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 : 

A= s+1 * 2 / (s+1) (s+2)  where s=-1

A= 2/-1+2

A= 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-pk = 1/ 1-e pkTs. Z-1

1/ Fs = 1/5 = 0.5 sec = Ts

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 p1Ts. Z-1    +  A2/ 1-e p2Ts. Z-1


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

H(Z) =  2/ 1-0.818Z-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.

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