Question

Exercise 1 (Transfer Function Analysis) MATLAB provides numerous commands for working with polynomials, ratios of polynomials, partial fraction expansions and transfer functions: see, for example, the commands roots, poly, conv, residue, zpk and tf (a) Use MATLAB to gener ate the continuous-time transfer function H5+15)( +26(s+72) s(s +56)2(s2 +5s +30) H(s) displaying the result in two forms: as (i) the given ratio of factors and (ii) a ratio of two polynomials. (b) Use MATLAB to calculate the partial fraction expansion of H(s) in part (a), obtaining a form suitable for table look-up to determine the impulse response h(t). You should observe the need to recombine a pair of complex-conjugate linear factors into their equivalent quadratic factor. Why might the MATLAB command real be important in this step? How does MATLAB represent the expansion of repeated poles? Verify using a plot that your calculation of h(t) indeed matches the output of MATLAB's impulse command (c) Repeat parts (a) and (b) but for the discrete-time transfer function displaying the two forms and computing the partial fraction expansion first in terms of z (as given) and then equivalently in terms ofz-1. How does the command residuez help? How useful is the impulse command if the system was known to be stable?

Please do part C only, thank you.

Answer 1

MATLAB:

clc;clear all;close all;
b=[1 0 0 0]%Numerator
a=[1 13/6 1/6 -1/3]%denominator

%Impulse response using residuez
[z,p,k]=residuez(b,a)

figure;
subplot(121)
h=impz(b,a,20)
stem([0:19],h,'r')
xlabel('n');ylabel('h(n)');title('Impulse response from partial fraction expansion')

%Impulse response from impz command
subplot(122)
n=0:1:19
%combine partial fraction coeffs and poles in the follwoing way
H=(z(1)*(p(1).^n)).+(z(2)*(p(2).^n)).+(z(3)*(p(3).^n))
%k is an empty vector. Hence its not included
stem(n,H,'b')
xlabel('n');ylabel('h(n)');title('Impulse response from impz command')

%TF in Z and interms of z^-1
sys1=tf(b,a,0.1)
sys2 = tf(b,a,0.1,'Variable','z^-1')
%To prove that H(z) in z and z^-1 are same.
figure;
bode(sys1);
figure;
bode(sys2);

Command window:

>> z
z =

0.057143
1.142857
-0.200000

>> p
p =

0.33333
-2.00000
-0.50000

>> k
k = [](0x0)

>>poly(p)
ans =

1.00000 2.16667 0.16667 -0.33333

Transfer function 'sys1' from input 'u1' to output ...

z^3
y1: -----------------------------------
z^3 + 2.167 z^2 + 0.1667 z - 0.3333

Sampling time: 0.1 s
Discrete-time model.

Transfer function 'sys2' from input 'u1' to output ...

1
y1: ------------------------------------------
1 + 2.167 z^-1 + 0.1667 z^-2 - 0.3333 z^-3

Sampling time: 0.1 s
Discrete-time model.

plots:

Impulse response from partial fraction expansion 400000 Impulse response from impz command 400000 200000 200000 -200000 -200000 400000 -400000 -600000 -600000 20 0 15 20 15

Bode Diagram 4 2 -6 -8 -10 10 10 150 100 50 10 Frequency [rad/s]

Bode Diagram 2 -6 -8 -10 10 10 150 100 50 10 Frequency [rad/s]