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Theorem: AWGN Channel Capacity

The capacity of a bandlimited AWGN channel with bandwidth $B$ (Hz), average transmit power $P$ (W), and one-sided noise power spectral density $N_0$ (W/Hz) is

$C = B \log_2\!\left(1 + \frac{P}{N_0 B}\right) \quad \text{bits/s}$

Equivalently, with $\text{SNR} = P/(N_0 B)$ :

$C = B \log_2(1 + \text{SNR})$

The capacity-achieving input distribution is Gaussian: $X \sim \mathcal{N}(0, P)$ .

The AWGN capacity formula captures two fundamental resources: bandwidth $B$ (number of independent signalling dimensions per second) and SNR (energy per dimension relative to noise). Each independent dimension contributes $\log_2(1 + \text{SNR})$ bits, and there are $2B$ real dimensions per second (by Nyquist), but the complex baseband formulation gives $B$ complex dimensions, leading to $C = B\log_2(1 + \text{SNR})$ .

Proof

Channel model

The continuous-time AWGN channel $r(t) = x(t) + w(t)$ with bandwidth $B$ is equivalent to $2B$ real-valued samples per second (Nyquist sampling theorem), or $n = 2BT$ samples in $T$ seconds:

$r_i = x_i + w_i, \quad w_i \sim \mathcal{N}(0, N_0/2), \quad i = 1,\ldots,n$

Power constraint: $\frac{1}{n}\sum_{i=1}^n x_i^2 \leq P/(2B) = P_{\text{dim}}$ .

Mutual information computation

For input $X \sim \mathcal{N}(0, P_{\text{dim}})$ :

$h(Y) = \frac{1}{2}\log_2(2\pi e(P_{\text{dim}} + N_0/2))$

$h(Y|X) = h(W) = \frac{1}{2}\log_2(2\pi e \cdot N_0/2)$

Per-dimension mutual information:

$I = \frac{1}{2}\log_2\!\left(1 + \frac{2P_{\text{dim}}}{N_0}\right) = \frac{1}{2}\log_2(1 + \text{SNR})$

Optimality of Gaussian input

Among all distributions with variance $\leq P_{\text{dim}}$ , the Gaussian maximises $h(Y)$ (maximum entropy property). Since $h(Y|X) = h(W)$ is fixed, the Gaussian input maximises $I$ .

Total capacity

With $n = 2BT$ dimensions in $T$ seconds:

$C = \frac{n}{T} \cdot \frac{1}{2}\log_2(1+\text{SNR}) = B\log_2(1+\text{SNR}) \quad \text{bits/s}$

$\blacksquare$

, ,

Definition:
Shannon Limit

The Shannon limit is the minimum $E_b/N_0$ required for reliable communication at any positive rate. As the spectral efficiency $\eta = C/B \to 0$ (infinite bandwidth regime):

$\frac{E_b}{N_0}\bigg|_{\min} = \ln 2 = -1.59 \text{ dB}$

This follows from $C = B\log_2(1 + E_b C/(N_0 B))$ . As $B \to \infty$ with $P$ fixed:

$C \to \frac{P}{N_0 \ln 2}$

so $E_b/N_0 = P/(C N_0) \to \ln 2$ .

No communication system — regardless of coding, modulation, or signal processing — can operate reliably below $-1.59$ dB.

,

AWGN Channel Capacity vs SNR

Capacity $C = B\log_2(1 + \text{SNR})$ as a function of SNR for different bandwidths. The Shannon limit at $E_b/N_0 = -1.59$ dB is shown as a horizontal reference. Observe the logarithmic growth: doubling SNR adds only one bit per dimension.

Parameters

Bandwidth B (MHz)

Show Shannon limit (

E_b/N_0 = -1.59

dB)

Practical Codes: Gap to AWGN Capacity

Compare the performance of practical coding families against the AWGN capacity. Each point represents the achieved spectral efficiency at a given $E_b/N_0$ for BER $= 10^{-5}$ . Observe how modern codes (turbo, LDPC, polar) approach the Shannon boundary within 1-2 dB.

Parameters

Show turbo codes

Show LDPC codes

Show polar codes

Show uncoded modulation

Example: AWGN Capacity at SNR = 10 dB

Compute the channel capacity and spectral efficiency for an AWGN channel with bandwidth $B = 1$ MHz and SNR $= 10$ dB.

Solution

Convert SNR

$\text{SNR} = 10^{10/10} = 10$ .

Compute capacity

$C = 1 \times 10^6 \times \log_2(1 + 10) = 10^6 \times \log_2(11)KATEXPLACEHOLDER0END= 10^6 \times 3.459 = 3.459 \text{ Mbps}$ $

Spectral efficiency

$\eta = C/B = 3.459 \text{ bits/s/Hz}$ $To achieve this with uncoded QAM, we would need 16-QAM ($ \eta = 4 $bits/s/Hz) derated by a code rate of$ 3.459/4 = 0.865 $.$ \blacksquare$

Example: Bandwidth-Power Trade-off

A communication system must deliver $C = 1$ Mbps. Compare the required SNR for bandwidths $B = 0.5$ , $1$ , and $10$ MHz. What happens as $B \to \infty$ ?

Solution

Required SNR

From $C = B\log_2(1+\text{SNR})$ : $\text{SNR} = 2^{C/B} - 1$ .

$B = 0.5$ MHz: $\text{SNR} = 2^2 - 1 = 3 = 4.77$ dB.

$B = 1$ MHz: $\text{SNR} = 2^1 - 1 = 1 = 0$ dB.

$B = 10$ MHz: $\text{SNR} = 2^{0.1} - 1 = 0.0718 = -11.4$ dB.

Required power

$P = \text{SNR} \times N_0 B$ . With $N_0 = 10^{-20}$ W/Hz:

$B = 0.5$ MHz: $P = 3 \times 10^{-20} \times 5 \times 10^5 = 1.5 \times 10^{-14}$ W.

$B = 1$ MHz: $P = 1 \times 10^{-20} \times 10^6 = 10^{-14}$ W.

$B = 10$ MHz: $P = 0.0718 \times 10^{-20} \times 10^7 = 7.18 \times 10^{-15}$ W.

Infinite bandwidth limit

As $B \to \infty$ : $P \to C N_0 \ln 2 = 10^6 \times 10^{-20} \times 0.693 = 6.93 \times 10^{-15}$ W.

Increasing bandwidth further yields diminishing returns — the power converges to the Shannon limit. $\blacksquare$

Sphere-Packing Interpretation of AWGN Capacity

Animated visualisation of the sphere-packing argument. A large signal sphere of radius

\sqrt{nP}

is packed with noise spheres of radius

\sqrt{nN_0/2}

. The number of non-overlapping noise spheres gives the maximum number of distinguishable codewords

M \approx (1+\text{SNR})^{n/2}

.

The number of non-overlapping noise spheres packed inside the signal sphere determines the channel capacity.

Sphere-Packing Interpretation

The AWGN capacity has an elegant geometric interpretation. In $n = 2BT$ dimensions, the noise vector $\mathbf{w}$ has squared norm concentrated around $n N_0/2$ (by the law of large numbers). Each codeword creates a "noise sphere" of radius $\sqrt{n N_0/2}$ . The transmitted signal is constrained to a sphere of radius $\sqrt{nP/(2B)}$ . The maximum number of non-overlapping noise spheres is

$M \approx \frac{\text{Vol}(\text{signal sphere})}{\text{Vol}(\text{noise sphere})} = \left(\frac{P/(2B) + N_0/2}{N_0/2}\right)^{n/2} = (1 + \text{SNR})^{n/2}$

The rate $R = \frac{1}{n}\log_2 M = \frac{1}{2}\log_2(1+\text{SNR})$ bits per dimension, giving $C = 2B \times R = B\log_2(1+\text{SNR})$ .

,

Quick Check

An engineer claims to have designed a coding scheme that achieves BER $= 10^{-6}$ at $E_b/N_0 = -2$ dB over an AWGN channel. Is this possible?

Yes, with sufficiently long block length

No, it violates the Shannon limit of $-1.59$ dB

Yes, if the bandwidth is large enough

It depends on the modulation scheme used

Correction:

No, it violates the Shannon limit of

-1.59

dB

Correct. The minimum $E_b/N_0$ for reliable communication is $\ln 2 = -1.59$ dB, regardless of bandwidth, block length, or coding scheme. $-2$ dB $< -1.59$ dB, so this claim is impossible.

Common Mistake: $C = B\log_2(1+\text{SNR})$ Requires Gaussian Input

Mistake:

Applying $C = B\log_2(1+\text{SNR})$ to compute the capacity of a system using discrete constellations (QAM, PSK) and claiming this is the achievable rate.

Correction:

The formula $C = B\log_2(1+\text{SNR})$ is the capacity with Gaussian input — the maximum over all input distributions. With a finite constellation (e.g., 16-QAM), the achievable rate is the constellation-constrained capacity, which saturates at $\log_2 M$ bits/symbol for high SNR and is strictly less than the Gaussian capacity. For 16-QAM, the maximum is 4 bits/symbol regardless of SNR.

⚠️Engineering Note

Finite Block Length Penalty in Real Systems

Shannon's capacity formula assumes infinite block length — in practice, codes have finite block length $n$ , introducing a rate penalty. The normal approximation for the maximum achievable rate at block length $n$ and error probability $\epsilon$ is

$R^*(n, \epsilon) \approx C - \sqrt{\frac{V}{n}} Q^{-1}(\epsilon)$

where $V$ is the channel dispersion (for AWGN: $V = \frac{1}{2}\!\left(1 - \frac{1}{(1+\text{SNR})^2}\right) (\log_2 e)^2$ ) and $Q^{-1}$ is the inverse Q-function.

At 5G NR block lengths ( $n = 1000\text{-}8000$ for data channels), this penalty is 0.5-1.5 dB relative to the asymptotic capacity. For ultra-reliable low-latency communication (URLLC) with $\epsilon = 10^{-5}$ and short blocks ( $n = 200$ - $500$ ), the penalty can exceed 3 dB — a significant design consideration.

Practical Constraints

•
5G NR data: $n = 1000$ - $8448$ (LDPC), penalty ~0.5-1 dB
•
5G NR control: $n = 54$ - $1024$ (polar), penalty ~1-3 dB
•
URLLC: $n = 200$ - $500$ , $\epsilon = 10^{-5}$ , penalty ~2-4 dB

AWGN Capacity

The maximum achievable rate over a bandlimited additive white Gaussian noise channel: $C = B\log_2(1 + P/(N_0 B))$ bits/s. Achieved by Gaussian signalling.

Shannon Limit

The minimum energy per bit required for reliable communication: $E_b/N_0 = \ln 2 = -1.59$ dB. This is the ultimate power efficiency limit, approached as bandwidth goes to infinity.

Spectral Efficiency (Capacity)

The capacity per unit bandwidth: $\eta = C/B = \log_2(1+\text{SNR})$ bits/s/Hz. Represents the maximum number of bits that can be reliably transmitted per second per Hertz of bandwidth.

AWGN Channel Capacity

Theorem: AWGN Channel Capacity

Channel model

Mutual information computation

Optimality of Gaussian input

Total capacity

Definition: Shannon Limit

AWGN Channel Capacity vs SNR

Parameters

Practical Codes: Gap to AWGN Capacity

Parameters

Example: AWGN Capacity at SNR = 10 dB

Convert SNR

Compute capacity

Spectral efficiency

Example: Bandwidth-Power Trade-off

Required SNR

Required power

Infinite bandwidth limit

Sphere-Packing Interpretation of AWGN Capacity

Sphere-Packing Interpretation

Quick Check

Common Mistake: C=Blog⁡2(1+SNR)C = B\log_2(1+\text{SNR})C=Blog2​(1+SNR) Requires Gaussian Input

Finite Block Length Penalty in Real Systems

AWGN Capacity

Shannon Limit

Spectral Efficiency (Capacity)

Definition:
Shannon Limit

Common Mistake: $C = B\log_2(1+\text{SNR})$ Requires Gaussian Input