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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 zeroinput 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 eroInput R esponse + 6D + 9) yet) = (3D + 5 )f(t) let us determine yo(t), t he zeroinput 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 zeroinput 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 eroinput 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 zeroinput 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 eroinput
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 zeroinput 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 zeroinput 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 secondorder polynomial can be determined by using the formula
in Sec. B.710 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.
 Spring '13
 Bayliss
 Signal Processing, The Land

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