The bit error probability for GMSK is given by
A. $${P_e} = \frac{1}{2}\left\{ {\sqrt {\frac{{2v{E_b}}}{{{N_0}}}} } \right\}$$
B. $${P_e} = Q\left\{ {\sqrt {\frac{{2v{E_b}}}{{{N_0}}}} } \right\}$$
C. $${P_e} = \sqrt {\frac{{2v{E_b}}}{{{N_0}}}} $$
D. $${P_e} = {Q^2}\left\{ {\sqrt {\frac{{2v{E_b}}}{{{N_0}}}} } \right\}$$
Answer: Option B
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