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BER for Square M-QAM

BER for Square M-QAM. EE 241 Ronell B. Sicat December 12, 2009. Challenge: Compute the Probability of Symbol/Bit Error for 64-QAM. Finding SER can be ok, but how about BER?. Why. ... BER? Because ultimately we are sending 1’s and 0’s.

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BER for Square M-QAM

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  1. BER for Square M-QAM EE 241 Ronell B. Sicat December 12, 2009

  2. Challenge: Compute the Probability of Symbol/Bit Error for 64-QAM • Finding SER can be ok, but how about BER?

  3. Why... • ... BER? Because ultimately we are sending 1’s and 0’s. • ... do we want to find a generalized expression for BER? Because computing it is not easy. How... • ... is the BER computation done? Similar to SER computations we did before (can be tedious). • ... do we find the BER expression? By induction.

  4. Outline • Describe system model and assumptions • Derive BER expressions • M = 16 (in detail) • M = 64 and M = 256 (use provided results) • Find generalized BER expression • Verify result and discuss observations

  5. System Model and Assumptions • Square M-ary QAM (M = 2N where N is even) • M symbols mapped to k = log2M bits • Each symbol is represented by • where Ac and As are represented by log2M level amplitudes which take values of either where d is half of the minimum distance (unlike what we use in most of our examples).

  6. System Model and Assumptions • Half distance d • Gray code is used to minimize bit error. • Assumptions: • Equiprobable symbols • Noise is AWGN, zero mean, variance = No/2 • No error due to synchronization

  7. BER Computation (M = 16) Signal constellation for square 16-QAM

  8. BER Computation (M = 16) • Each symbol is represented by four bits (i1, q1, i2, q2) with i1, i2 as corresponding to I and q1, q2 corresponding to Q. • Consider two cases: • i1, q1 only (first bits of I and Q) • i2, q2 only (second bits of I and Q) • Let’s represent the probability of bit error for case 1 as Pb(1) and Pb(2) for case 2. • Pb(k) = probability that the kth bits of I and Q are in error (average error for ik and qk), e.g.

  9. BER Computation (M = 16) Signal constellation for case 1

  10. BER Computation (M = 16) Divide and Conquer! • Next we find • But and Note: AWGN channel

  11. BER Computation (M = 16)

  12. BER Computation (M = 16) Rotating the constellation by 90 degrees.

  13. BER Computation (M = 16) • By symmetry • Finally, the bit error rate for case 1 is:

  14. BER Computation (M = 16) • We now express this in terms of erfc( ) using we get

  15. BER Computation (M = 16) • We now consider case 2 where we disregard i1,q1 to get Pb(2). • Using a similar method we get

  16. BER Computation (M = 16) Signal constellation for case 2.

  17. BER Computation (M = 16) • We now have • But

  18. BER Computation (M = 16) .. ..

  19. BER Computation (M = 16) • Going back and using the results, we get and by rotation, Thus, by averaging we get

  20. BER Computation (M = 16) • Expressing this in terms of erfc( ) • Recall, • Finally, the BER for M = 16 is

  21. BER Computation (M = 64) • Next is to find BER for M = 64 and M = 256. • Do we want to do this? • Use these steps • Decompose problem into k = (log2M)/2 cases and get Pb(k) for each case. • Express Pb(k) in terms of erfc( ) and r = Eb/N0. • BER is the average of the Pb(k) of the k cases.

  22. BER Computation (M = 64) • We are provided with the following for M = 64

  23. BER Computation • Before we move on let us define the following • For example, for M = 16

  24. BER Computation (M = 256) • We are provided with the R,S pairs for 256-QAM

  25. BER Computation (M = 256) • As reference we are given, • Finally we can get,

  26. Generalized BER expression • Now that we have the BER expressions for M = 16, 64 and 256, let’s try to look for a pattern. • 1st observation: Each Pb(k) has a factor of • For the next observations, let us look at R,S tables

  27. Generalized BER expression R,S components for Pb(1) WHAT PATTERN DO YOU SEE?

  28. Generalized BER expression • For k = 1, R is always 1 • For k = 1, S starts with 1 and is increasing by 2 as 1, 3, 5, 7... • There are terms.

  29. Generalized BER expression

  30. Generalized BER expression • S is the same as in k = 1 • R is decreasing (absolutely) and changing signs until -1.

  31. Generalized BER expression • Number of terms per case (conditional probability) = • If we combine all our observations, we come up with

  32. Generalized BER expression • Finally, to get the BER expression, we take the average of all the conditional probabilities • Now you can get the expression for BER for any square QAM given M.

  33. Verify result . • Example for M = 4 (QPSK) using conventional method

  34. Verify result .

  35. Verify result • Example for M = 4 (QPSK) using BER expression

  36. Verify result VERIFIED for M = 4.

  37. Thank You for your attention. =) Shukran!

  38. Generalized BER expression Other values of k

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