1
$\begingroup$

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 $$H(j\omega) = \left\{ \begin{array}{ll} 1 & \mbox{if } |\omega| \leq \frac{\omega_0}{2} \\ 0 & \mbox{if } |\omega| \geq \frac{\omega_0}{2} \end{array} \right.$$ 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 $t>0$ 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.

Improve this question
$\endgroup$
13
  • $\begingroup$ You can either solve in the frequency domain by taking Laplace transform of y(t) = Y(s) and then multiply with H(s), then take inverse Laplace transform, OR you can solve in the time domain by noting that h(t) is a sinc function, and convolving that sinc function with y(t). $\endgroup$
    – Dan Boschen
    Commented Jul 3, 2016 at 2:38
  • $\begingroup$ Is $y(t)$ defined for $-\infty<t<\infty$? $\endgroup$
    – Matt L.
    Commented Jul 3, 2016 at 8:33
  • $\begingroup$ @DanBoschen: The problem with the suggested Laplace transform approach is that $Y(s)$ doesn't exist, neither does $H(s)$. A solution could be to use the Fourier transform instead (we already have $H(\omega)$), but we still have a problem with $Y(\omega)$, unless the OP forgot to add a step function to the definition of $y(t)$. In the latter case, the problem can be solved using the Fourier transform, but it still can't be solved using the Laplace transform (I've assumed that $\lambda\neq 0$). $\endgroup$
    – Matt L.
    Commented Jul 3, 2016 at 8:37
  • $\begingroup$ The square in the term will give $y(t)$ a DC and a component at $2 \omega_0$. With $e^{-x}$ in, we get a damped cosine with an offset. The ideal lowpass $H(\omega)$ would take out the component at $2 \omega$ and leave the $e^{-x}$ envelope. I would go down the Fourier route too. @MattL. It could be assumed that they "start" at the same time (step at $t=0$) (?) $\endgroup$
    – A_A
    Commented Jul 3, 2016 at 9:01
  • $\begingroup$ @A_A: Well, that's the question I was asking the OP. Note that you also assume that $\lambda$ is real-valued and positive. It shouldn't be us assuming all those things, but the OP should clarify the question. $\endgroup$
    – Matt L.
    Commented Jul 3, 2016 at 10:04

1 Answer 1

Reset to default
0
$\begingroup$

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:

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

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.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.