29Nyquist Limit on Bandwidth Find the highest data rate possible for a given bandwidth, BBinary data (two states)Zero noise on channelExample shown with band from 0 Hz to B Hz (Bandwidth B) Maximum frequency is B HzPeriod = 1/B1Nyquist: Max data rate is 2B (assuming two signal levels)Two signal events per cycle
30Nyquist Limit on Bandwidth (general) If each signal point can be more than two states, we can have a higher data rateM states gives log2M bits per signal pointPeriod = 1/B4 signal levels:2 bits/signal10001101General Nyquist: Max data rate is 2B log2MM signal levels, 2 signals per cycle
31Practical LimitsNyquist: Limit based on the number of signal levels and bandwidthClever engineer: Use a huge number of signal levels and transmit at an arbitrarily large data rateThe enemy: NoiseAs the number of signal levels grows, the differences between levels becomes very smallNoise has an easier time corrupting bits2 levels - better margins4 levels - noise corrupts data
32Characterizing Noise Noise is only a problem when it corrupts data Important characteristic is its size relative to the minimum signal informationSignal-to-Noise RatioSNR = signal power / noise powerSNR(dB) = 10 log10(S/N)Shannon’s Formula for maximum capacity in bpsC = B log2(1 + SNR)Capacity can be increased by:Increasing BandwidthIncreasing SNR (capacity is linear in SNR(dB) )SNR in linear formWarning: Assumes uniform (white) noise!
33Shannon meets Nyquist From Nyquist: From Shannon: Equating: or M is the number of levels needed to meet Shannon LimitSNR is the S/N ratio needed to support the M signal levelsExample: To support 16 levels (4 bits), we need a SNR of 255 (24 dB)Example: To achieve Shannon limit with SNR of 30dB, we need 32 levels