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I have a signal that is described below

$x(t) = \begin{cases} -1, & t<0 \\ 2t-1, & 0\leq t<1 \\ 2-t, & 1\leq t<2 \\ 0, & t\geq 2 \end {cases}$

$x(-t) = \begin{cases} -1, & t>0 \\ -2t+1, & -1\leq t<0 \\ t-2, & -2\leq t<-1 \\ 0, & t\leq -2 \end {cases}$

and I want to find the odd and even signal that $x_0(t)+x_e(t) = x(t)$

and then I have to find $x_e(t) = \frac 1 2 (x(t)+x(-t))$ $x_o(t) = \frac 1 2 (x(t)-x(-t))$

How to do it? Also I don't know if what I did is correct. Can anyone help?

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I don't understand why you define $g$ from $x$, but we can find odd and even parts of $g$. Your $g(t)$ should allow $t$ to range over $(-\infty,\infty )$ as $x(t)$ does, not $[0,3)$. For example, you should add a line to $g(t)$ saying $g(t)=-2$ if $t \lt 0$. Then you would have values of $g(t)$ and $g(-t)$ at all times and you could substitute them into your last two equations.

Added: the calculation of $g(t)$ is not correct. It should be $g(t) = \begin{cases} -2 & t\le 0 \\ 2t-2, & 0\leq t<1 \\ 1-t, & 1\leq t<2 \\ -1, & 2\leq t \end {cases}$

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    I will try to do it and I' ll post the result.2012-03-21