Unformatted text preview: Domain Analysis of DiscreteTime Systems
y[k + 2] = y[k + 1]  + J[k + 2 ] 2f[k + 1]
(9.5)
S etting k =  2 a nd t hen s ubstituting y [I] = 2, y [2] = 1, frO] = f [I] = 0 (recall t hat
0.24y[k] f [k] = k s tarting a t k = 0), we o btain
vIol =2 0.24(1) +0  0 S etting k =  1 i n Eq. (9.5) a nd t hen s ubstituting vIOl
w e o btain y[l] = 1.76  0.24(2) = 1.76 = 1. 76, Y [I] = 2, f [l] = 1, frO] = 0, +I  0 = 2.28 = 1.76, Y[I] 2(1) = 1.8576 S etting k = 0 i n Eq. (9.5) a nd t hen s ubstituting vIOl
= I yields = 2.28, J[2] =2 a nd I P] y[2]
a nd so on. = 2.28  0.24(1.76) +2 • 9.1 577 DiscreteTime System equations We shall see in t he f uture t hat t he s olution o f a difference equation o btained
in this direct (iterative) way is useful in many situations. Despite t he m any uses
of this method, a closedform solution o f a difference equation is far more useful
in s tudy o f t he s ystem behavior a nd i ts dependence on t he i nput a nd t he various
system parameters. For t his r eason we s hall develop a systematic procedure t o
a nalyze discretetime systems along lines similar t o t hose used for continuoustime
systems. Operational Notation
I n difference equations i t is convenient t o use operational n otation s imilar t o
t hat used in differential equations for t he sake o f c ompactness. I n c ontinuoustime systems we used t he o perator D t o denote t he o peration of differentiation.
For discretetime systems we s hall use the o perator E t o denote t he o peration for
advancing t he sequence by one t ime u nit. T hus Note carefully t he i terative or recursive n ature o f t he c omputations. From t he E f[k] == f [k n i nitial conditions (and input) we o btained y[O] first. T hen, u sing this value of
y[O] a nd t he previous n  1 initial conditions (along with t he i nput), we found y[l]. E2 f [k] == f [k N ext, using y[O], Y[l] along with the initial conditions a nd i nput, we o btained y[2],
a nd so on. This m ethod is general a nd c an b e a pplied t o a difference e quation o f
a ny order. T he i terative procedure cannot be applied t o c ontinuoustime systems.
I t is i nteresting t hat t he h ardware realization o f Eq. (9.3a) depicted in Fig. 8.17
( with'Y = 0.5) g enerates t he solution precisely in this (iterative) fashion. + 1]
+ 2]
(9.6) T he firstorder difference e quation o f t he s avings account problem was found
t o b e [see Eq. (8.25b)] + 1 ] ay[k] + 1] 6 . E xercise E 9.l
Using t he i terative m ethod, find t he first t hree t erms o f y[k] i f
y[k + 1]  2y[k] = i [k]
T he i nitial c ondition is y [I] = 10 a nd t he i nput I[k] = 2 s tarting a t k = o.
A nswer: vIOl = 20,Y[I] = 42, andy[2] = 86. 'V Using t he o perational notation, we c an e xpress t his e quation as o or C omputer E xample C 9.1
Solve E xample 9.2 using MATLAB.
i .(c91.m) Y =[l 2]; Y =Y';
k =2:2; k =k';
f=[O 0 0]';
f or m =1:length(k}2
y =Y(m+l}0.24*Y(m)+f(m+2}2*f(m+l};
Y =[Y;y];
F =m;f=[f;F];
e nd
s tem(k,Y}
[ kY] y[k ( E  a)y[k] = E f[k] (9.8)...
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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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