PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

552 Practical MATLAB® Applications for Engineers

0

2.3000

-6.3250

54.6006

-192.4085

38.0520

****************************

Coefficients of the denominator polynomial of H(z)

1.0000

-2.1700

1.8366

-0.1757

-0.4300

0.1169

***************************

The first 11 coefficients of the impulse response h(n) are:

***************************

coef. num. h(n)

***************************

ans =

0 0

1.0000 2.3000

2.0000 -1.3340

3.0000 47.4817

4.0000 -86.5192

5.0000 -236.1448

6.0000 -346.0348

7.0000 -311.8158

8.0000 -125.3485

9.0000 148.4517

10.0000 376.3726

***************************

Note that the system transfer function is given by

Hz

zz z z z

()

.....

.




  



2 36 325 54 6006192 408538 0520

12

12 3 4 5

117 zz zzz  12 34 5 1 8366....0 1757 0 43 0 1169

Note also that the corresponding system impulse response h(n) = Z−^1 [H(z)] is given by

h(n) = 0 + 2.3δ(n − 1) – 1.3403δ(n − 2) + 47.4817δ(n − 3) – 86.5192δ(n − 4) – 236.1448δ(n−5) −

346.0348δ(n−6) – 311.8158δ(n − 7) – 155.3485δ(n − 8) + 148.4517δ(n − 9) + 376.3726δ(n − 10)

Example 5.17

Let the system difference equation be given by

g(n) = 0.5g(n − 1) + 2f(n)

Create the script fi le diff_eq_res, that simulates the given difference equation and returns

the response plots for n ≤ 10, for each of the following inputs:

a. f(n) = u(n) (step response)

b. f(n) = δ(n) (impulse response)