In QPSK the average probability of bit error in AWGN channel is obtained as
A. $${P_{eQPSK}} = Q\left[ {\sqrt {\frac{{2{E_b}}}{{{N_o}}}} } \right]$$
B. $${P_{eQPSK}} = {Q^2}\left[ {\sqrt {\frac{{2{E_b}}}{{{N_o}}}} } \right]$$
C. $${P_{eQPSK}} = Q\left[ {\sqrt {\frac{{{E_b}}}{{{N_o}}}} } \right]$$
D. $${P_{eQPSK}} = Q\left[ {\frac{1}{2}\sqrt {\frac{{2{E_b}}}{{{N_o}}}} } \right]$$
Answer: Option A
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