu = 9.75$ , $\mu = 2.4375$ .

Check: $P_3 = 2.4375 - 4.0 = -1.5625 < 0$ . Sub-channel 3 is turned off.

Recompute with 3 active sub-channels

$3\mu - (0.25 + 1.0 + 0.5) = 4$

$3\mu = 5.75$ , $\mu = 1.9167$ .

$P_1 = 1.9167 - 0.25 = 1.667$ $P_2 = 1.9167 - 1.0 = 0.917$ $P_3 = 0$ (off) $P_4 = 1.9167 - 0.5 = 1.417$

Check: $\sum P_k = 1.667 + 0.917 + 0 + 1.417 = 4.0$ . Correct.

Compute capacity

$C_1 = \log_2(1 + 4 \times 1.667) = \log_2(7.667) = 2.94$ bits

$C_2 = \log_2(1 + 1 \times 0.917) = \log_2(1.917) = 0.94$ bits

$C_3 = 0$ bits

$C_4 = \log_2(1 + 2 \times 1.417) = \log_2(3.833) = 1.94$ bits

$C_{\text{total}} = 2.94 + 0.94 + 0 + 1.94 = 5.82 \text{ bits/channel use}$

Compare with equal power ( $P_k = 1$ each): $C_{\text{equal}} = \log_2(5) + \log_2(2) + \log_2(1.25) + \log_2(3) = 5.49$ bits.

Water-filling gain: $5.82 - 5.49 = 0.33$ bits. $\blacksquare$

Water-Filling Algorithm

The water-filling solution can be computed by a simple iterative algorithm:

Sort sub-channels by noise floor $N_0/|g_k|^2$ in ascending order
Start with all $K$ sub-channels active
Compute $\mu = (P + \sum_{k \in \mathcal{A}} N_0/|g_k|^2)/|\mathcal{A}|$ where $\mathcal{A}$ is the active set
If $P_k = \mu - N_0/|g_k|^2 < 0$ for any $k \in \mathcal{A}$ , remove that sub-channel from $\mathcal{A}$ and go to step 3
Terminate when all active sub-channels have $P_k \geq 0$

This converges in at most $K$ iterations.

Quick Check

In water-filling power allocation, what happens to a sub-channel whose gain $|g_k|^2$ is very small (deep fade)?

It receives the most power to compensate for the poor channel

It receives no power — the transmitter stays silent on that sub-channel

It receives equal power as all other sub-channels

It receives power proportional to its channel gain

Correction:

It receives no power — the transmitter stays silent on that sub-channel

Correct. When $|g_k|^2 < N_0/\mu$ (the noise floor exceeds the water level), the water-filling solution allocates zero power. It is more efficient to redirect that power to sub-channels with better gains.

OFDM Converts Frequency-Selective into Parallel Flat Channels

Orthogonal Frequency Division Multiplexing (OFDM) is the practical realisation of the parallel sub-channel decomposition. By dividing the wideband channel into $K$ narrowband sub-carriers (each experiencing flat fading), OFDM transforms the frequency-selective channel into $K$ independent flat-fading channels.

With a cyclic prefix longer than the channel delay spread, each OFDM sub-carrier sees a scalar channel $y_k = H_k x_k + w_k$ , exactly matching the parallel channel model. Water-filling across OFDM sub-carriers approaches the frequency-selective channel capacity. This is why OFDM is the dominant waveform in 4G LTE, 5G NR, Wi-Fi, and DVB.

Common Mistake: Water-Filling Requires CSI at the Transmitter

Mistake:

Assuming water-filling gains are always available. In FDD systems, the downlink channel is not directly observed by the transmitter.

Correction:

Water-filling requires the transmitter to know $|H(f_k)|^2$ for each sub-carrier. In TDD systems, channel reciprocity provides this. In FDD systems, the receiver must quantise and feed back the channel state, introducing overhead and delay. With imperfect CSIT, the water-filling gain is reduced, and equal power allocation can be near-optimal at high SNR.

Why This Matters: OFDM and Water-Filling in 5G NR

5G NR uses OFDM with up to 3300 sub-carriers (at 30 kHz spacing) or 400 sub-carriers (at 240 kHz spacing for mmWave). While full water-filling across all sub-carriers is not implemented due to feedback overhead, 5G NR performs sub-band power allocation:

The bandwidth is divided into sub-bands of 4-16 PRBs
The UE reports per-sub-band CQI
The scheduler adapts the MCS per sub-band

This coarse-grained adaptation captures most of the water-filling gain while keeping the feedback overhead manageable.

See full treatment in Peak-to-Average Power Ratio (PAPR)

Parallel Channels

A set of independent sub-channels that can be used simultaneously. A frequency-selective channel decomposes into parallel flat-fading sub-channels via OFDM. The total capacity is the sum of individual sub-channel capacities, optimised by water-filling.

Water-Filling (Frequency)

The optimal power allocation across frequency sub-channels that maximises the total capacity under a sum-power constraint. More power is allocated to sub-channels with higher gain; sub-channels in deep fades are turned off.

Capacity of Frequency-Selective Channels