BER for M array orthogonal signal

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i need help in calculating the number of bit errors made per decision for my simulation.

M=8 for my case

For example;

if i send symbol X1, and i receive symbol Y1, no error is made

if i send symbol X1, and i receive symbol Y2, an error is made <--- how to calculate the number of bit errors

if i send symbol X1, and i receive symbol Y6, an error is made <-- whats the respective number of bit errors?

Thanks in advance =)
 

It depends on the the "symbol to bit" encoding. For example with Gray coding, adjacent symbols differ by one bit.

For orthogonal signaling with equiprobable symbols, a symbol error is equiprobable in all (M-1) directions. So any bit to symbol encoding will suffice.

Precisely for orthogonal signalling, Prob of bit error=(2^(k-1)/(2^k-1))*Prob of symbol error

which is roughly 0.5* Prob of symbol error.


I should also mention that this holds only for AWGN channels.

Added after 59 seconds:

Forgot k=log2(M)

M= no of orthogonal signals
 

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    tomatokid

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Hi

Thanks for your help.. I have got the correct simulation results...



Since the system is of 8 equal energy, equally likely orthogonal signals, so for same SNR value, i should get the same BER for every signal right?

Is it necessary to use gray coding for my case?
 

any random bit to symbol encoding will suffice because there is no nearest neighbor concept in orthogonal constellations unlike mqam or mpsk.
 

How will u prove that BER for msb bit is less than BER for lsb bits in mqam....
 

darock said:
any random bit to symbol encoding will suffice because there is no nearest neighbor concept in orthogonal constellations unlike mqam or mpsk.
Click to expand...

Sorry... I dun really understand... Can explain in more details?
 

@sohilitri


Well, it is not neccesarily true that MSB error prob is less than LSB error probability.

Consider QPSK or 4-QAM as shown in the following.

00 01

10 11

Here both LSB and MSB error are equiprobable.

Then what is the significance of the symbol to bit encoding?

Well, to making error in both the bits is less probable to making error in one of the bits. Therefore symbol to bit encoding can be chosen to minimize the total number of bits in error.
 

Thanks Darock,

But u see in M-QAM ...LSB bits are coded with ASK and MSB bits are coded with PSK...so there is lot of diff....as we know that PSK has min probability error....
 

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