
#1
Apr1807, 07:14 AM

P: 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
Apr1807, 07:27 AM

P: n/a

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
Apr1807, 12:42 PM

P: 133

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


#4
Apr1807, 04:45 PM

P: n/a

even and odd signals
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
Apr1907, 01:06 AM

P: 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
Apr1907, 10:20 AM

P: n/a

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?



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