--- title: "Shannon-Hartley theorem; capacity limit 1.44S/N₀" subject: "Communication Systems" chapter: 6 key: "semester_6/communication_systems:shannon-hartley-1-44" source: https://www.bctengineer.com/q/semester_6/communication_systems/shannon-hartley-1-44 site: BCT Engineering --- # Shannon-Hartley theorem; capacity limit 1.44S/N₀ Communication Systems · Chapter 6: Baseband Digital Data Transmission - Asked in: 2080 Baisakh (CS II), 6+4 marks; 2075 Asoj (CS II), 6+4 marks (asked 2 times) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/shannon-hartley-1-44 - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-6 ## Question Explain Shannon Hartley channel capacity theorem and its implication and theoretical limits. Show that channel capacity is 1.44S/N₀, when channel bandwidth tends to infinity. ## Short answer Shannon-Hartley gives C = B log2(1+S/N) b/s, N=N0B; error-free transmission is possible if R C$, errors cannot be made arbitrarily small, whatever coding is used. **Implications** - **Bandwidth–SNR trade-off:** the same capacity can be reached with a wide band and low SNR (e.g. spread spectrum) or a narrow band and high SNR (e.g. M-ary QAM). - Capacity grows **linearly with B** but only **logarithmically with S/N**. Doubling the power adds only about $B$ b/s at high SNR. - It gives a benchmark: practical modems and codes (turbo, LDPC) are judged by how close they come to $C$. - It proves that error-free communication is possible over noisy channels with proper coding. It does not say how to build the code. **Theoretical limits** - **Ideal system:** $R = C$. Writing $S = E_b C$: $$ \frac{C}{B} = \log_2\!\left(1 + \frac{E_b}{N_0}\frac{C}{B}\right) \Rightarrow \frac{E_b}{N_0} = \frac{2^{C/B} - 1}{C/B} $$ - As $C/B \to 0$ (infinite bandwidth), $E_b/N_0 \to \ln 2 = 0.693$, i.e. **−1.6 dB**. This is the **Shannon limit**: below it, no error-free transmission is possible at any bandwidth. - Increasing bandwidth alone cannot give unlimited capacity, because noise power $N_0 B$ also grows (shown below). #### Capacity with infinite bandwidth $$ C = B \log_2\!\left(1 + \frac{S}{N_0 B}\right) $$ Let $x = \dfrac{S}{N_0 B}$, so $B = \dfrac{S}{N_0 x}$. As $B \to \infty$, $x \to 0$: $$ \begin{aligned} C &= \frac{S}{N_0} \cdot \frac{1}{x} \log_2(1 + x) = \frac{S}{N_0} \log_2 (1+x)^{1/x} \\ C_\infty &= \lim_{x \to 0} \frac{S}{N_0} \log_2 (1+x)^{1/x} = \frac{S}{N_0} \log_2 e \\ &= \frac{S}{N_0} \times 1.4427 \approx 1.44 \frac{S}{N_0} \end{aligned} $$ using $\lim_{x \to 0} (1+x)^{1/x} = e$ and $\log_2 e = 1/\ln 2 = 1.4427$. So even with infinite bandwidth, the capacity stays finite at $C_\infty \approx 1.44\, S/N_0$ b/s. Capacity can be raised only by raising the signal power or lowering the noise density. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the Shannon-Hartley formula C = B log2(1 + S/N) give? A. The minimum bit rate of a channel B. The highest rate with arbitrarily small error over a band-limited AWGN channel C. The average error probability D. The bandwidth needed for any code Answer: B. If R <= C, suitable coding gives arbitrarily small error probability; if R > C it cannot. (Key point: Capacity formula) 2. What is the capacity as B tends to infinity? A. Infinite B. Zero C. 1.44 S/N0, which is finite D. 0.693 S/N0 Answer: C. C_inf = (S/N0) log2 e = 1.44 S/N0, because noise power N0 B grows with B. (Key point: Infinite bandwidth) 3. Which limit is used in deriving C_inf = 1.44 S/N0? A. lim x->0 (1 + x)^(1/x) = e B. lim x->inf (1 + x)^x = 1 C. lim x->0 sin(x)/x = 0 D. lim x->inf log2 x = 0 Answer: A. With x = S/(N0 B), C = (S/N0) log2 (1 + x)^(1/x), which tends to (S/N0) log2 e. (Key point: Derivation step) 4. What is the Shannon limit on Eb/N0 for reliable communication? A. ln 2 = 0.693, i.e. -1.6 dB B. 1.44, i.e. 1.6 dB C. 0 dB exactly D. 3 dB Answer: A. As C/B tends to 0, Eb/N0 tends to ln 2, about -1.6 dB; below it, no error-free transmission is possible. 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