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Unformatted text preview: C (An 107 + ... + alA + ao = 0 (2.5a) This result m eans t hat ceAt is indeed a solution of Eq. (2.4), provided t hat A satisfies
Eq. (2.5a). N ote t hat t he polynomial in Eq. (2.5a) is identical to t he polynomial
Q (D) in Eq. (2.4b), with A replacing D. Therefore, Eq. (2.5a) can be expressed as Q(A) = 0 (2.5b) In this case t he r oot A r epeats twice. Observe t hat t he c haracteristic modes in this
case are eAt a nd teAt. C ontinuing this p attern, we c an show t hat for the differential
equation
(2.8)
(D  Wyo(t) = 0
t he c haracteristic modes are e )'t, W hen Q(A) is e xpressed in factorized form, Eq. (2.5b) can be represented as
YO (t) =
(2.5c)
Clearly, A has n solutions: AI, A2, . .. , An. Consequently, Eq. (2.4) has n possible
solutions: q eA,t, c 2e A2t , . " , c ne Ant , w ith c t, C2, . .. , C as a rbitrary c onstants. We
n
can readily show t hat a general solution is given by t he sum of these n s olutions,t
so t hat
yo(t) = c leA,t + C2e).,t + ... + c ne Ant
(2.6)
where C l, C2, . .. , C n a re a rbitrary c onstants determined by n c onstraints (the
auxiliary conditions) on t he solution.
t To prove this assertion, assume that YI (t), Y2(t), . .. , Yn(t) are all solutions of Eq. (2.4). Then te)'t, t 2e A
t, ... , t rle).t, a nd t hat t he s olution is (c t + C2t + ... + crt r l) e ).t (2.9) Q (D)YI(t) = 0
Q (D)Y2(t) = 0
Q (D)Yn(t) = 0
,Cn, respectively, and adding them together yields
Q (D) [ qYI (t) + C2Y2(t) + ... + cnYn(t)] = 0
This result shows that q YI(t) + C2Y2(t) + ... + cnYn(t) is also a solution of the homogeneous
equation (2.4).
:j: The term eigenvalue is German for characteristic value. MUltiplying these equations by C I,C2, . .. 108 2 T imeDomain A nalysis o f C ontinuousTime S ystems 2.2 C onsequently, f or a s ystem w ith t he c haracteristic p olynomial t he c haracteristic m odes a re e Att , t eAt!, . .. , t r1e At !, eAr+1t,
s olution is yo(t) = ( Cl ( 2.llb) . .. , e Ant a nd t he S etting t = 0 in Eqs. (2.l1a) and ( 2.llb), a nd substituting t he initial conditions yo(O) = (}
a nd lio(O) =  5 we o btain
0 =Cl+C2 + C2t + ... + c rtrl)e Att + c r +le Ar +1 t + ... + c ne Ant  5 =  Cl  2C2 Complex Roots
T he p rocedure for h andling c omplex r oots is t he s ame a s t hat for r eal r oots.
F or c omplex r oots t he u sual p rocedure l eads t o c omplex c haracteristic m odes a nd
t he c omplex f orm o f s olution. H owever, i t is possible t o a void t he c omplex f orm
a ltogether b y s electing a r eal f orm o f s olution, as d escribed b elow.
F or a r eal s ystem, c omplex r oots m ust o ccur i n p airs o f c onjugates i f t he coefficients o f t he c haracteristic p olynomial Q(.>.) a re t o b e r eal. T herefore, i f a: + j /3
is a c haracteristic r oot, a:  j/3 m ust a lso b e a c haracteristic r oot. T he z eroinput
r esponse c orresponding t o t his p air o f c omplex c onjugate r oots is Solving these two simultaneous equations in two unknowns for C1 a nd C2 yields
C1 Therefore F or a r eal s ystem, t he r esponse yo(t) m ust a lso b e r eal. T his is p ossible o nly i f Cl
a nd C2...
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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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