Zero mean Gaussian noise of variance N is applied to a half wave rectifier. The mean squared value of the rectifier output will be
(a) Zero (b) N/2
(c) N \sqrt{z} (d) N
Question 2.2.48: Zero mean Gaussian noise of variance N is applied to a half ...
Half wave rectifier is given as
y = x for x ≥ 0
= 0 for x< 0
\begin{aligned}E\left(y^{2}\right) &=\int_{0}^{\infty} y^{2}+(y) d y \\&=\int_{0}^{\infty} y^{2}\left[\frac{1}{2} \delta(y)+\frac{1}{\sqrt{2 \pi N}} e^{\frac{-y^{2}}{2 N}}\right] d y \\&=0+\int_{0}^{\infty} \frac{y^{2}}{\sqrt{2 \pi N}} e^{\frac{-y^{2}}{2 N}} d y\end{aligned}
\begin{aligned}&\text { Let that } \frac{y}{\sqrt{N}}=t \\&d y=\sqrt{N} d t \\&E\left[y^{2}\right]=\frac{1}{\sqrt{2 \pi N}} \int_{0}^{\infty} N t^{2} e^{\frac{-t^{2}}{2}} \sqrt{N} d t \\&=\frac{N}{\sqrt{2 \pi}} \int_{0}^{\infty} t^{2} e^{\frac{-t^{2}}{2}} d t \\&E\left[y^{2}\right]=\frac{N}{2}\end{aligned}
Hence, the correct option is (b)
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