Question

20 points] Q1. The Stately State Transition Matrix, Ф 16 2 3 13 Consider a state transition matrix, Ф, for a SISO LTI system A1. A- 5 11 10 8 9 76 12 Please determine and justify: (a) Ф(t) for this system A 4 14 15 1 ] (b) Ф(s) for this system A (c) System A's characteristic polynomial (d) Ф(z) for this system A via Tustin's method (ie, trapazoid-rule) (e) A difference equation assuming: (1) a step input at the first step (u(k)), followed by (2) ZOH sampling, with (3) H-I, and Г as given. [Hint: What is z-transform of the ZOH, i.e., G(E) for a system G(s)?]

Answer 1

Consider the matrix A 13 16 2 3 11 10 5 8 7 6 12 14 25 1 State transition matrix is D()=[sl-A Write the MATLAB code to determine the matrix, D(t) A= [16 2 3 13; 5 11 10 8; 9 7 6 12;4 14 25 1]; I-eye (4); syms s fi-inv (s*I-A) fi t-ilaplace (fi)

fi 2 3 2 2 2 2 s 189 s 1078 309 s - 1678 - 13 s 169 s136 S 18 s399 s - 896 3 s | 1 1 1 #1 2 3 2 2 2 8 s+9 s - 408 5 s87 s 10 s45 s - 1854 261 s - 1974 528 S23 s #1 #1 #1 #1 2 2 3 2 2 s 28 s 29 s 1224 12 s 151 s408 - 9 s+ 25 s 408 7 s 67 s 1224 #1 #1 #1 #1 2 2 2 3 2 - 14 s 125s 1348 4 s 227 s 776 25 s 523 s 2068 S 33 s 231 s 136 #1 #1 #1 #1

Here =-s4 = # 1 34s3200s2 - 4230s 4080 (a) Thus, the matrix (i) is the inverse Laplace transform of D(s) 1 A +34s200 s -4230s4080 -S -13s2 169s+ 3 -S 2 2 3s2 309s +18s s 189s 1078 1678 399s-896 136 10s2 2 - 23s 829s 5s 87s 45s D(s) S 1854 528 261s 1974 408 12s2 7s267s -9s225s s - 28s + 151s 1224 408 29s 1224 408 S 25s2-523s 4s2227s -14s2125s s3 33s 23 1s + + 776 1348 2068 136

1 -s34s200s (-4230s +4080 (-2s2-189s -3s2-309s -13s2169s -318s +136 399s-896 -1078 +1678 -323s2 -5s2-87s -10s2-45s -8s2-9s ) 261s -1974 -528 +1854 +408 -7s2-67s -12s2151s -9s25s -s28s +408 1224 29s 1224 -408 -25s2523s -4s2-227s -333s -14s2125s +776 +1348 -2068 231s 136

(b) Thus, the matrix D(s) is, 1 D(s) -s34s200s -4230s 4080 (-2s2-189s -3s2 -309s -13s2 169s -s* +18s2 399s 896 -1078 +1678 +136 -s3 +23s2 -10s2-45s -5s87s -8s-9s -St +1854 +408 -528 261s 1974 -9s225s -7s2-67s -s328s2 -12s2151s +408 1224 29s -1224 -408 -4s2 227s -14s125s -25s2523s -s33s 231s 136 -2068 +776 1348 (c) Write the MATLAB code to obtain the characteristic polynomial A [16 2 3 13;5 11 10 8;9 7 6 12;4 14 25 1]; syms s CP det (s*I-A) CP = 34 s 3-200*s^2 4230*s - 4080 s^4 Thus, the characteristic polynomial of A is, s-34s3-200s2 +4230s -4080.

(d) Consider D(s) 1 DS)(-s34s3 200s2 X -4230s 4080 -3s2-309s -2s2189s -13s2169s -s318s2 399s 896 +136 -1078 +1678 -323s -5s2-87s -10s-45s -8s2-9s + -528 261s 1974 +1854 +408 -9s225s -7s2-67s+ -12s2151s 3 - 28s2 +408 1224 29s 1224 -408 -4s2 227s -25s2523s -14s2125s -s3+33s2 231s 136 -2068 +776 +1348 for s Substitute z-1 Tz 1

1 D(z) 34s3+200s -4230s4080 -2s2-189s -3s2309s -13s2169s s3 18s 399s 896 -1078 +1678 +136 -5s2-87s -8s2-9s 3 -10s2 -45s -s +23s2+ 261s 1974 +1854 +408 -528 -7s2 67s -9s225s -328s2 -12s2 151s +408 -408 1224 29s 1224 -4s2-227s -14s2125s -s333s2 -25s523s 231s 136 +776 +1348 -2068