Question

Consider a BPSK signal as follows:
$s\left(t\right)=\{\begin{array}{c}+10\varphi \left(t\right);\text{for logic-1}\\ -10\varphi \left(t\right);\text{for logic-0}\end{array}$

where, $\varphi \left(t\right)=\{\begin{array}{c}\sqrt{\frac{2}{T_b}}cos\left(2\pi f_{c}t\right);0\le t\le T_b\\ 0;\text{otherwise}\end{array}$

Here, $T_b$ is the bit duration which is equal to $200\mu s$ and $f_c$ is the carrier frequency which is equal to $2MHz$ . This $BPSK$ signal is transmitted through an $AWGN$ channel and the output signal is processed by two correlator receivers separately.
${\text{Receiver-1 with a local oscillator carrier of}}^{\prime \prime }cos(4\pi \times {10}^{6}t{)}^{\prime \prime }$ and
${\text{Receiver-2 with a local oscillator carrier of}}^{\prime \prime }100cos(4\pi \times {10}^{6}t{)}^{\prime \prime }$
Both the receivers are identical in all other aspects (including optimum threshold value).
If the probability of bit errors (BER) of receiver-2 is $Q\left(1\right)$, then that of receiver-2 will be

• A. $Q\left(1\right)$
• B. $Q\left(0.1\right)$
• C. $Q\left(100\right)$
• D. $Q\left(1\right)$

Answer 1

The correct option is A $Q\left(1\right)$

Variation in the amplitude of local oscillator carrier does not have any impact on the BER
So, both the receiver will have same BER.