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I have a output signal $Y(t)= \sum_{n=-\infty}^{\infty} Z_n \delta (t-n\tau) h(t)$ and $Z_n$ is a random variable with equal probability to be $+-1$ and h(t) is $sinc(\frac t {\tau})$

and the Expected value of $Y(t)$ is

$E(Y(t)) = E[\sum_{n=-\infty}^{\infty} Z_n \delta (t-n\tau) h(t)]$

$=\sum_{n=-\infty}^{\infty}h(n\tau) E[Z_n]$

$=0$

But if I calculated in other way

$\int_{-\infty}^\infty t Y(t) dt$

$=\int_{-\infty}^\infty t \sum_{n=-\infty}^{\infty} Z_n \delta (t-n\tau) h(t) dt$

$= \int_{-\infty}^\infty \sum_{n=-\infty}^{\infty} tZ_n \delta (t-n\tau) h(t) dt$

$= \int_{-\infty}^\infty \sum_{n=-\infty}^{\infty} n\tau Z_n h(n\tau) dt$

$= \sum_{n=-\infty}^{\infty} [n\tau h(n\tau)\int_{-\infty}^\infty Z_n dt]$

Why I have two different result with same signal? Thanks for any help.

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  • $\begingroup$ To the Dilip's Question: Actually, This is a tutorial question from my class. The original setting is convolution between $\delta(t-n\tau) and h(t)$, as @Dilip say. I just curious why I can have two different set of solution when I time two signal together. Now I know this is because the second one is calculated the joint probability. Thanks Ixop and Dilip. $\endgroup$ – Samuel Mar 19 '13 at 6:49
  • $\begingroup$ So it is a convolution as my answer suggested, and so what $Y(t)$ works out to be is $$Y(t) = \sum_{n=-\infty}^{\infty} Z_n h(t-n\tau) = \sum_{n=-\infty}^{\infty} Z_n\operatorname{sinc}\left(\frac{t-n\tau}{\tau}\right).$$ In other words, both of the calculations you present in your question are nonsensical since you are not even looking at the correct $Y(t)$, and the fact that the two calculations give different answers is really not of much interest. $\endgroup$ – Dilip Sarwate Mar 19 '13 at 16:53
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Your second equation $E[Y(t)] = \int_{-\infty}^\infty tY(t)dt$ is wrong; that won't calculate the expected value of $Y(t)$. The equation you are thinking of is $E[Y(t)] = \int_\Omega Y(t)P[Y(t)]$, where you would determine $P[Y(t)]$ by a joint probability of the $Z_n$'s. That's why they don't match.

In any case, @Dilip-Sarwate has an answer showing that you can simplify this particular equation to a state where the answer becomes trivial (and is equal to the first answer you found).

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If $h(t)$ is continuous at $a$, then $h(t)\delta(t-a) = h(a)\delta(t-a)$. Since the $\operatorname{sinc}$ function is continuous everywhere, $$\begin{align} Y(t) &= \sum_{n=-\infty}^{\infty} Z_n\delta(t-n\tau)h(t)\\ &= \sum_{n=-\infty}^{\infty} Z_n\delta(t-n\tau)h(n\tau)\\ &= \sum_{n=-\infty}^{\infty} Z_n\delta(t-n\tau)\operatorname{sinc}(n)\\ &= Z_0\delta(t) \end{align}$$ since $\operatorname{sinc}(n) = \begin{cases}1,&n = 0,\\0,& n\neq 0.\end{cases}$ You (or your textbook's authors) probably meant to write something else. You sure you are not missing a $\star$ indicating a convolution between that $\delta(t-n\tau)$ and $h(t)$?

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