Question 11.15.10: Consider a discrete-time system with frequency response H(ω)...
Consider a discrete-time system with frequency response
H(ω) = \frac{0.1\ +\ 0.1e^{-jω}\ +\ 0.18e^{-2jω}\ +\ 0.18e^{-3jω}\ +\ 0.09e^{-4jω}\ +\ 0.09e^{-5jω}}{1 – 1.5e^{-jω}\ +\ 2.2e^{-2jω}\ -\ 1.5e^{-3jω}\ +\ 0.8e^{-4jω}\ -\ 0.18e^{-5jω}}.
a. Plot the frequency response of the system.
b. Compute and plot the impulse response and the step response of the system.
The frequency response of the discrete-time system.
a.
num=[.1 .1 .18 .18 .09 .09];
den=[1 -1.5 2.2 -1.5 0.8 -0.18];
freqz(num,den)
b.The system transfer function H(z) = \frac{0.1z^5\ +\ 0.1z^4\ +\ 0.18z^3\ +\ 0.18z^2\ +\ 0.09z\ +\ 0.09}{z^5\ -\ 1.5z^4\ +\ 2.2z^3\ -\ 1.5z^2\ +\ 0.8z\ -\ 0.18} is derived from the frequency response H(v) by substituting e^{jω} with z and multiplying the numerator and the denominator by z^5. Now, the impulse response and the step response of the system can be computed by the commands filter, dstep, and dimpulse.
delta=[1 zeros(1,39)];
h=filter(num,den,delta);
stem(0:39,h)
legend('Impulse response')
h2=dimpulse(num,den,40)
stem(0:39,h2)
legend('Impulse response')
u=ones(1,40);
s=filter(num,den,u);
stem(0:39,s)
legend('Step response')
s2=dstep(num,den,40);
stem(0:39,s2);
legend('Step response')
