Even and odd part of a signal

Any signal can be split as x[n] = xe[n] + xo[n], where

• even part: xe[n] = ½{x[n] + x[−n]} (symmetric, xe[−n] = xe[n])
• odd part: xo[n] = ½{x[n] − x[−n]} (antisymmetric, xo[−n] = −xo[n])

Given: x[n] = 1 for −4 ≤ n ≤ 0 and x[n] = 2 for 1 ≤ n ≤ 4 (zero elsewhere). The folded signal x[−n] is 1 for 0 ≤ n ≤ 4 and 2 for −4 ≤ n ≤ −1.

Even part

xe[n] = 1.5 for −4 ≤ n ≤ −1
      = 1   for n = 0
      = 1.5 for 1 ≤ n ≤ 4

As a sequence: xe = {1.5, 1.5, 1.5, 1.5, 1, 1.5, 1.5, 1.5, 1.5} (n = −4 … 4).

Odd part

xo[n] = −0.5 for −4 ≤ n ≤ −1
      =  0   for n = 0
      = +0.5 for 1 ≤ n ≤ 4

As a sequence: xo = {−0.5, −0.5, −0.5, −0.5, 0, 0.5, 0.5, 0.5, 0.5}.

Check: xe[n] + xo[n] gives 1 for n ≤ 0 and 2 for n ≥ 1, which is x[n].