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% Aufgabe 2
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 A=[ 0  1;
    -1 -1]
 b=[0 1]'
 c=[1 0]
 d=0

 % (a)
 sys=ss(A,b,c,d)

 % (b)
 step(sys)
 impulse(sys)

 % (c)
 x0=[0 0]
 t=0:0.01:20;
 u=sin(0.5 * pi * t);
 lsim(sys, u, t, x0) % zeichet u und y
 grid on
 % Verstärkung für große t aus Plot: 0.465
 gainDb = 20 * log10(0.465)
 % Phasen-Verzögerung für große t aus Plot: 16 - 17.5
 phaseShiftRad = (16 - 17.5) * 0.5 * pi
 phaseShiftDeg = phaseShiftRad * 180 / pi 

 % (d)
 figure  % neues Plot-Fenster oeffnen
 bode(sys)
 grid on
 % aus dem Bode-Diagramm können die Verstärkung und die Phase für
 % die Frequenz pi / 2 = 1.571 abgelesen werden.

 % (e) Bem.: System ist in RNF gegeben => G(s) ablesbar.
 Gs = tf(sys)