We say that a continuous signal x(t) is even if x(t)=x(−t) for all t. Similarly, x(t) is odd if x(t)=−x(−t) for all t.
Note that if x(t) is odd, x(0)=0.
Some common even signals you will be familiar with are x(t)=cos(t) and x(t)=t2. Some common odd signals you will be familiar with are x(t)=sin(t) and x(t)=t3.
There are exactly analogous definitions for discrete signals.
x[n] is even if x[n]=x[−n] for all n. x[n] is odd if x[n]=−x[−n] for all n.
Many signals, of course, are neither even nor odd. x(t)=Aebt, the exponential, is neither even nor odd. Neither are most time-shifted functions.
However, we are going to make use of the interesting fact that any signal x[t] that is defined for all t, can be expressed as the sum of an even signal and an odd signal.
This feels like quite a strong result. Suppose
x(t)=xE(t)+xO(t), where xE(t) and xO(t) are even and odd signals respectively.
Then x(−t)=xE(−t)+xO(−t).
But xE(−t)=xE(t), and xO(−t)=−xO(t).
So from that we can see that
x(−t)=xE(t)−xO(t).
Adding that to the equation we started with, we get:
x(t)+x(−t)=2xE(t)+xO(t)−xO(t)=2xE(t).
xE(t)=x(t)+x(−t)2
If instead we subtract the two equations, we get:
x(t)−x(−t)=2xO(t)
xO(t)=x(t)−x(−t)2
We're not quite done, because we started by assuming these signals exist. But we know that x(t) is defined for all t, so the odd and even components are well-defined, and it's clear that
x(t)+x(−t)2 is always even and
x(t)−x(−t)2 is always odd.