# 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.,

𝑥(𝑡) = 𝑥(−𝑡); for all 𝑡     … continuous time signal

𝑥(𝑛) = 𝑥(−𝑛); 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.,

𝑥(−𝑡) = −𝑥(𝑡); for all 𝑡    … continuous time signal

𝑥(−𝑛) = −𝑥(𝑛); 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.,

𝑥(𝑡) = 𝑥𝑒 (𝑡) + 𝑥𝑜 (𝑡)     … (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,

𝑥(−𝑡) = 𝑥𝑒 (−𝑡) + 𝑥𝑜 (−𝑡)

⟹ 𝑥(−𝑡) = 𝑥𝑒 (𝑡) − 𝑥𝑜 (𝑡)     … (2)

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

𝑥(𝑡) + 𝑥(−𝑡) = 𝑥𝑒 (𝑡) + 𝑥𝑜 (𝑡) + 𝑥𝑒 (𝑡) − 𝑥𝑜 (𝑡) = 2𝑥𝑒 (𝑡)

∴ 𝑥𝑒 (𝑡) =12[𝑥(𝑡) + 𝑥(−𝑡)] … (3)

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

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

𝑥(𝑡) − 𝑥(−𝑡) = [𝑥𝑒 (𝑡) + 𝑥𝑜 (𝑡)] − [𝑥𝑒 (𝑡) − 𝑥𝑜 (𝑡)]

⟹ 𝑥(𝑡) − 𝑥(−𝑡) = 𝑥𝑒 (𝑡) + 𝑥𝑜 (𝑡) − 𝑥𝑒 (𝑡) + 𝑥𝑜 (𝑡) = 2𝑥𝑜 (𝑡)

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

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,

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

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

## Numerical Example 1

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

### Solution

Given signal is,

𝑥(𝑡) = 𝑒𝑗4𝑡

∴ 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,

$$\mathrm{\therefore x_{0}(t)=\frac{1}{2}\left [ x(t)-x(-t) \right ]=\frac{1}{2}\left ( e^{j4t}-e^{-j4t} \right )=j\sin4t }$$

## 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,

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,

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

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

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