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"O" system, the steady-state error for a unit-ramp function is infinity; for a type
"1" system, it is finite; and for systems of type lL2" or higher, it is zero.
Similarly, the steady-state error to a unit-parabolic input function, r(t) = t2 or
r (s) = 1/S3, can be obtained as
e(oo) = ~
(5,92)
g
lr
i
.
N
E
q
l
d
456 PERFORMANCE, STABILITY, DYNAMICS, AND CONTROL
where
= lrrri_[S2G(S)]
s~ or
(5.93)
We observe that for a given system to have zero steady-state error to aunit-parabolic
input, we must have at least have three integrators in the forward path. Therefore,
the steady-state error to unit-parabolic input oftype "0" and type "1" systems is infi-
nite; for type "2" systems,it is finite; and for systems oftype "3" or higher,itis zero.
5.6 FrequencyResponse
In steady-state; a sinusoidal input to a linear system generates a sinusoidal
response (output) of the same frequency. However, the magnitude and phase angles
of the response are generally different from those of theinput and also var}r with the
frequency of the applied input.ln the following, we will determine the steady-state
response (magnitude and phase angle) to sinusoidalinputs.2
In general, a sinusoid input function can be represented as
r (t) = A cos cot + B sin c.ot
- Mi cos(cot + 41)
(5.94)
(5.95)
Here, Mt and ~/ are the magrutude and phase angle of theinput sinusoid function
and are given by the following expressions:
In phasor notation,
r(t) : Mt 24t
(5.96)
(5.97)
(5.98)
Furthermore, we assume that we can represent r (t) as iumber, r (t) -.
A - jB so that A - jB = N.tteid and A + jB = M,ae_?4,pTfakrnug the Laplace
transform of Eq. (5.94), we get
f(s)=(AXB )
The response to a sinusoidalinput is given by
y(s) = (Al:_B )G(s)
As + Beo
. 'J ))G(s)
~s+-]co) .s~
ki k2 +...
= s+ lcn + s
'- Jco
(5.99)
(5.100)
(5.101)
(5.102)
Because we are interested in only the steady-state response, we have ignored the
terms corresponding to G(s), which generate the transient response. Iriecall that
LINEAR SYSTEMS, THEORY, AND DESIGN:A BRIEF REVIEW 457
the steady-state response comes from the poles because of:input function, which
in this case are at s = :+: jco. Using partial fraction method, we get
k,= A +B G(s)ls=_jto
s- ja
A+
= +2- jBG(_jc,o)
.
J
(5.103)
(5.104)
Because G( jc.o)is acomplex number,we can write G( jto) = M8e j48 or G(- jco) =
Mge-jb,. With A + j B = M,e~idf , we have
Similarly,
ki - ~M,e-)*i Mge-i*8
_ Mt Mg e-j(41+48)
2
/
k2 = M2M e/(4f +~)
~*
~l
where * denotes the complex conjugate. Then,
Yoo(S) = M2M [
e-j(41+4g) ej(4f +4,)
~s+-]co) + ~s co)
(5-105)
(5.106)
(5.107)
(5.108)
(5.109)
where the suffix oo denotes the steady-state value (t -* oo) and Mg and 4g are
the magnitude and phase angle of the transfer function G(s) (with s = jr.o) and
are given by
M8 - IG( jto)l
Z~g = ZG(jw)
Taking the inverse Laplace transforms in Eq. (5.109), we get
Yoo(t) = M2M [e-/M +*r +ou) + ej~ +48 +wr)]
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