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4! 58 θr, then the state equations are in the form of CCF. 22  ( sI − A ) = −1 For a unit-step function input, u ( t ) s =1 / s. 479 e  (c) Characteristic equation: (d) ∆( s ) = s 2 + 80 . 65 s + 322 From the state equations we see that whenever there is to increase the effective value of Ra by (1 + K ) Rs . =0 . 58 Ra there is ( 1 + K ) Rs . Thus, the purpose of R is s This improves the time constant of the system. 18 θr. The equations are in the form of CCF with v as the input. 1434 e  −1 (b) (c) Characteristic equati on: 2 .

D) Eigenvalues of A: −1, −1, −1 The matrix A is already in Jordan canonical form. Thus, the DF transformation matrix T is the identity matrix I. 9597  5-11 (a) 1 −2  0 0  S = [B AB] = S = B  1 −1 1  2 AB A B  =  2 −2 2     3 −3 3  S = [B AB] = S is singular. (b) S is singular. (c)  2   2 2 +2 2  2+ 2   S is singular. (d) 45 −1 0 1  1 −3   1 −2 4  AB A B  =  0 0 0    1 −4 14  S = B 5-12 (a) 2 S is singular. Rewrite the differential equations as: d θm 2 dt 2 B d θm 2 =− J dt 2 − K J θm + State variables: x = θ , x m 1 Ki 2 J = dia ia dθ =− Kb dθ m dt m x , dt 3 La dt − Ra La ia + K a Ks (θ La r −θm ) = ia State equations: Output equation:  dx1    dt   0     dx2  =  − K  dt   J  dx   K K  3  − a s  dt   La 1 − B J − Kb La      0    x1    Ki    x2 +  0  θ r J     x  K K  Ra  3   a s  −   La  La  0 y = 1 0 0 x (b) Forward-path transfer function:  s  Θm ( s ) K G ( s) = = [1 0 0 ] J E (s )  0  −1 s+ B J Kb La  0   Ki  − J   Ra  s+ L a  −1    0    KiK a  0 =  K  ∆ o ( s)  a  La  ∆ o ( s ) = J La s + ( BLa + Ra J ) s + ( KLa + Ki Kb + Ra B) s + KRa = 0 3 2 Closed-loop transfer function:   s  Θm ( s ) K M ( s) = = [1 0 0 ]   J Θr ( s )   KaKs  La = −1 s+ B J Kb La K i Ka K s  0   Ki  − J   R s+ a La  −1    0    K s G( s )  0 =  K K  1 + K s (s )  a s  La  JLa s + ( BLa + Ra J ) s + ( KLa + Ki Kb + Ra B) s + K i K a K s + KRa 3 2 5-13 (a) 46 = x1 A=  0 1  −1 0  A = 2 − 1 0   0 −1  A = 3  0 − 1 1 0  A = 4 1 0  0 1  (1) Infinite series expansion: 3 5 t t  t2 t4  1 − + − L t − + − L  1 2 2 2!

Taking the inverse Laplace transform −1 37 on both sides of the equation gives the desired relationship for 5-3 (a) Characteristic equation: Eigenvalues: s ∆( s ) = = −0 . 5 − j 1. 323 , φ( t ) . j 1. 333e −4 t −t ∆ ( s ) = ( s + 3) = 0 2 −4 t −4 t Eigenvalues: = −3, − 3 s State transition matrix:  e −3 t φ ( t) =  0 (d) Characteristic equation: ∆( s ) = −3 t − 9 = 0 Eigenvalues: 2 s   e  0 s = −3 , 3 s = − State transition matrix:  e3 t φ ( t) =  0 (e) Characteristic equation:   e  0 −3 t ∆ ( s ) = s + 4 = 0 Eigenvalues: 2 j2, j2 State transition matrix:  cos2 t  − sin2 t φ ( t) = (f) Characteristic equation: ∆( s ) = s 3 s i n 2t  cos2t  + 5 s + 8 s + 4 = 0 Eigenvalues: 2 s = − 1, − 2 , −2 State transition matrix:  e− t  φ ( t) = 0   0 (g) Characteristic equation: ∆( s ) = s 3 0 e 0 + 15 e  φ ( t) = 0   0 −5 t 5-4 State transition equation: x (t ) = φ (t )x( t ) + −2 t + 75 s + 125 = 0 2 s   te  −2t e  0 −2 t te e −5 t −5 t 0 Eigenvalues: s = − 5, − 5, −5   te  −5 t e  0 −5 t ∫ φ (t − τ )Br (τ )d τ t 0 (a) 38 φ (t ) for each part is given in Problem 5-3.