# New PDF release: kuo and golnaragh automatic control systems 8ed - solutions

By Benjamin Kuo

Read Online or Download kuo and golnaragh automatic control systems 8ed - solutions manual PDF

Similar mathematics books

Visual Thinking in Mathematics by Marcus Giaquinto PDF

Visible pondering - visible mind's eye or conception of diagrams and image arrays, and psychological operations on them - is omnipresent in arithmetic. is that this visible considering basically a mental reduction, facilitating snatch of what's accumulated by means of different potential? Or does it even have epistemological services, as a way of discovery, figuring out, or even evidence?

New PDF release: The Joy of x: A Guided Tour of Math, from One to Infinity

Many folks take math in highschool and rapidly disregard a lot of it. yet math performs a component in all of our lives all the time, no matter if we all know it or now not. within the pleasure of x, Steven Strogatz expands on his hit long island instances sequence to give an explanation for the large principles of math lightly and obviously, with wit, perception, and outstanding illustrations.

Additional info for kuo and golnaragh automatic control systems 8ed - solutions manual

Sample text

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.