As shown in Figure (b). The output of the low pass filter is $$y(t)$$.
Let $$E$$ be the expectation operator and consider the following statements :
$$\left( {\rm I} \right)$$ $$E\left( {X\left( t \right)} \right) = E\left( {Y\left( t \right)} \right)$$
$$\left( {{\rm I}{\rm I}} \right)$$ $$\,\,\,\,\,\,\,\,E\left( {{X^2}\left( t \right)} \right) = E\left( {{Y^2}\left( t \right)} \right)$$
$$\left( {{\rm I}{\rm I}{\rm I}} \right)\,$$ $$\,\,\,\,\,\,E\left( {{Y^2}\left( t \right)} \right) = 2$$
Select the correct option:
For the closed loop system shown, the root locus for 0 < K < $$\infty$$ intersects the imaginary axis for K = 1.5. The closed loop system is stable for
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