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me352_hw5_soln

# me352_hw5_soln - ME 352 System Dynamics and Control...

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1 ME 352 System Dynamics and Control Homework #5 – Solutions 1. (a) Type 0 (b) Type 1 (c) Type 3 (d) Type 0 2. Input Error Constants Steady-state Error ________________________________________________________________________________ u t s ( ) K K p = ) 1 ( 1 K + (a) tu t s ( ) K v = 0 t u t s 2 2 ( ) / K a = 0 The above results are valid if the value of K corresponds to a stable closed-loop system. ________________________________________________________________________________ u t s ( ) K p = ∞ 0 (b) tu t s ( ) 667 . 1 = v K 6 . 0 t u t s 2 2 ( ) / K a = 0 ________________________________________________________________________________ (c) The closed-loop system is unstable. It is meaningless to conduct a steady-state error analysis. ________________________________________________________________________________ u t s ( ) 10 = p K 0.909 (d) tu t s ( ) 0 = v K t u t s 2 2 ( ) / 0 = a K 3. The following MATLAB code was used. % ME352 hw#5, question #3 % define plant and controllers Gs = tf(1,[1 1 0]); Ds1 = tf(5*[1 3],[1 8]); Ds2 = tf(5*conv([1 3],[1 0.1]), conv([1 8],[1 0.01])); % compute step responses sys1 = feedback(series(Gs,Ds1),1); sys2 = feedback(series(Gs,Ds2),1); figure(1) step(sys1, '-' ,sys2, '-.' ) legend( 'system-i' , 'system-ii' ); % compute control effort ctrl1 = feedback(Ds1,Gs); ctrl2 = feedback(Ds2,Gs); figure(2) step(ctrl1, '-' ,ctrl2, '-.' ) legend( 'control effort-i' , 'control effort-ii' ); (a) Both systems eliminate steady-state error. Step responses are shown below. Overshoot and rise-times are

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me352_hw5_soln - ME 352 System Dynamics and Control...

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