Homework 6 Solutions
Problem 1:
(c) Yes, both poles are inside the unit circle.
Problem 2:
Problem 3
1) The stability boundary in s-plane is the jω axis. Replacing s in z = esT with jω gives
That is, the stability boundary is mapped to the unit circle in z-plane.
2) Assume a point
in the very vicinity of s = 0 such that
Using z = esT, the relative locations of the points near z = 1 are given as
where the fact that cosωdT ≈ 1 and sin ωdT ≈ 0 for small ωdT has been used. Hence, the points
around s = 0 are mapped to points around z = 1 by factor of T.
3) An arbitrary location in the s-plane is represented by
where ωn is in rad/sec. Thus, s-plane locations are in terms of frequency. Using z = esT = es2π/ωs,
the corresponding z-plane location is
Since the ωn/ωs is nondimensional, the information contained in z is normalized to sample rate.
4) The locations in s-plane (s = −σ ± jω) are mapped to z-plane locations through
Given that z on the negative real axis,
Indeed, if the above expression is true then z = −e−σT represents the negative real axis with a
damped frequency ωd.
5) An arbitrary vertical line in the left half s-plane is given by
Using z = esT, the line is mapped to
Thus, vertical lines in s-plane correspond to circles in z-plane.
6) An arbitrary horizontal line in the s-plane is given by
Using z = esT, the line is mapped to
Thus, horizontal lines in s-plane correspond to radial lines in z-plane.
7) Let –ωs ≤ ω ≤ –ωs , where –ωs = 2π/T. Then, ω1 = ω + ωs/2 maps to
 =  (ω + ωs /2) =  ωT  ωs T/2 =  ω  π = − ωT
Hence, frequencies separated by ωs/2 maps back into the frequencies within –ωs ≤ ω ≤ –ωs.
Problem 4:
(a)
(b)
Figure E13.13
Problem 5:
Problem 6:
Figure 13.8
Problem 7:
Problem 8:
Matlab script:
close all
clear all
num = [1.7 1.7*0.46];
den = [1 1 0.5];
sys = tf(num,den,0.1)
figure(1)
step(sys);
xlabel('samples');
ylabel('step response output');
title('discrete time step response');
sysc = d2c(sys)
figure(2)
step(sysc);
xlabel('time');
ylabel('step response output');
title('Continuous time step response');
Output:
Transfer function:
1.7 z + 0.782
------------z^2 + z + 0.5
Sampling time: 0.1
Transfer function:
13.37 s + 563.1
--------------------s^2 + 6.931 s + 567.2
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