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    ELEN90055 Control Systems: Worksheet 4 - Solutions To Starred Problems

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    Mingyue Wang

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    © All Rights Reserved
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    ELEN90055 Control Systems: Worksheet 4 - Solutions To Starred Problems

    Uploaded by

    Mingyue Wang
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    ws4-21_solutions

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    31 views4 pages

    ELEN90055 Control Systems: Worksheet 4 - Solutions To Starred Problems

    Uploaded by

    Mingyue Wang

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 4

    ELEN90055 Control Systems

    Worksheet 4 - Solutions to Starred Problems


    Semester 2, 2021

    1. The transfer function of the second order system is usually written as


    ωn2 ωn2
    H(s) = =
    s2 + 2ζ ωn s + ωn2 (s + ζ ωs )2 + ωn2 (1 − ζ 2 )
    So the characteristic equation is
    (s + ζ ωn )2 + ωn2 (1 − ζ 2 ) = 0
    and for 0 ≤ ζ < 1, the root of the characteristic equation are
    q
    s = − ζ ωn ± j ωn 1 − ζ 2 .
    |{z} | {z }
    σ ωd

    × 𝑗𝜔𝑑
    𝜔𝑛 𝜃 sin 𝜃 = 𝜁

    −𝜁𝜔𝑛

    × −𝑗𝜔𝑑

    (a)
    4.6
    ts = ≤ 9.2 ⇒ ζ ωn ≥ 0.5
    ζ ωn

    −0.5

    Figure 1: σ

    1
    If the overshoot is less than or equal to 0.17 then ζ ≥ 0.5 and θ = sin−1 ζ = 30◦).

    30°

    −0.5

    Figure 2: ζ

    From Franklin, Powel &


    1.8 Emami-Naeini“Feedback
    tr ≈ ≤ 0.6 ⇒ ωn ≥ 3 control of dynamic systems
    ωn
    (slide 7, L7-21.pdf).

    𝑗3 See also Goodwin, Graebe


    and Salgado eq. (4.8.15) in
    30° sec. 4.8 for the exact formula
    relating overshoot to
    −0.5 damping coefficient.

    −𝑗3

    Figure 3: ωn

    So the region in the complex s-plane for which the poles could be placed are shown
    below

    𝑗3

    30°

    −0.5

    −𝑗3

    Figure 4:

    2
    (b) • If the settling time is 9.2sec, then ζ ωn = 0.5 and therefore the poles should be
    exactly on the line ζ ωn = 0.5.
    • To make the rise time small, we need to make ωn large as tr = 1.8ωn .
    • The overshoot condition is satisfied if the poles are located in the highlighted
    area in Figure 2.
    The location of the poles that satisfy the above conditions are shown in Figure 5.

    𝑗3

    × 30°

    −0.5
    ×

    −𝑗3

    Figure 5: Problem 1(b).

    2.
    1.8
    tr = ≤2 ⇒ ωn ≥ 0.9.
    ωn

    M p ≤ 10% ⇒ ζ ≥ 0.6 ⇒ θ ≥ 36.8◦

    The region of the complex plane where all the dominant poles of the closed-loop transfer
    function should be located is highlighted in Figure 6

    0.9𝑗

    36.8°

    −0.9𝑗

    Figure 6: Problem 2.

    3. Rise time is a measure of how fast the system reacts to a change in its input, and settling
    time is a measure of how fast the system can settle down to steady state with a desired miss

    3
    ratio (that is, how fast the system’s transient decays). So depending on the application,
    both rise time and settling time can be used as measures of how fast the response is. In
    other words, depending on how we define a faster response, the controller can be designed
    such that we have a lower rise time or a lower settling time.
    In this specific problem, we have two responses in which the rise times are close to each
    other (0.5sec and 1sec) while the settling times are very different (4sec and 10sec). Because
    of this big difference in the settling times, we say that the response with tr = 1sec and
    ts = 4sec is faster. See the following step responses to compare the responses of these two
    systems.

    Step Response
    2

    1.5
    Amplitude

    0.5

    0
    0 2 4 6 8 10 12
    Time (seconds)

    Figure 7: Problem 3a.

    Step Response
    2

    1.5
    Amplitude

    0.5

    0
    0 2 4 6 8 10 12
    Time (seconds)

    Figure 8: Problem 3b.

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