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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.

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  • $\begingroup$ 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 Jul 3 '16 at 2:38
  • $\begingroup$ Is $y(t)$ defined for $-\infty<t<\infty$? $\endgroup$ – Matt L. Jul 3 '16 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. Jul 3 '16 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 Jul 3 '16 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. Jul 3 '16 at 10:04
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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.

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