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    System Response

    This module is concern with the response ofLTI system.

    L.T. is used to investigate the response of first

    and second order systems. Higher ordersystems can be considered to be the sum ofthe response of first and second order system.

    Unit step, ramp, and sinusoidal signal playimportant role in control system analysis. It istherefore we will investigate this signals.

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    (1)

    2. Convolution

    Output of LTI system is the convolution

    of its input and its impulse response:

    h(t)r(t) c(t)

    t

    dthrthtrtc )()()(*)()(

    t

    dhtr )()(

    Taking the L.T. of (1) yields

    C(s) = R(s)H(s) (2)Where C(s)is the L.T. of c(t)

    R(s) is the L.T. of r(t)

    H(s) is the L.T. of h(t)

    H(s) is called the transfer function (T.F)

    3. Derivative

    If the input is r(t) then the output is c(t)

    wherer(t) denotes the derivative ofr(t)

    Review of some LTI properties

    We will express system as in figure below,system input is r(t), output is c(t), and

    impulse response is h(t)

    4. Integral

    If the input is r(t)dtthen the output isc(t)dt

    1. Impulse response

    Impulse response, denoted by h(t), is the

    output of the system when its input is

    impulse (t). h(t) is called the impulse

    response of the system or the weighting

    function5. Poles and ZeroT.F.isusually rational and therefore can be

    expressed as N(s)/D(s). Poles is the values

    ofs resulting T.F to be infinite. Zeroes is

    the values ofs resulting T.F to be zero

    System Response

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    Poles, Zeros, and System Response The output response of a system is the sum of two

    responses: theforced response andthe natural response The poles of a transfer function are (1) the values of

    the Laplace transform variable, s, that cause the

    transfer function to become infinite or (2) any roots of

    the denominator of the transfer function that are

    common to roots of the numerator

    The zeros of a transfer function are (1) the values of the

    Laplace transform variable, s, that cause the transferfunction to become zero, or (2) any roots of the

    numerator of the transfer function that are common to

    roots of the denominator.

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    Pole dan Zero Pada System Orde Satu

    Fungsi alih sistem orde 1,

    Untuk mencari A dan B,

    Persamaan respon sistemnya,

    Secara grafis,

    INTRO

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    Time Response of the First Order Systems.

    We can found the differential equation

    first we write (1) as

    The diff. Eq. is the inverse L.T. of (2)

    Now we take the L.T of (3) and include

    the initial condition term to get

    1)(

    )()(

    s

    K

    sR

    sCsG (1)

    )(

    )(

    1sR

    KsCs

    (2)

    )()(1

    )( tr

    Ktc

    tc (3)

    )(

    )(

    1)0()( sR

    KsCcssC (4)

    Solving for C(s) yields

    )/1(

    )R(s)/(

    )/1(

    )0()(

    s

    K

    s

    csC (5)

    Note that the initial condition as an

    input has a Laplace transform ofc(0),which is constant.

    The inverse L.T of a constant is impulse

    (t). Hence the initial condition appearsas the impulse function

    Here we can see that the impulse

    function has a practical meaning, eventhough it is not a realizable signal

    1

    s

    K

    1

    1

    s

    R(s)

    c(0)

    +

    +

    C(s)

    Here we will investigate the time response

    of the first order systems.

    The transfer function of a general first

    order system can be written as:

    The eq. can be shown in the block diagram as

    shown in the figure bellow.

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    Time Response of the First Order Systems.

    Unit step response

    For unit step input R(s)=1/s, then

    Since we usually ignore the initial

    condition in block diagram, we use the

    system block diagram as shown bellow.

    )()/1(

    )/()( sR

    s

    KsC

    (1)

    Suppose that the initial condition is

    zero then

    1

    s

    KR(s) C(s)

    /1s

    1

    )/1(

    )/()(

    s

    K

    s

    K

    s

    KsC (2)

    Taking the inverse L.T of (2) yields

    )1()( t

    eKtc (3)

    The first term originates in the pole of input

    R(s) and is called the forced response or steady

    state response The second term originates inthe pole of the transfer function G(s) and is

    called the natural response Figure below plot

    c(t)

    t

    c(t)

    K1/

    )1( t

    eK

    The final value or the steady state value ofc(t)

    is K that is

    lim c(t)= Kt

    c(t) is considered to reach final value after

    reaching 98% of its final value. The parameter is called the time constant. The smaller the time

    constant the faster the system reaches

    the final value.

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    Time Response of Second Order System

    22

    2

    2)(

    )()(

    nn

    n

    sssR

    sCsG

    (1)

    The standard form second order system is

    Where = damping ratio, n = naturalfrequency, or undamped frequency.

    Consider the unit step response of this system

    sssRsGsC

    nn

    n

    )2()()()(

    22

    2

    (2)

    Case 1: 0

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    General Second-Order System

    Natural Frequency,n

    The natural frequency of a second-order system is the frequency of

    oscillation of the system without damping

    Damping Ratio,

    So,

    = =

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    PROBLEM:

    Given the transfer function of Eq. , find and n

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    Time Response Specification in Design

    The typical of unit step response of a system

    is as in figure below

    nt

    c(t)

    Mpt

    1.0

    0.9

    0.1

    Tr Tp

    1+ d

    1d

    css

    Ts

    Rise time Tr

    is the time required for the

    response to rise from 10% to 90% of the final

    value css. Mpt is the peak value, and Tpis the

    time required to reach Mpt.

    Mptcsspercent overshoot =100%

    css

    Settling time, Ts,is the time required for the

    output to settle within a certain percent of its

    final value. Common values are 5% and 2%.

    Settling time is proportional to the ,

    kTs = k =

    n

    If 2% is used to specify the settling value thenk=4. T

    r

    , Ts

    ,and css

    , are equally meaningful for,over, critically, or under-damped cases, whileTp,and Mp,are meaningful only for under-damped cases.

    Under-damped cases.

    To find Tp,we differentiate c(t) and equalizing

    the result to zero and solve the equation for t.

    We will find that2

    1/1

    1

    eMT p

    n

    p

    %100shootover%2

    1/ e

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    Time Response Specification in DesignRecall that

    1

    or

    1

    pn

    n

    p TT

    and

    %100shootover%2

    1/ eBoth %over-shootand nTp are functionsonly of and can be plotted as in thefollowing figure

    %over-shoot

    n

    Tp

    100

    80

    60

    40

    20

    0

    5.0

    4.6

    4.2

    3.8

    3.4

    3.00.2 0.4 0.6 0.8 1

    %over-s

    hoot

    nTp

    Example:

    Servomotor is used to control the position of plotter

    pen as in the following figure.

    a

    a

    a

    a

    Kss

    K

    ssK

    ssK

    sT 5.02

    5.0

    )2(/5.01

    )2(/5.0

    )( 2

    ann K5.0and222

    Here we have

    Suppose that we want to have =1, the fastest

    response with no overshoot thenn= 1 and Ka=2.

    Note that the settling time is

    Tp= 4/ n= 4sIt is not fast enough. If we want the faster response

    First we have to choose a different motor, then

    Redesign the compensator.

    0.5/s(s+2)

    motorR(s)

    Kp

    Amplifier

    +

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    TRANSIENT RESPONS ANALYSIS WITH MATLAB(2)

    Example 2 : find the unit impuls- respons for following equation,

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    Sumber Pustaka

    Norman S. Nise, Control Systems Engineering, Fourth Edition,

    2004, Wiley & Sons..