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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.

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    $\begingroup$ Welcome on DSP SE Michael, what have you tried so far? $\endgroup$
    – Gilles
    Sep 26, 2018 at 6:00
  • $\begingroup$ @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.... $\endgroup$
    – Michael
    Sep 26, 2018 at 6:10
  • $\begingroup$ 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. $\endgroup$
    – hulappa
    Sep 26, 2018 at 7:29
  • $\begingroup$ @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 .... $\endgroup$
    – Michael
    Sep 26, 2018 at 8:27
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    $\begingroup$ This and this might be useful to you. $\endgroup$
    – jojeck
    Sep 26, 2018 at 8:35

1 Answer 1

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Substitute the last two equations into the first and it should be clear.

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.)

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  • $\begingroup$ 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? $\endgroup$ Sep 26, 2018 at 18:26
  • $\begingroup$ @robertbristow-johnson this is good to know, but I don't think the original question requires this proof. I just put that last part in as a throwaway line $\endgroup$
    – Robert L.
    Sep 26, 2018 at 18:31

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