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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 tried to simplify it, then do what I need and in the end add 1 again to get the result. –  place_gpon Mar 21 '12 at 17:47
    
@place_gpon: I don't see the simplification. There seems to be some confusion whether you are shifting in signal level or time as $x$ has no breakpoint at $3$ –  Ross Millikan Mar 21 '12 at 17:53
    
Maybe it was a bad idea. I removed the g things. Still don't know how to add them. –  place_gpon Mar 21 '12 at 18:00
    
@place_gpon: Now you have steps at $\{-2,-1,0,1,2\}$. For example $x_e(t)=\frac 12 (-1 +0)=\frac{-1}2$ for $t \le -2$. You just need to work it out at each step. –  Ross Millikan Mar 21 '12 at 18:05
    
I will try to do it and I' ll post the result. –  place_gpon Mar 21 '12 at 18:07
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