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Unformatted text preview: matched. Therefore, Ao = bn a nd h (t) = bn8(t) + characteristic modes (2.79) To determine t he c haracteristic mode terms in t he above equation, let us consider
a system So whose i nput f (t) a nd t he c orresponding o utput x (t) a re related by Q (D)x(t) = f (t) (2.80) 162 2 T imepomain Analysis of ContinuousTime Systems Observe t hat b oth t he systems S a nd So have t he same characteristic polynomial;
namely, Q(>.), a nd, consequently, t he s ame characteristic modes. Moreover, So is
t he s ame as S w ith P(D) = 1, t hat is, bn = O. Therefore, according t o Eq. (2.79),
t he impulse response of So consists of characteristic mode terms only without a n
impulse a t t = O. L et us denote this impulse response of So b y Yn(t). Observe
t hat Yn (t) c onsists of characteristic modes of S, a nd therefore may be viewed as a
zeroinput response of S. Now Yn(t) is t he response of So t o i nput 8(t). Therefore,
according t o Eq. (2.80) 2.9 S ummary 163 T he righthand side of Eq. (2.84) is a linear combination of t he derivatives of Yn(t)u(t). Evaluating these derivatives is clumsy and inconvenient because of the
presence of u(t). T he derivatives will generate an impulse and its derivatives at the
origin. Fortunately when m :s; n [Eq. (2.77)], we c an avoid this difficulty by using
t he observation in Eq. (2.79), which asserts t hat a t t = 0 ( the origin), h(t) = bno(t).
Therefore, we need not bother t o find h (t) a t t he origin. This simplification means
t hat i nstead of deriving P(D)[Yn(t)u(t)], we c an derive P(D)Yn(t) and add t o it the
term bno(t), so t hat (2.81a) Q(D)Yn(t) = 8(t)
or = bno(t) + [P(D)Yn(t)]u(t) (2.85) (2.81b)
or
(2.81c)
where y~k) (t) r epresents t he k th derivative of Yn(t). T he r ighthand side contains a
single impulse t erm 8(t). T his is possible only if y~nl) (t) has a u nit j ump discontinuity a t t = 0, so t hat y~n) (t) = 8(t). Moreover, t he lowerorder t erms c an not have
any j ump d iscontinuity because t his would mean t he presence of t he derivatives of
8(t). Suppose, f or instance, Yn(t) h as a jump discontinuity, t hen i ts derivative Yn(t)
c ontains a n i mpulse 8(t), a nd i ts second derivative Yn(t) contains t he first derivative
of t he impulse 8 (t), a nd so on. B ut t his is impossible because t he r ighthand side of
Eq. (2.81c) consists of only 8(t). For this reason only y~nl) (t) c an have a unit j ump
d iscontinuity so t hat y~n)(t) is 8(t). T here c an be no j ump discontinuities in any
of t he r emaining variables because this would give rise t o higherorder derivatives
of 8(t) o n t he l efthand side. Therefore Yn(O) = y~l) (0) = '" = y~n2) (0) = 0 (no
discontinuity a t t = 0). Therefore, t he n initial conditions on Yn(t) a re
y~nl)(o) = 1 Yn(O) = y~l)(O) = . .. = y~n2)(0) =0 (2.82) T his discussion m eans t hat Yn(t) is t he z eroinput response of t he s ystem S s ubject
t o i nitial conditions (2.82).
We now show t hat for t he same i nput f (t) t o b oth systems, S a nd So, t...
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This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.
 Spring '13
 Bayliss
 Signal Processing, The Land

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