Home / Signals and Systems / Discrete Time Convolution Properties | Discrete Time Signal

# Discrete Time Convolution Properties | Discrete Time Signal

Want create site? Find Free WordPress Themes and plugins.

Discrete-Time Convolution

Convolution is such an effective tool that can be utilized to determine a linear time-invariant (LTI) system’s output from an input and the impulse response knowledge.

Given two discrete time signals x[n] and h[n], the convolution is defined by

$x\left[ n \right]*h\left[ n \right]=y\left[ n \right]=\sum\limits_{i=-\infty }^{\infty }{{}}x\left[ i \right]h\left[ n-i \right]~~~~~~~~~~~~~~~~~~~~~~~\left( 1 \right)$

The summation on the right side is called the convolution sum.

It should be noted that the convolution sum exists when x[n] and h[n] are both zero for all integers n<0.

If x[n] and  h[n] are zero for all integers n<0, then  x[i]=0 for all integers i<0 and h[n-i] =0 for all integer n-i<0 (or n<i) . Thus the summation on i in (1) may be taken from i=0 to i=n, and the convolution operation is given by,

$x\left[ n \right]*h\left[ n \right]=\left\{ \begin{matrix} \begin{matrix}\begin{matrix}0, & {} & {} \\\end{matrix} & n=-1,-2,\ldots \\\end{matrix} \\\begin{matrix}\sum\limits_{i=0}^{n}{x\left[ i \right]h\left[ n-i \right]} & {} & {} & n=0,1,2,\ldots \\\end{matrix} \\\end{matrix} \right.\text{ (2)}$

Since the summation in (2) is over a finite range of integers (i=0 to i=n), the convolution sum exists. Hence any two signals that are zero for all integers n<0 can be convolved.

To compute the convolution (1) or (2)

1. First, change the discrete-time index n to i in the signals x[n] and h[n].
2. Flip the signals h[i] to obtain h[-i] (it is called folding).
3. For each output index n, shift by n to get h[n-i] .(Positive value of n gives right shift.)
4. The product x[n]*h[n] is formed and y[n] is computed by summing the values of x[i]*h[n-i] as i ranges over the set of integers.

## Discrete-Time Convolution Example

Suppose that x[n]=anu[n] and h[n]= anu[n] Where u[n] is a discrete-time unit-step function and a and b are fixed non zero  real numbers.

Step 1:

Change discrete time signal index n to i in both signals:

$x\left[ i \right]={{a}^{i}}u\left[ i \right]$

$h\left[ i \right]={{b}^{i}}u\left[ i \right]$

Step 2:

Flip h[i] to get h[-i]:

$h\left[ -i \right]={{b}^{-i}}u\left[ -i \right]$

Step 3:

Shift by n to get h[n-i]:

$h\left[ n-i \right]={{b}^{n-i}}u\left[ n-i \right]$

Step 4:

Find y[n] by summing the product x[i]h[n-i] over a finite range of i:

$y\left[ n \right]=\sum\limits_{i=-\infty }^{\infty }{{}}x\left[ i \right]h\left[ n-i \right]~~~~~~~~~~~\left( 3 \right)$

If x[n] is zero for all integers n<0, then x[i]=0 for all integers i<0, So,

$u\left[ i \right]=\left\{ \begin{matrix}\begin{matrix} 1 & i\ge 0 \\\end{matrix} \\\begin{matrix} 0 & i<0 \\\end{matrix} \\\end{matrix} \right.$

Similarly, If h[n] is zero for all integers n<0, then h[n-i]=0 for all integers n-i<0 (or n<i), So,

$u\left[ n-i \right]=\left\{ \begin{matrix}\begin{matrix}1 & i\le n \\\end{matrix} \\\begin{matrix}0 & i>n \\\end{matrix} \\\end{matrix} \right.$

Thus the summation on i in (3) may be taken from i=0 to i=n and the convolution operation is given by,

$y\left[ n \right]=\sum\limits_{i=0}^{n}{{}}x\left[ i \right]h\left[ n-i \right]$

$y\left[ n \right]=\sum\limits_{i=0}^{n}{{}}{{a}^{i}}{{b}^{n-i}}$

$y\left[ n \right]={{b}^{n}}\sum\limits_{i=0}^{n}{{}}{{\left( \frac{a}{b} \right)}^{i}}$

If a=b,

$y\left[ n \right]={{b}^{n}}\sum\limits_{i=0}^{n}{{}}{{\left( 1 \right)}^{i}}$

If a≠b, then standard math relation gives

$\sum\limits_{i=0}^{N-1}{{}}{{\left( r \right)}^{i}}=\frac{1-{{\left( r \right)}^{N}}}{1-r}~~~~~~~~r\ne 1$

Whereas,

$y\left[ n \right]=\left\{ \begin{matrix}\begin{matrix}n+1 & ~~~~~~~~~~~~~~~~~a=b \\\end{matrix} \\\begin{matrix}\frac{1-{{\left( {}^{a}/{}_{b} \right)}^{n+1}}}{1-\left( {}^{a}/{}_{b} \right)} & a\ne b \\\end{matrix} \\\end{matrix} \right.$

## Discrete-Time Convolution Properties

The convolution operation satisfies a number of useful properties which are given below:

Commutative Property

If x[n] is a signal and h[n] is an impulse response, then

$\fn_jvn&space;\large&space;\large&space;x[n]*h[n]=h[n]*x[n]$

Associative Property

If x[n] is a signal and h1[n] and h2[n] are impulse responses, then

$\fn_jvn&space;\large&space;(x[n]*h_1&space;[n])*h_2&space;[n]=x[n]*(h_1&space;[n]*h_2&space;[n])$

Distributive Property

If x[n] is a signal and h1[n] and h2[n] are impulse responses, then

$\fn_jvn&space;\large&space;x[n]*(h_1&space;[n]+h_2&space;[n])=x[n]*h_1&space;[n]+x[n]*h_2&space;[n])$

Did you find apk for android? You can find new Free Android Games and apps.