# Second order system impulse response generation

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This module describes the recursive generation of the impulse response of a second order system.

## Introduction

This module examines the recursive generation of the impulse response of a second order system. This analysis can be used to determine how to generate a cosine or sine function recursively.

## Second order systems

The transfer function of a general second order system is given as:

$H\left(z\right)=\frac{{b}_{0}{z}^{2}+{b}_{1}z+{b}_{2}}{{z}^{2}+{a}_{1}z+{a}_{2}}$

The Direct Form II structure for the implementation of the second order system is shown in the following figure.

The difference equation for the second order system is

$y\left(n\right)+{a}_{1}y\left(n-1\right)+{a}_{2}y\left(n-2\right)={b}_{0}x\left(n\right)+{b}_{1}x\left(n-1\right)+{b}_{2}x\left(n-2\right)$
$y\left(n\right)=-{a}_{1}y\left(n-1\right)-{a}_{2}y\left(n-2\right)={b}_{0}x\left(n\right)+{b}_{1}x\left(n-1\right)+{b}_{2}x\left(n-2\right)$

To recursively determine the impulse response of the system start with the following:

$h\left(-1\right)=h\left(-2\right)=0,x\left(n\right)=\delta \left(n\right)$

Then using Equation 3 the impulse response can be found recursively by:

$h\left(0\right)=-{a}_{1}h\left(-1\right)-{a}_{2}h\left(-2\right)+{b}_{0}\delta \left(0\right)+{b}_{1}\delta \left(-1\right)+{b}_{2}\delta \left(-2\right)={b}_{0}$
$h\left(1\right)=-{a}_{1}h\left(0\right)-{a}_{2}h\left(-1\right)+{b}_{0}\delta \left(1\right)+{b}_{1}\delta \left(0\right)+{b}_{2}\delta \left(-1\right)$
$h\left(1\right)=-{a}_{1}{b}_{0}+{b}_{1}$
$h\left(2\right)=-{a}_{1}h\left(1\right)-{a}_{2}h\left(0\right)+{b}_{0}\delta \left(2\right)+{b}_{1}\delta \left(1\right)+{b}_{2}\delta \left(0\right)$
$h\left(2\right)=-{a}_{1}\left(-{a}_{1}+{b}_{1}\right)-{a}_{2}{b}_{0}+{b}_{2}$

For the values of n >2 the recursive equation reduces to

$h\left(n\right)=-{a}_{1}h\left(n-1\right)-{a}_{2}h\left(n-2\right)$

because the values of $x\left(n\right),x\left(n-1\right)$ and $x\left(n-2\right)$ will all be zero for n >2.

So, if you know the values of $h\left(0\right)$ and $h\left(1\right)$ , then the Equation 10 can be used to find future values of $h\left(n\right)$ .

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