484 hppcndix A Solutioiis to the Exercises o = arctan(-) B = I ele , y(t) = + (!/(I P - f 'a4sd in M 53" M cos(41 - 53") - 1.2 e -?' 1E l l ) exteriial part internal part S o ~ ~ ~7.4 ion Kq (7.15): ~ r l y ( l )+ ~ i g ~ (= t ) For nnilateral signals that start at t = with (4.34): ?;(t) (t) YY(.) -0 - y(0) X(S) Puttirig these pairs into (7.15), yields (7.16) (LL[SY(S)- S O 21) y(0)] ~ 7.5~ + O O Y ( S ) = p l i s X ( s ) - r(0)]t I9(,X(S) t ~ ~ ~ Initial conditions from the block dingram: y(&) = ,(0) ( - ) = -2zo {/(OS)= ?J(O ) I- r(O+).0.5 + = +2 -4-0 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 2R H ( s ) =z _ , Re{s} '> - from t>heblock diagram, direct form 11 -4 G ( S ) = ~ s +1 X ( s ) - C(E(1) - k ( t ) (t - I)&(t- 1)) + + Y ( s ) = H ( s ) Xjs) t G(S)z ( ) 120 - S+l 1- z0 = il = + -F{eP3-l) s +1 I) I_ $ ( A T + [-s + -](P-5-1) sf -2 - S i l g ( l ) ;= [4(1 - X { ) ) P 2 s - + -p-p ii I ' - 2; s ( t ) t- _ P + s +1 2p -d'-11] E(f - 1) - 4x0 -s+ I