** Baud **
"Baud" is the name for an information "symbol." Baud are usually sent at a
rate known as the "baud rate" (B). The inverse of the "baud rate" is the
time it takes to send one baud, the "baud period" (T).
B = 1/T (Baud/second)
Baud is from the name Baudot. M. Baudot invented the "Baudot code", a five-
bit code used on early teletype machines to send letters and numbers.
A Baud has a certain number of recognizable states (L). Binary numbers can
be assigned to these states. The number of bits (n) that can be carried by a
Baud ("number of bits per Baud") is just the maximum number of bits in the
set of binary numbers (codes) that be assigned to the states.
n = log2(L) or 2^n = L
Almost all standards use a number of states that correspond to an even number
of bits. Typical values are:
n L codes
(bits) (levels)
1 2 0,1
2 4 00,01,10,11
3 8 000,001,010,011,100,101,110,111
4 16 0000,0001,0010,0011,0100,0101,0110,0111,1000,...,1111
5 32 00000, ... ,11111
The states in a simple "baseband" case would be voltage or current values
(called "levels").
For "broadband" modulation the states could represent discrete values of
amplitude and/or phase. The "states" must be far enough apart so that
"noise" does not cause one state to be confused with another by the receiver
(i.e., cause transmission errors).
For example, 0 volts could be transmitted to represent the data bit "0" and 5
volts could represent the data bit "1". At the receiver the average value
of the voltage during a baud period would be measured and the data value
would be interpreted as a "0" if the voltage was less than 1.0 volt, or a "1"
if the voltage were greater than 1.0 volts. This scheme would allow the
signal voltage to decrease by 60% and noise voltages of up to +/- 0.99 volts
to be present without causing errors.
Note in the example above that the receiver must have a "clock oscillator"
that is synchronized with the baud rate and time of arrival so it knows when
to start and stop averaging the voltage. More complicated coding schemes
like RTZ and Manchester are used to make is easier for the receiver to
synchronize its "bit clock."
** Bandwidth (Nyquist)
The Nyquist theorem says the maximum bit rate (Rmax, in bits/second) for a
channel with bandwidth (H, in Hz) is:
Rmax <= 2 H log2(L)
Since log2(L) is the number of bits per Baud, so the maximum Baud per second, B;
Bmax <= 2 H
One can approach this maximum Baud rate only with baseband encoding schemes
that have only have one voltage change per Baud (NRZ, NRZI). Broadband schemes
require twice the bandwidth (B = fmax - fmin) of the basebase scheme
(B = fmax) that provides the modulation. A more useful equation is:
Bmax <= H/F (F = Baud Complication Factor, cycles per Baud)
A table of F values for the coding schemes we have discussed:
Scheme F Max. Changes per Baud
(cycles per Baud)
NRZ 0.5 1
NRZI 0.5 1
RTZ 1.0 2
Manchester 1.0 2
Broadband 1.0 1
Note that the Baud Complication Factor (F) is independent of the number of
levels. It only depends on the maximum number of times the voltage (current,
phase, or amplitude) changes during a Baud period. The Broadband factor is
twice as large as NRZ because changing the carrier amplitude or phase at a
rate B generates sidebands at f-B/2 and f+B/2 (f is the carrier frequency),
so the bandwidth is ((f+B/2) - (f-(B/2)) which is just equal to the Baud rate B.
The Bit Rate (R) of a channel is the numbers of useful bits per second that
are transmitted.
R = n B Bit Rate = (bits/Baud) * (Baud/second)
and the generalized equation is
R <= n * H / F
Max. R (bits/sec) = (bits/Baud) * (Bandwidth, cycles/sec) / (cycles/Baud)
** Effect of Noise (Shannon)**
Since the bandwidth (H) is always limited and expensive, it would seem
desirable to use encoding schemes with as many "bits per Baud" (n) as
possible. On magnetic and optical recording media there are only two levels
possible, hence one bit per baud is used. Even here, every other potential
bit is "wasted" when Manchester coding is used and the even potential bits
are always the inverse of the preceding bit (so the Baud has two levels and a
transition in the middle, but carries only one data bit of information).
Detecting multiple levels increases the complication of the receiver
circuitry. Each additional bit added to a multi-level code doubles the
number of levels and halves the voltage difference between levels (assuming
the maximum signal voltage is fixed).
The ultimate limit is due to noise. The noise level (N) is defined as the
minimum spacing between voltage levels before random variations due noise
will cause the error rate to exceed some acceptable value (e.g., 1 error in
10E12 bits). The maximum number of levels is then maximum signal level (S)
divided by N, plus 1. Note: lower case n is Bits/Baud, upper case is Noise.
L <= sqrt((S/N)+1) (L is usually a power of two (2,4,8,16,32,64,128,...))
and
n <= 0.5 log2((S/N)+1)
R <= (H/F) 0.5 log2((S/N)+1)
bits/sec <= (cycles/sec/(cycles/Baud) (max. bits per Baud)
If we note that the minimum value of F is 0.5, then we have Nyquist's
Theorem, but the form above is more useful since it shows the effect of Baud
Complication Factor (F), as well as the signal power (S) and the noise
power (N).
** Exercises **
Given a channel with bandwidth H, a signal to noise ratio (S/N), and a given
encoding scheme (RTZ, Manchester, QAM, ...), find the maximum bit rate (R).
Given a value for (S/N), what is the maximum number of levels and Bits/Baud
(n).
Given a Baud period (1/B) and (S/N), what is the maximum bit rate (R).
(1/20/99)