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c annot b e e xpressed as a n e xplicit function o f t he i nput. T hese l imitations make i t
useless in t he t heoretical s tudy o f systems. Problems
9.11 Solve iteratively (first three terms only):
( a) y[k + 1 ] 0.5y[k]
(b)y[k + 1] = 0, with y [l] + 2y[k] = J[k + 1], = 10 with J[k] = eku[k] and y [l] =0 9.12 Solve the following equation iteratively (first three terms only):
y [k] 0.6y[k  1] 0.16y[k  2] = 0 with y [l] =  25, y [2] =0 9.13 Solve iteratively the secondorder difference Eq. (8.26b) in chapter 8 (first three terms
only), assuming y [l] = y [2] = 0 and f [k] = 100u[k].
9.14 Solve the following equation iteratively (first three terms only):
y[k + 2] + 3y[k + 1]
with f [k] = (3)k u [k], y [l] + 2y[k] = J[k + 2] + 3J[k + 1] + 3f[k]
= 3, and y [2] = 2 9.15 Repeat Prob. 9.14 if
y[k]
with f [k] + 2y[k  1] + y[k  2] = 2 f[k] f[k  1] = (3)ku[k],y[1] = 2, and y [2] = 3. 9.21 Solve
y[k + 2] T he s tability criterion in t erms o f t he l ocation of characteristic roots of t he
s ystem c an be summarized as follows:
1. An LTID system is asymptotically stable if a nd only if all t he c haracteristic
roots are inside t he u nit circle. T he r oots may b e r epeated o r u nrepeated. + 3y[k + 1] + 2y[k] = 0 if y [l] = 0, y [2] = 1 + 2] + 2y[k + 1] + y[k] = 0 if y [l] = 1, y [2] = 1 if y [l] = 1, y [2] = 0 9.22 Solve
9.23 Solve 2. An LTID system is unstable if a nd o nly if e ither one or b oth o f t he following
tThere is a possibility of an impulse ark] in addition to characteristic modes. y[k y[k + 2 ] 2y[k + 1] + 2y[k] =0 9.24 Find v[k], t he voltage at the kth node of the resistive ladder depicted in Fig. P8.55
in Chapter 8, if V = 100 volts and a = 2. 9 6 12 T imeDomain A nalysis o f D iscreteTime S ystems 613 P roblems Hint: T he difference e quation for v[k] is given i n P rob. 8.45. T he a uxiliary conditions
a re v[O] = 100 a nd v[N] = O.
9 .31 y[k
9 .32 + 1] + 2y[k] = F ind t he t otal r esponse of a s ystem specified by t he e quation y[k = E f[k] 9 .47 R epeat P rob. 9.31 i f = J[k + 1] 9 .48 I n t he s avings a ccount p roblem d escribed i n E xample 8 .5 ( chapter 8 ), a p erson deposits $500 a t t he b eginning o f e very m onth, s tarting a t k = 0 w ith t he e xception a t
k = 4, w hen i nstead o f d epositing $500, s he w ithdraws $1000. F ind y[k] i f t he i nterest
r ate is 1 % p er m onth ( r = 0 .01).
H int: Because t he d eposit s tarts a t k = 0, t he i nitial condition y [l] = O. W ithdrawal
is a d eposit o f n egative a mount. 9 .49 T o p ayoff a l oan o f M d ollars in N p ayments u sing a fixed m onthly p ayment o f P
d ollars, show t hat ao = a l = a2 = . .. = a nl = 0
t he r esulting e quation is called a n onrecursive difference e quation.
( a) Show t hat t he i mpulse response o f a s ystem d escribed b y t his e quation is + bn l6[k  1] + ... + bl6[k  n R epeat P rob. 9.41 i f t he i mpulse response h[k] = (0.5)ku[k], a nd t he i nput f [k] is
( a) 2ku[k] ( b) 2(k3)u[k] ( c) 2ku[k  2].
H...
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 Spring '13
 Bayliss
 Signal Processing, The Land

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