Question

A data is transmitted using BPSK and BFSK, modulation scheme whose signal space diagram is shown below.



The probability of error in BPSK signal in Q(x) 'x' is constatns. Assume the signals are received at the receivers which are perfectly synchronized, then the probability of BFSK signal is equal to

• A. the value of $P_{e\left(FSK\right)}<P_{e\left(PSK\right)}$
• B. the value of $P_{e\left(FSK\right)}>P_{e\left(PSK\right)}$
• C. the value of $d_1<d_2$
• D. the value of $d_1>d_2$

Answer 1

Answer: (B), (D)

the value of $d_1>d_2$



$P_e=Q\left(\frac{d}{\sqrt{2N_o}}\right)$
$d_1$ for BPSK $=2\sqrt{E_b}$
$d_2$ for BPSK $=\sqrt{2E_b}$
$P_{e\left(QPSK\right)}=Q\left(\sqrt{\frac{2E_b}{N_o}}\right)=Q\left(x\right)$
$P_{e\left(FSK\right)}=Q\left(\sqrt{\frac{E_b}{N_o}}\right)$
$P_{e\left(FSK\right)}=Q\left(\frac{x}{\sqrt{2}}\right)$