Signal Processing and Linear Systems-B.P.Lathi copy

# an a re distinct i f t here a re repeated roots same

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 , . &quot; , 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 r-le).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 ime-Domain A nalysis o f C ontinuous-Time 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 r-1e 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 rtr-l)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(.&gt;.) 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 ero-input 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...
View Full Document

## This note was uploaded on 04/14/2013 for the course ENG 350 taught by Professor Bayliss during the Spring '13 term at Northwestern.

Ask a homework question - tutors are online