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  • 7/29/2019 Asignacion Control 1 SH

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    Universidad de Puerto Rico

    Recinto Universitario de Mayagez

    Mayagez, Puerto Rico

    Computer Homework #1

    Ennio A. Gaud Figueroa (802-07-2972)

    Rubn J. Prez Rivera (843-06-6160)

    INEL 4505

    Prof. Shawn David Hunt

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    We plotted in MATLAB the step response of a 3rd

    order system with finite zeros

    ()

    ( )

    We varied this equation and plotted the results to see how changed with each parameter. We seta step response control and we will be evaluating based on that step response.

    Step Response Control

    >> w=500;

    >> z=2;

    >> zeta=.75;

    >> alpha=5;

    >> b=[(w*w)/(z*alpha)*[1 z]];

    >> a=[conv([1 2*zeta*w w*w],[1 alpha])];

    >> sys=tf(b,a);

    >> step(sys);

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    Step Response (Change in Wn)

    >> w=100;

    >> z=2;

    >> zeta=.75;

    >> alpha=5;

    >> b=[(w*w)/(z*alpha)*[1 z]];

    >> a=[conv([1 2*zeta*w w*w],[1 alpha])];

    >> sys=tf(b,a);

    >> step(sys);

    0 0.2 0.4 0.6 0.8 1 1.20

    0.02

    0.04

    0.06

    0.08

    0.1

    0.12Step Response

    Time (sec)

    Amplitude

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    Here we can see comparing this graph with the step response control graph that the amplitude

    change as we change the natural frequency of the system. Also we can see that the system gets

    to its final value a bit late.

    Step Response ( Change in z)

    >> w=500;

    >> z=5;

    >> zeta=.75;

    >> alpha=5;

    >> b=[(w*w)/(z*alpha)*[1 z]];

    >> a=[conv([1 2*zeta*w w*w],[1 alpha])];

    >> sys=tf(b,a);

    0 0.2 0.4 0.6 0.8 1 1.20

    0.01

    0.02

    0.03

    0.04

    0.05

    0.06

    0.07

    0.08

    0.09

    0.1Step Response

    Time (sec)

    Amplitude

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    >> step(sys);

    We can see that putting a zero on the system, the amplitude changed; and the system got to itsfinal value faster. It also presents an overshoot in the rise as the system tries to get to steady

    state.

    0 0.005 0.01 0.0150

    0.005

    0.01

    0.015

    0.02

    0.025

    0.03

    0.035

    0.04

    0.045Step Response

    Time (sec)

    Amplitude

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    Step Response (Change in )

    >> w=500;

    >> z=2;

    >> zeta=100;

    >> alpha=5;

    >> b=[(w*w)/(z*alpha)*[1 z]];

    >> a=[conv([1 2*zeta*w w*w],[1 alpha])];

    >> sys=tf(b,a);

    >> step(sys);

    0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.60

    0.005

    0.01

    0.015

    0.02

    0.025

    0.03

    0.035

    0.04

    0.045Step Response

    Time (sec)

    Amplitude

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    With the change in , we can see how the pole affect the response of the system and we see how

    with an addtional pole we can make the system get faster to its final value although this can

    make the system not stable.