Impulse invariant method example

Step 1 : Analog frequency transfer function H(s) will be given. If it not given then obtain expression of H(s) from the given specification

Step 2 : If required H(s) by using fraction expansion

Step 3 : Obtain Z transform of each PFE term using in-variance transformation equation

Step 4 : Obtain H(z) this is required digital IIR filter

Find out H(Z) using impulse in-variance method at 5 Hz sampling frequency from H(s) as given below :

H(s) = 2 / (s+1) (s+2)

Step 1 : Given analog transfer function is, 

 H(s) = 2 / (s+1) (s+2)

Step 2 : We will expand H(s) using partial fraction expansion as :

H(s) = A₁/s+1 + A₂/s+2

p₁= -1 and p₂ =-2

Step 3 : 

A₁ = s+1 * 2 / (s+1) (s+2)  where s=-1

A₁ = 2/-1+2

A₁ = 2

Same way 

A₂= s+2 * 2 / (s+1) (s+2)  where s=-2

A₂= 2/-2+1

A₂= -2

H(s) = 2/s+1 -  2/s+2

Step 3 : Obtain Z transform of each PFE term using in-variance transformation equation

1 / s-p_k = 1/ 1-e ^p_k^T_s. Z^-1

^1/ F_s = 1/5 = 0.5 sec = ^T_s

_{1/ s+1  →} 1/ 1-e ^-1(0.2) . Z^-1

_{1/ s+1  →} 1/ 1-e ^-0.2 . Z^-1

_{1/ s+2 →} 1/ 1-e ^-2(.2) . Z^-1

_{1/ s+2 →} 1/ 1-e ^-0.4 . Z^-1

Step 4 : Obtain H(z) this is required digital IIR filter

H(Z) = A₁/ 1-e ^p₁^T_s. Z^{-1    +}  A₂/ 1-e ^p₂^T_s. Z^-1

^{H(Z) =} 2/ 1-e ^-0.2 . Z^{-1    -}  2/1-e ^-0.4 . Z^-1

^{H(Z) =  2/ 1-0.818}Z^-1 - 2/ 1-0.67Z^-1

^{H(Z) = 2Z / Z-0.818 - 2Z/ Z-0.67}

H(Z) = 2Z(Z-.67-2Z (Z-0.818) / (Z-0.818)(Z-0.67)

^{H(Z) = 0.29 Z /} Z²-1.488Z+0.54

This require transfer function for digital IIR filter.