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16.06 Principles of Automatic Control Lecture 5 Dynamic Response: Usually, we find the response of a system using Laplace techniques. Often, do within Matlab. Example: DC Motor. Suppose: J 0.01 kg¨m 2 ; b 0.001 N-m-sec K t K e 1 n-M/A 1 V/(rad/sec) R a 10Ω,L 1 H Then Θ 100 psq“ V a s 3 ` 10.1s 2 ` 101s Ω sΘ 100s psq“ psq“ V a V a s 3 ` 10.1s 2 ` 101s 100 s 2 ` 10.1s ` 101 100 Gpsq“ ps ` 5.05 ` j 8.6889qps ` 5.05 j 8.6889q What is the step response of the motor? That is, what is the velocity of the motor as a function of time, if v a ptq“ σptq? By hand, would do: 1

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  • 16.06 Principles of Automatic Control Lecture 5

    Dynamic Response:

    Usually, we nd the response of a system using Laplace techniques. Often, do within Matlab.

    Example: DC Motor.

    Suppose:

    J 0.01 kgm2; b 0.001 N-m-sec Kt Ke 1 n-M/A 1 V/(rad/sec) Ra 10, L 1 H Then

    100 psq Va s3 ` 10.1s2 ` 101s s 100s psq psq Va Va s3 ` 10.1s2 ` 101s

    100 s2 ` 10.1s ` 101

    100 Gpsq ps ` 5.05 ` j8.6889qps ` 5.05 `j8.6889q

    What is the step response of the motor? That is, what is the velocity of the motor as a function of time, if vaptq ptq?

    By hand, would do:

    1

  • j1

    gsptq L 1 Gpsqs

    1 100 Gpsq

    s sps ` a ` jbqps ` a jbqr1 r2 r3 ` ` s s ` a ` jb s ` a jb

    Would nd r1, r2, r3 by partial fraction expansion. Then nd L 1of each term, add together, and simplify. A lot of work.

    Instead, use MATLAB:

    num=[0 0 100]; den=[1 10.1 101]; sysg=tf(num,den); t=0:0.01:5; y=step(sysg,t); plot(t,y);

    The above code produces the following gure:

    0 1 2 3 4 50

    0.2

    0.4

    0.6

    0.8

    1

    1.2

    1.4

    Figure 1: Velocity of the motor.

    The above system was an open-loop system. Would do the same for a closed-loop system, after nding the transfer function.

    2

  • Example:

    The transfer function from aileron input (a) to roll angle () is given by

    k psq a sps ` 1q

    where k steady roll-rate per unit of aileron deection roll subsidence time constant

    I M9

    Suppose a is measured in % of full deection, so a 1 is full right aileron, a 1 if full left one. Then a typical set of parameters might be

    k 100 deg/sec 0.5 sec

    100 Gpsq

    sp0.5s ` 1q

    Suppose we implement the following control law:

    K G(s)-

    r e a

    What is the transfer function of a closed-loop system?

    Kk KGpsq sps`1q

    Hpsq Kk 1 ` KGpsq 1 `

    sps`1q Kk

    s2 ` s ` Kk Suppose we take K 0.1/deg.

    3

  • Then:

    10 Hpsq

    0.5s2 ` s ` 10 20

    Hpsq s2 ` 2s ` 20

    Find step response via MATLAB:

    num=[0 0 20]; den=[1 2 20]; sysg=tf(num,den); t=0:0.01:5; y=step(sysg,t); plot(t,y); xlabel(Time, t (sec)); ylabel(Roll angle, \phi (deg));

    0 1 2 3 4 50

    0.5

    1

    1.5

    Time, t (sec)

    Rol

    l ang

    le,

    (de

    g)

    Figure 2: Roll angle vs time.

    The result (shown in Figure 2 is NOT very good. Oscillatory!

    More on this later.

    4

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    16.06 Principles of Automatic ControlFall 2012

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