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# Second Order System Impulse Response Generation

Module by: David Waldo. E-mail the author

Summary: 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 ( z ) = b 0 z 2 + b 1 z + b 2 z 2 + a 1 z + a 2 H ( z ) = b 0 z 2 + b 1 z + b 2 z 2 + a 1 z + a 2 size 12{H $$z$$ = { {b rSub { size 8{0} } z rSup { size 8{2} } +b rSub { size 8{1} } z+b rSub { size 8{2} } } over {z rSup { size 8{2} } +a rSub { size 8{1} } z+a rSub { size 8{2} } } } } {}
(1)

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 ( n ) + a 1 y ( n 1 ) + a 2 y ( n 2 ) = b 0 x ( n ) + b 1 x ( n 1 ) + b 2 x ( n 2 ) y ( n ) + a 1 y ( n 1 ) + a 2 y ( n 2 ) = b 0 x ( n ) + b 1 x ( n 1 ) + b 2 x ( n 2 ) size 12{y $$n$$ +a rSub { size 8{1} } y $$n - 1$$ +a rSub { size 8{2} } y $$n - 2$$ =b rSub { size 8{0} } x $$n$$ +b rSub { size 8{1} } x $$n - 1$$ +b rSub { size 8{2} } x $$n - 2$$ } {}
(2)
y ( n ) = a 1 y ( n 1 ) a 2 y ( n 2 ) = b 0 x ( n ) + b 1 x ( n 1 ) + b 2 x ( n 2 ) y ( n ) = a 1 y ( n 1 ) a 2 y ( n 2 ) = b 0 x ( n ) + b 1 x ( n 1 ) + b 2 x ( n 2 ) size 12{y $$n$$ = - a rSub { size 8{1} } y $$n - 1$$ - a rSub { size 8{2} } y $$n - 2$$ =b rSub { size 8{0} } x $$n$$ +b rSub { size 8{1} } x $$n - 1$$ +b rSub { size 8{2} } x $$n - 2$$ } {}
(3)

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

h ( 1 ) = h ( 2 ) = 0, x ( n ) = δ ( n ) h ( 1 ) = h ( 2 ) = 0, x ( n ) = δ ( n ) size 12{h $$- 1$$ =h $$- 2$$ =0,x $$n$$ =δ $$n$$ } {}
(4)

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

h ( 0 ) = a 1 h ( 1 ) a 2 h ( 2 ) + b 0 δ ( 0 ) + b 1 δ ( 1 ) + b 2 δ ( 2 ) = b 0 h ( 0 ) = a 1 h ( 1 ) a 2 h ( 2 ) + b 0 δ ( 0 ) + b 1 δ ( 1 ) + b 2 δ ( 2 ) = b 0 size 12{h $$0$$ = - a rSub { size 8{1} } h $$- 1$$ - a rSub { size 8{2} } h $$- 2$$ +b rSub { size 8{0} } δ $$0$$ +b rSub { size 8{1} } δ $$- 1$$ +b rSub { size 8{2} } δ $$- 2$$ =b rSub { size 8{0} } } {}
(5)
h ( 1 ) = a 1 h ( 0 ) a 2 h ( 1 ) + b 0 δ ( 1 ) + b 1 δ ( 0 ) + b 2 δ ( 1 ) h ( 1 ) = a 1 h ( 0 ) a 2 h ( 1 ) + b 0 δ ( 1 ) + b 1 δ ( 0 ) + b 2 δ ( 1 ) size 12{h $$1$$ = - a rSub { size 8{1} } h $$0$$ - a rSub { size 8{2} } h $$- 1$$ +b rSub { size 8{0} } δ $$1$$ +b rSub { size 8{1} } δ $$0$$ +b rSub { size 8{2} } δ $$- 1$$ } {}
(6)
h ( 1 ) = a 1 b 0 + b 1 h ( 1 ) = a 1 b 0 + b 1 size 12{h $$1$$ = - a rSub { size 8{1} } b rSub { size 8{0} } +b rSub { size 8{1} } } {}
(7)
h ( 2 ) = a 1 h ( 1 ) a 2 h ( 0 ) + b 0 δ ( 2 ) + b 1 δ ( 1 ) + b 2 δ ( 0 ) h ( 2 ) = a 1 h ( 1 ) a 2 h ( 0 ) + b 0 δ ( 2 ) + b 1 δ ( 1 ) + b 2 δ ( 0 ) size 12{h $$2$$ = - a rSub { size 8{1} } h $$1$$ - a rSub { size 8{2} } h $$0$$ +b rSub { size 8{0} } δ $$2$$ +b rSub { size 8{1} } δ $$1$$ +b rSub { size 8{2} } δ $$0$$ } {}
(8)

h ( 2 ) = a 1 ( a 1 + b 1 ) a 2 b 0 + b 2 h ( 2 ) = a 1 ( a 1 + b 1 ) a 2 b 0 + b 2 size 12{h $$2$$ = - a rSub { size 8{1} } $$- a rSub { size 8{1} } +b rSub { size 8{1} }$$ - a rSub { size 8{2} } b rSub { size 8{0} } +b rSub { size 8{2} } } {}
(9)

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

h ( n ) = a 1 h ( n 1 ) a 2 h ( n 2 ) h ( n ) = a 1 h ( n 1 ) a 2 h ( n 2 ) size 12{h $$n$$ = - a rSub { size 8{1} } h $$n - 1$$ - a rSub { size 8{2} } h $$n - 2$$ } {}
(10)

because the values of x(n),x(n1)x(n),x(n1) size 12{x $$n$$ ,x $$n - 1$$ } {} and x(n2)x(n2) size 12{x $$n - 2$$ } {} will all be zero for n>2.

So, if you know the values of h(0)h(0) size 12{h $$0$$ } {} and h(1)h(1) size 12{h $$1$$ } {}, then the Equation 10 can be used to find future values of h(n)h(n) size 12{h $$n$$ } {}.

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