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)
= A

_{1}/s+1 + A_{2}/s+2
p

_{1}= -1 and p_{2}=-2
Step 3 :

A

_{1 }= s+1 * 2 / (s+1) (s+2) where s=-1
A

_{1 }= 2/-1+2
A

_{1 }= 2
Same way

A

_{2}= s+2 * 2 / (s+1) (s+2) where s=-2
A

_{2}= 2/-2+1
A

_{2}= -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) = A

_{1}/ 1-e^{p}_{1}^{T}_{s}. Z^{-1 + }A_{2}/ 1-e^{p}_{2}^{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 / }Z

^{2}-1.488Z+0.54

This require transfer function for digital IIR filter.

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)
= A

_{1}/s+1 + A_{2}/s+2
p

_{1}= -1 and p_{2}=-2
Step 3 :

A

_{1 }= s+1 * 2 / (s+1) (s+2) where s=-1
A

_{1 }= 2/-1+2
A

_{1 }= 2
Same way

A

_{2}= s+2 * 2 / (s+1) (s+2) where s=-2
A

_{2}= 2/-2+1
A

_{2}= -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) = A

_{1}/ 1-e^{p}_{1}^{T}_{s}. Z^{-1 + }A_{2}/ 1-e^{p}_{2}^{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 / }Z

^{2}-1.488Z+0.54

This require transfer function for digital IIR filter.