Ferkans — Interactive Telecom Tutor

Connecting Continuous-Time to Discrete-Time

So far we have analyzed the discrete-time AWGN channel. But real communication happens in continuous time: a transmitter sends a waveform $x(t)$ , it propagates through a channel, and the receiver observes $y(t) = x(t) + z(t)$ . How do we bridge the gap?

The answer is the Nyquist-Shannon sampling theorem, which shows that a band-limited signal of bandwidth $W$ Hz is completely described by $2W$ samples per second. This converts the continuous-time channel into a discrete-time one, and the capacity per second becomes $C = W\log(1 + \text{SNR})$ . This section makes this connection rigorous.

Definition:
Continuous-Time Band-Limited AWGN Channel

The continuous-time AWGN channel with single-sided bandwidth $W$ Hz is

$y(t) = x(t) + z(t), \quad t \in [0, T],$

where:

$x(t)$ is the transmitted waveform, band-limited to $[-W, W]$ , with average power $\frac{1}{T}\int_0^T x^2(t)\,dt \leq P_x$ ,
$z(t)$ is white Gaussian noise with two-sided power spectral density $N_0/2$ .

The noise power in the band is $N = N_0W$ (for a real baseband channel with bandwidth $W$ ).

Theorem: Capacity of the Band-Limited AWGN Channel

The capacity of the continuous-time band-limited AWGN channel with bandwidth $W$ Hz, signal power $P_x$ , and noise spectral density $N_0/2$ is

$C = W\log_2\!\left(1 + \frac{P_x}{N_0W}\right) \quad \text{bits per second}.$

This is the celebrated Shannon-Hartley formula.

By the sampling theorem, the band-limited channel produces $2W$ independent real samples per second. Each sample sees an i.i.d. AWGN channel with signal power $P_x/(2W)$ and noise power $N_0/2$ , giving capacity $\frac{1}{2}\log(1 + P_x/(N_0W))$ bits per sample. Multiplying by $2W$ samples per second yields the result.

Proof

Apply the sampling theorem

By the Nyquist-Shannon theorem, a signal band-limited to $W$ Hz is completely described by samples at rate $2W$ per second. Over a time interval $T$ , this gives $n = 2WT$ samples.

Convert to discrete-time

The sampled channel is $Y_i = X_i + Z_i$ where $Z_i \sim \mathcal{N}(0, N_0/2)$ are i.i.d. The signal power per sample is $\mathbb{E}[X_i^2] = P_x/(2W)$ , so the per-sample SNR is $\text{SNR} = P_x/(N_0W)$ .

Compute the capacity in bits/s

The per-sample capacity is $\frac{1}{2}\log(1 + \text{SNR})$ bits per real sample. With $2W$ samples per second:

$C = 2W \times \frac{1}{2}\log\!\left(1 + \frac{P_x}{N_0W}\right) = W\log\!\left(1 + \frac{P_x}{N_0W}\right) \text{ bits/s.}$

,

Shannon-Hartley formula

The capacity of the band-limited AWGN channel: $C = W\log_2(1 + P/(N_0W))$ bits/s. Named for Shannon (information-theoretic derivation) and Hartley (earlier work on the log relationship between bandwidth and capacity).

Definition:
Passband Channel Capacity

For a passband channel centered at carrier frequency $f_0 \gg W$ with single-sided bandwidth $W$ , the complex baseband equivalent has bandwidth $W/2$ Hz and sample rate $W$ complex samples per second.

Each complex sample sees the channel $Y = X + Z$ , $Z \sim \mathcal{CN}(0, N_0)$ , with power constraint $\mathbb{E}[|X|^2] \leq E_s$ and $\text{SNR} = E_s/N_0$ .

The capacity per complex symbol is $\log(1 + \text{SNR})$ bits, and the capacity in bits/s is

$C = W \log_2(1 + \text{SNR}) \quad \text{bits/s},$

which is the same as the real baseband result — both conventions give the same bits/s.

Example: Shannon Capacity of a 5G NR Channel

A 5G NR cell operates at carrier frequency $f_0 = 3.5$ GHz with bandwidth $W = 100$ MHz and received $\text{SNR} = 15$ dB. What is the maximum achievable data rate?

Solution

Convert SNR to linear

$\text{SNR} = 10^{15/10} \approx 31.6$ .

Apply the Shannon-Hartley formula

$C = 100 \times 10^6 \times \log_2(1 + 31.6) = 100 \times 10^6 \times 5.02 \approx 502 \text{ Mbit/s.}$ $

Compare with practical throughput

In practice, 5G NR achieves about 60-80% of the Shannon limit due to cyclic prefix overhead (~7%), reference signal overhead (~15-20%), control channel overhead, and coding gap. A realistic throughput is approximately $300$ – $400$ Mbit/s for this configuration.

Bandwidth vs. Power Tradeoff

The Shannon-Hartley capacity curve

C = W\log_2(1 + P/(N_0W))

shows two regimes: bandwidth-limited at high SNR and power-limited at low SNR. The capacity saturates at

P/(N_0\ln 2)

as

W \to \infty

.

Capacity vs. Bandwidth

Explore how capacity scales with bandwidth for fixed power $P$ . At small bandwidth, capacity grows almost linearly with $W$ (bandwidth-limited regime). At large bandwidth, capacity saturates at $P/(N_0\ln 2)$ (power-limited regime).

Parameters

Transmit power (dBm)23

N_0

(dBm/Hz)-174

Maximum bandwidth (MHz)500

Two Operating Regimes

The Shannon-Hartley formula reveals two fundamentally different operating regimes:

Bandwidth-limited regime ( $W$ small, $\text{SNR}$ high): Capacity grows approximately as $W\log_2(\text{SNR})$ . Adding more bandwidth is very valuable. This is the regime of fiber optics and high-SNR short-range wireless links.

Power-limited regime ( $W$ large, $\text{SNR}$ small): Capacity saturates at $P/(N_0\ln 2)$ regardless of bandwidth. The bottleneck is energy, not spectrum. This is the regime of satellite communication, deep-space links, and IoT devices.

Most practical systems operate between these extremes, and the spectral efficiency $\eta = C/W$ indicates which regime dominates: $\eta \gg 1$ means bandwidth-limited, $\eta \ll 1$ means power-limited.

🔧Engineering Note

Thermal Noise Floor

The noise spectral density $N_0$ has a fundamental physical origin: at temperature $T$ Kelvin, the thermal noise power spectral density is $N_0 = k_B T$ , where $k_B = 1.381 \times 10^{-23}$ J/K is Boltzmann's constant.

At room temperature ( $T = 290$ K): $N_0 = 4.00 \times 10^{-21}$ W/Hz $= -174$ dBm/Hz.

This is the ultimate noise floor for any receiver operating at room temperature. In practice, the receiver noise figure $F$ (typically 3-8 dB) increases the effective noise spectral density to ${N_0}_{\text{eff}} = F \cdot k_B T$ .

Practical Constraints

•
Thermal noise floor at 290 K: $-174$ dBm/Hz
•
Receiver noise figure adds 3-8 dB in practice
•
Cryogenic receivers (radio astronomy) can achieve $F < 0.1$ dB

Historical Note: From Hartley to Shannon

Ralph Hartley published "Transmission of Information" in 1928, establishing the logarithmic relationship between the number of distinguishable signal levels and the amount of information. His formula $C = W\log_2 M$ (for $M$ signal levels) captures the bandwidth dependence but misses the noise.

Shannon's 1948 breakthrough was to incorporate noise into the picture, yielding $C = W\log_2(1 + \text{SNR})$ . The "+1" inside the logarithm — which seems like a minor detail — is actually the crucial difference: it says that even with infinite bandwidth and infinite signal levels, noise limits what you can communicate. Hartley's formula gives $C \to \infty$ as $M \to \infty$ ; Shannon's formula correctly gives a finite limit.

Quick Check

A system with bandwidth $W$ and power $P$ achieves capacity $C_{1}$ . If the bandwidth is doubled to $2W$ (keeping power $P$ fixed), the new capacity $C_{2}$ satisfies:

$C_{2} = 2C_{1}$

$C_{2} < 2C_{1}$

$C_{2} > 2C_{1}$

$C_{2} = C_{1}$ (bandwidth does not help)

Correction:

C_{2} < 2C_{1}

Doubling bandwidth increases noise power ( $N = N_0W$ ) while keeping signal power fixed, so the SNR halves. By the concavity of $x \mapsto x\log(1 + a/x)$ , we get $C_{2} < 2C_{1}$ . The capacity increases but by less than a factor of two — this is the diminishing returns of bandwidth in the power-limited regime.

Band-Limited Channels and the Sampling Theorem