Every digital communication system has to answer two questions: how much spectrum do I need? (occupied bandwidth) and how fast can I theoretically transmit through it? (channel capacity). The first answer depends on the modulation scheme; the second depends on the signal-to-noise ratio. Nyquist gave the lower bound on bandwidth; Shannon gave the upper bound on capacity.
Nyquist’s bandwidth theorem
For a noiseless channel, the minimum bandwidth required to transmit at symbol rate Rs without inter-symbol interference is Rs/2 Hz (baseband) or Rs Hz (passband). This is the "Nyquist bandwidth." Real systems can't use ideal brick-wall filters, so they add excess bandwidth via a roll-off factor α (typically 0.20–0.35), giving occupied bandwidth B = Rs · (1 + α).
Shannon’s channel capacity
Shannon's 1948 formula: C = B · log2(1 + SNR), where SNR is in linear scale (not dB). This is the upper bound on error-free bit rate for any modulation, any coding. At SNR = 20 dB (100×), Shannon η = log2(101) ≈ 6.66 bits/s/Hz — meaning a 1 MHz channel could theoretically carry up to 6.66 Mbps. Real systems get close (LTE, WiFi 6 reach ~80% of Shannon η in good conditions) but never exceed it.
Bits per symbol — the modulation tradeoff
Higher-order modulations pack more bits per symbol: BPSK = 1, QPSK = 2, 16-QAM = 4, 256-QAM = 8, 4096-QAM = 12. Bandwidth stays the same per symbol — you get more bits without more spectrum. But the constellation points get closer together, so you need higher SNR to distinguish them. Rule of thumb: each extra bit/symbol needs ~3 dB more SNR. 1024-QAM (WiFi 6) needs ~30 dB SNR; 4096-QAM (WiFi 7) needs ~36 dB.
ASK, FSK, PSK, QAM
- ASK (Amplitude Shift Keying): encodes data in signal amplitude. OOK (on-off keying) is the simplest. Vulnerable to amplitude noise.
- FSK (Frequency Shift Keying): encodes data in carrier frequency. Constant-envelope (immune to amplitude noise), used in Bluetooth Classic, garage door openers, low-rate IoT. Wider spectrum than PSK/QAM. FSK bandwidth depends on the modulation index
h = 2Δf / Rs; this tool's 2-FSK/MSK/4-FSK values assume h = 0.5 (the most spectrally efficient case). For larger h, the actual bandwidth is wider (Carson's rule: B ≈ 2Δf + Rs).
- PSK (Phase Shift Keying): encodes data in carrier phase. BPSK (2 phases), QPSK (4 phases), 8-PSK (8 phases). Constant envelope. Limited to k ≤ ~3 before SNR demand becomes prohibitive.
- QAM (Quadrature Amplitude Modulation): encodes data in BOTH phase and amplitude. The constellation is a 2D grid of points. Far more bits per symbol than PSK at the same SNR. Used in LTE, WiFi 4+, DOCSIS, DVB.
What's the difference between Nyquist bandwidth and occupied bandwidth?
Nyquist bandwidth is the THEORETICAL minimum needed to transmit at symbol rate Rs with no inter-symbol interference — equal to Rs at passband (or Rs/2 baseband). Occupied bandwidth is what your signal ACTUALLY uses after pulse shaping. Real systems use raised-cosine (or root-raised-cosine) filters with roll-off α > 0, giving occupied B = Rs · (1 + α). For α = 0.25, the occupied bandwidth is 25% wider than Nyquist. The smaller α, the more efficient — but the filter gets harder to build and more sensitive to timing errors.
Why can't real systems reach the Shannon limit?
Shannon proved capacity exists with arbitrarily small error probability, but achieving it requires (a) infinitely long codewords (infinite latency), (b) infinite-complexity decoders, and (c) Gaussian-distributed signal (continuous, not discrete constellations). Real systems use finite-length error-correction codes (LDPC, polar, turbo), discrete QAM constellations, and have practical hardware impairments. Modern systems get 70–85% of Shannon η; the last few % requires exotic coding and is often not worth the complexity. 5G NR LDPC + 256-QAM in good conditions reaches ~80%.
What is Eb/N0 and why does it matter?
Eb/N0 (read "E-b over N-naught") is the energy per bit divided by noise power spectral density — a normalized SNR that lets you compare systems with different data rates and bandwidths fairly. Relationship: Eb/N0 = SNR × B / Rb = SNR / η. Lower Eb/N0 means more efficient use of energy per bit. Each modulation has a minimum Eb/N0 to achieve a target bit-error rate (BER). BPSK needs ~9.6 dB for 10⁻⁵ BER (uncoded); 256-QAM needs ~24 dB. Error-correction codes can reduce required Eb/N0 by 3–10 dB ("coding gain").
Why does QAM beat PSK for high bit rates?
At high orders, QAM's 2D constellation packs points more efficiently than PSK's 1D ring. Consider 16-QAM (4×4 grid) vs 16-PSK (16 points on a circle). The 16 points on a circle are very close together — small noise pushes you to a neighbor. The 16-point grid has more breathing room because the constellation expands in both dimensions. For a given average power, QAM's minimum point spacing is larger than PSK's for M ≥ 16. Below M = 8, the difference is minor; above it, QAM dominates. That's why no one uses 16-PSK or 32-PSK in practice.
What's a typical real-world example of these numbers?
5G NR with 100 MHz channel: typical operating point is 256-QAM with rate-3/4 LDPC at ~25 dB SNR. Symbol rate ~100 Mbaud × spectral-efficiency overhead ≈ 90 Mbaud net. Bit rate = 90 Mbaud × 8 bits/symbol × 0.75 (code rate) ≈ 540 Mbps. Shannon at 25 dB, 100 MHz = ~830 Mbps. So 5G achieves ~65% of Shannon η — a typical real-world figure. The remaining 35% is overhead for control channels, pilot signals, guard bands, and coding inefficiency.
Should I use a smaller roll-off α to save bandwidth?
Tradeoffs. Smaller α (closer to 0) means closer to Nyquist minimum — saves spectrum. But: (a) the filter needs more taps to achieve sharper roll-off, increasing DSP complexity and latency; (b) the time-domain pulse has bigger sidelobes, more sensitive to timing jitter; (c) the peak-to-average power ratio (PAPR) grows, requiring more linear amplifiers. Modern systems pick α based on the regulatory mask and hardware budget. LTE uses α ≈ 0.22, DVB-S2X allows 0.05–0.35. Don't go below 0.10 unless you have a very stable clock and high-quality DSP.