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In the problem below, I'm not understanding how they came up with the formula for X(t) used to calculate the autocorrelation $E\Big[X(t)X(t+\tau)\Big]$. All they say about X(t) is that it has equal probability of switch polarity from +A to -A and vica versa... but, the picture they show of X(t), looks more like its a discrete signal X[n] instead of a continous signal X(t)....

example

pico
  • 941

2 Answers2

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On base of the fact that $X\left(t\right)$ and $X\left(t+\tau\right)$ only take values in $\left\{ A,-A\right\} $ we are allowed to conclude that: $$\Omega=\left\{ X\left(t\right)=X\left(t+\tau\right)\right\} \cup\left\{ X\left(t\right)\neq X\left(t+\tau\right)\right\} $$

Moreover: $$\left\{ X\left(t\right)=X\left(t+\tau\right)\right\} \cap\left\{ X\left(t\right)\neq X\left(t+\tau\right)\right\} =\varnothing$$

This justifies the equality: $$\mathbb{E}\left[X\left(t\right)X\left(t+\tau\right)\right]=$$$$\mathbb{E}\left[X\left(t\right)X\left(t+\tau\right)\mid X\left(t\right)=X\left(t+\tau\right)\right]P\left(X\left(t\right)=X\left(t+\tau\right)\right)+$$$$\mathbb{E}\left[X\left(t\right)X\left(t+\tau\right)\mid X\left(t\right)\neq X\left(t+\tau\right)\right]P\left(X\left(t\right)\neq X\left(t+\tau\right)\right)$$

Under condition $X\left(t\right)=X\left(t+\tau\right)$ we find that $X\left(t\right)X\left(t+\tau\right)=A^{2}$.

Under condition $X\left(t\right)\neq X\left(t+\tau\right)$ we have $X\left(t\right)X\left(t+\tau\right)=-A^{2}$.

So we can extend the equality with: $$\cdots=A^{2}P\left(X\left(t\right)=X\left(t+\tau\right)\right)+\left(-A^{2}\right)P\left(X\left(t\right)\neq X\left(t+\tau\right)\right)$$

drhab
  • 151,093
  • Thanks makes sense.... Incidentally, do you know at which step the requirement for $\tau > 0$ enters into the derivation...the range for $\tau$ is implicitly $\tau > 0$.... but, then, at (9.91) they say, something like auto-correlation is even symmetric, therefore we can put the absolute value around $|\tau|$ and drop requirement for $|\tau| >0$ and just make it for all $\tau$... but, really, i have no idea why there's a restriction on $\tau>0$ in the first place. Seems to me if you write $R(t, t+\tau) = E[x(t) x(t+\tau)]$, then $\tau$ could actually be negative.. – pico Jul 14 '20 at 14:07
  • unless there's a step, that restricts $\tau$ to being only positive? – pico Jul 14 '20 at 14:11
  • Dunno. Maybe it is forbidden that $t+\tau<0$ so that there is a restriction on $\tau$ that depends on $t$. In that case $\tau<0$ is possible but is not unlimited. – drhab Jul 14 '20 at 14:44
  • Ok, noevermind, I think its because the poison distribution adds this restriction because it requires $(\alpha \tau) > 0$... and we already know $\alpha \in [0, \infty]$ so that means $\tau$ needs to be positive. $\tau > 0$ – pico Jul 14 '20 at 14:45
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\begin{align} E[X(t) X(t+ \tau)] &= E[X(t) X(t+ \tau)| X(t) \textrm{ and } X(t+\tau) \textrm{ have same signal} ] P(X(t) \textrm{ and } X(t+\tau) \textrm{ have same signal}) + \\ &+ E[X(t) X(t+ \tau)| X(t) \textrm{ and } X(t+\tau) \textrm{ have different signals} ] P(X(t) \textrm{ and } X(t+\tau) \textrm{ have different signals}) \\ &= A^2 P(X(t) \textrm{ and } X(t+\tau) \textrm{ have same signal}) - \\ &-A^2 P(X(t) \textrm{ and } X(t+\tau) \textrm{ have different signals}) \\ &= A^2 P(\textrm{number of crossings is even}) - \\ &-A^2 P(\textrm{number of crossings is odd}) \\ \end{align}

Daniel S.
  • 823