+ Post New Thread
Results 1 to 13 of 13

26th March 2009, 12:41 #1
 Join Date
 Dec 2006
 Location
 Iraq
 Posts
 1,142
 Helped
 406 / 406
 Points
 15,510
 Level
 30
Explanatio of 16QAM and QPSK statements
can anybody explaine the following please:
16 QAM with code rate of 1/2 coding to give 2 bits per symbol.
and
QPSK with code rate of 3/4 coding to give 1.5 bit per symbol.
also if you know the reference, please advise me
thanks in advance.

26th March 2009, 12:41

26th March 2009, 14:53 #2
 Join Date
 Dec 2008
 Posts
 76
 Helped
 23 / 23
 Points
 1,362
 Level
 8
qpsk coding rate
Hi,
16 QAM produces 4 bits/symbols with an (1/2) rate (bits/symbol) coding will produce
4*(1/2)=2 bits/symbols
QPSK with code rate of 3/4 coding to give 1.5 bit per symbol.
2 * (3/4) = 1.5bit/ symbol
whether the explantion is enough for you or are u expecting some other reply.
Happy Learning
1 members found this post helpful.

26th March 2009, 15:08 #3
 Join Date
 Dec 2006
 Location
 Iraq
 Posts
 1,142
 Helped
 406 / 406
 Points
 15,510
 Level
 30
16qam
My Friend,
If i have 128 binary digits and need to use the QPSK modulation, then the number of output symbols will be 64, in each symbol there are 2 bits. Now if i used the QPSK with 3/4 coding, what is the outbut will be?

26th March 2009, 17:00 #4
 Join Date
 Mar 2009
 Posts
 12
 Helped
 12 / 12
 Points
 824
 Level
 6
qam 16 3/4
may i know the clarification also?
Thank you
1 members found this post helpful.

26th March 2009, 17:00

26th March 2009, 18:00 #5
 Join Date
 Dec 2008
 Posts
 76
 Helped
 23 / 23
 Points
 1,362
 Level
 8
16 qam bitrate
At first what is the your bit rate(source bit rate=R_s)?????
if there are k=128 bits for (1/2) rate coder, there are 100% redundant bits (nk)
i.e., 128 bits. ; n=128+128=256.
so code rate=r=k/n=128/256=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.
If R_s is the bit rate which is incoming bits to the encoder.
the o/p of the Mary modulator in your case QPSK (M=4, in this case)
At the o/p of the Mary modualtor: symbol rate R_sym
R_sym= R_c / log2 (M) = 2*R_s / log2 (4) = R_s in your case.
Happy learning
1 members found this post helpful.

26th March 2009, 18:00

26th March 2009, 18:28 #6
 Join Date
 Dec 2006
 Location
 Iraq
 Posts
 1,142
 Helped
 406 / 406
 Points
 15,510
 Level
 30
qam bit rate
My friend,
Thank you for your help, but i am asking about the number of output symbols in case of QPSK with code rate of 3/4 coding if the input is 128 bit.
please rectify me if i am asking something wrong. In OFDM i need to know what is the data that i have before i process it by the IFFT block.
Many Thanks

26th March 2009, 19:30 #7
 Join Date
 Dec 2008
 Posts
 76
 Helped
 23 / 23
 Points
 1,362
 Level
 8
16 qam coding rate
If the i/p bit is transferred at the rate of 128 bits/ sec.
then in case of of QPSK (M=4) with code rate of 3/4 coding
Rc= (3/4)*128 = 96
NO. OF SYMBOLS TRANSFERRED AT THE O/P QPSK MODUALTOR IS : 96/log2 (4) = 96/2 = 48 symbols/sec
i think this will help you to further proceed in your learning.
Happy learning
Added after 5 minutes:
hVE YOU GOT CONVINCED OR ARE YOU EXPECTING ANOTHER BETTER REPLY.
WHETHER YOU CHECKED WITH MY ANSWER ???? I HAVE SHARED MY KNOWLEGE WITH YOU.
KINDLY CHECK WITH THE ANSWER AND DO REPLY ME.
1 members found this post helpful.

26th March 2009, 19:55 #8
 Join Date
 Dec 2006
 Location
 Iraq
 Posts
 1,142
 Helped
 406 / 406
 Points
 15,510
 Level
 30
qam code rate
I donate 50 points for you. Your discussion was very clear and good.
Many thanks.
by the way, there are about 10 new members from Malaysia joined our group all are students of PhD and M.s.c in communications. Most of them are Iraqi students. So that, Please rewrite your answer in the group as this very nice from you

27th March 2009, 08:07 #9
 Join Date
 Dec 2008
 Posts
 76
 Helped
 23 / 23
 Points
 1,362
 Level
 8
qpsk code rate
Hi Muntather,
"Please rewrite your answer in the group as this very nice from you "
i can't follow this line.
what do you mean by that.....
Happy learning
1 members found this post helpful.

27th March 2009, 11:56 #10
 Join Date
 Dec 2006
 Location
 Iraq
 Posts
 1,142
 Helped
 406 / 406
 Points
 15,510
 Level
 30
qam coding rate
just to put some force in the group. May be this will help some of the members. I do not mean other thing. My Friend,
My goal is to let every one learn.
I am sorry if you understand it in a wrong way.

27th March 2009, 11:56

2nd April 2009, 07:27 #11
 Join Date
 Oct 2008
 Location
 Sindh, Pakistan
 Posts
 895
 Helped
 180 / 180
 Points
 9,408
 Level
 23
16 qam symbol rate
Originally Posted by rramya
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.
Secondly, the idea about the 1.5 bits/symbol ... I think this is related to information theory, we dont have a 0.5 bit practically. I am completely convinced by the math that you have provided but this shows that in one symbol after encoding there would be equivalent information to 1.5 bits. Even though the symbol would really be carrying 2 bits i.e. 2 bits/symbol
Do you agree? Y/N

2nd April 2009, 10:46 #12
 Join Date
 Dec 2008
 Posts
 76
 Helped
 23 / 23
 Points
 1,362
 Level
 8
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.
1 members found this post helpful.

1st September 2009, 20:21 #13
 Join Date
 Sep 2009
 Posts
 1
 Helped
 0 / 0
 Points
 585
 Level
 5
16 qpsk
Hi, Im looking for an answer with a problem that I have with my wireless connection.
The antena outside my house has very high signal but the modulation fluctuates too much. For example:
QAM64 3/4
QAM16 1/2
QAM16 3/4
QPSK 3/4
Data ranges for download and upload of 29+
It seems that this fluctuations are affecting my internet connection. I'm experiencing also fluctuations on connection timeouts (Sporadic connection timeout). This is affecting my downloads and other communication that requires stable connection.
Do anyone knows if its possible to fix a wireless antena with just one modulation?
Thanks for any help
QLands
+ Post New Thread
Please login