Output of lowpass filter with damped sine wave input The random process $$Y(t)=\cos(\omega_0t)\cos(\omega_0t+\pi N(t))$$where $N(t)$ is a Poisson process of parameter $\lambda$ enters a lowpass filter with transfer function $$H(j\omega) =
\left\{
 \begin{array}{ll}
  1  &amp; \mbox{if } |\omega| \leq \frac{\omega_0}{2} \\
  0  &amp; \mbox{if } |\omega| \geq \frac{\omega_0}{2}
 \end{array}
\right.$$ If $Z(t)$ is the output of the filter, find the expectation and variance of both $Y(t)$ and $Z(t)$.


So, first we have that:

$$\mathbb{E}[Y(t)]=\cos(\omega_0t)\mathbb{E}[\cos(\omega_0t+\pi N(t))]=\cos(\omega_0t)\mathbb{E}[\cos(\omega_0t)\cos(\pi N(t))-\sin(\omega_0t)\sin(\pi N(t))]]=\cos^2(\omega_0t)\mathbb{E}[\cos(\pi N(t))]$$

Doing that expectation by definition (with the summation) we get that $$\mathbb{E}[Y(t)]=\cos^2(\omega_0t)e^{-2\lambda t}$$

With a similar procedure we find that

$$\mathbb{V}[Y(t)]=\cos^4(\omega_0t)\cdot(1-e^{-4\lambda t})$$

First I tried to find $\mathbb{E}[Z(t)]=h(t)*\mathbb{E}[Y(t)]$

So using the property that states that the transform of $x(t)y(t)$ is $\frac{1}{2\pi}X(j\omega)Y(j\omega)$ we get that

$$\mathcal{F}(\cos^2(\omega_0t))=\frac{\pi}{2}\delta(\omega+2\omega_0)+\frac{\pi}{2}\delta(\omega-2\omega_0)+\pi\delta(\omega)$$

But it is known that the transform of the exponential multiplied by the step function (the random process involved is valid only for $t&gt;0$ by the definition of the Poisson process probability function) is $\frac{1}{2\lambda+j\omega}$ so

$$\mathcal{F}(y(t))=Y(j\omega)=\frac14\frac{1}{2\lambda+j(\omega+2\omega_0)}+\frac14\frac{1}{2\lambda+j(\omega-2\omega_0)}+\frac12\frac{1}{2\lambda+j\omega}$$

So what I need would be to antitransform $H(j\omega)Y(j\omega)$, and now I'm stuck. How could this be done? And regarding $\mathbb{V}[Z(t)]$, I have no idea of how to even begin to solve it.
I've finally figured this out thanks to a friend of mine. The thing is that $N(t)$ can be even or odd, and in both cases, the signal we get is easy to deal with.

We can calculate the desired expectation this way:

$$\mathbb{E}[Z(t)]=\mathbb{E}[h(t)*Y(t)]=\mathbb{E}[h(t)*Y(t)|N(t) =2n]\mathbb{P}[N(t)=2n]+\mathbb{E}[h(t)*Y(t)|N(t) =2n+1]\mathbb{P}[N(t)=2n+1]$$

But when $N(t)$ is even or odd, $Y(t)$ becomes simple:

$$Y(t) =
\left\{
 \begin{array}{ll}
  \cos^2(\omega_0t)  &amp; \mbox{if } N(t)\mbox{ is even } \\
  -\cos(\omega_0t)\sin(\omega_0t)    &amp; \mbox{if } N(t)\mbox{ is odd }
 \end{array}
\right.$$

So

$$\mathbb{E}[Z(t)]=\mathbb{P}[N(t)=2n]\cdot h(t)*\cos^2(\omega_0t)-\mathbb{P}[N(t)=2n+1]\cdot h(t)*(\cos(\omega_0t)\sin(\omega_0t))$$

And those two probabilities are easy to calculate, as well as the two convolutions left.