A Gaussian channel has SNR = P/N = 15 dB (approximately 31.6 in linear scale). What is the channel capacity?
AC = 15 bits per channel use
BC = (1/2) log2(1 + 31.6) ≈ (1/2)(5.02) ≈ 2.51 bits per channel use
CC = log2(15) ≈ 3.91 bits per channel use
DC = 31.6 bits per channel use
First convert dB to linear: 15 dB means P/N = 10^(15/10) = 31.6. Then C = (1/2) log2(1 + 31.6) = (1/2) log2(32.6) ≈ (1/2)(5.03) ≈ 2.51 bits per channel use. The 1/2 factor comes from the real-valued channel; a complex channel (as in modern wireless systems) would give log2(1 + SNR) without the 1/2. Each 3 dB of SNR increase adds approximately 0.5 bits of capacity — this logarithmic scaling means gains get harder as SNR increases.
Question 2 Multiple Choice
As SNR approaches infinity, the capacity of the Gaussian channel grows without bound. As SNR approaches zero, capacity approaches zero linearly. Which regime is more relevant for modern wireless communications?
AThe high-SNR regime, because modern systems operate at very high power levels
BThe low-SNR (bandwidth-rich, power-limited) regime, because technologies like spread-spectrum and IoT devices often operate below 0 dB SNR by using wide bandwidth and sophisticated coding
CNeither — modern systems operate exactly at the Shannon limit
DBoth regimes are equally relevant
Many modern systems (GPS, spread-spectrum, IoT sensors, deep-space communication) operate at very low SNR by spreading the signal across wide bandwidth. At low SNR, C ≈ (P/N) * (1/(2*ln2)) bits per channel use, which is linear in SNR. The bandwidth-power tradeoff C = W*log2(1+P/(NW)) shows that as bandwidth W increases with fixed power, capacity approaches P/(N*ln2) — a finite limit determined by power alone. This 'ultimate Shannon limit' of -1.59 dB per bit is a key benchmark for power-efficient communication design.
Question 3 Short Answer
Explain why the Gaussian distribution is the capacity-achieving input distribution for the AWGN channel, connecting this to the maximum-entropy property of the Gaussian.
Think about your answer, then reveal below.
Model answer: The capacity C = max I(X;Y) = max [h(Y) - h(Y|X)] = max h(Y) - h(Z), since h(Y|X) = h(Z) is fixed (the noise is independent of the input). So we need to maximize h(Y) subject to the power constraint. Y = X + Z, and the variance of Y is at most P + N (with equality when X has variance P). Among all distributions with variance P + N, the Gaussian maximizes differential entropy. This is achieved when X is Gaussian with variance P, since the sum of independent Gaussians is Gaussian. The maximum entropy property of the Gaussian is thus directly responsible for the Gaussian input being optimal.
This is a beautiful example of how information-theoretic properties (maximum entropy of the Gaussian) yield engineering results (the capacity-achieving strategy). The same reasoning extends to vector Gaussian channels: the capacity-achieving input is always Gaussian, with covariance structure determined by the channel and power constraints.