1. Sep 6, 2011

### tina_singh

1)-why is x(t)+x(-t) always even??..no matter if x(t) even or odd???

2)-when we talk about unit step function...u(t)..and we add..u(t)+u(-t)..the value of both is 1 at t=0..so does'n't that gets added twice??..and it becomes 2 at t=0...

3)when we have x(-t) and we time shift it say x(-t-3) it shifts toward the -ve t axis.. where as x(t-3) the function is shifted on the + axis..why is it so??

i would be really greatful if you can help me out with the above 3 doubts..

Last edited by a moderator: Sep 28, 2011
2. Sep 28, 2011

### jvclark2

Re: even and odd signals..plzz help

1) x(t)+x(-t) = x(t)+x(t) for even functions and if x(t) is even then x(t) + x(t) would be even.
As for odd functions the definition of an odd function is for an f(x), -f(x)=f(-x) and therefore x(t)+x(-t) = 0 for an odd function which is technically even.

2) Yes I believe so assuming you define your step function as u(t) = 1 if tâ‰¥0 and 0 otherwise

3) For x(t), a shift to x(t-3) would be a shift to the right. Likewise if you wanted to shift x(-t) to the right you would need to have x(-(t-3)) =x(-t+3) and NOT x(-t-3). Just be careful and use parentheses because when you shift you are substituting the independent variable not just throwing a "-3" in there somewhere.

3. Sep 28, 2011

### LCKurtz

Re: even and odd signals..plzz help

Define F(t) = x(t) + x(-t). F will be even if F(-t) = F(t). Does that work? Does it matter what the formula for x(t) is?

The value at a single point usually doesn't matter because u(t) is usually used in integration. Sometimes u(t) isn't even defined at 0 because of this; it is just defined as u(t) = 0 for t < 0 and u(t) = 1 for t > 0.
Both x(t-3) and x((-t) - 3) are shifted to the right.