umardar27
Junior Member level 1
Hi,
I am new to MATLAB and was trying to simulate BPSK and QPSK in matlab.
I had a queiry regarding noise scaling .Why the noise is devided by sqrt(1/ebno) and sqrt(2) . Here is sample programme
ebno_index=1:7
ebno=10.^(ebno_index/10)
....
noise=sqrt(1/ebno)*randn(1,100) % for BPSK
...
noise=sqrt(1/ebno)*(randn(1,100)+j*randn(1,100))/sqrt(2) % for QPSK
tahanks in advance.
umar
---------------------------------------------------------------------------------------------
ebno=10.^(ebno_index/10)
is not convert noise power in dB to normal value
noise=sqrt(1/ebno)*randn(1,100) % for BPSK
this assumes that your signal power is 1. sqrt is because noise variance (=power) becomes k^2 if the signal is multplied by k.
in the complex case (QPSK), the power has to be equally split between the 2 dims,
and so 1/sqrt(2) is required, so that total noise power is ebno.
HTH,
-b
-------------------------------------------------------------------------------------------------
Thanks a lot for the response.
If i correctly understood ur answer then thats mean if there is QAM16 and not QPSK then the noise has to divided by sqrt(8).
Another quiery is if we assume that our signal power is not 1 and is say 2 then will our noise become sqrt(2/ebno)*randn(1,100)
There is one more quiery and i.e. is the channel treated the same way. e.g if we assume channel taps to be (randn+j*randn) then we are to divide it by sqrt(2) and what will happen if channel taps are incresed from one to say 3 taps.
Thanking you in advance.
P.S it will be appreciated if you can send some basic material (tutorials/papers) in this regard for my better understanding.
-----------------------------------------------------------------------------------------------
I am new to MATLAB and was trying to simulate BPSK and QPSK in matlab.
I had a queiry regarding noise scaling .Why the noise is devided by sqrt(1/ebno) and sqrt(2) . Here is sample programme
ebno_index=1:7
ebno=10.^(ebno_index/10)
....
noise=sqrt(1/ebno)*randn(1,100) % for BPSK
...
noise=sqrt(1/ebno)*(randn(1,100)+j*randn(1,100))/sqrt(2) % for QPSK
tahanks in advance.
umar
---------------------------------------------------------------------------------------------
ebno=10.^(ebno_index/10)
is not convert noise power in dB to normal value
noise=sqrt(1/ebno)*randn(1,100) % for BPSK
this assumes that your signal power is 1. sqrt is because noise variance (=power) becomes k^2 if the signal is multplied by k.
in the complex case (QPSK), the power has to be equally split between the 2 dims,
and so 1/sqrt(2) is required, so that total noise power is ebno.
HTH,
-b
-------------------------------------------------------------------------------------------------
Thanks a lot for the response.
If i correctly understood ur answer then thats mean if there is QAM16 and not QPSK then the noise has to divided by sqrt(8).
Another quiery is if we assume that our signal power is not 1 and is say 2 then will our noise become sqrt(2/ebno)*randn(1,100)
There is one more quiery and i.e. is the channel treated the same way. e.g if we assume channel taps to be (randn+j*randn) then we are to divide it by sqrt(2) and what will happen if channel taps are incresed from one to say 3 taps.
Thanking you in advance.
P.S it will be appreciated if you can send some basic material (tutorials/papers) in this regard for my better understanding.
-----------------------------------------------------------------------------------------------