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

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

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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(.) &quot; [ :] 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: &lt;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...
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