# Even and odd signals

1. ### dervast

133
Hi i am stuck with something really simple :(
I know that we can express a signal with the even and odd signal
x(t)=xe(t)+xo(t)
(xe(t) means even signal and xo(t) means odd signal)

x(-t)=x(t) for even signals and (1)
x(-t)=-x(t) for odd signals (2)

where even signal is
xe(t)=1/2[x(t)+x(-t)] (3) and the odd one is
xo(t)=1/2[x(t)-x(-t)] (4)
i can validate that x(t)=xe(t)+xo(t) if i use equations 3 and 4
x(t)=1/2[x(t)+x(-t)]+1/2[x(t)-x(-t)]= 1/2x(t)+1/2x(t)+1/2x(-t)-1/2x(-t)= x(t) done

My problem arise when i try to use (1)+(2) to (3)+(4) to prove what i want
using (1) to (3) we have xe(t)=1/2[x(t)+x(t)] =2/2x(t)
using (2) to (4) we have xo(t)=1/2[x(t)-(-x(t))] =2/2x(t) and that means that i have proved that x(t)=4x(t)

P.S Plz tell me where i am wrond and correct my bad english mathematical phrases

2. ### vsage

0
Why would you consider equations 1 and 2 at all? A generic function x(t) will not have those properties always. The only function for which both of those equations can be true is x(t) = 0, which is consistent with your result x(t) = 4x(t).

3. ### dervast

133
I cant understand what u are saying me . Plz try to clarify where i am wrong.
Thx a lot

4. ### vsage

0
Ok here is what makes your thinking wrong: You try to use x(-t) = x(t) and x(-t) = -x(t), two VERY SPECIFIC conditions, to show something for any x(t) which could have any sort of shape. What you meant to use for equations 1 and 2 was:

xe(-t) = xe(t) (1)
xo(-t) = -xo(t) (2)

If you assume equations 1 and 2 are true then you are implying x(t) = 0 for all t

I hope I made myself more understandable.

5. ### dervast

133
Yes thx a lot i have clearly understand my wrong. But if someone ask me to prove that
xe(t)=1/2[x(t)+x(-t)] how shouldi think to prove that?

6. ### vsage

0
I would use the definition of an even function xe(t) = xe(-t), but recall that xe(t) = 1/2[x(t) + x(-t)], so what would xe(-t) be?