# Even odd functions

#### 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..

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#### micromass

What did you try to answer this??

For the first, fill in -a in x(t)+x(-t) and see if

$$x(-a)+x(-(-a))=x(a)+x(-a)$$

For the second one. It actually never matters what the unit step function is in 0. So saying that u(t)+u(-t)=2 in 0, is correct, but it doesn't matter.
Note that a lot of people choose that the unit step function is 0, or 1/2 in 0.

The third one. We actually have a function y(t)=x(-t). Then x(-t-3)=y(t+3). So it makes sense that it gets shifted to the other side.

#### tina_singh

okh..thanks for answering..i got the first 2 parts...
buh i m still a little confuse about the third...see when we say there are 2 functions x(t) and x(t-2) it means x(t-2) is delayed by 2 sec with respect to x(t) therefore it shifts in the positive x direction.. does'n't the same apply for x(-t) and x(-t-2) the second function is time delayed by 2 secs with respect to the first so even it should shift to the positive x axis..isn't it????

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#### micromass

The - in front of the t reverses the direction. So shifting towards the positive axis becomes shifting towards the negative axis and vice versa.

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