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c haracteristic r oot a t 'Y =  3. T he c haracteristic m odes a re ( _3)k a nd k ( _ 3)k. H ence,
t he z eroinput response is j =chl k c os (f3k + 0) ( 9.22) We c an d etermine t he a rbitrary c onstants Cl a nd C2 f rom t he i nitial c onditions following
t he p rocedure i n p art ( a). I t is left a s a n e xercise for t he r eader t o s how t hat Cl = 4 a nd
C2 = 3 s o t hat w here c a nd 0 a re a rbitrary c onstants d etermined f rom t he a uxiliary c onditions.
T his i s t he s olution i n r eal f orm, w hich a voids d ealing w ith c omplex n umbers .
• york] E xample 9 .3
( a) F or a n L TID s ystem d escribed by t he difference e quation y[k + 2J  0.6y[k + IJ  0.16y[kJ = 5J[k + 2J ¥, (E2  0.6E  0.16)y[kJ = 5E2 J[kJ 'Y2  0.6'Y  0.16 = (')' + 0.2)(,,(  0.8) = ( E + 3)f[k] york] = c(0.9)k cos (2Jk C
2 =cos
0.9 (9.24) + 0) (7r ) =0 [v'23
1.]
 +0
cosO+smO
6
.9
2 T he c haracteristic r oots a re 'Yl =  0.2 a nd 'Y2 = 0.8. T he z eroinput response is
e
1 =cos
(0.9)2 (9.25)
T o d etermine a rbitrary c onstants Cl a nd C2, we s et k =  1 a nd  2 i n Eq. (9.25), t hen
s ubstitute y o[IJ = 0 a nd y o[2J =
t o o btaint ¥ T herefore find t he z eroinput r esponse o f a n L TID T o d etermine t he a rbitrary c onstants C a nd 0, we s et k =  1 a nd  2 i n t his e quation a nd
s ubstituting t he i nitial conditions y o[I] = 2, yo[2] = 1, we o btain T he c haracteristic e quation is ('Y + 0.2)('Y  0.8) = 0 US w hen t he i nitial conditions a re y o[I] = 2 a nd yo[2] = 1.
T he c haracteristic p olynomial is (')'2 1.56'Y+0.81) = ( ')'0.78 j0.45)('Y0. 7 8+j0.45).
T he c haracteristic r oots a re 0.78 ± jO.45; t hat is, 0 .ge±jii. T hus, [')'[ = 0 .9 a nd f3 = 7r/6,
a nd t he z eroinput response, according t o E q. (9.22), is given by (9.23b) T he c haracteristic polynomial is 0=5Cl+~C2} (4 + 3 k)(  3l (E2  1.56E + 0.81)y[k] (9.23a) f ind t he t otal r esponse i f t he i nitial conditions a re y [IJ = 0 a nd Y[2J =
a nd i f t he
i nput f[kJ = 4  ku[kJ. I n t his e xample we shall d etermine t he z eroinput c omponent yo[kJ
o nly. T he z erostate c omponent is d etermined l ater i n E xample 9.7.
T he s ystem e quation i n o perational n otation is 25 = ( c) F or t he case o f c omplex r oots, let
s ystem d escribed by t he e quation C (7r) =0  (12
v'3.]
+0
cosO+smO
3
.81
2
C or ~ e cos 0 + /8 c sin 0 = 2 1.~2 c cos 0 + ; 12 e sin 0 = 1 = ;. 25 " 4 = 25cr + TIi C2 T hese a re t wo simultaneous e quations in two unknowns c cos 0 a nd e s in
t hese e quations yields
(9.26)
c cos 0 =2.308 t The initial conditions Y[I] a nd y[2] are the conditions given on the total response. B ut because
t he input does not s tart until k = 0, the zerostate response is zero for k < O. Hence, a t k =  1
a nd  2 t he total response consists of only the zeroinput component, so t hat y[  1] = YO [ 1] and
y[2] = Yo[2]. c sin 0 =  0.397
Dividing c sin 0 b y e cos 0 yields o. S olution o f 5 82 9 T imeDomain A nalysis o f D iscreteTime S ystems t an () =  0.397 =  0.172
2.308
1 9 .3 5 83 9.3 The Unit Impuls...
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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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