University:
Eastern Kentucky University
Course:
CSC 310 | Data Structures
Academic year:

2020
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305

Pages:

5
Author:

Bidoggia5pt

Chapter 5 5.1 The transfer function from u to y is hat hat g (s) = (s * 1/3)/(s + 1/5) = s/(s ^ 2 + 1) If ucts = sint, den hat y (s)= hat g(s) * G(s) = s/(s ^ 2 + 1) * 1/(s ^ 1) = s/((s ^ 2 + 1) ^ 2) wned y(t) = 0, 5t * sin t which is not bounded. Thus the network is not BIBO stable. 52 hat y (s)= int a ^ infty overline g(z) * e ^ (- s * epsilon) dr Jer sigma = sigma + j*omega If 030, chu |e ^ (- j * t)| = |e ^ (- sigma * t)||e ^ (- j * omega*t)| = e ^ (- tau * tau*tau) < =, for all t. If one system is BIBO stable, CAM integrate |f(t)| < ∞ dt from x to te \mathcal{S} >= emptyset Thus we have for | hat g (s)|< = int 0 ^ infty | hat g (t)| |e ^ (- s * t)| * dt < = int 0 ^ infty | hat g (t)|dt< infty 5.3 integrate |g(t)| dt from 0 to ∞ = integrate 1/(1 + E) dE from 0 to ∞ =f 11 (i+E)| 0 ^ infty zdx Thus the system in not BIBO table. Ar f(t) = t * e ^ (- c) we love hat y(s) = f * \{y(e)\} = 1/((s(t)) ^ 3) all ih poles have negative runt part thus the eyetem B180 stable. 5.4 f^ 2 (x)= (e ^ (-2x))/(s + 1) indZ crad /pun 0 Z e mu,f ^ 2 s aligned g(t)= L ^ - 1 [ hat g (t)]=[ matrix e^ -(t-2) &fr< z\\ 0&fr ne z\\ 0&fr ne z matrix ] int 0 ^ infty |f(t)| 1 dt= integrate 2 ^ (- (t - 2)) dp from 0 to ∞ = integrate xi ^ (- t) depsilon from 0 to ∞ =-e^ - pi | t = 0 ^ (e*pi) = - [0 - i] = 1 Thas the system is BIBO stable. 5.50 a y(y) = w(x - 1) If r( pm)= delta(phi) then g_{y}(epsilon) = y(t) = a*delta(t - 1) - a ^ 2 * delta(t - lambda) a ^ 3 * delta' * (e - 3) - a ^ 4 * delta(e - delta) +*** T int 0 ^ infty | hat g vec F (c)|d tau=[a| tau|a|^ 2 + |a| ^ 3 +*** = |a| * sum alpha = 0 to ∞ |a| ^ alpha =[ |a| i-|a| &i xi|a|< I\\ infty&i xi|z|z| matrix Thus the feedback system in BIBO stable if and only if |e| < 1 For a = 1 we have the following pair r(6) (८) 45 The bounded input excites an unbounded output. 5.6 hat g(s) = (J - 2)/(delta * r') if mu(e) = beta , then y(x) hat f(0) * 3 = - 6 of mu(t) = sin 2 ) , then hat g (t) | hat f(jz) |4 hat n (xt+4 hat f (jz)) 1.26 (22+1.25) hat g(1) = [[- 2, 3]] * [[5 + 1, - i * 0], [0, 5 - 1]] ^ - 1 * [[- 1], [0]] = [[- 2, 3]] * [[1/(J + i), (tau ^ 2 - 1)/(J ^ 2 - 1)], [0, i/(S - 1)]] * [[- 2], [0]] = 4/(J + 1) . It is BIBO stable. 5.8 g[k] = phi(0.6)^ * f = 0, 1, 2 oint (y) = 3/3 * [g(b)] = (0.8j)/((j - 0.8) ^ 2) Its poles lie inside the unit circle, thus the system is BIBO stable. 5.4 The mutarx has segenvalues and thus the equation is not asymptotically stable ror murginally stable. 5.10 The matini has eigenvalues -1,0,0; Thus the equation is not asymptotically stable. If the upeated eigenvalue o is a suple root of the minunal polynomial or, equivalently, has only Jordan blocks of order 1, chan the equation is marginally stable, we compute the eigenvectors associated with lambda = 0 : (A-1)= 0 Loo V=0 which yields two linearly incteperdent ergonvectors (or of and [101]. Thus the fordan form of A is hat A = [[- 1, 0, 0, 0, 0], [- 1, 1, 0, 1, 0], [0, 1, 0, 1, 0], [0, 0, 0, 1, 0]] and the equation is marginally stable 5.11 The metine has ergewalues -1,0,0; chus the equation is not asymptotically stable. we comprite the eigenectors. associated with lambda = 0 (a-a mp) \mathcal{V} = [[- i, 0, i, 0], [0, 0, i, 0], [0, 0, 0, i]] * \mathcal{nu} = 0 It has only one cornearly independent -21-

Linear System Theory and Design Chapter 5

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