So what relation Baud and Bit rate ?

After we see relation bandwidth and baud rate / bit rate in previous article, we will discuss further about that relation more deeply in this section. Relationship between the bandwidth of the baud rate derived using sinc pulse concept.
Sinc pulse have zero crossing in interval 1/(2W). With Fourier analysis we can show that the sinc pulse has no frequency components higher than W as seen below :

If the transmission channel is an ideal lowpass filter with a bandwidth higher than W , the channel will be suitable to be used for sending sinc pulse which have first zero crossing at t=1/2 W without distortion. The output pulse shape will remain because of all the components in the output frequency will remain the same as the input.
The nature of the sinc pulse which has a periodic zero crossing  pulse every 1/2 W ( pulse  with a maximum frequency W )  can be used to send the next pulse right on t = 1/2 W.
The periodic time we know as T = 1/2 W.  So we conclude that bit rate = 1/ T = 2W.

If we increase data rate become W –> B, we have interval time each pulse T–> 1/2 B, it means  data rate r –> 1/T = 2B.
This value gives us the maximum rate theoretically for symbol transmission.
From here we get the relation become :

r <= 2B atau B >= r/2

In fact, there is no such thing as sinc pulses so that our analysis produces a maximum symbol rate in a channel lowpass and used in a similar pulse with sinc pulses which has wide 1.5 until 2 times from sinc pulse.

If we can make some symbol amplituda different with each bit representing a carry bit, we can increase the data rate while maintaining symbol rate in constant value.

In general the number of symbols determined by the number of bits of information represent : M = 2^k
So the relation between number symbol and bit rate is : Bit rate = rb = r log2 M [bps].

If we have 2 bit information, we have symbol M = ^k = 2^ 2 = 4
So bit rate is = rb = r. log2 M = r log2 4 = 2 bps.  It means if we have data 1 kilobaud, we will have 2 kbps.
Here r is baudrate, and rb is bitrate.

So with a certain baud rate we can continue to raise the bit rate by adding the number of symbols, or in other words increase the number of bits carried by one symbol.
So just increase number bit in one symbol in order to increase bit rate . Is that true ?
If the focus only in bandwidth we can say : true !
But there’s another factor which make we say it FALSE.  It is noise factor.

The greater number of symbols will make detection more difficult and the effects will be more significant noise, because it  can changes in the level of symbol. Noise will reduce analog communication quality and raises error in digital communication. We state relativity between signal and noise in :

(S/N)dB = 10 log (S/N) [dB]

Shannon stated that an error free bit rate  ( bit error without errors) on a transmission channel can not exceed the maximum capacity C.

C = B log2(1+S/N)

Where C = maximum information data rate in bit per second, and B =  bandwidth line  in hertz, and S/N = Signal to noise Ratio.

For example if a transmission channel has bandwidth = 4Khz, we have symbol rate maximum r <= 2B = 8 kbauds.
It means we can send 8000 signal (symbol) per second.
But if we have  S/N = 28 dB or = 631 we just have
C = B log2(1 + S/N) = 4.000 log2(632) = 37.2 Kbps.

The question is how many bit per symbol / signal can be transfer for that ?
Just combine formula above :
bit rate = r log2 M
37,2 kbps = 8 kbauds * log2 2^k
We get k =4. It means we just have 4 bit maximum for one signal or symbol in that transmission line.

Another great posts :

• Baud Rate not Bit Rate ??