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) = 𝑥_𝑒 (𝑡) + 𝑥_𝑜 (𝑡)     … (1)

Where,

• 𝑥_𝑒 (𝑡) is the even component of the signal, and

• 𝑥_𝑜 (𝑡) is the odd component of the signal.

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

𝑥(−𝑡) = 𝑥_𝑒 (−𝑡) + 𝑥_𝑜 (−𝑡)

⟹ 𝑥(−𝑡) = 𝑥_𝑒 (𝑡) − 𝑥_𝑜 (𝑡)     … (2)

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

𝑥(𝑡) + 𝑥(−𝑡) = 𝑥_𝑒 (𝑡) + 𝑥_𝑜 (𝑡) + 𝑥_𝑒 (𝑡) − 𝑥_𝑜 (𝑡) = 2𝑥_𝑒 (𝑡)

∴ 𝑥_𝑒 (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) = [𝑥_𝑒 (𝑡) + 𝑥_𝑜 (𝑡)] − [𝑥_𝑒 (𝑡) − 𝑥_𝑜 (𝑡)]

⟹ 𝑥(𝑡) − 𝑥(−𝑡) = 𝑥_𝑒 (𝑡) + 𝑥_𝑜 (𝑡) − 𝑥_𝑒 (𝑡) + 𝑥_𝑜 (𝑡) = 2𝑥_𝑜 (𝑡)

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

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) = 𝑒^𝑗4𝑡.

Solution

Given signal is,

𝑥(𝑡) = 𝑒^𝑗4𝑡

∴ 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,

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

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

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