Assignment 5

Problem 2

g)

$$\begin{array}{rl}{\bar{\sigma }}_y^2& =\underset{t\to \infty }{\lim }E[y(t{)}^2]\\ {\bar{\sigma }}_y^2& =\underset{t\to \infty }{\lim }E\left[{\int }_0^{t}g({\tau }_1)w(t-{\tau }_1))d{\tau }_1{\int }_0^{t}g({\tau }_2)w(t-{\tau }_2))d{\tau }_2\right]\\ {\bar{\sigma }}_y^2& =\underset{t\to \infty }{\lim }E\left[{\int }_0^t{\int }_0^{t}g({\tau }_1)g({\tau }_2)w(t-{\tau }_1)w(t-{\tau }_2)d{\tau }_{1}d{\tau }_2\right]\\ {\bar{\sigma }}_y^2& =\underset{t\to \infty }{\lim }{\int }_0^t{\int }_0^{t}g({\tau }_1)g({\tau }_2)R_w({\tau }_2-{\tau }_1)d{\tau }_{1}d{\tau }_2\\ {\bar{\sigma }}_y^2& =\underset{t\to \infty }{\lim }4{\int }_0^{t}g({\tau }_2{)}^{2}d{\tau }_2\\ {\bar{\sigma }}_y^2& =\underset{t\to \infty }{\lim }4\left(-\frac{1}{2}{e}^{-8t}+2e^{-6t}-\frac{9}{4}{e}^{-4t}+\frac{3}{4}\right)\\ {\bar{\sigma }}_y^2& =3\end{array}$$