Even and Odd Components of a Signal



Even Signal

A signal is said to be an even signal if it is symmetrical about the vertical axis or time origin, i.e.,

$$\mathrm{x(t) = x(-t);\: \text{ for all t   ..... continuous time signal}}$$

$$\mathrm{x(n) = x(-n);\: \text{ for all n   ..... discrete time signal}}$$

Odd Signal

A signal is said to be an odd signal if it is anti-symmetrical about the vertical axis, i.e.,

$$\mathrm{x(-t) \:=\: -x(t);\: \text{ for all t   ..... continuous time signal}}$$

$$\mathrm{x(-n) \:=\: -x(n);\: \text{ for all n   ..... discrete time signal}}$$

Determination of Even and Odd Components of a Signal

Continuous-time Case

Every signal need not be either purely even signal or purely odd signal, but the signal can be expressed as the sum of even and odd components, i.e.,

$$\mathrm{x(t) \:=\: x_e (t) \:+\: x_o \:(t) \:\: \dotso \: (1)}$$

Where,

  • xe (t) is the even component of the signal, and
  • xo (t) is the odd component of the signal.

By the definition of even and odd signals, we have,

$$\mathrm{x(-t) \:=\: x_e (-t) \:+\: x_o (-t)}$$

$$\mathrm{\Rightarrow \: x(-t) \:=\: x_e (t) \:-\: x_o (t) \:\: \dotso \: (2)}$$

Adding eqns. (1) & (2), we get,

$$\mathrm{x(t) \:+\: x(-t) \:=\: x_e (t) \:+\: x_o (t) \:+\: x_e (t) \:-\: x_o (t) \:=\: 2x_e (t)}$$

$$\mathrm{\therefore \: x_e (t) \:=\: 12[x(t) \:+\: x(-t)] \:\: \dotso \: (3)}$$

$$\mathrm{\therefore \: x_{e}(t) \:=\: \frac{1}{2}\left[ x(t) \:+\: x(-t) \right] \:\: \dotso \: (3)}$$

Again, subtracting eqn. (2) from eqn. (1), we get,

$$\mathrm{x(t) \:-\: x(-t) \:=\: [x_e (t) \:+\: x_o (t)] \:-\: [x_e (t) \:-\: x_o (t)]}$$

$$\mathrm{\Rightarrow \: x(t) \:-\: x(-t) \:=\: x_e (t) \:+\: x_o (t) \:-\: x_e (t) \:+\: x_o (t) \:=\: 2x_o (t)}$$

$$\mathrm{\therefore \: x_{0}(t) \:=\: \frac{1}{2}\left[x(t) \:+\: x(-t) \right]\:\: \dotso \: (4)}$$

Thus, the equations (3) and (4) gives the even and odd components of a continuous-time signal respectively.

Discrete-time Case

The even and odd components of a discrete-time signal x(n) are given by,

$$\mathrm{\therefore \: x_{e}(n) \:=\: \frac{1}{2}\left[ x(n)\:+\:x(-n) \right ]} \:\: \dotso \:(5)$$

$$\mathrm{\therefore \: x_{0}(n) \:=\: \frac{1}{2}\left[ x(n)\:-\:x(-n) \right ]} \:\: \dotso \:(6)$$

Numerical Example 1

Find the even and odd components of the continuous-time signal x(t) = ej4t.

Solution

Given signal is,

$$\mathrm{x(t) \:=\: e^{j4t}}$$

$$\mathrm{\therefore \: x(-t) = e^{-j4t}}$$

The even component of the signal is,

$$\mathrm{\therefore \: x_{e}(t) \:=\: \frac{1}{2}\left[ x(t)\:+\:x(-t) \right]\:=\:\frac{1}{2}\left( e^{j4t}\:+\:e^{-j4t} \right ) \:=\: \cos 4t}$$

And, the odd component of the signal is,

$$\mathrm{\therefore \: x_{0}(t) \:=\: \frac{1}{2}\left[ x(t) \:-\: x(-t) \right] \:=\: \frac{1}{2}\left( e^{j4t} \:- \: e^{-j4t} \right) \:=\: j\:\sin 4t }$$

Numerical Example 2

Find the even and odd components of the discrete-time signal x(n), where,

$$\mathrm{x(n) \:=\: \begin{Bmatrix} 5,6,3,4,1\ \uparrow \ \end{Bmatrix}}$$

Solution

The given discrete time sequence is,

$$\mathrm{x(n) \:=\: \begin{Bmatrix} 5,6,3,4,1\ \uparrow \ \end{Bmatrix}}$$

Here,

$$\mathrm{n \:=\: 0,\: 1,\: 2,\: 3,\: 4}$$

$$\mathrm{\therefore \: x(-n) \:=\: \begin{Bmatrix} 1, 4, 3, 6, 5\ \uparrow \ \end{Bmatrix}}$$

Hence, the even component of the sequence is,

$$\mathrm{x_{e}(n) \:=\: \frac{1}{2}\left[ x(n) \:+\: x(-n) \right ] \:=\: \frac{1}{2}\left[5,\: 6,\: 3,\: 4,\: 1 + 1,\: 4,\: 3,\: 6,\: 5 \right ]}$$

$$\mathrm{\Rightarrow\: x_{e}(n)\:=\: \frac{1}{2}\left[5 + 1,\: 6 + 4,\: 3 + 3,\: 4 + 6,\: 1 + 5 \right ]\:= \:\frac{1}{2} \left [6,\: 10,\: 6,\: 10,\: 6 \right ]}$$

$$\mathrm{\therefore \: x_{e}(n) \:=\: \begin{Bmatrix} 3, 5, 3, 5, 3\ \uparrow \ \end{Bmatrix}}$$

And the odd component of the sequence is,

$$\mathrm{ x_{0}(n) \:=\: \frac{1}{2}\left[ x(n) \:-\: x(-n) \right ] \:=\: \frac{1}{2}\left[5,\: 6,\: 3,\: 4,\: 1 - 1,\: 4,\: 3,\: 6,\: 5 \right ]}$$

$$\mathrm{\Rightarrow \: x_{0}(n) \:=\: \frac{1}{2}\left [5-1,\:6-4,\:3-3,\:4-6,\:1-5\right ] \:=\: \frac{1}{2}\left [4,\:2,\:0,\:-2,\:-4\right ]}$$

$$\mathrm{\therefore \: x_{0}(n) \:=\: \begin{Bmatrix} 2, 1, 0, -1, -4\ \uparrow \ \end{Bmatrix}}$$

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