m-ary and nyquist formula
I am really sorry for the mislead information.
Thank you very much for pinpointing.
well back to your question.
let me rewrite your question
WHY IS SIGNAL BANDWIDTH IS HALF THE SIGNALLING RATE?
answer:
first the formula :
maximum rate of information transmission of PCM systems. (SIGNALLING RATE)
R_b =2W log2 L bits/sec
if L=2, (i.e., L==> No. of representation level of the quantizer)
Then, R_b =2W bits/sec
which obeys nyquist theorem.
note:
As per nyquist theorem, sampling freq should be atleast twice the maximum content of the signal freq
suppose if L=M^n which represents the possible code words.
then R_b =2W log2 (M^n) bits/sec
R_b =2Wn log2 (M) bits/sec
this formula applied to M-ary PCM syatem, where the parameter average power transmitted(P),noise power (No) in the particular bandwidth B
are included reduces to final eqn
R_b = B log2 ( 1+ 12P/(k^2 No B ) )
where
No B = variance of the channel noise measured in bandwidth B
This eqn can be compared with Shannon's channel cpacity theorem.
C= B log2 ( 1+ P/ No B)
Note:
here the channel capacity is the maximum signalling rate (R_b)
i am very very sorry in my previous message i have wrritten Signalling rate as bandwidth. It is wrong.
Actualy maximum signalling rate is te channel capacity in digital comm systems.
Happy learning