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Entropy and information rate; Shannon limit as bandwidth → ∞

Short answer

Entropy H is the average information per symbol in bits/symbol. For a source emitting r symbols per second, information rate R = r H bits/s. As bandwidth B → ∞, Shannon capacity approaches C∞ = (S/N0) log2 e ≈ 1.44 S/N0 bits/s, with a required Eb/N0 of at least -1.6 dB.

State and derive the relation between entropy and information rate. Derive the expression which shows the limit of Shannon's channel capacity theorem when bandwidth tends to infinity.

Relation between entropy and information rate

Entropy HH is the average information per symbol, in bits/symbol. Information rate RR is the average information produced per second, in bits/s.

Derivation: let a discrete memoryless source emit rr symbols per second from an alphabet {s1,…,sM}\{s_1, \dots, s_M\} with probabilities p1,…,pMp_1, \dots, p_M.

  1. In a long time TT seconds the source emits N=rTN = rT symbols.
  2. Symbol sks_k occurs about Npk=rTpkN p_k = rTp_k times, and each occurrence carries log⁡2(1/pk)\log_2(1/p_k) bits.
  3. Total information in TT seconds:
IT=∑k=1MrT pklog⁡21pk=rT HI_T = \sum_{k=1}^{M} rT\,p_k\log_2\frac{1}{p_k} = rT\,H
  1. Information per second:
R=ITT=r H bits/sR = \frac{I_T}{T} = r\,H\ \text{bits/s}

where rr is the symbol rate (symbols/s) and HH the entropy (bits/symbol). Example: r=1000r = 1000 symbols/s and H=1.75H = 1.75 bits/symbol give R=1750R = 1750 bits/s. For equally likely symbols, Rmax=rlog⁡2MR_{max} = r\log_2 M.

Limit of Shannon capacity when B→∞B \to \infty

Shannon–Hartley theorem for an AWGN channel (noise PSD N0N_0, noise power N=N0BN = N_0B):

C=Blog⁡2(1+SN0B) bits/sC = B\log_2\left(1+\frac{S}{N_0B}\right)\ \text{bits/s}

Rewrite with x=SN0Bx = \dfrac{S}{N_0B}, so B=SN0xB = \dfrac{S}{N_0 x}:

C=SN0⋅1xlog⁡2(1+x)=SN0log⁡2(1+x)1/x\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} \end{aligned}

As B→∞B \to \infty, x→0x \to 0, and lim⁡x→0(1+x)1/x=e\lim_{x \to 0}(1+x)^{1/x} = e:

C∞=SN0log⁡2e=SN0⋅1ln⁡2≈1.44 SN0 bits/sC_\infty = \frac{S}{N_0}\log_2 e = \frac{S}{N_0}\cdot\frac{1}{\ln 2} \approx 1.44\,\frac{S}{N_0}\ \text{bits/s}

Meaning:

  • Capacity does not become infinite when bandwidth becomes infinite, because noise power N0BN_0 B also increases. It approaches a finite limit set by S/N0S/N_0.
  • At this limit, with energy per bit Eb=S/CE_b = S/C:
EbN0=SN0C∞=ln⁡2=0.693  ⇒  −1.6 dB\frac{E_b}{N_0} = \frac{S}{N_0 C_\infty} = \ln 2 = 0.693 \;\Rightarrow\; -1.6\ \text{dB}

This is the Shannon limit: no coding scheme can give reliable transmission with Eb/N0E_b/N_0 below −1.6 dB.

 C
 |           ......................  1.44 S/N0
 |       ..'
 |    .'
 |  .'
 | /
 +-------------------------------->  B

✓ Answer checked by Claude Opus 5.5 (high effort)

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