# Show that any continuous-time signal $x(t)$ can be represented as $x(t)= x_e(t) + x_o(t)$

Show that any continuous-time signal $$x(t)$$ can be represented as $$x(t)= x_e(t) + x_o(t)$$

where $$x_e(t) =\frac{1}{2}[x(t) + x(-t)]$$ and $$x_o(t) = \frac{1}{2} [x(t) − x(-t)]$$ are even and odd functions, respectively.

• Welcome on DSP SE Michael, what have you tried so far? Sep 26, 2018 at 6:00
• @Gilles Nothing...sorry for my weak math foundation, i don't know how to start it. Maybe you can give a direction to me first so i can try to do something.... Sep 26, 2018 at 6:10
• You are trying to show that the equation is true. Perhaps you should try to see what happens if you plug the expressions given for $x_e(t)$ and $x_o(t)$ to it. Sep 26, 2018 at 7:29
• @hulappa i tried that,x(t)=1/2 x(t)*2 = x(t) So this is the answer? Is that simple? Maybe i thought too complicated .... Sep 26, 2018 at 8:27
• This and this might be useful to you.
– jojeck
Sep 26, 2018 at 8:35

iI.e., if you have $$x_e(t), x_o(t)$$ given to you. Substitute these into the equation for $$x(t)$$ and you should see that both sides are equal to each other.
(Note that this proves the equality, but doesn't prove that $$x_e(t)$$ is even or that $$x_o(t)$$ is odd.)
• we define $x_e(t)$ to be even symmetry, i.e. $$x_e(-t)=x_e(t)$$ we define that. and we define $x_o(t)$ to be odd symmetry, to wit: $$x_o(-t)=-x_o(t)$$. then we posit that $$x(t) = x_e(t) + x_o(t)$$ and then ask, for that to be true, what must $x_e(t)$ and $x_o(t)$ be? Sep 26, 2018 at 18:26