# Even and Odd Components of a Signal

Electronics & ElectricalElectronDigital Electronics

## Even Signal

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

š„(š”) = š„(−š”); for all š”     … continuous time signal

š„(š) = š„(−š); for all š     … 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.,

š„(−š”) = −š„(š”); for all š”    … continuous time signal

š„(−š) = −š„(š); for all š     … 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.,

š„(š”) = š„š (š”) + š„š (š”)     … (1)

Where,

• š„š (š”) is the even component of the signal, and

• š„š (š”) is the odd component of the signal.

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

š„(−š”) = š„š (−š”) + š„š (−š”)

ā¹ š„(−š”) = š„š (š”) − š„š (š”)     … (2)

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

š„(š”) + š„(−š”) = š„š (š”) + š„š (š”) + š„š (š”) − š„š (š”) = 2š„š (š”)

∴ š„š (š”) =12[š„(š”) + š„(−š”)] … (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,

š„(š”) − š„(−š”) = [š„š (š”) + š„š (š”)] − [š„š (š”) − š„š (š”)]

ā¹ š„(š”) − š„(−š”) = š„š (š”) + š„š (š”) − š„š (š”) + š„š (š”) = 2š„š (š”)

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

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 š„(š”) = šš4š”.

### Solution

Given signal is,

š„(š”) = šš4š”

∴ š„(−š”) = š−š4š”

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,

š = 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,

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

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

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