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Channel Capacity: Nyquist and Shannon Limits. Based on Chapter 3 of William Stallings, Data and Computer Communication, 8 th Ed. Kevin Bolding Electrical Engineering Seattle Pacific University. Find the highest data rate possible for a given bandwidth, B Binary data (two states)

Channel Capacity: Nyquist and Shannon Limits

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Channel Capacity:Nyquist and Shannon Limits

Based on Chapter 3 of William Stallings, Data and Computer Communication, 8th Ed.

Kevin BoldingElectrical EngineeringSeattle Pacific University

Find the highest data rate possible for a given bandwidth, B

Binary data (two states)

Zero noise on channel

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Example shown with bandfrom 0 Hz to B Hz (Bandwidth B)Maximum frequency is B Hz

Period = 1/B

- Nyquist: Max data rate is 2B (assuming two signal levels)
- Two signal events per cycle

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

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Period = 1/B

4 signal levels:

2 bits/signal

- General Nyquist: Max data rate is 2B log2M
- M signal levels, 2 signals per cycle

4 levels - noise corrupts data

2 levels - better margins

- Nyquist: Limit based on the number of signal levels and bandwidth
- Clever engineer: Use a huge number of signal levels and transmit at an arbitrarily large data rate

- The enemy: Noise
- As the number of signal levels grows, the differences between levels becomes very small
- Noise has an easier time corrupting bits

- Noise is only a problem when it corrupts data
- Important characteristic is its size relative to the minimum signal information

- Signal-to-Noise Ratio
- SNR = signal power / noise power
- SNR(dB) = 10 log10(S/N)

- Shannon’s 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) )

SNR in linear form

Warning: Assumes uniform (white) noise!

From Nyquist:

From Shannon:

Equating:

or

SNR is the S/N ratio needed tosupport the M signal levels

M is the number of levelsneeded to meet Shannon Limit

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

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- The Nyquist Limit requires two signaling events per Hertz
- C=2B log2M
- This must be achieved using waveforms with frequency components <= B

Period = 1/B

“Corners” require higher-frequency components

- We need a way to represent a ‘1’ with a pulse that has no components greater than B
- Must be able to overlap two pulses per Hertz without loss of information

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- The Sinc Pulse is defined as sin(x)/x
- Sinc pulse at frequency f requires bandwidth f
- sin(x 2f)/(x 2f)

- Note that the sinc pulse is zero at all multiples of 1/2f except for the singular pulse

- Pulses can overlap as long as each one is centered on a multiple of 1/2f

- When the pulses are summed, checking the waveform at each multiple of 1/2f gives the orignal data