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Channel Capacity No. 1Seattle Pacific University
Channel Capacity:Nyquist and Shannon Limits
Based on Chapter 3 of William Stallings, Data andComputer
Communication, 8th Ed.
Kevin BoldingElectrical Engineering
Seattle Pacific University
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Channel Capacity No. 2Seattle Pacific University
Nyquist Limit on Bandwidth Find the highest data rate possible
for a given bandwidth, B
Binary data (two states) Zero noise on channel
1 0 1 0 0 0 1 0 1 1 0 1 00 0
Period = 1/B
Nyquist: Max data rate is 2B (assuming two signal levels) Two
signal events per cycle
Example shown with bandfrom 0 Hz to B Hz (Bandwidth B)
Maximum frequency is B Hz
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Channel Capacity No. 3Seattle Pacific University
Nyquist Limit on Bandwidth (general) If each signal point can be
more than two states, we can have
a higher data rate
M states gives log2M bits per signal point
10 00 11 00 00 00 11 01 10 10 01 00 0000 11
Period = 1/B
General Nyquist: Max data rate is 2B log2M M signal levels, 2
signals per cycle
4 signal levels:2 bits/signal
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Channel Capacity No. 5Seattle Pacific University
Characterizing Noise
Noise is only a problem when it corrupts data
Important characteristic is its size relative to theminimum
signal information
Signal-to-Noise Ratio
SNR = signal power / noise power SNR(dB) = 10 log10(S/N)
Shannons Formula for maximum capacity in bps
C = B log2
(1 + SNR)
Capacity can be increased by:
Increasing Bandwidth
Increasing SNR (capacity is linear in SNR(dB) )
Warning: Assumes
uniform (white) noise!
SNR in linear form
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Channel Capacity No. 6Seattle Pacific University
Shannon meets NyquistFrom Nyquist: MBC 2log2
From Shannon: )1(log 2SNRBC
Equating: )1(loglog2 22 SNRBMB
)1(loglog2 22 SNRM
)1(loglog 22
2 22SNRM
SNRM 12
SNRM 1 12 MSNRor
M is the number of levels
needed to meet Shannon Limit
SNR is the S/N ratio needed to
support the M signal levels
Example: 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
)1(loglog 22
2 SNRM
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Achieving the Nyquist Limit
The Nyquist Limit requires two signaling events per Hertz
C=2B log2M This must be achieved using waveforms with
frequency
components
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Sinc (Nyquist) Pulses The Sinc Pulse is defined as
sin(x)/x
Sinc pulse at frequency frequires bandwidth f
sin(x 2f)/(x 2f)
Note that the sinc pulse iszero at all multiples of 1/2fexcept
for the singular pulse
Pulses can overlap as longas each one is centered on amultiple
of 1/2f
-5 -4 -3 -2 -1 0 1 2 3 4 5-0.5
0
0.5
1
1.50 1 0 0 0 0 1 1 10 0 0 1 1 1 0 0 1 0
When the pulses aresummed, checking thewaveform at each multiple
of1/2f gives the orignal data
-5 -4 -3 -2 -1 0 1 2 3 4 5-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
-5 -4 -3 -2 -1 0 1 2 3 4 5
0 1 0 0 0 0 1 1 10 0 0 1 1 1 0 0 1 0
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
