Signal Processing and Linear Systems-B.P.Lathi copy

C onsequently w e s hould b e a ble t o o btain d

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: inverse L aplace t ransform o f t his e quation yields (1- - ~e-4' + -9') 1 [ 1 [ (_¥e-4' + ¥eX1(t) • 2+ [ a nd (13.33a) + BF(s)] 801 13.3 Solution o f S tate E quations E xample 1 3.5 F ind t he s tate v ector x (t) for t he s ystem whose s tate e quation is given by 36 1 36 e 45 20 9t X2(t) x= A x+Bf u (t) w here f (t) = u (t) a nd t he i nitial conditions are Xl (0) F rom E q. (13.33b), we h ave = 2, X(s) = L et us first find ~(s). X2(O) o = 1. ~(s)[x(O) + B F(s)] ( Sl-A)=S[: ~]-[=:: a nd ~(s) = ( sl - A)- 1 = [ 12 ! J=[s:6 s-+%J (S+:n~+9) 2 /3 ( 8+4)(8+9) ] - 36 ( 8+4)(8+9) 8 +12 ( 8+4)(8+9) C omputer E xample C 13.2 Solve E xample 13.5 using MATLAB. Caution: S ee c aution i n E xample C I3.1. A =[-12 2 /3;-36 -1]; B =[l/3; 1]; C =[O 0); D =O; x O=[2;1]; t =0:.01:3; t =t'; f =ones(length(t),l); [ y,x)=lsim(A,B,C,D,f,t,xO); p lot(t,x) 0 We have (13.36a) T he Output Now, x(O) is given as T he o utput e quation is given by x(O) = [ :] Also, F (s) y = ~, a nd B F(.) " [ :] Therefore H7] = C x+Df a nd Y (s) = C X(s) (13.36b) ) U(t) + D F(s) T he s ubstitution of Eq. (13.33b) into this equation yields • 13 802 Y (s) = C {.(s)[x(O) S tate-Space A nalysis 13.3 + BF(s)]} + D F(s) S olution o f S tate E quations H(s) = C+(s)B + D (13.37) 1 0] [o v z ero-state r esponse z ero-input response 1 T he z ero-state response ( that is, t he r esponse Y (s) w hen x (O)=O), is given b y s t3 1 [ 2 Y (s) = [ C.(s)B + DJF(s) ( 13.38a) N ote t hat t he t ransfer f unction o f a s ystem is defined u nder t he z ero-state c ondition [see Eq. (6.53)J. T he m atrix C .(s)B + D is t he t ransfer f unction m atrix H (s) o f t he s ystem, which relates t he r esponses VI, Y2, . .. , Yk t o t he i nputs f l, 12, . .. , f j: <S+1)~+2) ( 8+1)(8+2) (8+~R!+2) Y (s) = H (s )F(s) (13.39) 5 +2 r 2 (8 2 ) ( 8+1)(8+2) = [0 1] [Xl] + [1 0] [11] h -2 -3 1 X2 (13.40a) 1 and an output equation [:] [J [:} [::J [::] Remember t hat t he i jth element of the transfer function matrix in Eq. (13.42) represents the transfer function t hat relates the o utput Yi(t) t o t he input f i(t). Thus, t he transfer function t hat relates the o utput Y3 t o t he input h is H32(S), where and A)-l = s _1]-1 [ 2 s +3 s +3 ( 8+1)(s+2) [ (S+1~(8+2) ] -2 ( S+1)(8+2) ( s+1)(s+2) Hence, the transfer function matrix H (s) is given by + 5s + 2 + l )(s + 2) s2 H32(S) = (s • C omputer E xample C 13.3 A =[O 1 j-2 - 3]j B =[l Oj1 l ]j C =[l 0 ;1 1 ;0 2]; D =[O 0 ;1 OjO 1 ]; [ num1,den1]=ss2tf(A,B,C,D,1) [num2,den2]=ss2tf(A,B,C,D,2) 0 Characteristic Roots (Eigenvalues) o f a Matrix (13.40c) = ( sl - = H (s)F(s) (13.40b) I n this case, +(8) 8 +58+2 ( 8+1)(8+2) Solve Example 13.6 using MATLAB. Caution: T he common factor (s + 1) in two of the transfer functions in Eq. (13.42) are canceled. T he MATLAB answer gives transfer function with common factor. Let us consider a system with a s tate equation Xl] [X2 (13.42) 2 Y (s) o E xample 1 3.6 1 s +2 a nd the zero-state response is T he m atrix H (s) is a k x j m atrix (k is t he n umber o f o utputs a nd j is t he n umber o f inputs). T he...
View Full Document

Ask a homework question - tutors are online