Outline

Nyquist Theorem
Shannon Theorem
Serial Lines:
- Asynchronous Communication
- Encoding
- Baud Rates, Maximum Data Rates

Nyquist Theorem

Nyquist Sampling Theorem:
- if all significant frequencies of a signal are less than bandwidth B
- and if we sample the signal with a frequency 2B or higher,
- we can exactly reconstruct the signal.
- any sampling rate less than 2B will lose information
formulated by Nyquist, proven by Shannon in 1949

with a signal for which the maximum frequency is higher than B, the reconstructed signal may not resemble the original signal

assume samples are encoded digitally using K symbols
for example, there are 16 possible symbols, 0..9A..EF, encoding 16 possible sample values
for example, a symbol can be a voltage level on a wire
then, the maximum data rate is D = 2B log₂K bits/s
for example, with 32 symbols and a bandwidth B=1MHz, the maximum data rate is 2*1MHz*log₂32 bits/s or 10Mb/s
a symbol can be encoded as a unique signal level (AM), or a unique phase (PM), or a unique frequency (FM), or a combination of these (modern modems)
the rate of symbol transmission is also known as the Baud rate but this term is frequently mis-used (as bits/second)
in theory, we could have a very large number of symbols, allowing very high transmission rate without high bandwidth
in practice, we cannot use a high number of symbols because we cannot tell them apart: all real circuits suffer from noise

it is impossible to reach very high data rates on bandlimited circuits in the presence of noise
given a signal power S, noise power N, signal-to-noise ratio S/N
a decibel level dB is dB = 10 log₁₀ S/N
for example S/N = 20dB means the signal is 100 times more powerful than the noise
Shannon's theorem: the capacity C of a channel with bandwidth B Hz is C = B log₂(1+S/N) b/s
for example if S/N = 20dB, and the channel has bandwidth B = 1MHz,
- C = B log₂(1+S/N) b/s
- C = 1MHz log₂(1+100) b/s
- 6Mb/s < C < 7Mb/s
if S/N = 0dB (signal power and noise power are the same), C = 1Mb/s
this can be very hard to achieve in practice!

a given telephone circuit has a bandwidth of 3000 Hz and a S/N ratio of 30dB (factor of 1000)
Shannon limit gives us C = 3000 log₂(1+1000) b/s =~ 30,000 b/s
to encode information on this channel, we can send 3000 samples / second, and use about 1,000 different symbols:
- each symbol can encode 10 bits
- the 30dB S/N ratio means we have 1,000 times more power in the signal than in the noise, and so (intuitively) we may be able to distinguish about that many symbols
- the Nyquist theorem says
  - if we try to send any more than 3,000 symbols per second,
  - and we are bandlimited to 3,000 Hz
  - the higher frequencies will be lost
  - symbols will be lost
  - so we should not use more than 3,000 symbols per second
  - to decode, we should look for samples at least 6,000 times per second
in-class exercise:
- what channel capacity can we get if the S/N ratio is 10dB and the bandwidth is 1,000 Hz?
- with S/N ratio 50dB?
- how many different symbols are we going to use?

RS-232 standard, later augmented with RS-422 standard
asynchronous communication: receiver must be able to decode sender's signal without knowledge of the sender's clock (with synchronous communication, the sender sends the clock as well as the data)
serial communication: one bit is sent on the wire before the next bit is sent (opposite of parallel communication)

possible encoding: no voltage is no bit (idle), positive voltage is 1, negative voltage is 0 bit
RS-232 encoding: positive voltage is 0, negative (or zero) voltage is 1, idle sends 1
two sides of RS-232 must agree on the number of data bits sent (e.g. 5, 8 -- 4 shown below) and on the bit time (e.g. 9600 baud)
sender sends the start bit (always 0, so high), the n data bits (LSB first), and 1 or more stop bits (always 1, so low)
receiver starts an accurate clock at the center of the start bit, uses it to read the data bits
two signal wires plus ground give bidirectional communication (full duplex -- both sides can send at the same time)

BREAK is a long sequence of zero bits -- multiple hundreds of milliseconds
if a bit signal is corrupted (e.g. by lightning), the signal on the wire may not be what was intended to be sent
if the start bit is distorted (by noise or bad wiring), the receiver may interpret the successive bits at the wrong time
if the two endpoints fail to agree on bit time or number of bits, the communication will fail and random-looking characters will be received

A Baud is the number of changes per second in the signal (named after Baudot)
for RS-232 and many other systems, the number of Baud is the same as the number of bits per second
but 10Mb/s Manchester-encoding Ethernet runs at 20MBaud
56Kb/s modems run at much lower baud rates (9600 baud?), but use many symbols