Question

Which of the following causal analog transfer functions is used to design causal IIR digital filter with transfer function? $$ H(z) = \frac{0.05z}{z-e^{-0.42}} + \frac{0.05z}{z-e^{-0.2}} $$ Assume impulse invariance transformation with \( T = 0.1 \, \text{s} \). 

• A. \( H(S) = \frac{0.5}{S+4.2} + \frac{0.5}{S+2} \)
• B. \( H(S) = \frac{0.5}{S+2.1} + \frac{0.5}{S+4} \)
• C. \( H(S) = \frac{0.5}{S-2.1} + \frac{0.5}{S-4} \)
• D. \( H(S) = \frac{0.5}{S-4.2} + \frac{0.5}{S-2} \)

Hint:

The impulse invariance method maps analog poles to digital poles using the relation \(z = e^{sT}\) . Understand the mapping between the s plane and z plane for different transformation methods.

Answer 1

The Correct Option is A

Step 1: To find the analog transfer function, we need to consider the impulse invariance transformation which uses the relation \(z = e^{sT}\) . 
Step 2: For the first term of H(z), \( z - e^{-0.42} \), the s plane pole is obtained by the following relation: \[ e^{-0.42} = e^{s \times 0.1} \] So, \(s = -0.42/0.1= -4.2 \). Similarly, for \( z - e^{-0.2} \), \(s = -0.2/0.1= -2 \) 
Step 3: The digital filter transfer function with poles \(z = e^{-0.42}\) and \(z = e^{-0.2}\) , has corresponding poles at \(s = -4.2\) and \(s = -2 \) in analog domain. Thus, the transfer function is: \[ H(S) = \frac{0.5}{S+4.2} + \frac{0.5}{S+2} \]