and the initial coidit,ions are not needed.
Trang 1484 hppcndix A Solutioiis to the Exercises
4
3
B = I e l e , o = arctan(-) M in M 53"
y(t) = (!/(I P - 3 f + 2 cos(41 - 53") - 1.2 e -?' 1 E l l )
'a4sd
part
S o ~ ~ ~ i o n 7.4
Kq (7.15): ~ r l y ( l ) + ~ i g ~ ( t ) =
For nnilateral signals t h a t start at t = 0 with (4.34):
( L L [ S Y ( S ) - y(0)] + O O Y ( S ) = p l i s X ( s ) - r(0)] t I9(,X(S)
Puttirig these pairs into (7.15), yields (7.16)
21) Initial conditions from the block dingram:
y(&) = ,(0) ( - 2 ) 0 5 4 = -2zo
{/(OS) = ?J(O ) I- .r(O+) 0.5 + 4 = -4-0 + 2
b) Sirice the initial s t a t r has beeii given, (7.21) is suitable f i x t h e > solution and the initial coidit,ions are not needed
H ( s ) =z _ ,
2 R
6 + 1 Re{s} '> - 1 from t>he block diagram, direct form 11 -4
s + 1
G ( S ) = ~
X ( s ) - C(E(1) - k ( t ) + (t - I)&(t - 1))
+ - F { e P 3 - l ) -
s + I
Y ( s ) = H ( s ) Xjs) t G(S) z ( 0 ) = I_
s + 1 $ ( A T I )
2 120 2 - 2
P +
= il - - + - p - p I _
S i l ii s s + 1
-
- + [- + -](P-5-1)
g ( l ) ;= [4(1 - X { ) ) P ' - 2; s ( t ) t- 2 p - d ' - 1 1 ] E ( f - 1)