Question

Consider the system below :



Given : T = 0.1 m sec; A = 0.2 mV; $N_0={10}^{-10}$ Watt/Hz. Which of the following is/are correct?

• A. Maximum SNR output when h(t) = S(T-t) is 0.16.
• B. $S_1\left(t\right)=-S_2(t-T)$
• C. Minimum probability of error at the output is Q[0.2].
• D. $E_d=0$

Answer 1

Answer: (A), (B), (C)

Minimum probability of error at the output is Q[0.2].

$E_d=E\left[S_1\right(t)-S_2(t\left)\right]$
$=2A^{2}T$

Max SNR at output of matched filer when $h\left(t\right)=S(T-t)is\frac{E_d}{N_0/2}$


$(SNR{)}_{O/P}=\frac{2A^{2}T}{N_{0/2}}=16\times {10}^{-2}=0.16$

$\Rightarrow (P_c{)}_{min}=Q\left[\sqrt{\frac{E_d}{2N_0}}\right]=Q\left[\frac{2A^{2}T}{2\times {10}^{-10}}\right]$

$=Q\left[\sqrt{\frac{4\times {10}^{-8}\times {10}^{-4}}{{10}^{-10}}}\right]=A\left[0.2\right]$