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

# cnynt 0 this result shows that q yit c2y2t

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Unformatted text preview: a re c onjugates. L et a nd T his y ields = - 5, yo(t) = _ 5e- t C2 =5 + 5 e- 2t T his is t he zero-input component of yet) for t :2: O. ( b) Similar procedure may b e followed for repeated roots. For instance, for a system specified by (D2 (2.10a) Yo(t) = 109 S ystem R esponse t o I nternal C onditions: Z ero-Input R esponse + 6D + 9) yet) = (3D + 5 )f(t) let us determine yo(t), t he zero-input component of the response if t he initial conditions are yo(O) = 3 a nd lio(O) = - 7. T he characteristic polynomial is A2 + 6A + 9 = (A + 3)2, and its characteristic roots are A1 = - 3, A2 = - 3 ( repeated roots). Consequently, t he characteristic modes of t he system are e - 3t a nd t e- 3t . T he zero-input response, being a linear combination of t he characteristic modes, is given by :'ei° e(o+j{3)t + : 'e- jO e (a- j {3)t 2 2 = ~eot [ei({3t+O) = ce ot c os (/3t + e -i({3HO)] + 9) ( 2.10b) T herefore, t he z ero-input r esponse c orresponding t o c omplex c onjugate r oots a :±j/3 c an b e e xpressed i n a c omplex f orm ( 2.lOa) o r a r eal f orm (2.10b). T he l atter is m ore c onvenient f rom a c omputational v iewpoint b ecause i t a voids d ealing w ith c omplex n umbers. • E xample 2 .1 (a) F ind yo(t), t he zero-input component of t he response for an LTI system described by t he following differential equation: (D2 + 3 D + 2) yet) = D f(t) when t he i nitial conditions are yo (0) = 0, lio(O) = - 5. Note t hat yo(t), b eing t he z ero-input component ( J(t) = 0), is t he s olution of (D2 + 3 D + 2)yo(t) = O. T he c haracteristic polynomial of t he s ystem is A2+ 3A + 2. T he c haracteristic equation of t he s ystem is therefore A2 + 3A + 2 = (A + 1)(A + 2) = O. T he characteristic roots of t he s ystem are A1 = - 1 a nd A2 = - 2, a nd t he c haracteristic modes of t he s ystem are e - t a nd e - 2t. C onsequently, t he zero-input component of t he loop current is (2.l1a) To determine t he a rbitrary constants C, a nd C2, we differentiate Eq. (2.l1a) to obtain We can find t he a rbitrary constants C1 a nd C2 from t he initial conditions yo(O) = 3 a nd lio(O) = - 7 following t he procedure in p art ( a). T he reader can show t hat C1 = 3 a nd C2 = 2. Hence, yo(t) = (3 + 2 t)e- 3t ( c) For t he case of complex roots, let us find t he zero-input response of an LTI system described by t he equation: (D2 + 4D + 40) yet) = (D + 2 )f(t) with initial conditions yo(O) = 2 a nd lio(O) = 16.78. T he characteristic polynomial is A2 + 4A + 40 = (A + 2 - j6)(A + 2 + j 6). T he characteristic roots are - 2 ± j 6.t T he solution can b e w ritten either in the complex form A2t (2.lOa) or in t he real form (2.10b). T he complex form is yo(t) = Cle Att + c2e , where Al = - 2 + j 6 a nd A2 = - 2 - j6. Since Q = - 2 a nd (3 = 6, t he real form solution is [see Eq. (2.l0b)] (2.12a) Yo(t) = ce - 2t cos (6t + II) tThe complex conjugate roots of a second-order polynomial can be determined by using the formula in Sec. B.7-10 or by expressing the polynomial as a sum of two squares. The latter can be accomplished by completing the square with the first two terms, as sh...
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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.

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