How to Obtain Discrete-Time Impulse Response from Transfer Function

How to Obtain Discrete-Time Impulse Response h(1),h(2),...,h(n) from Transfer Function G(s)

1. Find the inverse Laplace transform of G(s) . It will be the system’s continuous-time impulse response h(t) . h(t)  L1 (G(s)) 1 For example: G(s)   h(t)  e t s 1

2. Compute h(1),h(2),...,h(n) from h(t)

kT h(kT)  h(kT  T) h(k)   h()d  T (k 1)T 2 where T is the sampling period

1 h(kT)  h(kT  T) e kT  e(k1)T For example: G(s)   h(k)  T  T s 1 2 2 3. How to select T : approximately 1/5 to ½ of the dominant time constant

1 For example: G(s)   time constant=1 second sampling period 0.2 to 0.5 second s 1 e 0.5k  e 0.5(k 1) Hence, if we use T=0.5 second  h(k)  0.5  0.25(e 0.5k  e 0.5(k 1) ) 2

4. How to select n ? Let nT to be 5 times of the dominant time constant

1  for G(s)  with T=0.5 second  n  10 s 1  for k=1:10;h(k)=0.5*(exp(-0.5*k)+exp(-0.5*(k-1)));end;h'

0.8033 0.4872 0.2955 0.1792 0.1087 0.0659 0.0400 0.0243 0.0147 0.0089  h= [0.8033 0.4872 0.2955 0.1792 0.1087 0.0659 0.0400 0.0243 0.0147 0.0089]

1 5. How to handle the delay

1 For example G(s)  e0.5s  delay = 0.5 second  1 step (delay=1 T) s 1  h= [0 0.8033 0.4872 0.2955 0.1792 0.1087 0.0659 0.0400 0.0243 0.0147 0.0089]

6. Estimate hmax and hmin

How to Obtain Discrete-Time Impulse Response h(1),h(2),...,h(n) from Impulse Transfer Function G(z) (1) h(k)  Z 1 (G(z)) (2) Select the order n so that h(k) is very small for all. (3) Estimate the ranges of h(k) for k  n .

1 HW: The process is given by G(s)  . Assume c is uncertain but in the range from s 2  2s  c 2 to 3.5. Design one interval controller and one PI controller and then simulate the step response of the closed-loop system for the cases when c  2, 2.5, 3, and 3.5 using the design interval and PI controller. Discuss the results and draw some conclusions. (Suggestions- For PI Controller: use c=2.75 for the design. For Interval Control: Calculate the discrete-time impulse responses corresponding to c=2.0, 2.1, …, and 3.5 and find the envelope in order to determine hmin and hmax.)