Wiley signals and systems e book TLFe BO 544

Apperidix A . Solutioris to t,lir Exerriscs

+

y[3] = 0 - 2 3 ( - 2 )
y[4] 0 2 . (-4) 3
’

+

-

529

= -4

5

:

g[k] ( 1)A ( k 4l)&!k]
b) y[k] = 1,- 1 , 2 > - 2 , 3 , - 3 .
Solution 14.4
y[k] = J [ k ]- g[1,

-

!),a]= 4 2 1

=

~

y[O]

21
1

y 31 = 431 - y [ l ] = -6
] 0
0[4] 1*[1]- ~ [ 2 =
y[S] = .“5] - y[3] = 5

Solution 14.5
Intcrrial part:

1eax-c oiit iiiput S i g d

External part:

y,,+[O]

=2

sct initial state> t o zero

1 =2

.+[ I ] = 2 1 + 0.5 . (1

+ .vrxc [OJ) = 3.2

ycxt[2] = 2 ’ 1 + 0.5 . ( 1 + 3 . 5 ) = 4.25
y,,+[3] = 2 . 1

+ 0.5 . ( I + 4.23) = 1.625

Iriilial condition

Solution 14.6
a)
0

h

iiiiique assigizmcnt of iiiii,ial states is orily possible with ca,rioriic:;tlfolms.

DF I:

Begin wit,li the fcedback brmch.