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
This Question Belongs to Electronics And Communications Engineering >> Digital Communication
A. Only 800 Hz component
B. 800 Hz and 900 Hz components
C. 800 Hz and 1000 Hz components
D. 800 Hz, 900 Hz and 100 Hz components
Increased pulse width in the flat top sampling, leads to
A. Attenuation of high frequencies in reproduction
B. Attenuation of low frequencies in reproduction
C. Greater aliasing errors in reproduction.
D. No harmful effects in reproduction
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