qam 16 1/2
comm_eng wrote:
How can the (data) rate change? Doesn't the channel encoder just add redundency at the same data rate of the incoming stream.
2*R_s shows that after suppose convolutional encoding the data rate of the encoded symbols is greater than the data rate of the incoming bit stream.
I think the data rate remains same, only that now you have half of the user bits flowing in the same amount of time.
ANS: YOU R CORRECT:
i never talk abt data rate change!!!!
all i talk abt
the channel encoder rate ... because of adding redundant bits....
eg in case: for convolutional encoder rate of (1/2)
channel encoder bit rate is given by = R_c= (n/k)*R_s = (256/128)*R_s =2*R_s
where R_s is the bit rate generated by the source.
which means: we require double the bandwidth of the uncoded Systems.....and hence the capacity of the channel needs to be doubled for the . hence R_c=2*R_s
INTREPRETATION:
since BW is a constrainst,
if we cant double the bandwidth of the single user channel: automatically,
ONLY HALF OF THE USER BITS ARE TRANSMITTED in the given channel , utilizing the available BW.
Secondly, the idea about the 1.5 bits/symbol :
suppose if R_s = 3 bit/sec.... encoder rate = (3/4)
R_c=(4/3)R_s =4 bit/sec
Here , we have original bits = 3 bits
redundant bit = 1 bit
At the Modulator o/p , the symbol rate is R_Mod_sym = R_c/log2(M)
for QPSK M=4.
R_Mod_sym = R_c/log2(M) = ( 4 ) / log2(4) = 4/( 2) = 2 symbols /sec
now,
bits/symbol = 3 (bit/sec) / (2 ) (symbols /sec) = (3/2) bits / symbol...
= 1.5 bits/symbol
INTREPRETATION:
we have 3 bits as original bits there are 2 symbols (which has got 2 bits/symbol, in case of QPSK) we get at the o/p of the modulator ......
bits/symbol = 1.5 bits/symbol ..............
Happy Learning.