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You have the definition of the random process as : $$X(t) = a \sin(\omega_0 t + \Theta) \tag{1}$$ where $a$ and $\omega_0$ are deterministic constants and $\Theta$ is a continuous R.V. uniformly distributed in $[0,2\pi]$. According to the indexed-set of RV interpretation of a R.P., for each index $t$ you have a new R.V denoted as $X_t$ which has its own ...

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You have a single random variable $\Theta$ in this example, so taking the expectation with respect to that random variable results in a single integral. The formula for $R_X(\tau)$ in your question is rather messy and full of mistakes; it should be as follows: \begin{align}R_X(\tau)&=E[X(t+\tau)X(t)]\\&=E[a\sin(\omega_0t+\omega_0\tau+\Theta)\cdot ... 1 the PDF of X can be written as f_X(x) = \begin{cases} \frac{1}{4} \text{ for } -2 \leq x < 0 \\ \frac{1}{2} \text{ for } 0 \leq x < 1 \\ 0 \text{ otherwise,} \end{cases} but for the edges it remains unclear whether the intervals should or should not include their borders as Dilip pointed out. Now the function $Y(X)$ applies an offset of $\pm 1$ ...

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There is no impulse because there is zero probability mass at $0$. As the problem itself shows, it is perfectly all right for a pdf to make sudden jumps in value; note that the graph of $f_X$ (yes, I do mean $X$, not $Y$) jumps in value at $0$ and the value of $f_X(0)$ cannot be determined from the graph. $f_X(0)$ could have value $\frac 14$ or $\frac 12$ ...

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