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    4.4 The initial and final value theorems

    In system analysis, the investigator often needs to know only the

    steady-state value of the output for a preliminary study. In such

    cases, the final value theorem is frequently useful. Ify(t)and

    dy/dtpossess Laplace transforms, if the following limit exists,

    and ify(t)approaches a definite value as t , this theorem

    states that

    )]!lim)"

    ssYys

    = #.#-$)

    %he theorem&s conditions are satisfied if the system model is

    linear with constant coefficients !i.e., transforma'le so that Y(s)( T(s) V(s)] and if all the roots of the denominator ofsY(s)lie in

    the left-hand plane. %he most common failure of the theorem is

    the case where the input vis a pure sinusoid. %his introduces

    purely imaginary roots and a steady-state sinusoidal response.

    %hus,y(t)does not approach a definite value as t .

    %he companion theorem is the initial value theorem.

    )]!lim)" ssYys

    = #.#-)

    %he conditions under which the theorem is valid are that the

    limit and the transforms of y and dy*dt exist.

    If the ramp response of dy*dt( ry +'v were not availa'le, the

    final value theorem could have 'een used to find the steady-state

    difference 'etween the input and output quickly. or v(t) =mt,

    we o'tain

    =

    =++

    +==

    $

    $,)

    $

    $lim

    )$

    lim)])!lim))

    "

    ""

    bm

    b

    s

    bs

    s

    m

    s

    m

    s

    b

    s

    mssYsVsyv

    s

    ss

    #.#-)

    %hus, the output does not follow the input unless $=b .

    %he initial and final value theorems are especially useful when

    the system is represented only in terms of its transfer function.

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    %he theorems can thus 'e directly applied without working

    through the solution of a differential equation.

    Example4.8

    $) erive formulas for the steady-state difference

    'etween the input V(s)and the output Y(s)for a unit-step

    input and a unit-ramp input. %he system diagram is given

    in igure #.$

    ) /pply the results to the case where G(s) = K/s.

    igure #.$ / unity feed'ack system. %he error ise(t)= v(t) -y(t).

    0olution1

    a) %he difference 'etween V(s)and Y(s)isE(s)in the figure,

    where

    E(s) = V(s) - Y(s) = V(s) - G(s)E(s) #.#-#)

    0olve forE(s)in terms of V(s).

    )$

    ))

    sG

    sVsE

    +=

    %he steady-state difference from the final value theorem is

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    )$

    )lim)lim

    "" sG

    ssVssEe

    ssss

    +==

    #.#-2)

    or a unit-step input, V(s)( $*s, and #.#-2) gives

    )"$

    $

    )$

    $lim

    " GsGe

    sss +

    =+

    = #.#-3)

    where )lim)" " sGG s=

    or a unit-ramp input, V(s)( $*s2and

    )lim

    $

    )

    $lim

    "

    " ssGssGse

    s

    sss

    =+

    = #.#-4)

    ') or G(s) =K/s, =

    sKGs

    *lim)"" , and "sse . %hus, this

    particular system has a 5ero steady-state difference 'etween the

    output and the step input.

    or a unit-ramp input, KssGs = )lim" , and ess($*K, which is a

    finite 'ut non5ero difference.

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    4.5 Impulse response

    6esides the step function, the pulse function and its

    approximation, the impulse, appear quite often in the analysis

    and design of dynamic systems. In addition to 'eing an

    analytically convenient approximation of an input applied for

    only a very short time, the impulse is also useful for estimating

    the system&s parameters experimentally. %he impulse is an

    a'straction that does not exist in the physical word 'ut can 'e

    thought of as the limit of a rectangular pulse whose duration %

    approaches 5ero while maintaining its strength /. %he strength

    of an impulse or pulse is the area under its time curve igure

    #.$#).

    igure #.$# Impulse and rectangular pulse. a) Impulse of

    strength /. ') 7ectangular pulse. %he impulse is the limit of the

    pulse as "T with / held constant.

    %he impulse response of the first-order model bvrydt

    dy+= can 'e

    o'tained 'y the Laplace transform method. rom ta'les, the

    transform of an impulse )tv of strengthAis

    V(s) =A

    rom #.$-8),

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    rs

    bAyA

    rs

    b

    rs

    ysY

    +=

    +

    =

    )")")

    In the time domain, this 'ecomes

    rtebAyty ])"!) +=

    %hus, the impulse can 'e thought of as 'eing equivalent to an

    addition initial condition of magnitude bA.

    4.6 Pulse response

    %he response due to a pulse 'y using the step response to find

    y(T),which is then used as the initial condition for a 5ero inputsolution. /lternatively, the shifting property of Laplace

    transforms can 'e applied. 9ith this viewpoint, the pulse in

    igure #.$#' is taken to 'e composed of a step input of

    magnitudestarting at t(", followed at t=T'y a step input of

    magnitude -) see igure #.$2).

    igure #.$2 7ectangular pulse as the superposition of two step

    functions.

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    %he pulse input can now 'e expressed as follows.

    ))) Tt!t!tv ss = #.3-$)

    Its transform is

    )$$$

    )]!)]!) sTsTss es

    se

    sTt!"t!"sV ===

    #.3-)

    /ssume that the system is sta'le and the initial condition is 5ero.

    %he pulse response is found from #.$-8) with *$=r and a

    partial fraction expansion.

    sTsTsT

    es

    #e

    s

    #

    s

    #

    s

    #

    s

    e

    s

    bsY

    +

    ++

    =

    +

    = $$$$

    $

    $)

    %he transform has 'een expressed as the sum of elementary

    transforms. :ross multiplication gives

    bs#s# =++ )$$

    or

    b#

    b#

    ==

    $

    In the time domain, we o'tain

    )))) *)$

    *$ Tt!#Tt!e#

    t!#e#

    ty ssTt

    s

    t+=

    or $%t%T,

    *

    *$) tt beb#e#

    ty =+= #.3-)

    or Tt ,

    **

    *)$*$

    )$)

    tTTtt

    eeb##e

    #

    e

    #

    ty

    =+=

    #.3-#)

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    %his response is shown in igure #.$3. %he previous equation,

    when written in terms of the pulse strengthA=T, is

    **

    )$)tT ee

    T

    bAty =

    #.3-2)

    If the strengthAis kept constant as Tapproaches 5ero,

    L&;opital&s rule gives

    *

    ")lim

    t

    TbAety

    =

    %his is the same as the impulse response when y") is 5ero.

    igure #.$3

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    a) =se the following 'lock diagram to write the system&s

    differential equation model directly in state varia'le form. %he

    state varia'les arex$andx.

    ') =se 'lock diagram reduction to derive the transfer function

    $s)*s)

    / torque Tis applied to a load of inertia*. %he damping is

    negligi'le so that )) sTs*s = , where is the speed of the load.

    or a sinusoidal torque tAtT +sin) = , plot the frequency

    response curves withA**( for a two-decade range centered at

    $=

    +

    rad*unit time.

    =se the final value theorem to compute the steady-state error e

    'etween the input and the output for the system shown in the

    following figure with the input functions as follows1

    a)s) ( $*s ')s) ( $*s

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