This preview shows page 1. Sign up to view the full content.
Unformatted text preview: n) (9.2c) T his e quation shows t hat y[k], t he o utput a t t he k th i nstant, is c omputed from 2 n+l
pieces of information. These are t he p ast n values of t he o utput: y[k  1], y[k 2], . .. , y [kn], t he p ast n values o fthe i nput: f [kl]' f [k2]' . .. , f [kn]' a nd the
p resent value o f t he i nput f[k). I f t he i nput is causal, t hen f [IJ = f [2] = . .. =
f [  n] = 0, a nd we need only n initial conditions y (I], y [2], . .. , y [n]. T his result
allows us to compute iteratively or recursively t he o utput y[O], y[l], y[2], y[3], . .. ,
a nd so o n.t For instance, to find y[O] we s et k = 0 in Eq. (9.2c). T he l efthand
s ide is y[O), a nd t he righthand side contains t erms y [I], y [2), . .. , y [n), a nd
t he i nputs frO), f [I), f [2), . .. , f [n]. Therefore, t o begin with, we m ust know
t he n initial conditions y [IJ, y [2J, . .. , y [n]. Knowing these conditions and t he
i nput f[k), we c an iteratively find t he response y[O), y[IJ, y[2], . .. , a nd so on. The
following examples demonstrate this procedure. This method basically reflects the
m anner in which a computer would solve a difference equation, given t he i nput and
i nitial conditions.
t For this reason Eq. (9.2) is called a r ecursive d ifference e quation. However in Eq. (9.2), if
a l = a 2 = ... = a nl = 0, then, according to E q. (9.2c), determination of the present
o utput y[kJ does not require the past values y[k  1], y[k  2], . .. , etc. For this reason, when
a i = 0, (i = 0 ,1, . .. , n  1), the difference E q. (9.2) is n onrecursive. This classification is
i mportant in designing and realizing digital filters. In this chapter, however, this classification
is n ot important. The analysis techniques developed here (and in Chap. 12) apply t o general ao = recursive and nonrecursive systems. Observe t hat a nonrecursive s ystem is a special case o f a
recursive system with ao = a l = . .. = a nl = O. Design o f recursive and nonrecursive s ystems is discussed in Chapter 12. +0 = 8 Now, s etting k = 1 i n E q. ( 9.3b) a nd u sing t he value y[O] = 8 ( computed i n t he first s tep)
a nd f[IJ = (1)2 = 1, we o btain y[I) = 0.5(8) + (1)2 = 5
Next, s etting k = 2 i n Eq. (9.3b) a nd u sing t he value y[l] = 5 ( computed i n t he p revious
s tep) a nd f[2) = (2)2, we o btain y[2] = 0.5(5) + (2)2 = 6.5
C ontinuing in this way iteratively, we o btain + (3)2 = 12.25
0.5(12.25) + (4)2 = 22.125 y(3) = 0.5(6.5)
y[4] = T he o utput y[k] is d epicted in Fig. 9.1. • We now present one more example o f i terative s olutionthis t ime f?r a.secondorder equation. Iterative method can be applied t o a difference e quatIon I II d elay
o r advance operator form. In Example 9.1 we considered t he former. Let us now
apply t he i terative method t o t he advance o perator form.
• E xample 9 .2
S olve i teratively y[k + 2 ] y[k + 1] + 0.24y[k] = f[k + 2 ] 2f[k + IJ (9.4) w ith i nitial conditions y [I] = 2, y[2] = 1 a nd a c ausal i nput f[k] = k ( starting a t
k = 0). T he s ystem e quation c an b e e xpressed a s 5 76 9 T ime...
View
Full
Document
 Spring '13
 Bayliss
 Signal Processing, The Land

Click to edit the document details