- 2076 Asoj (CS II) · 8 marks
Implications and limitations of Shannon-Hartley theorem
Short answer
Capacity C = B log2(1+S/N) sets an upper error-free rate for AWGN channels. It implies a bandwidth–SNR trade-off, limit examples such as 30.9 kbps for 3.1 kHz at 30 dB, and the asymptotic limit 1.44 S/N0 requiring Eb/N0 ≥ -1.6 dB. Limitations: assumes only AWGN, is non-constructive, needs infinite delay, and Gaussian signaling.
Discuss with examples, the implications and limitations of Shannon Hartley Channel capacity theorem.
Shannon–Hartley theorem: the capacity of a channel of bandwidth Hz, disturbed by additive white Gaussian noise, with average signal power and noise power , is
If the information rate , there exists a coding scheme giving an arbitrarily small error probability; if , errors cannot be made small.
Implications
- Upper bound on rate: it gives the maximum error-free rate of any channel, a benchmark for real systems. Example: telephone line, kHz, dB (1000):
This is why voice-band modems stopped near 33.6 kbps. 2. Bandwidth–SNR trade-off: the same capacity can be obtained with large bandwidth and low SNR, or small bandwidth and high SNR. To keep 30.9 kbps in half the bandwidth (1550 Hz) we need (60 dB), much more power. Wideband systems (spread spectrum, FM, UWB) use bandwidth to save power. 3. Noise does not forbid error-free communication; it only limits the rate. Good channel coding (turbo, LDPC) approaches the limit. 4. Bandwidth limit: as , , and reliable transmission needs dB (Shannon limit). 5. Bandwidth efficiency plane: separates possible from impossible regions for choosing modulation (e.g. 64-QAM needs high SNR).
Limitations
- It assumes AWGN only; it does not directly apply to impulse noise, interference, fading and multipath channels.
- It is an existence theorem: it does not tell how to build the code or modulator that reaches capacity.
- Reaching capacity needs very long code words, so infinite delay and complexity; practical systems operate below .
- It assumes a Gaussian-distributed signal, average power constraint and a linear, band-limited channel; practical signals use finite constellations (e.g. 16-QAM), which reach less.
- Increasing bandwidth gives only a limited gain (capacity saturates at ), and increasing SNR gives only a logarithmic gain: doubling at high SNR adds only about bits/s.
Example of the logarithmic limitation: with kHz, raising from 1000 to 2000 increases from 30.9 kbps to only about 34.0 kbps.
✓ Answer checked by Claude Opus 5.5 (high effort)
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