Answered: Problem #5 Find the transfer function…

Problem #5 Find the transfer function of a system whose state equations are given as: 1 1 -1] 4 3 0 |-2 1 10 y=[20 30 10]x

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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[1 1 1] 024. The system of equations | 0 0 1x=| b, | is solvable if |0 0 1 b, (c) b, = b, (d) b, = 0 (e) none (a) h, = b, = 0 (b) b, = b, # 0 025. If A = B+C and B= B' and C' =-C, then %3D %3D (1) C = A– A" »C=÷(4- A') (6) C=;(4+A") (4) C = A+ A" (e) none c=-(4-A') (0) C = (A+ A°) («)C= A+ A°
Consider the following real-world problem: Suppose that the population of Memphis is 22,400, and 60% of the population are susceptible to contract a serious illness from a very contagious bacteria that was introduced as a form of biological warfare. The remaining people have no reaction to the bacteria. Assume the susceptible people are equally mixed throughout the population. On January 1, 2022 (time t=0), the number N(t) of people who had contracted the disease was 180 and was increasing at the rate of 39 per day. Assume that N’(t) is proportional to the product of the numbers of those who have caught the disease and of those who have not. Write the differential equation that models the physical situation and solve for the exact answer.
X1 1 -2 3 1 X2 10 Solve the system 2 -4 8. 3 10 X3 3 -6 10 X4 27 X5 Select one: a. x1 = 5 + 2s – 3t, x2 = s, x3 = -3 – 2t, x4 =7+ 4t, x5 = t %D %3D b. x1 = 5 + 3t, x2 = s, x3 = 3 – 2t, x4 = 7+4t, x5 = t %3D Oc.(x1,x2, x3, x4, x5) = (-1,–-2, 1, 1,0) O d. (x1, x2, x3, X4, X5) = (1, –2, 1, 1,0) No colution 6
Match b and a system representation to its corresponding difference equation y[n] = -1.46x[n] - 3.53x[n-1] + 3.93x[n-2] - 2.86y[n-1] - 4.1y[n-2] -0.65y[n-3] y[n] = -1.46x[n] - 3.53x[n-1] + 3.93x[n-2] + 2.86y[n-1] + 4.1y[n-2] + 0.65y[n-3] y[n] = 1.46x[n] + 3.53x[n-1] - 3.93x[n-2] + 2.86y[n-1] + 4.1y[n-2] + 0.65y[n-3] y[n] = -1.46x[n] - 3.53x[n-1] - 3.93x[n-2] + 2.86y[n-1] + 4.1y[n-2] + 0.65y[n-3] y[n] = 1.46x[n] + 3.53x[n-1] - 3.93x[n-2] - 2.86y[n-1] - 4.1y[n-2] -0.65y[n-3] 1. 2. 3. 4. 5. b = [-1.46 -3.53 3.93]; a = [1 2.86 4.1 0.65] b = [-1.46 -3.53 -3.93]; a = [1 -2.86 -4.1 -0.65] b = [-1.46 -3.53 3.93]; a = [1 -2.86 -4.1 -0.65] b = [1.46 3.53 -3.93]; a = [1 2.86 4.1 0.65] b = [1.46 3.53 -3.93]; a = [1 -2.86 -4.1 -0.65]
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If f: X→ Y and B, C e P (Y), then -f-" (B) f1 (~ B) and f-1 (B - C) = f1 (B) ~f-1 (C)
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