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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.

Figure 1:
Figure 1 (graphics1.png)

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(n−1)x(n),x(n−1) size 12{x \( n \) ,x \( n - 1 \) } {} and x(n−2)x(n−2) 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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