Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 0 down vote accepted

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.