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8/13/2019 Chapter 4-2.doc
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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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