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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