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Lösung 6 Aufgabe 1 a) b) c) Aufgabe 2 Aufgabe 3 a) 5 4 2 6 4 20 9 2 ...

Lösung 6 Aufgabe 1 a) b) c) Aufgabe 2 Aufgabe 3 a) 5 4 2 6 4 20 9 2 ...

Lösung 6 Aufgabe 1 a) b) c) Aufgabe 2 Aufgabe 3 a) 5 4 2 6 4 20 9 2

Institut für Chemie und Bioingenieurwissenschaften Dr. Robert N. Grass Regelungstechnik FS 12 Lösung 6 Aufgabe 1 y G 1 a) G2 r 1 G1 b) c) y r y r Aufgabe 2 Y R G 7 G 1 1 7 G G G G 1GG1GG G G1G2 G3 1 G H G H Aufgabe 3 a) X s 6 1 1 2 3 4 G1G2G3 G4G 1 G 2 1 G 1GHGH 1 1 2 1 1 1 G G 2 H 6s2 2 s9s20s4 6 2 3 4 5 5 4 6s 2 s4s5s4 Applying the initial value theorem we get: 6ss2 s4 5 x0lim 0 2 s s Since X(s) has all poles with negative real part we can use the final value theorem 6ss2 s4 5 xlim s0 2 s 0 x(t) is converging. Finally x(t) is smooth because all the poles are real. b) X s 10s 2 3 2 s6s10s2 s32is32is2 - 1 - 10s Applying the initial value theorem we get 2 3

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