Correct Answer - Option 3 : 2.5 μW
Concept:
1) PSD and Autocorrelation function form a Fourier pair
2) \(PS{D_{output}} = {\left| {H\left( \omega \right)} \right|^2}PS{D_{input}}\)
Calculation:
\(H\left( {j\omega } \right) = \frac{{\frac{1}{{j\omega C}}}}{{R + \frac{1}{{j\omega C}}}}\)
\(= \frac{1}{{1 + j\omega RC}}\)
output PSD
\({S_y}\left( \omega \right) = {\left| {H\left( \omega \right)} \right|^2}{S_x}\left( \omega \right)\)
\(= \frac{1}{{{R^2}{C^2}{\omega ^2} + 1}}{S_x}\left( \omega \right)\)
\({S_x}\left( \omega \right) = \frac{{{N_o}}}{2}\) (input PSD)
\(= \frac{1}{{{R^2}{C^2}{\omega ^2} + 1}}\frac{{{N_o}}}{2}\)
The Inverse Fourier Transform will give ACF.
Autocorrelation Function:
\({R_y}\left( \tau \right) = \frac{{{N_o}}}{{4RC}}\;{e^{ - \;\frac{\tau }{{RC}}}}\)
Average output power: is obtained from ACF by subsisting τ = 0
R = 1 kΩ and C = 0.1 μF
\({R_y}\left( {\tau = 0} \right) = \frac{{{N_o}}}{{4\left( {RC} \right)}} = \frac{{{{0.001\times10}^{ - 6}}}}{{\begin{array}{*{20}{c}} {4\times10^3\times0.1 \times {{10}^{ - 6}}} \end{array}}}\)
⇒ 2.5 μW
option 3 is correct.
