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 Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1 of 106   Table of Contents for Linear Linear System Theory and  Search document  Solutions Manual LINEAR LINEAR SYSTEM SYSTEM THEO THEORY RY,, 2/E Wilson Wilson J. Rugh Department of Electrical and Computer Engineering Johns Hopkins University Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month. Read Free Foron 30this Days Sign up to vote title   Not useful Cancel anytime. Useful     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1 of 106   Table of Contents for Linear Linear System Theory and  Search document  PREFACE With some lingering ambivalence about the merits of the undertaking, but with a bit more dedication Theory . Rou the first time around, I prepared this Solutions Manual for the second edition of Linear System Theory. 40% of the exercises are addressed, including all exercises in Chapter 1 and all others used in developments in text. This coverage complements the 60% of those in an unscientific unscientific survey who wanted a solutions manual perhaps does not overly overly upset the 40% who voted no. (The main contention between the two groups involved inevitable appearance of pirated student copies and the view that an available solution spoils the exercise.) I expect that a number of my solutions could be improved, and that some could be improved using techniques from the text. Also the press of time and my flagging enthusiasm for text processing impeded crafting of economical solutions—some solutions may contain too many steps or too many words. Howev hope that the error rate in these pages is low and that the value of this manual is greater than the price paid. Please send comments and corrections to the author at rugh@jhu.edu or ECE Department, Johns Hop University, Baltimore, MD 21218 USA. Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month. Read Free Foron 30this Days Sign up to vote title   Not useful Cancel anytime. Useful     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1 of 106   Table of Contents for Linear Linear System Theory and  Search document  PREFACE With some lingering ambivalence about the merits of the undertaking, but with a bit more dedication Theory . Rou the first time around, I prepared this Solutions Manual for the second edition of Linear System Theory. 40% of the exercises are addressed, including all exercises in Chapter 1 and all others used in developments in text. This coverage complements the 60% of those in an unscientific unscientific survey who wanted a solutions manual perhaps does not overly overly upset the 40% who voted no. (The main contention between the two groups involved inevitable appearance of pirated student copies and the view that an available solution spoils the exercise.) I expect that a number of my solutions could be improved, and that some could be improved using techniques from the text. Also the press of time and my flagging enthusiasm for text processing impeded crafting of economical solutions—some solutions may contain too many steps or too many words. Howev hope that the error rate in these pages is low and that the value of this manual is greater than the price paid. Please send comments and corrections to the author at rugh@jhu.edu or ECE Department, Johns Hop University, Baltimore, MD 21218 USA. Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month. Read Free Foron 30this Days Sign up to vote title   Not useful Cancel anytime. Useful     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1 of 106   Table of Contents for Linear Linear System Theory and  Search document  CHAPTER 1 Solution 1.1 (a) For k = 2, (A + B )2 = A 2 + AB + BA + B 2 . If AB = BA, the then (A + B )2 = A 2 + 2AB + B 2 . In ge k AB = BA, then the k-fold product ( A + B ) can be written as a sum of terms of the form A j B k − j , j = 0, . . . , k number of terms that can be written as A j B k − j is given by the binomial coefficient . Therefore AB j implies ( A + B )k = k Σ j =0 k j A j B k− j (b) Write det [λ I − A (t )] = λn + an −1 (t )λn −1 + . . . + a 1 (t )λ + a 0 (t ) where invertibility of A (t ) implies a 0 (t ) ≠ 0. The Cayley-Hamilton theorem implies A n (t ) + an −1 (t )A n −1 (t ) + . . . + a 0 (t )I = 0 Master your semester with Scribd − vote Free 30this Days a (t )I − . . . − a − (t )A −Read A (t ) −up (tFor ) on Sign to title _− ________________________________ A − (t ) = a (t )  Useful  Not useful & The New York Times for all t . Multiplying through by A −1 (t ) yields 1 1 n 1 0 n 2 n 1 Cancel anytime.   Special offer forfor students: Onlya$4.99/month. all t . Since 0 (t ) = det [ −A (t )], a 0 (t ) = det A (t ). Assume ε > 0 is such that  det A (t ) ≥ ε for all t A (t ) ≤ α we have aij (t ) ≤ α, and thus there exists a γ such that a j (t ) ≤ γ for all t. Then, for all t   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1 of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions __________ _ det (λ I − A ) = det (λ I − A ) = det (λ I − A ) _ H H (e) α A has eigenvalues α λ1 , . . . , αλn since Ap = λp implies ( α A )p = (αλ)p. (f) Eigenvalues of A T A are not nicely related to eigenvalues of A. Consider the example A= 0 α 0 0 , A TA = 0 0 0 α where the eigenvalues of A are both zero, and the eigenvalues of A T A are 0, α. (If A is symmetric, then applies.) Solution 1.3 (a) If the eigenvalues of A are all zero, then det ( λ I − A ) = λn and the Cayley-Hamilton theorem shows that nilpotent. On the other hand if one eigenvalue, say λ1 is nonzero, let p be a corresponding eigenvector. T A k p = λ k1 p ≠ 0 for all k ≥ 0, and A cannot be nilpotent. _ (b) Suppose Q is real and symmetric, and λ is an eigenvalue of Q. Then λ also _ _is_ an eigenvalue. From H H Qp = p p Qp = p p Qp = λ p, and eigenvalue/eigenvector equation we get . Also λ λ _ _ _ transposing H H H p Qp = λ p p. Subtracting the two results gives ( λ − λ)p p = 0. Since p ≠ 0, this gives λ = λ, that is, λ is real (c) If A is upper triangular, then λ I − A is upper triangular. Recursive Laplace expansion of the determinant a the first column gives det (λ I − A ) = (λ − a 11 ) . . . (λ − ann ) You're Reading a Preview which implies the eigenvalues of A are the diagonal entries a 11 , . . . , ann . Unlock full access with a free trial. Solution 1.4 Download With Free Trial (a) A= 0 0 1 0 implies A T A = (b) Master your semester with Scribd 3 1 A= implies 1 3 & The New York Times 1 0 0 0 implies A  = 1 10 6Free For 30 Days Read A T A = Sign up to vote on this title 6 10  Then Special offer for students: Only $4.99/month.  Not useful Cancel anytime. Useful det (λI − A T A ) = (λ − 16)(λ − 4)     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition. Khalil - Nonlinear Systems Slides     Save Embed Share Print 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E Solution 1.6  Solutions By definition of the spectral norm, for any α ≠ 0 we can write A  = max A x  = max x  = 1 = max α x  = 1 Since this holds for any x  = 1 A x  _______ x  αA x  A α x  __________ ________ = max x  = 1/α αx  α x  α ≠ 0, A x  A x  _______ _______ = max x≠0 x  x  ≠ 0 x  A  = max Therefore A  ≥ for any x A x  _______ x  ≠ 0, which gives A x  ≤ A x  Solution 1.7 By definition of the spectral norm, AB  = You're max Reading (AB )x  =a max A (Bx ) Preview x  = 1 x  = 1 max { by trial. Exercise 1.6 ≤ Unlock Aaccess Bxwith } a, free full x  = 1 = A  max Bx  = A B  x  = 1 Download With Free Trial If A is invertible, then A A −1 = I and the obvious I  = 1 give 1 = A A −1  ≤ A A −1  Therefore Master your semester with Scribd 1 _____ A −  ≥ A  & The New York Times 1 Special offer for students: Only $4.99/month. Read Free Foron 30this Days Sign up to vote title   Not useful Cancel anytime. Useful     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Upload Sign In Join RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music  This is the solution manual of Rugh W.J Linear system theory. 2 edition. Khalil - Nonlinear Systems Slides     Save Embed Share Print 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions A= 0 A 12 0 0 then partitioning the vector x similarly we see that max A x  = max A 12 x 2  = A 12  x  = 1 Solution 1.9 x 2  = 1 By the Cauchy-Schwarz inequality, and x T  = x , x T A x  ≤ x T A x  = A T x x  ≤ A T x 2 = A x 2 This immediately gives x T A x ≥ − A x 2 If λ is an eigenvalue of A and x is a corresponding unity-norm eigenvector, then λ = λx  = λ x  = A x  ≤ A x  = A  You're Reading a Preview Since Q = Q T , Q T Q = Q 2 , and the eigenvalues of Q 2 are λ 21 , . . . , Solution 1.10 λ 2n . Therefore access2 with a free trial. Q  = full Unlock (Q ) = max λi  √λmax 1≤i≤n For the other equality Cauchy-Schwarz gives Download With Free Trial x T Qx | ≤ x T Q x  = Qx x  ≤ Q x 2 = [ max λi  ] x T x Master your Scribd x . Choosing x Thereforesemester | x Qx | ≤ Q  forwith all unity-norm the eigenvalue that yields max λ  gives ≤ ≤ & The New York Times T 1 i Special offer for students: Only $4.99/month. Thus  T Qx  Q  n i 1≤i≤n a as a unity-norm eigenvector of Q correspondi  Read Free Foron 30this Days Sign up to vote title Useful  Not useful Cancel anytime. x Ta Qxa  = x Ta [ max λi  ] xa = max λi  1≤i≤n  1≤i≤n    Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E Solutions max x T A T A x = x  = 1 so we have A  = Solution 1.12  λmax (A T A ) √λmax (A T A ) . Since A T A > 0 we have λ i (A T A ) > 0, i = 1, . . . , n, and (A T A )−1 > 0. Then by Exercise 1.11 A −1 2 = λmax ((A T A )−1 ) = n 1 _________ λmin (A T A ) Π λ i (A T A ) = i =1 __________________ λmin (A T A ) . det (A T A ) = A 2(n −1) _________ (det A )2 [λmax (A T A )]n −1 ≤ _____________ 2 (det A ) Therefore A −1  ≤ A n −1 ________ det A  You're Reading a Preview Solution 1.13 Assume A ≠ 0, for the zero case is trivial. For any unity-norm x and y, Unlock full access with a free trial. y T A x  ≤ y T A x  ≤ y  A Free x  =Trial A  Download With Therefore max y T A x  ≤ A  Master your semester x be suchwith Now let unity-norm that A xScribd  = A , and let Ax _____ & The New York Times y = x , y  = 1 a a a Special offer for students: Only $4.99/month. Then ya  = 1 and a A  Read Free Foron 30this Days Sign up to vote title   Not useful Cancel anytime. Useful     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1 of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions example A (t ) = t 0 0 t2 , A (t ) = t, 0≤t ≤1 t2 , 1 < t < ∞ Clearly the time derivative of A (t ) is discontinuous at t = 1. (This overlaps Exercise 1.18 a bit.) Also the eigenvalues of continuously-differentiable A (t ) are not necessarily continuously differentiable, consi A (t ) = 0 1 −1 −t An easy computation gives the eigenvalues √t 2 − 4 t λ(t ) = __ ± _______ 2 2 Thus . t 1 ________ λ(t ) = __ ± 2 2 √t 2 − 4 and this function is not continuous at t = 2. Solution 1.15 You're Reading a Preview ≠ 0, 0 < λmin (Q ) x T x ≤ x T Q x ≤ λ max (Q ) x T x Clearly Q is positive definite, and by Rayleigh-Ritz if x Unlock full access with a free trial. λ min (Q ) (respectively, λ max (Q )) shows that these inequalitie Choosing x as an eigenvector corresponding to tight. Thus Download With Free Trial ε1 ≤ λmin (Q ) , λmax (Q ) ≤ ε2 Therefore 1 1 λmin (Q −1 ) = _______ ≥ ___ λmax (Q ) ε2 Read Free Foron 30this Days Sign up to vote title 1 1 ___ _______ 1 − ≤ ε Useful  Not useful λmax (Q ) = λmin (Q )  1 Cancel anytime. Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month. Thus Rayleigh-Ritz for the positive definite matrix Q −1 gives     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E Solution 1.17  Solutions Using the product rule to differentiate A (t ) A −1 (t ) = I yields . d ___ A −1 (t ) = 0 A (t ) A −1 (t ) + A (t ) dt which gives . d ___ A −1 (t ) = −A −1 (t ) A (t ) A −1 (t ) dt Solution 1.18 Assuming differentiability of both x (t ) and x (t ), and using the chain rule for s functions,  d d ___ ___ x (t ) x (t )2  = 2x (t ) dt dt = 2x (t ) d ___ x (t ) dt Also we can write, using the product rule and the Cauchy-Schwarz inequality,  .T . . d T d ___ ___ x (t ) x (t ) =  x (t ) x (t ) + x T (t ) x (t ) = 2x T (t ) x (t ) x (t )2  =  dt dt . a Preview ≤ 2x (tYou're )x (t )Reading  Unlock full access For t such that x (t ) ≠ 0, comparing these expressions gives with a free trial. d ___ . x (t ) ≤ x (t )  Download With Free Trial dt If x (t ) = 0 on a closed interval, then on that interval the result is trivial. If x (t ) = 0 at an isolated point, continuity arguments show that the result is valid. Note that for the differentiable function x (t ) = t, x (t ) is not differentiable at t = 0. Thus we must make the assumption that x (t ) is differentiable. (While inequality is not explicitly used in the book, the added differentiability hypothesis explains why  we T Read Free For 30 Days 2 Sign up to vote on this title differentiate x (t ) = x (t ) x (t ) instead of x (t ).)  Master your semester with Scribd & The New York Times  Useful  Not useful Cancel anytime. Special offer for students: Only $4.99/month. Solution 1.19 To prove the contrapositive claim, suppose for each i, j there is a constant βij such that   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download   of 106 Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E Solution 1.20  Solutions If λ(t ), p (t ) are a pointwise-in- t eigenvalue/eigenvector pair for A −1 (t ), then A −1 (t ) p (t ) = λ(t ) p (t ) = λ(t )p (t ) Therefore, for every t , λ(t ) = A −1 (t ) p (t ) _____________ p (t ) −1 A (t )p (t ) ≤α ≤ _______________ p (t ) Since this holds for any eigenvalue/eigenvector pair, det A (t ) = 1 1 _________________ ___________ = . 1 − λ1 (t ) . . λn (t ) det A (t ) 1 >0 ≥ ___ n α for all t . Solution 1.21 Using Exercise 1.10 and the assumptions Q (t ) ≥ 0, tb tb tb ta ta ∫ Q (σ) d σ = ∫ λ max [Q ( ≥ ta , tb tb ta ta σ)] d σ ≤ ∫ tr [Q (σ)] d σ = tr ∫ Q (σ) d σ Note that tb You're Reading ≥ 0Preview Q (σ ) d σ a ∫ ta Unlock full access with a free trial. since for every x tb tb ta ta Download With Free Trial x ∫ Q (σ ) d σ x = ∫ x T Q (σ ) x d σ ≥ 0 T Thus, using a property of the trace on page 8 of Chapter 1, we have Master your semester with d σvote ) d σ ≤ tr ∫ Q (σ) d σ ≤ n Read ) to Foron σFree ∫ Q (σScribd ∫ Q (up 30this Days Sign title & The New York Times  Useful  Not useful Finally, Special offer for students: Only $4.99/month. tb tb tb ta ta ta Cancel anytime. tb ∫     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload Join RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music  This is the solution manual of Rugh W.J Linear system theory. 2 edition. Khalil - Nonlinear Systems Slides     Save Embed Share Print 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document  CHAPTER 2 Solution 2.3 . The nominal solution for ũ (t ) = sin (3t ) is ỹ (t ) = sin t. Let x 1 (t ) = y (t ), x 2 (t ) = y (t ) to writ state equation . x (t ) = − x 2 (t ) 3 (4/ 3)x 1 (t ) (1/ 3)u (t ) − Computing the Jacobians and evaluating gives the linearized state equation You're Reading a Preview 0 0 1 x δ (t ) + u δ (t ) 2 −4 sin −1trial. /3 0 Unlock full taccess with a free . x δ (t ) = y δ (t ) = where x δ (t ) = x (t ) − sin t cos t , 1 0 x δ (t ) Download With Free Trial u δ (t ) = u (t ) − sin (3t ) , y δ (t ) = y (t ) − sin t , x δ (0) = x (0) − Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title & The New York Times Solution 2.5 For ũ = 0 constant nominal solutions are solutions  of Useful  Not useful Special offer for students: Only $4.99/month. Cancel anytime. 0 = x̃ 2 − 2x̃ 1 x̃ 2 = x̃ 2 (1−2x̃ 1 ) 0 1     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1 of 106   Table of Contents for Linear Linear System Theory and  Search document  rank A = rank [ A b ]. Also, x̃ is a constant nominal with c x̃ = 0 if and only if 0 = A x̃ + bũ 0 = c x̃ that is, if and only if −bũ A x̃ = c 0 As above, this holds if and only if A c rank = rank A b c 0 Finally, x̃ is a constant nominal with c x̃ = ũ if and only if 0 = A x̃ + bũ = ( A + bc ) x̃ and this holds if and only if x̃ ∈ Ker [ A + bc ] (If A is invertible, we can be more explicit. For any ũ the unique constant nominal is x̃ = −A −1 bũ. Then ỹ Reading −1.) ũ ≠ 0 if and only if c A −1 b = 0, and ỹ = ũ ifYou're and only if c A −1 ba=Preview Unlock full access with a free trial. Solution 2.8 (a) Since Download With Free Trial A B C 0 is invertible, for any K Master your semester with Scribd A + BK B = 0 C & The New York is invertible. Let Times Special offer for students: Only $4.99/month. A + BK B A B C 0 R1 R2 I 0 Read Free Foron 30this Days Sign to vote title K I up  =  Not useful Cancel anytime. Useful I 0     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E Solution 2.10  Solutions For u (t ) = ũ, x̃ is a constant nominal if and only if 0 = (A + Dũ ) x̃ + bũ This holds if and only if bũ ∈ Im [ A + Dũ ], that is, if and only if rank ( A + Dũ ) = rank A+Dũ bũ If A +Dũ is invertible, then x̃ = −(A + Dũ )−1 bũ If A is invertible, then by continuity of the determinant det (A + Dũ ) ≠ 0 for all ũ such that ũ  is suffici small, and (+) defines a corresponding constant nominal. The corresponding linearized state equation is . x δ (t ) = (A + Dũ ) x δ (t ) + [ b − D (A + Dũ )−1 bũ ] u δ (t ) y δ (t ) = C x δ (t ) Solution 2.12 For the given nominal input, nominal output, and nominal initial state, the nominal solu satisfies 1 . 0 You're Reading x̃ (t ) = x̃ 1 (t ) − x̃ 3 (t ) a, Preview x̃ (0) = −3 −2 x̃ (t ) − 2 x̃ (t ) 2 3 Unlock full access with a free trial. 1 = x̃ 2 (t ) − 2 x̃ 3 (t ) Integrating for x̃ 1 (t ) and then x̃ 3 (t ) easilyDownload gives the nominal solution With Free Trialx̃ 1 (t ) = t, x̃ 2 (t ) = 2 t − 3, and x̃ 3 (t ) The corresponding linearized state equation is specified by A= 0 0 0 1 0 −1 0 1 −2 , B (t )= 0 t 0 , C= 0 1 −2 Master your semester with Scribd  Free Foron 30this Days It is unusual that the nominal input and nominal output are constants,Read but up the linearization istitle time varying. Sign to vote  & The New York Times  Useful  Not useful Special offer forSolution students: 2.14 Only $4.99/month. Compute Cancel anytime.   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition. Khalil - Nonlinear Systems Slides     Save Embed Share Print 1 Download   of 106 Table of Contents for Linear Linear System Theory and  Search document  CHAPTER 3 Solution 3.2 ∂ ___ ∂τ Differentiating term k +1 of the Peano-Baker series using Leibniz rule gives σ1 t σ2 σk ∫τ A (σ ) ∫τ A (σ ) ∫τ 1 ... 2 σ2 t ∫τ A (σ k +1 ) d σk +1 . . . d σ1 σk τ τ d _ . . . dσ a__ Preview t − A (τ) A (σ2 ) . . . d σk +1 . . . d σ2 A (t ) A (σ2 ) . . . A (σk +1You're ) d σk +1 Reading 2 dτ τ τ τ τ τ ∫ = ∫ ∫ ∫ ∫ Unlock full access with a free trial. t ∫τ + A (σ 1 ) t ∂ ___ ∫τ A (σ ) ∫τ ∂τ ∂ ___ σ2 σ1 2 d σk +1 . . . d σ1 ∫τ A (σk +1 ) With Download Free Trial ... σ2 σ1 ∂ τ ∫τ semester ∫τ = A (σ1 ) σk A (σ 2 ) ∫τ σk ... ∫τ A (σ k +1 ) Master your with Scribd Repeating this process k times gives & The New York Times σ σ t Special offer for students: Only $4.99/month. ∂ ___ ∂τ 1 1 σk 2 ∫ A (σ ) ∫ A (σ ) ∫ 2 d σk +1 . . . d σ1 ... ∫ A (σ k +1 ) Read Free Foron 30this Days Sign up to vote title   Not useful Cancel anytime. Useful d σk +1 . . . d σ1   _ d   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Upload Sign In RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition. Khalil - Nonlinear Systems Slides     Save Embed Share Print 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E Solution 3.6  Solutions Writing the state equation as a pair of scalar equations, the first one is . −t x (t ) ______ x 1 (t ) = 1 1 + t2 and an easy computation gives x 1 (t ) = x 1o _________ (1 + t 2 )1/ 2 Then the second scalar equation then becomes x 1o . −4t x (t ) + _________ ______ x 2 (t ) = 2 (1 + t 2 )1/ 2 1 + t2 The complete solution formula gives, with some help from Mathematica, . 1 ________ x 2o + x 2 (t ) = (1 + t 2 )2 t ∫ 0 (1 + σ2 )3/ 2 _________ d σ x 1o (1 + t 2 )2 1+t 2 (t 3 /4+5t/ 8)+(3/ 8) sinh−1 (t ) 1 _√ ____________________________ ________ x 1o x + = 2o (1 + t 2 )2 (1 + t 2 )2 If x 1o = 1, then as t →∞, x 2 (t ) → 1/ 4, not zero. You're Reading a Preview Solution 3.7 From the hint, letting Unlock full access with a free trial. t r (t ) = v (σ)φ(σ) d σ Downloadt∫o With Free Trial . we have r (t ) = v (t )φ(t ), and φ(t ) ≤ ψ (t ) + r (t ) Master your semester Scribd v (t ) gives Multiplying (*) through by thewith nonnegative Read Free Foron 30this Days Sign up to vote title v (t )φ(t ) ≤ v (t )ψ(t ) + v (t )r (t Useful & The New York Times )  Not useful Cancel anytime. Special offer foror students: Only $4.99/month. . () ( ) ( )≤ ( ) ( )     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions t ∫ v (τ) d τ t eo gives t ∫ v (τ) d τ t r (t ) ≤ ∫ v (σ)ψ(σ)e σ dσ to and using (*) yields the desired inequality. Solution 3.10 Multiply the state equation by 2 z T (t ) to obtain . d ___ z (t )2 2 z T (t ) z (t ) = dt n n Σ Σ 2 z (t )a (t ) z (t ) = i ij j i =1 j =1 n ≤Σ n Σ 2a (t )z (t )z (t ) , ij i j t ≥ to i =1 j =1 At each t ≥ to let You're Reading a Preview a (t ) = 2n 2 max aij (t ) 1 ≤ i, j ≤ n Unlock full access with a free trial. Note a (t ) is a continuous function of t, as a quick sample sketch indicates. Then, since zi (t ) ≤ z (t ), d Download ___ t )2 ,Trial t ≤ to z (t )2 ≤ aWith (t )z (Free dt Multiplying through by the positive quantity Master your semester with Scribde gives York Times & The New t − ∫ a (σ) d σ to t Special offer for students: Only $4.99/month. d ___ dt − ∫ a (σ) d σ e to z (t )2 Read Free Foron 30this Days Sign up to vote title  ≤0,  Not useful Cancel anytime. Useful t ≤ to     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download   of 106 Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions z (t ) = t t t to to ∫ A (σ) z (σ) d σ + ∫ σ∫ E (τ, σ) d τ z(σ) d σ t = t ∫ ∫σ A (σ) + E (τ, σ ) d τ to ∆ = z (σ) d σ t ∫ Â(t, σ) z (σ) d σ to Thus t t ∫ ∫ z (t ) =  Â (t, σ) z (σ) d σ ≤ Â (t, σ)z (σ) d σ to to By continuity, given T > 0 there exists a finite constant α such that Â (t, σ) ≤ α for to ≤ σ ≤ t ≤ to + T. Thus t ∫ z (t ) ≤ α z (t ) d σ , t ∈ [to , t o +T ] to and the Gronwall-Bellman inequality gives z (t ) = 0 for t than one solution. ∈ [to , to +T ], implying that there can be no m You're Reading a Preview Solution 3.13 From the Peano-Baker series, Unlock full access with a free trial. σ1 σk −1 t t . . . . . . . . . dσ Φ(t, τ) − I + A (σ1 ) dDownload + With A (σ1Free ) σ1 + 1 Trial A (σk ) d σk ∫τ σ1 t ∞ = ∫τ Σ ∫τ A (σ ) ∫τ ∫τ ∫τ σ j −1 ... ∫τ A (σ j ) d σ j . . . d σ1 Master your semester with Scribd  Read Free Foron 30this Days Sign up to vote title  For any fixed T > 0 there is a finite constant α such that A (t ) ≤ α for t ∈ [−T, T ], by continuity. Therefore & The New York Times Useful  Not useful  σ σ 1 j =k +1 ∞ t 1 Special offer for students: ... A (σ )  Only $4.99/month. Σ ∫τ j =k +1 1 ∫τ σj−1 ∫τ A (σ j ) d σ j . . . d σ1  ≤ ∞ t 1 Σ ∫τ A (σ ) ∫τ 1 j =k +1 Cancelσanytime. j−1 ... ∫τ A (σ j ) d σ j . . . d σ1    Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload Join RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download   of 106 Table of Contents for Linear Linear System Theory and   Search document Linear System Theory, 2/E Solutions ∞ α2T ) j _(_____ <ε j! Σ j =K +1 Using the hint, ∞ Σ j =k +1 ∞ (α2T )k +1+i α2T ) j __________ _(_____ = j! i =0 (k +1+i )! Σ ∞ (α2T )k +1 (α2T )i . ______ ≤ Σ ________ i i =0 (k +1)! k If k > α2T, then ∞ Σ j =k +1 α2T ) j _(_____ j! (α2T ) . ≤ ________ k +1 (k +1)! Because of the factorial in the denominator, given Solution 3.15 (α2T )k +1 1 __________________ _________ = (k −1)!(k +1)(k −α2T ) 1 − α2T/k ε > 0 there exists a K > α2T such that (*) holds. Writing the complete solution of the state equation at t f , we need to satisfy tf Ho x o + H f Φ(t f , t o ) xo + ∫ Φ(t f , σ ) f ( σ) d σ =h to Thus there exists a solution that satisfies the boundary conditions if and only if tf You're Reading a Preview to Unlock full access with a free trial. h − H f ∫ Φ(t f , σ ) f ( σ) d σ ∈ Im[ Ho + H f Φ (t f , t o ) ] There exists a unique solution that satisfies the boundary conditions if Ho + H f a solution x (t ) satisfying the boundary conditions: Download With Free Trial (1) Compute Φ(t, to ) for t (2) Compute Ho + H f Φ(t f , to ) is invertible. To com ∈ [to , t f ] Φ(t f , to ) Master your semester with Scribd (3) Compute ∫ Φ(t , σ) f ( σ) d σ & The New York Times (4) Solve (+) for x tf f to o Special offer for students: Only $4.99/month. t ∫ Read Free Foron 30this Days Sign up to vote title   Not useful Cancel anytime. Useful     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition. Khalil - Nonlinear Systems Slides     Save Embed Share Print Download 1 of 106   Table of Contents for Linear Linear System Theory and  Search document  CHAPTER 4 Solution 4.1 . An easy way to compute A (t ) is to use A (t ) = Φ(t, 0)Φ(0, t ). This gives A (t ) = This A (t ) commutes with its integral, so we can write − 2t − 1 1 − 2t Φ(t, τ) as the matrix exponential t −(t −τ) A (σ) d σ = exp ∫ (t −τ) You're Reading a Preview τ Φ(t, τ) = exp 2 −(t −τ) −(t −τ)2 Unlock full access with a free trial. Solution 4.4 A linear state equation corresponding to the n th -order differential equation is Download With Free Trial . x (t ) = ... ... . . . ... 0 0 −a 0 (t ) −a 1 (t ) . . . 0 0 . . . Master your semester with Scribd & The New York Times The corresponding adjoint state equation is Special offer for students: Only $4.99/month. 0 1 0 . . . ... 0 0 0 . . . 1 x (t ) −anRead Free Foron 30this Days Sign to vote title −1 (t ) up  Useful  Not useful Cancel anytime. a 0 (t )     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions . zn −2 (t ) = −zn −3 (t ) + an −3 (t ) zn (t ) gives . d2 d d3 ____ ___ ____ [ an −1 (t ) zn (t ) ] a t z t + [ ( ) ( ) ] − z t = z t ( ) ( ) n −2 n n n −2 dt dt 2 dt 3 = −zn −3 (t ) + an −3 (t ) zn (t ) − d2 d ____ ___ [ an −2 (t ) zn (t ) ] + 2 [ an −1 (t ) zn (t ) ] dt dt Continuing gives the n th -order differential equation d n −2 d n −1 dn _____ _____ ____ [ an −1 (t ) zn (t ) ] − n −2 [ an −2 (t ) zn (t ) ] zn (t ) = dt dt n −1 dt n d ___ [ a 1 (t ) zn (t ) ] + (−1)n +1 a 0 (t ) zn (t ) + . . . + (−1)n dt Solution 4.6 For the first matrix differential equation, write the transpose of the equation as (transpose differentiation commute) .T X (t ) = A T (t )X T (t ) , X T (to ) = XTo This has the unique solution X T (t ) = ΦA T (t )You're (t, to )X ToReading , so that a Preview T X (full t ) =access Xo Φ TAwith t o ) trial. (t ) (ta, free Unlock In the second matrix differential equation, let to verify (Leibniz rule) that a solution is Φk (t, τ) be the transition matrix for Ak (t ), k = 1, 2. Then it is Download With Free Trial t X (t ) = Φ1 (t, t o )Xo Φ T 2 (t, to) + ∫ Φ (t, σ)F (σ)Φ (t, σ) d σ 1 T 2 to Or, one can generate this expression by using the obvious integrating factors on the left and right sides of Master your semester with Scribd  Zon t )this differential. equation. (To show this is a unique solution, show that Read the (30 between Free For Days Signdifference up to vote title any two solu  satisfies Z (t ) = A (t )Z (t ) + Z (t )A (t ), with Z (t ) = 0. Integrate both sides and apply the Bellman-Gr & The New York Times Useful  Not useful  inequality to show Z (t ) is identically zero.) 1 Special offer for students: Only $4.99/month. T 2 o Cancel anytime.   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1   of 106 Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions Replacing τ as noted above gives Φ (t, = cos τ 0 0 cos τ = −sin τ −sin τ sin τ 0 cos τ sin τ cos τ 0). Solution 4.10 For sufficiency, suppose differentiable. Let z (t ) = T −1 (t ) x (t ) so that Φz (t, 0 + Φ x (t, 0) = T (t )e Rt . Then T (0) = I and T (t ) is continu 0) = T −1 (t )Φx (t, 0)T (0) = T −1 (t )T (t )e Rt = e Rt . Thus z (t ) = R z (t ). For necessity, suppose P (t ) is a variable change that gives . z (t ) = Ra z (t ) Then Φ z (t, 0) = e Ra t = P −1 (t )Φx (t, 0)P (0) that is, Φx (t, 0) = P (t )e Ra t P −1 (0) You're Reading a Preview Let T (t ) = P (t )P −1 (0) and R = P (0)Ra P −1 (0). Then Φ x (t, Unlock full access−1with free −trial. (0) RPa(0)t P 1 (0) 0) = T (t )P (0) e P = T (t )P (0) [ P −1 (0)e Rt P (0) ] P −1 (0) Download With Free Trial = T (t )e Rt Solution semester 4.11 Suppose Master your with Scribd Φ(t, 0) = e & The New York Times A1t e A2t  Then Special offer for students: Only $4.99/month. . Φ(t 0) d ___ e A1t e A2t =e Read Free Foron 30this Days Sign up to vote title A1t  Not useful Cancel anytime. Useful ( A +A ) e A2t     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download   of 106 Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E Solutions d ___ dt we have that Φ (t, Solution 4.13 0) = e  A1t A2t e e A1t e A2t = A (t )e A1t A2t e , e A10 e A20 =I . Writing ∂ ___ ∂t ΦA (t, τ) = A (t )ΦA (t, τ ) , Φ(τ, τ) = I in partitioned form shows that ∂ ___ ∂t Φ21 (t, τ) = A 22 (t )Φ21 (t, τ ) , Φ21 (τ, τ) = 0 Thus Φ21 (t, τ) is identically zero. But then ∂ ___ ∂t Φii (t, τ) = Aii (t )Φii (t, τ ) , Φii (τ, τ) = I for i = 1, 2, and ∂ ___ ∂t Φ12 (t, τ) = A 11 (t )Φ12 (t, τ ) + A 12 (t )Φ22 (t, τ) , Φ12 (τ, τ) = 0 Using Exercise 4.6 with F (t ) = A 12 (t ) Φ22 You're (t, τ) gives Reading a Preview t ∫τ with(σ a)free (t, σ ) A (σ, τ) d σ τ) = full Φ12 (t,Unlock Φ11access Φ22trial. 12 Download With Free Trial Solution 4.17 We need to compute a continuously-differentiable, invertible P (t ) such that t 1 1 t . 0 1 P (t ) − P −1 (t )P (t ) 2 2− t 2 t Master your semester with Scribd  Read Free Foron 30this Days Sign up to vote title  Multiplying on the left by P (t ), the result can be written as a dimension-4 linear state equation. Choos & The New York Times Useful  Not useful  initial condition corresponding to P (0) = I, some clever guessing gives = P −1 (t ) Cancel anytime. Special offer for students: Only $4.99/month. P (t ) = 1 0 t 1   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload Join RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music  This is the solution manual of Rugh W.J Linear system theory. 2 edition. Khalil - Nonlinear Systems Slides     Save Embed Share Print 1 Download   of 106 Table of Contents for Linear Linear System Theory and Linear System Theory, 2/E Solutions ∂ ___ ∂t Since Φ(−τ,   Search document Φ TA (−τ, −t ) = A T (−t )Φ TA (−τ, −t ) −τ) = I, we have F (t ) = A T (−t ). Or we can use the result of Exercise 3.2 to compute: ∂ ___ ∂t ∂ Φ (−τ, −t ) = Φ (−τ, −t )A (−t ) ΦA (−τ, −t ) =− _____ A ∂ (−t ) A This implies ∂ ___ ∂t Since Φ(−τ, Φ TA (−τ, −t ) = A T (−t )ΦA (−τ, −t ) −τ) = I, we have F (t ) = A T (−t ). Solution 4.25 We can write t+ σ Φ(t + σ, σ ) = I + ∫σ A (τ) d τ + ∞ τ1 t +σ Σ ∫σ τk −1 ∫σ ∫σ A (τ1 ) A (τ2 ) . . . k =2 A (τk ) d τk . . . d τ1 and _k _ ∞a Preview You're Reading 1 ___ At (σ)t k e = I + At (σ)t + _ At (σ)t Σ ! k =2 akfree Unlock full access with trial. Then _ R (t, σ) = Φ(t + σ, σ)Download − e At (σ)t  With Free Trial ∞ =  t +σ Σ ∫σ k =2 τ1 τk −1 ∫σ A (τ1 ) A (τ2 ) . . . _k 1 ___ At (σ)t k  A (τk ) d τk . . . d τ1 − k! σ ∫ Master your with Scribd t ) ≤ α and the triangle From A (semester inequality, Read Free Foron 30this Days Sign up to vote title ∞ 2 ∞ ___ − _t__ − α t  Not useful & The New York Times R (t, σ) ≤ 2 Σ α k ! = α t ΣkUseful ! k Special offer for students: Only $4.99/month. Using k =2 k k 2 2 2 k =2 k 2 Cancel anytime.     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition. Khalil - Nonlinear Systems Slides     Save Embed Share Print Download 1 of 106   Table of Contents for Linear Linear System Theory and  Search document  CHAPTER 5 Solution 5.3 Using the series definition, which involves talent in series recognition, A 2k +1 = 0 1 1 0 , A 2k = 1 0 0 1 , k = 0, 1, . . . gives e At =I+ = Using the Laplace transform method, 0 t3 t2 0 0 t 1 1 ___ ___ + + t3 0 0 t 2 a Preview t You're 0 3! 2! Reading + ... −t )/ 2 (e t −e −t )/ 2 (e t +e cosh t sinh t Unlock full access with=a free trial. t t t t − − (e −e )/ 2 (e +e )/ 2 sinh t cosh t Download With Free Trial (sI − A )−1 = s −1 −1 = −1 s 1 _____ 2 s −1 s _____ s 2 −1 s _____ 2 s −1 1 _____ 2 s −1 Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title which gives again & The New York Times  Useful  Not useful cosh t sinh t Special offer for students: Only $4.99/month. e At = Cancel anytime. sinh t cosh t Using the diagonalization method, computing eigenvectors for A and letting     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition. Khalil - Nonlinear Systems Slides     Save Embed Share Print 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions t Φ (t, ∫ A (σ) d σ 0) = e 0 t2/2 t t t2/2 = exp And since t2 / 2 0 0 t2/2 , 0 t t 0 t2/2 . exp commute, Φ(t, 1 0 0 1 0) = exp 0 1 1 0 t Using Exercise 5.3 gives Φ(t, Solution 5.7 2 0) = cosh t sinh t sinh t cosh t e t /2 0 2 0 e t /2 2 = 2 e t / 2 cosh t e t /2 sinh t 2 2 e t / 2 sinh t e t / 2 cosh t To verify that t You're A Reading e A σ d σ = eaAtPreview −I ∫ 0 Unlock full access with a free trial. note that the two sides agree at t = 0, and the derivatives of the two sides with respect to t are identical. If A is invertible and all its eigenvalues have negative real parts, then lim t → ∞ e At = 0. This gives Download ∞ With Free Trial A e Aσ d σ = − I ∫ 0 that is, Master your semester with Scribd ∞ − A =−∫e & The New York Times 1 0 Special offer for students: Only $4.99/month. 0 Aσ dσ = ∫ ∞ Free Foron 30this Days Sign to vote title e A σRead d σ up   Not useful Cancel anytime. Useful     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload Join RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1   of 106 Table of Contents for Linear Linear System Theory and   Search document Linear System Theory, 2/E Solutions t . x (t ) = bu (t ) + A t ∫e A (t −σ) D e ∫σ u (τ) d τ t t ∫ bu (σ) d σ + Du (t ) e A (t −σ) e 0 D ∫σ u (τ) d τ bu (σ) d σ 0 = A x (t ) + Dx (t )u (t ) + bu (t ) Solution 5.12 We will show how to define β0 (t ), . . . , βn −1 (t ) such that n −1 . n −1 n −1 βk (t )Pk = βk (t )APk , βk (0)Pk = I Σ Σ k =0 Σ k =0 k =0 which then gives the desired expression by Property 5.1. From the definitions, P 1 = AP 0 − λ1 I , P 2 = AP 1 − λ2 P 1 , . . . , Pn −1 = APn −2 − λn −1 Pn −2 Also Pn = (A −λn I )Pn −1 = 0 by the Cayley-Hamilton theorem, so APn −1 = λn Pn −1 . Now we equate coefficien like Pk ’s in (*), rewritten as n −1 . n −1 βk (t )Pk = βk (t )[Pk+1 + λk +1 Pk ] Σ k =0 to get equations for the desired βk (t )’s: Σ k =0 You're Reading a Preview . λ1 β0 (with P 0Unlock t ) =access t ) a free trial. : β0 (full . P 1 : β1 (t ) = β0 (t ) + λ2 β1 (t ) . Download With Free Trial . . . Pn −1 : βn −1 (t ) = βn −2 (t ) + λn βn −1 (t ) that is, Master your semester with Scribd . λ 0 β. (t ) & The New York Times β .(t ) = 1.. λ.. 1 0 2 1 Special offer for students: Only $4.99/month. . . . . . 0 0 ... ... . . . ... 0 0 . . . λ 1 0Read Foron 30this Days Sign up to vote title βFree 0 (t ) 0 Useful β1 (t )  Not useful  . Cancel anytime. . . . . . 0     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1   of 106 Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E Solutions Φ(t, t o ) = P −1 (t ) e P (t )SP o = Q (t, t o )e Solution 5.19  −1 (t ) (t −t ) o o P (to ) = P −1 (t )P (to ) e S (t −to ) P −1 (to )P (to ) S (t −to ) From the Floquet decomposition and Property 4.9, T det Φ(T, 0) = det e RT ∫ tr [A (σ)] d σ = e0 Because the integral in the exponent is positive, the product of eigenvalues of Φ(T, 0) is greater than unity, w implies that at least one eigenvalue of Φ (T, 0) has magnitude greater than unity.Thus by the argument follo Example 5.12 there exist unbounded solutions. Solution 5.20 Following the hint, define a real matrix S by e S 2T = Φ2 (T, 0) and set Q (t ) = Φ(t, 0)e −St Clearly Q (t ) is real and continuous, and You're Reading a Preview +2T ) Q (t +2T ) = Φ(t +2T,Unlock = Φ(with t +2aT,free T )Φ 0)e −S (tfull (T, 0)e −S 2T e −St access trial. = Φ(t +T, 0)Φ(T, 0)e −S 2T e −St = Φ(t +T, T )Φ2 (T, 0)e −S 2T e −St = Φ(t +T, TDownload )e −St = Φ(t, With 0)e −StFree Trial = Q (t ) That is, Q (t ) is 2T-periodic. (For a proof of the hint, see Chapter 8 of D.L. Lukes, Differential Equat Classical to Controlled , Academic Press, 1982.) Master your semester with Scribd  Read Free Foron 30this Days Sign up to vote title  & The New York Times  Useful  Not useful Solution 5.22 The solution will be T -periodic for initial state x if and only if x satisfies (see text equ Special offer for(32)) students: Only $4.99/month. o Cancel anytime. o   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download   of 106 Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions to +T 0= ∫ to +T z To Φ(to , σ ) f (σ) d σ = to ∫ z T (σ) f ( σ) d σ to completes the proof. Solution 5.24 Note A = −A T , and from Example 5.9, e At = cos t sin t −sin t cos t Therefore all solutions of the adjoint equation are periodic, with period of the form k 2π, where k is a pos integer. The forcing term has period T = 2π /ω, where we assume ω > 0. The rest of the analysis breaks d into 3 cases. Case 1: If ω ≠ 1, 1 / 2, 1 / 3, . . . then the adjoint equation has no T-periodic solution, so the condition (Exe 5.22) T ∫z T (σ) f ( σ) d σ = 0 0 holds vacuously. Thus there will exist corresponding periodic solutions. Case 2: If ω = 1, then T You're Reading a Preview T Aσ z f d = z ( ) ( ) σ σ σ ∫ ∫ o e f ( σ) d σ T T 0 full access with a free trial. Unlock 0 T T = −zo 1 ∫ sin (σ) d σ + zo 2 ∫ cos σ sin σ d σ Download With Free Trial 0 0 2 ≠0 Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title & The New York Times ∫ cos σ sin (σ/k ) d σ = ∫ sin σ sin  (σ/kUseful ) d σ = 0  Not useful so there is no periodic solution. Case 3: If ω = 1/k, k = 2, 3, . . . , then since Special offer for students: Only $4.99/month. T T 0 0 the condition (+) will hold, and there exist periodic solutions. Cancel anytime.     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download   of 106 Table of Contents for Linear Linear System Theory and  Search document  CHAPTER 6 Solution 6.1 If the state equation is uniformly stable, then there exists a positive the corresponding solution satisfies γ such that for any t o x (t ) ≤ γxo  , t ≥ to Given a positive ε, take δ = ε / γ. Then, regardless of to , xo  ≤ δ implies x (t ) ≤ γ δ = ε , t ≥ to You're Reading a Preview Conversely, given a positive ε suppose positive δ is such that, regardless of to , xo  ≤ δ implies x ( t ≥ to . For any ta ≥ to let xa be such that Unlock full access with a free trial. xa  = 1 , Φ(ta , to )xa  = Φ(ta , to ) With FreeatTrial Then xo = δ xa satisfies xo  = δ, and the Download corresponding solution t = ta satisfies x (ta ) = Φ(ta , to )xo  = δΦ(ta , to ) ≤ ε Therefore Master your semester with Scribd Φ(t , t ) ≤ ε / δ Read Free Foron 30this Days Sign up to vote title x can beTimes Such anYork selected for any t , t such that t ≥ t . Therefore Useful  Not useful & The New a a a o Special offer for students: Only $4.99/month. for all t and t with t ≥t , and we can take a o o Φ(t, t o ) ≤ ε / δ / δ to obtain Cancel anytime.     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition. Khalil - Nonlinear Systems Slides     Save Embed Share Print Download 1 of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions t σ1 t ∫τ ∫τ Φ(t, τ) = I + A (σ) d σ + A (σ1 ) t ∫τ A (σ ) d σ d σ 2 ∫τ 2 2 1 σ1 t ∫τ + . . .  ∫τ A (σ ) d σ d σ  + . . . = I  +  A (σ) d σ +  A (σ1 ) t 1 σ1 t ∫τ 2 ∫τ = 1 +  A (σ) d σ +  A (σ1 )  ∫τ A (σ ) d σ d σ  + . . . 2 2 1 (Be careful of t < τ.) Since A (t ) ≤ α for all t , t σ1 t ∫τ ∫τ ∫τ 1 d σ d σ  + . . . Φ(t, τ) ≤ 1 + α 1 d σ + α   2 = 1 + αt −τ + α2 2 1 t −τ2 _|_____ + ... 2! For | t −τ ≤ δ, Φ(t, τ) ≤ 1 + α δ + 2 2 δ _α ____ + ... 2! αδ =e You're Reading a Preview Unlock full access with a free trial. Solution 6.8 See the proof of Theorem 15.2. Download With Free Trial Solution 6.10 Write Re [λ] = −η, where η > 0 by assumption, so that t e λt  = t e −ηt , t ≥0 A simple maximization argument (settingScribd the derivative to zero) gives Master your semester with Read Free Foron 30this Days Sign up to vote title 1 ∆ − ≤ ___ β ≥ = t , 0 t e & The New York Times  Useful  Not useful ηe ηt Special offer forso students: Only $4.99/month. that Cancel anytime.     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload Join RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download   of 106 Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions ∞ ∫ 0 ∞ j +( j −1)+ . . . +1 j _2___________ e −(η / 2 )t dt t e  dt ≤ j (η e ) 0 . . . j +( j −1)+ +1 2j . ___ ≤ _2___________ η (η e ) j ... 22 j +( j −1)+ +1 _____________ = j e Re [λ] j +1 j ∫ λt Solution 6.12 By Theorem 6.4 uniform stability is equivalent to existence of a finite constant γ such e At  ≤ γ for all t ≥ 0. Writing e At = m σk ΣΣ Wkj k =1 j =1 where λ 1 , . . . , t j −1 λt ______ e k ( j −1)! λm are the distinct eigenvalues of A, suppose Re[λk ] ≤ 0 , k = 1, . . . , m Re[λk ] = 0 implies σ k = 1 λt λt Since t j −1 e k  is bounded if Re[ λk ] < 0 (for any j), and e k  = 1 if Re [λk ] = 0, it is clear that  You're Reading a Preview bounded for t ≥ 0. Thus (*) is a sufficient condition for uniform stability. A necessary condition for uniform stability is Unlock full access with a free trial. Re[λk ] ≤ 0 , k = 1, . . . , m For if Re[λk ] > 0 for some k, the proof of Theorem 6.2 shows that Trial e At  grows without bound as t Download With Free between this necessary condition and the sufficient condition is illustrated by the two cases A= 0 0 0 0 , A= → ∞. The 0 1 0 0 Both satisfy the necessary condition, neither satisfy the sufficient condition, and the first case is uniformly st Master your semester with Scribd  while the second case is not (unbounded solutions exist, as shown by easy computation oftitle the transition mat Read Free Foron 30this Days Sign up to vote  (It can be shown that a necessary and sufficient condition for uniform stability is that each eigenvalue of A & The New York Times  Useful  Not useful nonpositive real part and any eigenvalue of A with zero real part has algebraic multiplicity equal to its geom Special offer formultiplicity.) students: Only $4.99/month. Cancel anytime.   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions x (ta ) = Φ(ta , to )xa  = Φ(ta , t o ) ≤ γ e −λ(ta −to ) Since such an xa can be selected for any to and ta > to , we have Φ(t, τ) ≤ γ e −λ(t −τ) for all t , τ such that t ≥ τ, and the proof is complete. . The variable change z (t ) = P −1 (t ) x (t ) yields z (t ) = 0 if and only if . P −1 (t ) A (t )P (t ) − P −1 (t )P (t ) = 0 . for all t . This clearly is equivalent to P (t ) = A (t )P (t ), which is equivalent to Φ A (t, τ) = P (t )P −1 (τ). Now is a Lyapunov transformation, that is P (t ) ≤ ρ < ∞ and det P (t ) ≥ η > 0 for all t , then P (τ)n −1 __________ ΦA (t, τ) ≤ P (t )P −1 (τ) ≤ P (t ) det P (τ) Solution 6.18 ≤ ρn /η =∆ γ for all t and τ. Conversely, suppose ΦA (t, τ) ≤ γ for all t and τ. Let P (t ) = ΦA (t, 0). Then P (t ) ≤ γ and P −1 (t )n −1 ___________ 1 a −Preview = P (t )n −1 det P (t ) P (t ) ≤ You're−Reading 1 det P (t ) Unlock full access with a free trial. 1 for all t . Using P (t ) ≥ 1/P − (t ) gives 1 __________ det P (t ) With ≥ Free Download n P −1 (t )Trial and since P −1 (t ) = ΦA (0, t ) ≤ γ, det P (t ) ≥ 1 ___ γ Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title Thus P (t ) is a Lyapunov transformation, and clearly . & The New York Times Useful  Not useful P − (t ) A (t )P (t ) − P − (t )P ( t) = 0 1 Special offer for students: Only $4.99/month. for all t . n 1 Cancel anytime.     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition. Khalil - Nonlinear Systems Slides     Save Embed Share Print 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document  CHAPTER 7 Solution 7.3 Let Â = FA, and take Q = F −1 , which is positive definite since F is positive definite. Then sin is symmetric, T Â Q +QÂ = A T FF −1 + F −1 FA = A T + A < 0 This gives exponential stability by Theorem 7.4. You're a Preview a (t )Reading By our default assumptions, is continuous. Since Q is constant, symmetric, and pos definite, the first condition of Theorem 7.2 holds. Checking the second condition, Solution 7.5 Unlock full access with a free trial. −a (t ) −a (t )/ 2 A (t )Q +QA (t ) = −a (t ) / 2 −1 T ≤0 Download With Free Trial gives the requirements a (t ) ≥ 0 , 4a (t ) ≥ a 2 (t ) Thus the state equation is uniformly stable if a (t ) is a continuous function satisfying 0 Master your semester with Scribd Solution 7.6 With & The New York Times a (t ) 0 Q (t ) = Special offer for students: Only $4.99/month. 0 1 ≤ a (t ) ≤ 4 for all t. Read Free Foron 30this Days Sign up to vote title . Not useful  .Useful Cancelaanytime. (t ) 0 , A T (t )Q (t ) + Q (t )A (t ) + Q (t ) = 0 −4     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload Join RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download   of 106 Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions η ≤ a (t ) ≤ 1/ (2η) for all t . Then 2a ( t ) + 1 − η ≥ η + 1 > 1 (t )+1 _a ______ a (t ) and Q (t )−ηI 1 = 1+η > 1 − η ≥ 1 + ______ ≥ 0, for all t, follows easily. 1/ (2η) Similarly, with ρ = (2η+1)/ η we can show ρI −Q (t ) ≥ 0 using 1 2η+1 ρ − 2a (t ) − 1 ≥ _____ − 2 ___ − 1 = 1 η 2η 1 2η+1 a (t )+1 ____ ≥1 ρ − _______ ≥ _____ − 1 − η a (t ) a (t ) Next consider . A T (t )Q (t ) + Q (t ) A (t ) + Q (t ) = . 2a (t )−2a (t ) 0 0 . a (t ) _____ −2a (t )− 2 a (t ) ≤ −ν I This gives that for uniform exponential stability we also need existence of a small, positive constant . ν a 2 (t ) − 2a 3 (t ) ≤ a (t ) ≤ a (t )−ν/2 ν such th You're Reading a Preview for all t . For example, a (t ) = 1 satisfies these conditions. Unlock full access with a free trial. Solution 7.11 Suppose that for every symmetric, positive-definite M there exits a unique, sym Download With Free Trial positive-definite Q such that A T Q + QA + 2µQ = −M that is, Master your semester with Scribd  (A + µ I ) Q + Q (A + µ I ) = −M Read Free Foron 30this Days Sign up to vote title  Then byYork the argument above Theorem 7.11 we conclude that all eigenvalues have negative real p µ I useful & The New Times  Useful ofA +Not That is, if T Special offer for students: Only $4.99/month. Cancel anytime. 0 = det [ λI − (A +µ I ) ] = det [ ( λ − µ)I − A ]   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and   Search document Linear System Theory, 2/E Solutions ∞ ∫ ∞ T x Ta e A σ Me A σ xa dσ ≤ t ∫x e T A Tσ Me A σ xa a dσ 0 = x Ta Qxa ≤ λmax (Q ) = Q  Also, using a change of integration variable from σ to τ = σ − t, ∞ ∞ T T ATσ Aσ xa e Me xa d σ = x Ta e A (t + τ) Me A(t + τ) xa d τ ∫ t ∫ 0 T = x Ta e A t Qe At xa At e  ≥ λmin (Q )e At xa 2 = _______ −1 2 Q  Therefore e At 2 _______ Q −1  ≤ Q  Since t was arbitrary, this gives max e At  ≤ √Q Q −1  t≥0 Solution 7.17 You're Let F = A + (µ−ε)I. Then F  ≤Reading A +µ−aε,Preview all eigenvalues of F have real parts less tha and Unlock full access with a free trial. e Ft = e At e (µ−ε)t Download With Free Trial Thus e At  = e −(µ − ε)t e Ft  By Theorem 7.11 the unique solution of F T Q +QF = −I is ∞ T Q = e F σ e F σ d σ Read Free Foron 30this Days Sign up to vote title Master your semester with Scribd∫ & The New York Times For any n × 1 vector x , 0 Special offer for students: Only $4.99/month. d ___ T T x T e F σ e Fσ x = x T e F σ [ F T + F ] e F σ x   Not useful Cancel anytime. Useful     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download   of 106 Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions T x T e F t e Ft x ≤ 2 (A +µ−ε) x T Qx , t ≥0 which gives e Ft  ≤ √2 (A +µ−ε)Q  , t ≥ 0 Thus the desired inequality follows from (*). Solution 7.19 To show uniform exponential stability of A (t ), write the 1,2-entry of A (t ) as a (t ), an Q (t ) = q (t ) I, where 2+e −2t , t ≥ 1/ 2 q (t ) = q ⁄ (t ) , −1/ 2 < t < 1/ 2 1 2 3, t ≤ − 1/ 2 Here q ⁄ (t ) is a continuously-differentiable ‘patch’ satisfying 2 ≤ q ⁄ (t ) ≤ 3 for −1/ 2 < t < 1/ 2, and an condition to be specified below. Then we have 2 I ≤ Q (t ) ≤ 3 I for all t. Next consider . . −2q (t )+q (t ) a (t )q (t. ) ≤ −νI A T (t )Q (t ) + Q (t )A (t ) + Q (t ) = a (t )q (t ) −6q (t )+q (t ) 1 1 2 We choose ν = 1 and show that 2 You're Reading a Preview . −2q (t )+q (t )+1 a (t )q (t ) Unlock full access with a .free trial. ≤ 0 −6q (t )+q (t )+1 a (t )q (t ) . for all t . With t < −1/ 2 or t > 1/ 2 it is easy to show that q (t )−q (t )−1 ≥ 0, and a patch function can be sket Download With Free Trial such that this inequality is satisfied for −1/ 2 < t < 1/ 2. Then, for all t, . . −2q (t )+q (t )+1 ≤ −q (t ) ≤ 0 , −6q (t )+q (t )+1 ≤ −5q (t ) ≤ 0 . . [−2q (t )+q (t )+1][−6q (t )+q (t )+1] − a 2 (t )q 2 (t ) ≥ [5−a 2 (t )]q 2 (t ) ≥ 4q 2 (t ) ≥ 0 Master your semester with Scribd  Thus we have proven uniform exponential stability. Read Free Foron 30this Days Sign up to vote title  To show A (t ) is not uniformly exponentially stable, write the state equation as two scalar equation & The New York Times Useful  Not useful  compute T Cancel anytime. Special offer for students: Only $4.99/month. Φ T (t 0) = e −t 0 t ≥0   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Upload Sign In RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition. Khalil - Nonlinear Systems Slides     Save Embed Share Print 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document  CHAPTER 8 Solution 8.3 No. The matrix A= −2 0 √8 −1 has negative eigenvalues, but 8 −4a √Preview You're A + Reading AT = √8 Unlock full access with a free trial. has an eigenvalue at zero. Solution 8.6 −2 Download With Free Trial Viewing F (t )x (t ) as a forcing term, for any t o , xo , and t ≥ to we can write t x (t ) = ΦA +F (t, t o ) xo = ΦA (t, t o ) xo + ∫ ΦA (t, σ)F (σ) x (σ) d σ Master your semester with Scribd which gives, for suitable constants γ, λ > 0, & The New York Times − − Special offer for students: Only $4.99/month.x (t ) ≤ γ e λ(t to Read Free Foron 30this Days Sign up to vote title Useful  Not useful Cancel anytime. to ) xo  + γ e −λ(t −σ) F (σ)x (σ) d σ t ∫ to      Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition. Khalil - Nonlinear Systems Slides     Save Embed Share Print 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions t x (t ) ≤ γ e −λ(t −to ) ∫ γF (σ) d σ t xo  eo ∞ ≤ γ e −λ(t−t ) e o ∫ γF (σ) d σ to xo  ≤ γ e −λ(t−t ) e γ β xo  o and we conclude the desired uniform exponential stability. Solution 8.8 We can follow the proof of Theorem 8.7 (first and last portions) to show that the solution ∞ T Q (t ) = e A (t )σ e A (t )σ d σ ∫ 0 of A T (t )Q (t ) + Q (t ) A (t ) = −I is continuously-differentiable and satisfies, for all t, I ≤ Q (t ) ≤ ρ I You'reηReading a Preview where η and ρ are positive constants. Then with Unlock full access with a free . trial. F (t ) = A (t ) − 1⁄2Q −1 (t )Q (t ) an easy calculation shows Download With Free Trial . F T (t )Q (t ) + Q (t )F (t ) + Q (t ) = A T (t )Q (t ) + Q (t ) A (t ) = −I Thus . x (t ) = F (t ) x (t ) Master your semester with Scribd is uniformly exponentially stable by Theorem 7.4. & The New York Times Special offer for students: Only $4.99/month. Solution 8.9 As in Exercise 8.8 we have, for all t, Read Free Foron 30this Days Sign up to vote title   Not useful Cancel anytime. Useful     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106 Table of Contents for Linear   Linear System Theory and  Search document Linear System Theory, 2/E  Solutions Then the complete solution formula gives t x (t ) = ΦF (t, t o ) xo + . ∫ Φ (t, σ) ⁄ Q − (σ)Q (σ) x (σ) d σ 1 F 1 2 to and the result of Exercise 8.8 implies that there exists positive constants γ, x (t ) ≤ γ e −λ(t −to ) t ∫ xo  + γ e −λ(t −σ) to βρ2 ____ η λ such that, for any to and t ≥ to x (σ) d σ Therefore t e x (t ) ≤ γ e λt λto xo  + ____ λσ e x (σ) d σ ∫ _γβρ η 2 to and the Gronwall-Bellman inequality (Lemma 3.2) implies t ∫ γβρ /η d σ 2 e x (t ) ≤ γ e λt λto xo e to Thus x (t ) ≤ γ e −(λ−γβρ 2 /η)(t −to ) xo  You're Reading a Preview Now, writing the left side as ΦA (t, to )xo  and for any to and t ≥ to choosing the appropriate unity-norm x −(λ−γβρ 2 /η)(t −to ) Unlock t o )access ΦA (t,full  ≤ γ ewith a free trial. For β sufficiently small this gives the desired uniform exponential stability. (Note that Theorem 8.6 also ca . used to conclude that uniform exponential Download stability of x With t ) implies uniform exponential stability o (t ) = FFree (t ) x (Trial . . x (t ) = [ F (t ) + 1⁄2Q −1 (t )Q (t ) ] x (t ) = A (t ) x (t ) for β sufficiently small.) Master your semester with Scribd  Read Free 30 Days Sign up to vote title . Foron . this  t ) =Not A (tuseful With F (t ) = A (t ) + (µ / 2)I we have that F (t ) ≤ α Useful ), and the eigenvalu Solution 8.10 Times & The New York  + µ / 2, F ( F (t ) satisfy Re [λ (t )] ≤ −µ / 2. The unique solution of Cancel anytime. F Special offer for students: Only $4.99/month. F T (t )Q (t ) + Q (t )F (t ) = −I   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload Join RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1 of 106   Table of Contents for Linear Linear System Theory and   Search document Linear System Theory, 2/E Solutions ∞ −z T F T (t )τ e e F (t )τ z= d ∫τ ___ dσ zTe F T (t )σ e F (t )σ z dσ ∞ ≥ −(2α + µ) ∫ z T e F T (t )σ e F (t )σ z d σ T (t )σ e F (t )σ z d σ τ ∞ ≥ −(2α + µ) ∫ z T e F 0 ≥ −(2α + µ) z T Q (t ) z Thus eF T (t )τ e F (t )τ ≤ (2α + µ) Q (t ) , τ ≥ 0 and using e F (t )τ = e A(t )τ e (µ / 2) τ , τ≥0 gives e A(t )τ  ≤ √(2α+µ)ρ e (−µ / 2) τ , τ ≥ 0 You're Reading a Preview Solution 8.11 Unlock full access a free trial. u (t ) iswith Write (the chain rule is valid since a scalar) . q (t ) = −A −1 (u (t )) . . dA _db __ ___ (u (t ))u (t ) (u (t ))u (t ) A −1 (u (t ))b (u (t )) − A −1 (u (t )) Download With Free Trial du du . ∆ = −B̂ (t )u (t ) Then Master your semester with Scribd . Read Free Foron 30this Days Sign up to vote title x (t ) = A (u (t )) x (t ) + b (u (t )) & The New York Times = A (u (t )) [ x (t ) − q (t ) ] + A (u (t))q Useful t )) Not useful (t ) + b (u ( Special offer for students: Only $4.99/month. gives = A (u (t )) [ x (t ) − q (t ) ] Cancel anytime.     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload Join RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music  This is the solution manual of Rugh W.J Linear system theory. 2 edition. Khalil - Nonlinear Systems Slides     Save Embed Share Print 1 Download of 106   Table of Contents for Linear Linear System Theory and   Search document CHAPTER 9 Solution 9.7 Write B (A −βI )B (A −βI )2 B . . . = B AB −βB A 2 B −2βAB+ β2 B . . . 2 You're=Reading B AB aAPreview B ... Unlock full access with a free trial. Im 0 0 0 . . . −βIm β2 Im I m − 2β I m 0 0 . . . Im 0 . . . ... ... ... ... . . . Download solution is even easier using rank tests Clearly the two controllability matrices have the sameWith rank.Free ( TheTrial Chapter 13.) eigenvalues, Solution semester 9.8 Since A has negative-real-part Master your with Scribd ∞ Q = ∫ e BB e & The New York Times At Special offer for students: Only $4.99/month. is well defined, symmetric, and 0 T A Tt Read Free Foron 30this Days Sign up to vote title dt Useful  Not useful Cancel anytime.     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload Join RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music  This is the solution manual of Rugh W.J Linear system theory. 2 edition. Khalil - Nonlinear Systems Slides     Save Embed Share Print 1 Download of 106   Table of Contents for Linear Linear System Theory and   Search document Linear System Theory, 2/E Solutions 0= dj ___ dt j x T e At B = x TA j B t =0 for j = 0, 1, 2, . . . . But this implies B AB . . . A n −1 B xT =0 which contradicts the controllability hypothesis. Thus Q is positive definite. Solution 9.9 Suppose λ is an eigenvalue of A, and p is a corresponding left eigenvector. Then p p TA = This implies both ≠ 0, and λ pT _ p HA = λ p H , A T p = λp Now suppose Q is as claimed. Then _ p H AQp + p H QA T p = λ p H Qp + λ p H Qp = −p H BB T p that is, You're Reading a Preview 2Re [λ] p H Q p = −p H BB T p Unlock full access with a free trial. This gives Re [λ] ≤ 0 since Q is positive definite. Now suppose Re [λ] = 0. Then (*) gives p H B = 0. Also j = 1, 2, . . . , _ Download With Free _Trial j p H A j B = λ p H A j −1 B = . . . = λ p H B = 0 Thus p B AB . . . A − B = 0 Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title which contradicts the controllability assumption. Therefore Re [λ] < 0. & The New York Times  Useful  Not useful H Special offer for students: Only $4.99/month. Solution 9.10 Let n 1 Cancel anytime.     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download   of 106 Table of Contents for Linear Linear System Theory and  Search document L in e a r S y s t e m T h e o r y , 2 / E  S o lu t io n s Now suppose the state equation is output controllable on [ to , t f ], but that W y (to , t f ) is not invert invertible. ible. there exists a p × 1 vector ya ≠ 0 such that y aT Wy (to , t f )ya = 0. Using by now familiar familiar arguments, this gives y aT C (t f )Φ(t f , t )B (t ) = 0 , t ∈ [t o , t f ] Consider the initial state xo = Φ(to , t f )C T (t f )[ C (t f )C T (t f ) ]−1 ya which is well defined and nonzero since rank C (t f ) = p. There exists an input ua (t ) such that tf 0 = C (t f )Φ(t f , t o ) xo + ∫ C (t )Φ(t , σ)B (σ)u (σ) d σ f f a to tf = ya + ∫ C (t )Φ(t , σ)B (σ)u (σ) d σ f f a to Premultiplying by y aT gives 0= y aT ya This contradicts y a ≠ 0, and thus Wy (to , t f ) is invertible. The rank assumption on C (t f ) is needed in the necessity proof to guarantee that x o is well well define m = p = 1, invertibility of Wy (to , t f ) is equivalent to existence of a ta ∈ (to , t f ) such that C (t f )Φ(t f , t a )B (ta ) ≠ 0 That is, there exists a ta Solution 9.11 ∈ (to , t f ) such that the output response at t f to an impulse input at ta is nonzero. From Exercise 9.10, since rank C = p, the state equation is output controllable if and only i some fixed t f > 0, tf − ∫ Master your semester with Scribd ∆ Wy = Ce 0 A (t f t ) is invertible. We Times will will show this holds if and only if & The New York Special offer for students: Only $4.99/month. rank BB T e A T (t f −t ) C T dt Read Free Foron 30this Days Sign up to vote title  CB CAB CAB . . . CA n −1 B  Not useful Cancel anytime. Useful =p     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document L in e a r S y s t e m T h e o r y , 2 / E  S o lu t io n s y aT Ce A (t f −t ) B = y aT C n −1 Σ α (t − t ) A B = 0 , k k f t ∈ [0, t f ] k =0 Therefore y aT Wy ya = 0, which implies that Wy is not invertible. For m = p = 1 argue as in Solution 9.10 to show that a linear state equation is output controllable if only if its impulse response (equivalently, transfer function) is not identically zero. Solution 9.17 Beginning with y (t ) = c (t )x (t ) . . . y (t ) = c (t )x (t ) + c (t )x (t ) . = [c (t ) + c (t )A (t )]x (t ) + c (t )b (t )u (t ) = L 1 (t )x (t ) + L 0 (t )b (t )u (t ) it is easy to show by induction that y (k ) k −1 (t ) = Lk (t )x (t ) + Σ j =0 Now if k − j −1 __d____ __ ___ _ 2, . . . [ L j (t )b (t )u (t ) ] , k = 1, 2, k − j −1 dt __ −1 ∆ Ln (t )M = α0 (t ) α1 (t ) . . . αn −1 (t ) then n −1 Σ α (t )L (t ) = i i α0 (t ) . . . αn −1 (t ) i =0 L 0 (t ) . . . = Ln (t ) Ln −1 (t ) Thus we can write Master your semester with Scribd − − d −− _____ __ ____ __ Free For 30 Days Sign t ) ]to vote on this title (t )u (up [ L (t )bRead y (t ) − Σ α (t )y (t ) = L (t )x (t ) + Σ − − dt & The New York Times  Useful  Not useful (n ) n 1 i (i ) n 1 n i =0 Special offer for students: Only $4.99/month. j =0 n j 1 n j 1 j n −1 n −1 i −1 i 0 i 0 j 0 − Σ αi (t )Li (t )x (t ) − Σ αi (t ) Σ Cancel anytime. d i − j −1 ______ [ L j (t )b (t )u (t ) ] dt i − j −1     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1 of 106   Table of Contents for Linear Linear System Theory and  Search document  CHAPTER 10 Solution 10.2 We show show equivalence equivalence of full-rank full-rank failure failure in the respective respective controllability controllability and observab matrices, and thus conclude that one realization is controllable and observable (minimal) if and only if the oth controllable and observable observable (minimal). (minimal). First, rank if and only if there exits a nonzero, n B AB . . . A n −1 B <n × 1 vector q such that q T B = q T AB = . . . = q T A n −1 B = 0 This holds if and only if q T B = q T (A+BC )B = . . . = q T (A+BC )n −1 B = 0 which is equivalent to rank B (A+BC )B Similarly, semester with Scribd Master your & The New York Times Special offer for students: Only $4.99/month. rank . . . (A+BC )n −1 B C CA . . . <n Read Free Foron 30this Days Sign up to vote title  <n  Not useful Cancel anytime. Useful     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions C (t )B (σ) = H (t )F (σ) for all t , σ, picking an appropriate t o and t f > to , tf Mx ( t o , t f ) W x ( t o , t f ) = ∫C tf T (t )H (t ) dt to ∫ F (σ ) B (σ ) d σ T to where the left side is a product of invertible matrices by minimality. Therefore the two matrices on the right are invertible. Let tf P −1 = M −x 1 (to , t f ) ∫C T (t )H (t ) dt to Then multiply both sides of (*) by C T (t ) and integrate with respect to t to obtain tf Mx (to , t f )B (σ) = ∫C T (t )H (t ) dt F (σ) to for all σ. That is, B (σ) = P −1 F (σ) for all σ. Similarly, (*) gives You're Readingtfa Preview C (t )Wx (to , t f ) = H (t ) ∫ F (σ ) B T (σ ) d σ Unlock full access with to a free trial. that is, Download With Free Trial tf C (t ) = H (t ) ∫ F (σ)B T (σ) d σ W −x 1 (to , t f ) to Master your semester with Scribd − Read Free Foron 30this Days Sign up to vote title F (σ)B (σ) d σ W − (t , t ) = ∫ C (t )H (t ) dt M (t , t ) = P ∫ & The New York Times  Useful  Not useful But (**) then gives tf T to Special offer for students: Only $4.99/month. so we have 1 tf x 1 o T f to x o f Cancel anytime.     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition. Khalil - Nonlinear Systems Slides     Save Embed Share Print 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions d ___ X (t ) X (σ ) = X (t ) dt d ___ X (σ) dσ which implies d ___ X (t ) X (σ) dt d ___ X (σ) = X (−t ) dσ Integrate both sides with respect to t from a fixed to to a fixed t f > to to obtain tf ∫ d ___ X (t ) dt d ___ X (σ) = X (−t ) (t f − to ) dσ to dt X (σ) Now let 1 _____ A= t f −to tf ∫ X (−t ) d ___ X (t ) dt to dt to write d ___ X (σ) = A X (σ) , X (0) = I dσ This implies X (σ) = e A σ . (Of course thereYou're are quicker ways. a For example note that Reading Preview d ∂ X (t+σ) = X (t ) ___ ∂ X (t+σ) = ___ ___ X (σ) Unlock full access with a free trial. dσ ∂σ ∂t . . Evaluating at σ = 0 gives X (t ) = X (t )X (0), which implies . Download With FreeX. (0) Trial X (0)t t X (t ) = X (0)e =e Also the result holds for continuous solutions of the functional equation, though the proof is much more difficu Master your semester with Scribd  factorization unreviewed in the text Solution 10.12 If rank G = r we can write (admittedly using a matrix Read Free Foron 30this Days Sign up to vote title  & The New York Times G =CB  Useful  Not useful i i i i i Cancel anytime. Special offer forwhere students: C isOnly p × $4.99/month. r , B is r × m, and both have rank r . Then it is easy to check that i i i i i   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document  CHAPTER 11 Solution 11.4 Since rank b Ab = rank 1 1 −1 −1 =1 the state equation is not minimal. It is easy to compute the impulse response: σ) = C (t )e A (t −σ) B = (t 2 + 1) e −(t−σ) You'rerealization Reading a Preview Then a factorization is obvious, giving a minimal G (t, . x (tfull e t u (t with ) =access ) Unlock a free trial. y (t ) = (t 2 + 1)e −t x (t ) Download With Free Trial Solution 11.7 For the given impulse response, (t, σ ) = Master your semester withΓScribd  Sign up to vote on this title  & The New York Not useful It is easy to checkTimes that rank Γ (t, σ) = 2 for all t , σ, and a little more calculation that rank Γ (t,  Useful  shows 22 22 1+e 2t / 2+e 2σ / 2 e 2σ 0 Free For 30 Days e 2t Read Cancel anytime. a minimal realization is, using formulas in the proof of Theorem 11.3, Special offer forThen students: Only $4.99/month. 33   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1 of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions Γ= 1 1 1 . . . 1 1 1 . . . 1 ... 1 ... 1 ... . . . . . . and clearly the rank condition in Theorem 11.7 is satisfied with l = k = n = 1. Then, following the pro Theorem 11.7, F = F s = Fc = Fr = H 1 = H s1 = 1 and a minimal (dimension-1) realization is . x (t ) = x (t ) + u (t ) y (t ) = x (t ) For the truncated sequence, Γ= 1 1 1 0 . . . 1 1 0 0 . . . 1 0 0 0 . . . 0 0 0 0 . . . ... ... ... ... . . . You're Reading a Preview The rank condition in Theorem 11.7 is satisfied with l = k = n = 3. Taking Unlock full access with a free trial. 1 1 1 1 1 0 1 1 0 , Fs = H s3 = 1 0 0 Download With Free Trial0 0 0 1 0 0 F = H3 = Fc = 1 1 1 , Fr = Master your semester with Scribd gives a minimal realization specified by & The New York Times 0 1 0 Special offer for students: Only $4.99/month. A = Fs F −1 = 0 0 1 0 0 0 1 1 1 , B= Read Free Foron 30this Days Sign up to vote title 1 1 1   Not useful Cancel anytime. Useful , C= 1 0 0     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download   of 106 Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions Gi −1 Gi . . . = α0 Gn −2+i . . . G0 G1 . . . Gi −2 Gi −1 . . . + . . . + αi −2 Gn −1 . . . Gn −3+i . . . By ignoring the top entry, this linear combination shows that column i +1 is given by the same linear combin of the i −1 columns to its left, and so on. Thus by the rank assumption on Γ there cannot exist such an i , an first n columns of Γ are linearly independent. A similar argument shows that the first n columns of Γ n linearly independent, for every j ≥ 0, and thus that Γnn is invertible. It remains only to show that the given A, B, C provides a realization for G (s ), since minimality is immediate. Premultiplication by Γnn verifies Γ −nn1 Gk . . . = ek +1 , k = 0, . . . , n −1 Gn +k −1 Then, since A = Γ snn Γ−nn1 , A Gk . . . Gn +k −1 Now, CB = G 0 , and You're Reading a Preview Gk +1 . a free trial. Unlock access . = Γ snnfull ek +1 = with , k = 0, . . . , n −1 . Gn +k Download With Free Trial CA j B = CA j −1 A G0 . . . Master your semester with Scribd G − & The New York Times G = CA j −1 Gn up Read Free Foron 30this Days Sign to vote title n 1 Special offer for students: Only $4.99/month. = ... =C . . . G1 . . .  j  Not useful Cancel anytime. Useful = G j , j = 1, . . . , n     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Upload Sign In RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1 of 106   Table of Contents for Linear Linear System Theory and  Search document  CHAPTER 12 Solution 12.1 If the state equation is uniformly bounded-input, bounded-output stable, then it is clear from definition that given δ we can take ε = η δ. Now suppose the ε, δ condition holds. In particular we can take δ = 1 and assume ε is such that, for any u (t ) ≤ 1 , t ≥ to implies You're Preview (t ) ≤ ε , at ≥ to y Reading to with sup Now suppose u (t ) is any bounded input Unlock signal.full Given let µa = u (t ). Note µ > 0 can be assumed access free trial. t ≥ to otherwise we have a trivial case. Then u (t ) / µ ≤ 1 for all t ≥ to , and the zero-state response to u (t ) satisfie t Download With Free Trial y (t ) =  ∫ G (t, σ)u (σ) d σ to t = µ  ∫ G (t, σ)u (σ)/µ d σ Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title & The New York Times  Not useful ≤ µε = ε sup u (t ) ≥t , tUseful to Special offer for students: Only $4.99/month. Thus we have t ≥ to Cancel anytime. o     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions rank B AB . . . A n −1 B =n Then given δ we can take δ ε = λmin e −A δ ∫e Aτ T T BB T e A τ d τ e −A δ 0 For a time-varying example, take scalar a (t ) = 0, b (t ) = e −t /2 . Then W (t −δ, t ) = e −t (e δ −1) Given any δ > 0, W (t −δ, t ) > 0 for all t , but there exists no ε > 0 such that W (t −δ, t ) ≥ ε for all t . Solution 12.9 Consider a scalar state equation . x (t ) = b (t )u (t ) y (t ) = x (t ) where b (t ) is a ‘smooth bump function’ described as follows. It is a continuous, nonnegative function that is You're Reading a Preview for t ∉ [0, 1], and has unit area on [0, 1]. Then for any input signal the zero-state response satisfies Unlock full 1access with a free trial. y (t ) ≤ b (σ)u (σ) d σ ∫ 0 for any t. Thus for any to and any t ≥ to , Download With Free Trial 1 ∫ y (t ) ≤ b (σ) d σ . sup u (t ) t ≥ to Master your semester with Scribd ≤ sup u (t ) ≥ & The New York Times 0 t to Read Free Foron 30this Days Sign up to vote title   Not useful Cancel anytime. Useful   the state equation is uniformly bounded-input, bounded-output stable with η = 1. However if we consi Special offer forand students: Only $4.99/month. bounded input that is continuous and satisfies   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload Join RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download   of 106 Table of Contents for Linear Linear System Theory and   Search document Linear System Theory, 2/E Solutions u (t ) ≤ Let T = 2 max [T 1 , T 2 ]. Then for t ε ___ , t ≥ T2 2ρ ≥ T, t ∫ y (t ) ≤ G (t −σ)u (σ) d σ 0 t T ε ___ G (t −σ) d σ = µ G (t −σ) d σ + 2ρ T 0 ∫ ∫ Changing the variables of integration gives t ε ___ y (t ) ≤ µ G (τ) d τ + 2ρ t −T ∫ t −T ∫ G (τ) d τ 0 ε ε ___ ρ =ε + ≤ µ ___ 2µ This shows that y (t ) → 0 as t Solution 12.11 where u (t ) is n 2ρ → ∞. The hypotheses imply that given ε > 0 there exist δ1 , δ2 > 0 such that if You're Reading a Preview xo  < δ1 ; u (t ) < δ2 , t ≥ to × 1, then the solution of Unlock full access with a free trial. . x (t ) = A (t ) x (t ) + u (t ) , x (to ) = xo Download With Free Trial satisfies x (t ) < ε , t ≥ to In particular, with x = 0, this shows that if u (t ) < δ for t ≥ t , then the corresponding zero-state solutio Master your semester with Scribd  the state equation Read Free Foron 30this Days Sign up to vote title  . x (t ) = A (t ) x (t ) + u (t ) Useful  Not useful & The New York Times o 2 Special offer for students: Only $4.99/month. satisfies  ( ) for o Cancel anytime. y (t ) = x (t ) But this implies uniform bounded-input, bounded ut stabilit by Exe   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload Join RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and   Search document Linear System Theory, 2/E Solutions ∞ ∫ ∞ 0 ∞ t ∫ ∫ y (t )e −η t dt = 0 0 ∞ ∞ G (t −σ)e −λ σ d σ 0 ∞ ∫ ∫ G (t −σ)e −λ σ d σ 0 ∫ ∫ G (t −σ)e −η = e −η t dt = t e −η t dt 0 e −λ σ d σ dt 0 where all integrals are well-defined because of the stability assumption, and λ , η > 0. Changing the varia integration in the inner integral from t to γ = t −σ gives ∞ ∞ ∞ ηt − y (t )e dt = G (γ)e −η γ d γ e −η σ e −λ σ d σ ∫ ∫ ∫ 0 0 0 ∞ = G(s ) s = η ∫ e − η λ σ dσ ( + ) 0 = 1 _____ G(η) η+λ Without the stability assumption we can say that U (s ) = 1/(s+ λ) for Re [s ] > −λ, and the integral for converges for Re [s ] > Re [p 1 ], . . . , Re [pn ], where p 1 , . . . , pn are the poles of G (s ). Thus ∞ G (s ) _____ = a y (Preview t )e −st dt Y (s ) = Reading You're s+ λ 0 ∫ Unlock full access with a free trial. is valid for Re [s ] > −λ, Re [p 1 ], . . . , Re [p n ]. This implies that if then η > −λ, Re [p 1 ], . . . , Re [pn ] Download With Free Trial ∞ G (η) ∫ y (t )e −η dt = _____ η+λ t Master your semester with Scribd even though y (t ) may be unbounded. & The New York Times 0 Special offer forSolution students: 12.14 Only $4.99/month. Given u (t ), t v (t ) = y (t ) = C x (t ) we have ≥ 0, and Read Free Foron 30this Days Sign up to vote title   Not useful Cancel anytime. Useful   xo , suppose x (t ) is a solution of the given state equation. Then   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions we get t x (t ) − z (t ) = ∫ α (t −σ)Bu (σ) d σ 0 0 Then . w (t ) = −(CB )−1 CAP z (t ) − (CB )−1 CAB (CB )−1 C x (t ) + (CB )−1 C x (t ) = −(CB )−1 CAP z (t ) − (CB )−1 CAB (CB )−1 C x (t ) + (CB )−1 CA x (t ) + (CB )−1 CBu (t ) = −(CB )−1 CAP z (t ) + (CB )−1 CA [ −B (CB )−1 C + I ] x (t ) + u (t ) = (CB )−1 CAP [ x (t ) − z (t ) ] + u (t ) t = (CB )−1 CAP ∫ α (t −σ)Bu (σ) d σ + u (t ) 0 0 Again using PB = 0 gives w (t ) = u (t ) , t ≥0 To address stability, since PB = 0 we see that P is not invertible. Thus AP is not invertible, which implie second state equation is never exponentially stable. The scalar case with A = −1, B = C = 1 is unifo bounded-input, bounded-output stable, but the resulting You're Reading a Preview . z (t ) = −v (t ) Unlock full access with. a free trial. w (t ) = v (t ) + v (t ) is not, as the bounded input v (t ) = cos (e t )Download shows. With Free Trial Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month. Read Free Foron 30this Days Sign up to vote title   Not useful Cancel anytime. Useful     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document  CHAPTER 13 Solution 13.1 Suppose n = 2 and A has complex eigenvalues. Let A= a 11 a 12 a 21 a 22 b= , b1 b2 Then A has eigenvalues You're Reading a Preview a 11 +a 22 ± √(a 11 +a 22 )2 −4(a 11 a 22 −a 12 a 21 ) ____________________________________ Unlock full access 2 with a free trial. and since the eigenvalues are complex, Download With Free Trial (a 11 +a 22 )2 − 4(a 11 a 22 −a 12 a 21 ) = (a 11 −a 22 )2 + 4a 12 a 21 < 0 Supposing that det [ b Ab ] = 0, we will show that if b ≠ 0 we get a contradiction. For Ab ] = a b − a 0 = det [ bScribd Master your semester with implies & The New York Times (a −a ) b b = (a 21 Special offer for students: Only $4.99/month. If b 1 = 0, b 2 ≠ 0, then (**) implies 11 22 2 2 2 1 2 2 1 2 12 b 2 21 − (a 11 −a 22 )b 1 b 2 Read Free Foron 30this Days Sign up to vote title  Useful  Not useful b 2 −a b 2 )2 Cancel anytime. 1 12   2 a 12 = 0, which contradicts (*). If b 1 ≠ 0, b 2 = 0, then (**) implies a   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E Solution 13.4  Solutions We need to show that rank B AB . . . A n −1 B if and only if the ( n +p )-dimensional state equation . A 0 z (t ) = C 0 =n, A B C D rank B D z (t ) + = n+p u (t ) is controllable. First suppose (+) holds but (++) is not controllable. Then there exists a complex so such that so In −A 0 −C so Ip rank Since rank [ so I −A B D < n+p B ] = n, this implies rank −C so Ip D −A −C 0 B 0 D <p In turn, this implies so = 0, so that (*) becomes rank < n+p and this contradicts the second rank condition in (+). Conversely, supposing (++) is controllable, then You're Reading a Preview 2 ... B AB A B . . . = n+p D full CBaccess CAB Unlock with a free trial. rank This implies Download With n −1 Trial . . . Free B AB rank A B =n in other words, the first rank condition in (+) holds. Now suppose < n+p rank C D Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title Then & The New York Times  Useful  Not useful s I −A 0 B A B Special offer for students: Only $4.99/month. rank Cancel anytime. o n −C so Ip D < n +p so = 0     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E Solution 13.10  Solutions Since P −1 B (P −1 AP )P −1 B . . . (P −1 AP )n −1 P −1 B = P −1 B AB . . . A n −1 B and controllability indices are defined by a left-to-right linear independence search, it is clear that controllab indices are unaffected by state variable changes. For the second part, let rk be the number of linearly dependent columns in A k B that arise in the left-tocolumn search of [ B AB . . . A n −1 B ]. Note r 0 = 0 since rank B = m. Then rk is the number of controllab indices that have value ≤ k. This is because for each of the r k columns of the form A k Bi that are dependen have ρi ≤ k, since for j > 0 the vector A k +j Bi also will be dependent on columns to its left. Th k = 1, . . . , m, rk −rk −1 gives the number of controllability indices with value k. Writing BG ABG . . . A k BG = B AB . . . A k B 0 ... G ... . . . . . . . 0 0 .. G 0 . . . 0 0 . . . G and using the invertibility of G shows that the same sequence of rk ’s are generated by left-to-right column se in [ BG ABG . . . A n −1 BG ]. Solution 13.11 You're Reading a Preview For the time-invariant case, if Unlock full access with a free trial. p T A = p Tλ , p T B = 0 implies p = 0, then Download With Free Trial p T (A +BK ) = p T λ , p T B = 0 obviously implies p = 0. Therefore controllability of the open-loop state equation implies controllability of closed-loop state equation. In the time-varying case, suppose the open-loop state equation is controllable on [ to , t f ]. Thus g Read Free For 30 Days Sign up to vote on this title x (to ) = xo there exists an input signal ua (t ) such that the corresponding solution xa (t ) satisfies xa (t f ) =0. The closed-loop state equation  Useful  Not useful Cancel anytime. . Special offer for students: Only $4.99/month. z (t ) = [ A (t ) + B (t )K (t ) ] z (t ) + B (t )v (t ) Master your semester with Scribd & The New York Times   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload Join RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1 of 106   Table of Contents for Linear Linear System Theory and   Search document Linear System Theory, 2/E Solutions B̂b̂ = 0 0 . . . 0 1 Using × to denote various unimportant entries, set B̂b̂ = Bo Rb̂ = block diagonal 0 0 . . . ρi × 1 , i = 1, . . . , m 1 0 . . . . 0 1 × 1 . . . 0 0 0 0 ... ... . . . ... ... × × . . . b̂ = × 1 0 0 . . . 0 1 This gives a set of equations of the form m 0 = b̂ 1 + Σ × b̂ i i i =2 m 0 = b̂ 2 + Σ × b̂ i i i =3 You're ..Reading a Preview . Unlock full access with a free trial. 0 = b̂m −1 + × b̂m Download 1 = b̂m With Free Trial Clearly there is a solution for the entries of b̂, regardless of the × ’s. Now it is easy to conclude controllabili the single-input state equation by calculation of the form of the controllability matrix. Then changing to original state variables gives the result since controllability is preserved. In the original variables, take K and b = b̂. For an example to show that b alone does not suffice, take Exercise 13.11 with all ×’s zero.  Master your semester with Scribd & The New York Times Read Free Foron 30this Days Sign up to vote title   Not useful Cancel anytime. Useful  Supposing the rank of the controllability matrix is q, Theorem 13.1 gives an invertible P Special offer forSolution students:13.14 Only $4.99/month. that   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download   of 106 Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions P −1 (P−a 1 APa )P = Ã 11 Ã 13 Ã 21 Ã 22 Ã 23 0 CPa P = 0 0 , P −1 (P−a 1 B ) = Ã 33 B̃ 1 B̃ 2 0 ˜ 0 C˜ C 1 2 ˜ = Cˆ . It is easy to see that the state equation formed where Ã 11 is l × l, and in fact Ã 33 = Â 22 , C 2 2 ˜ C 1 , Ã 11 , B̃ 1 is both controllable and observable. Also an easy calculation using block triangular structure sh that the impulse response of the state equation defined by (*) is ˜ e C 1 A˜ 11 t B̃ 1 It remains only to show that l = s. Using the effect of variable changes on the controllability and observab matrices and the special structure of (*) give C CA . . . CA n −1 ˜ C 1 B AB . . . A n −1 B = ˜ Ã C 1 11 . . . ˜ Ã n −1 C 1 11 n −1 B̃ 1 Ã 11 B̃ 1 . . . Ã 11 B̃ 1 You're Reading a Preview Thus ˜ C Unlock full access with a free trial. 1 ˜ C 1 Ã 11 ˜ . . . A˜ n −1 B̃ . B˜ 1 A˜ 11 B 1 11 1 . Download With Free Trial . ˜ Ãn −1 C 1 11 rank =s But Master your semester withC˜ Scribd & The New York Timesrank C˜ Ã. = rank B̃ Read Free Foron 30this Days Sign up to vote title 1 1 Special offer for students: Only $4.99/month. 11 . . ˜ Ã C l −1 1 Ã 11 B̃ 1 Useful −1   Not useful . . . lCancel anytime. Ã11 B̃ 1 =l     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1 of 106   Table of Contents for Linear Linear System Theory and  Search document  CHAPTER 14 Solution 14.2 For any t f > 0, tf W= ∫ e− At BB T e −A T t dt 0 is symmetric and positive definite by controllability, and tf You're Reading a−AtPreview d ___ T −A T t T ∫ e BB e AW + WA = − dt 0 Unlock full access with a free trial. =−e −Atf BB T e −A T tf dt + BB T Download With Free Trial Letting K = −B T W −1 , we have (A + BK )W + W (A + BK )T = − ( e −Atf −A T tf Master your semester with Scribd  Read For 30 Days Sign up to vote this title Suppose λ is an eigenvalue of A +BK. Then λ is an eigenvalue of ( A+BK )Free , and weon let 0 be a correspon p ≠  eigenvector. ThenTimes & The New York  Useful  Not useful BB T e + BB T ) T Special offer for students: Only $4.99/month. Also, T (A + BK ) p = λp Cancel anytime.   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download   of 106 Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions Solution 14.5 (a) For any n × 1 vector x, x H (A + A T ) x = x H A x + x H A T x ≥ − 2αm x H x If λ is an eigenvalue of A, and x is a unity-norm eigenvector corresponding to λ, then _ H T A x = λ x , x A = λ xH and we conclude _ λ + λ ≥ −2 αm Therefore any eigenvalue of A satisfies Re [λ] ≥ −αm , and this implies that for α > αm all eigenvalues of have positive real parts. Therefore all eigenvalues of −(A T +α I ) = (−A −α I )T have negative real parts. (b) Using Theorem 7.11, with α > αm , the unique solution of Q (−A − α I )T + (−A − α I )Q = −BB T is ∞ Q= ∫ e− (A +α I )t T BB T e −(A +α I )t dt 0 Reading a Preview Clearly Q is positive semidefinite. If x QxYou're = 0, then T (A +α I )t Unlock access 0 , a tfree ≥ 0trial. x T e −full B =with and the usual sequential differentiation and evaluation at t = 0 gives a contradiction to controllability. Thus Download With Free Trial positive definite. (c) Now consider the linear state equation . z (t ) = ( A+ α I −BB T Q −1 )z (t ) Master your semester with Scribd Using (*) to write BB Q − gives Read Free Foron 30this Days Sign up to vote title . − z (t ) = −Q (A+ α I ) Q z (t ) Useful  Not useful & The New York Times T 1 T 1 Cancel anytime.   Q [ −( A + α$4.99/month. I )T ]Q −1 has negative-real-part eigenvalues, which proves that (**) is exponentially stable. Special offer forBut students: Only (d) Invoking Lemma 14.6 gives that   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload Join RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and   Search document Linear System Theory, 2/E Solutions A+BK +Bbk = A +B (K+bk ) has the specified characteristic polynomial. Thus for the original state equation, the feedback law u (t ) = (K+bk ) x (t ) yields a closed-loop state equation with specified characteristic polynomial. Solution 14.8 Without loss of generality we can assume the change of variables in Theorem 13.1 has performed so that A= where A 11 is q A 11 A 12 0 A 22 , B= B1 0 × q, and λ I −A 11 rank B1 = q for all complex values of λ . Then the eigenvalues of A comprise the eigenvalues of A 11 and the eigenvalu A 22 . Also, for any complex λ , rank λ I −A λ I −A 11 −A 12 B 1 0 λ I −A 22 0 = q + rank λ I −A 22 You're Reading a Preview B = rank Now suppose rank [λ I −A B ] = nUnlock for allfull nonnegative-real-part access with a free trial.eigenvalues of A . Then by (+) any eigenvalue must be an eigenvalue of A 11 , which implies that all eigenvalues of A 22 have negative real parts. we can compute an m × q matrix K 1 such that A 11 + B 1 K 1 has negative-real-part-eigenvalues. So se Download With Free Trial K = [ K 1 0 ] we have that A 11 +B 1 K 1 A 12 0 A 22 A + BK = Master your semester with Scribd has negative-real-part eigenvalues. On the other hand, if there exists a K = [K K ] such that & The New York Times 1 Special offer for students: Only $4.99/month. A + BK = 2 Read Free Foron 30this Days Sign up to vote title   Not useful Cancel anytime. Useful A 11 +B 1 K 1 A 12 +B 1 K 2 0 A 22     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions tr [A+BLC ] = tr [A ] + tr [BLC ] = tr [A ] + tr [CBL ] = tr [A ] >0 Thus at least one eigenvalue of A+BLC has positive real part, regardless of L. Solution 14.12 Write the k th -row of G(s ) in terms of the k th -row Ck of C as Ck (sI − A )−1 B = ∞ Σ C A Bs − k j ( j+1) j =0 The k th -relative degree κk is such that, since L Aj [Ck ](t )B (t ) = Ck A j B, κ −2 Ck B = . . . = Ck A k B = 0 Ck A κk −1 B≠0 Thus in the k th -row of G(s ), the minimum You're difference betweenathe numerator and denominator polynomial deg Reading Preview among the entries Gk1 (s ), . . . , Gkm (s ) is κk . Unlock full access with a free trial. Download With Free Trial Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month. Read Free Foron 30this Days Sign up to vote title   Not useful Cancel anytime. Useful     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1 of 106   Table of Contents for Linear Linear System Theory and  Search document  CHAPTER 15 Solution 15.2 The closed-loop state equation can be written as . x (t ) = Ax (t ) + BMz (t ) +BNv (t ) = Ax (t ) + BMz (t ) +BNC [Lz (t )+x (t )] . z (t ) = Fz(t ) + GC [Lz (t )+x (t )] You're Reading a Preview Making the variable change w (t ) = x (t )+Lz (t ) gives the description . w (t ) = Ax (t )Unlock + BMzfull BNCw LFztrial. (t ) + (t )a+free (t ) + LGCw (t ) access with = Ax (t ) + [BM+LF ]z (t ) + [BN+LG ]Cw (t ) Download With Free Trial = [A −HC ]w (t ) . z (t ) = Fz(t ) +GCw (t ) Thus the closed-loop state equation in matrix form is Master your semester with Scribd . A −HC w (t ) = . GC z (t ) & The New York Times Special offer forand students: Only the result is $4.99/month. clear. 0 F Read Free Foron 30this Days w (t ) up Sign to vote title ) z (t Useful   Not useful Cancel anytime.     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload Join RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and   Search document Linear System Theory, 2/E Solutions τ+δ ∫τ τ+δ B (σ )2 dσ ≤ γ 2 ∫τ Φ(τ, σ )B (σ)B T (σ)ΦT (τ, σ ) d σ ≤ γ 2 n ε1 =∆ β1 Now for any τ, and t ∈ [τ+k δ, τ+(k +1)δ], k = 0, 1, . . . , τ+(k +1)δ t ∫τ B (σ )2 ∫τ dσ ≤ B (σ)2 d σ k τ+( j +1)δ j =0 τ+j δ ≤Σ ∫ B (σ)2 d σ ≤ (k +1) β1 ≤ [1 + (t −τ) / δ ] β1 This bound is independent of k, so letting β 2 = β1 /δ we have t ∫τ B (σ) 2 d σ ≤ β 1 + β2 (t −τ) for all t, τ with t ≥ τ. (Of course this provides a simplification of the hypotheses of Theorem 15.5 bounded-A (t ) case.) You're Reading a Preview Unlock full access with a free trial. Write the given state equation in the partitioned form . za (t ) A 11 A 12 za (t ) B1 = + . Download Withz Free TrialB A A t ( ) 21 22 b 2 zb (t ) Solution 15.6 y (t ) = Ip c 2 u (t ) za (t ) zb (t ) 0 Master your semester with Scribd  and the reduced-dimension observer and feedback in the form Read Free Foron 30this Days Sign up to vote title  & The New Yorkz. (tTimes − Huseful + (Useful A −H A  A ] z (t ) ) = [ A − H A ] z (t ) + [ B − HB ]u (t ) + [ A  )HNot c 22 12 Special offer for students: Onlyẑ $4.99/month. b (t ) = zc (t ) + Hza (t ) 1 21 22Cancel anytime. 12 11 a   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download   of 106 Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions Y ( s ) = Ip 1 0 (sI −A −BK )− BN R(s ) which is the same as if a static state feedback gain K is used. Solution 15.9 Solution 15.10 Similar in style to Solution 14.8. Since u = Hz + Jv = Hz + JC 2 x + JD 21 r +JD 22 u ∆ we assume that I −JD 22 is invertible, and let L = (I −JD 22 )−1 to write u = LHz + LJC 2 x + L JD 21 r Then, substituting for u, . x = (A+BL JC 2 )x + BLHz + BLJD 21 r . z = (GC 2 +GD 22 LJC 2 )x + (F+GD 22 L H )z + (GD 22 +GD 22 L JD 21 )r y = (C 1 +D 1 L JC 2 )x + D 1 L Hz + D 1 L JD 21 r This gives the closed-loop coefficients You're Reading a Preview Â = A+BL JC 2 Unlock full BLH BLJD 21 access with a free trial. , B̂ = GC 2 +GD 22 L JC 2 F+GD 22 L H GD 22 +GD 22 L JD 21 ˆ= C C 1 +D 1 L JC 2 Download With = DFree L JDTrial D LH , D̂ 1 1 21 These expressions can be rewritten using L = (I − JD 22 )−1 = I + J (I −D 22 J )−1 D 22 Master your semester with Scribd which follows from Exercise 28.2 or is easily verified using the identity Exercise Read Free For28.1. 30this Days Signinup to vote on title & The New York Times  Useful  Not useful Special offer for students: Only $4.99/month. Cancel anytime.     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload Join RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1 of 106   Table of Contents for Linear Linear System Theory and  Search document  CHAPTER 16 Solution 16.4 By Theorem 16.16 there exist polynomial matrices X (s ), Y (s ), A (s ), and B (s ) such that N (s ) X (s ) + D (s )Y (s ) = Ip Na (s ) A (s ) + Da (s )B (s ) = Ip Since D −1 (s )N (s ) = D −a1 (s )Na (s ), Na (s ) = Da (s )D −1 (s )N (s ). Substituting this into (**) gives (s ) A (s ) +aDPreview Da (s )You're D −1 (s )NReading a (s )B (s ) = Ip that is, Unlock full access with a free trial. N (s ) A (s ) + D (s )B (s ) = D (s )D −a1 (s ) Similarly, N (s ) = D (s )D −a 1 (s )Na (s ), and substituting into (*) gives Download With Free Trial Na (s ) X (s ) + Da (s )Y (s ) = Da (s )D −1 (s ) Therefore D (s )D −a1 (s ) and [ D (s )D −a1 (s ) ]−1 both are polynomial matrices, and thus both are unimodular. Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title Solution 16.5 From the given equality, & The New York Times  Useful  Not useful N (s )D (s ) − D (s )N (s ) = 0 Special offer for students: Only $4.99/month. L L Cancel anytime. and since N (s ) and D (s ) are right coprime there exist polynomial matrices X (s ) and Y (s ) such that     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition. Khalil - Nonlinear Systems Slides     Save Embed Share Print 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions I 0 −[X (s )B (s )+Y (s )A (s )] I gives X (s ) Y (s ) NL (s ) DL (s ) D (s ) −D (s )[X (s )B (s )+Y (s )A (s )]+B (s ) − N (s ) N (s )[X (s )B (s )+Y (s )A (s )]+A (s ) =I That is X (s ) Y (s ) NL (s ) DL (s ) −1 = D (s ) −D (s )(X (s )B (s )+Y (s )A (s ))+B (s ) −N (s ) N (s )(X (s )B (s )+Y (s )A (s ))+A (s ) which is another polynomial matrix. Thus X (s ) Y (s ) NL (s ) DL (s ) is unimodular. Solution 16.7 The relationship −1 R Preview You're Reading (P ρ s+P 1 s+R 0 ρ−1 ) = a holds if R 1 and R 0 are such that Unlock full access with a free trial. I = (P ρ s+P ρ−1 ) (R 1 s+R 0 ) = P ρ R 1 s 2 + (P ρ R 0 +P ρ−1 R 1 )s + P ρ−1 R 0 Download With Free Taking R 0 = P −ρ−11 and R 1 = −P −ρ−11 P ρ P −ρ−11 , it remains to verify that PTrial ρ R 1 = 0. We have I = (P s ρ + . . . + P ) (Q s η + . . . + Q ) ρ η 0 0 = P ρ Q η s η+ρ + (P ρ Q η−1 +P ρ−1 Q η )s η+ρ−1 + . . . Master your with P − semester and Q − invertible.with ThereforeScribd Read Free Foron 30this Days Sign up to vote title 0 P Q = 0 , P Q − +P − Q = & The New York Times  Useful  Not useful ρ 1 η 1 ρ η ρ η 1 ρ 1 secondOnly equation gives Special offer forThe students: $4.99/month. P P Q Q −1 η Cancel anytime.     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload Join RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music  This is the solution manual of Rugh W.J Linear system theory. 2 edition. Khalil - Nonlinear Systems Slides     Save Embed Share Print 1 Download of 106   Table of Contents for Linear Linear System Theory and   Search document Linear System Theory, 2/E Solutions Dk (s ) = D̃ 1 (s )u 1,k (s ) + . . . + D̃J (s )uJ,k (s ) + . . . + D̃m (s )um,k (s ) , k = 1, . . . , m Using a similar column notation for D hc and D l (s ) gives D hc k s ck [D ] ˜] hc c 1 [D + D lk (s ) = [D̃ 1 s ˜ l hc c [D ] l +D̃ 1 (s )] u 1,k (s ) + . . . + [D̃ J s J +D̃ J (s )] uJ,k (s ) ˜ hc c [D ] l + . . . + [D̃ m s m +D̃ m (s )] um,k (s ) , k = 1, . . . , m We claim that c k [D ] = max j = 1, . . . , m { c j [D̃ ]+degree u j,k (s ) } hc hc This is shown by a an argument using linear independence of D̃ 1 , . . . , D̃ m as follows. Let c̃ = and let µ j,k be the coefficient of s term on the right side is ˜] c̃ −c j [D max j = 1, . . . , m { c j [D̃ ]+degree u j,k (s ) } in u j,k (s ). Then not all the m Σµ µ j,k are zero, and the vector coefficient of th hc j,k D̃ j j =1 By linear independence this sum is nonzero, which implies ck [D ] = c̃. Now, using the definition of J, You're Reading a Preview ck [D ] < cJ [D̃ ] ≤ . . . ≤ cm [D̃ ] , k = 1, . . . , J −1 Unlock full access with a free trial. and this implies uJ,k (s ) = . . . = um,k (s ) = 0. Thus U (s ) has the form Ua (s ) U b (s ) Download With Free Trial U (s ) = 0(m−J+1) × J Uc (s ) where U a (s ) is (J −1) × J, from which rank U (s ) ≤ m −1 for all values of s . This contradicts unimodularity, cJ [D ] = cJ [D̃ ]. The proof is complete since the roles of D (s ) and D̃ (s ) can be reversed. Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month. Read Free Foron 30this Days Sign up to vote title   Not useful Cancel anytime. Useful     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Upload Sign In RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document  CHAPTER 17 Solution 17.1 If . x (t ) = A x (t ) + Bu (t ) y (t ) = C x (t ) is a realization of G T (s ), then . T T z (t ) =Reading A x (t ) +aCPreview v (t ) You're w (t ) = B T z (t ) Unlock full access with a free trial. is a realization for G (s ) since T T ) ] = [ C (sIWith − A )−1Free G (s ) = [ G T (sDownload B] = B T (sI − A T )−1 C T Trial Furthermore, easy calculation of the controllability and observability matrices of the two realizations shows one is minimal if and only if the other is. Now, if N (s ) and D (s ) give a coprime left polynomial frac description for G (s ), then there exist polynomial matrices X (s ) and Y (s ) such that Master your semester with Scribd N (s ) X (s ) + D (s )Y (s ) =Read I Free Foron 30this Days Sign up to vote title Therefore & The New York Times  Useful  Not useful Special offer for students: Only $4.99/month. X T (s ) N T (s ) + Y T (s ) D T (s ) = I   Cancel anytime. which implies that N T ( ) and D T ( ) are right coprime. Also, since D ( ) is row reduced, D T ( ) is column redu   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions Therefore SB To [ sI − A To − Q −1 VBTo ] −1 = D −1 (s )ΨT (s ) Using the definition of N (s ), −1 D −1 (s )N (s ) = SBTo [ sI − (A To + Q −1 VBTo ) ] Q −1 B −1 = CQ [ sI − Q −1 AQ ] Q −1 B = C (sI − A )−1 B Note that D (s ) is row reduced since Dlr = S −1 , which is invertible. Finally, if the state equation is controllab well as observable, hence minimal, then it is clear from the definition of D (s ) that the degree of the polyno fraction description equals the dimension of the minimal realization. Therefore D −1 (s )N (s ) is a coprime polynomial fraction description. Solution 17.5 Suppose there is a nonzero h with the property that for each uo there is an xo such that t ∫ hCe xo + hCe A (t −σ) Buo e At so σ dσ = 0 , t ≥0 0 Suppose G (s ) = N (s )D −1 (s ) is a coprime right polynomial fraction description. Then taking Laplace transfo You're Reading a Preview gives −1 (sa)free access with (s ) D −so )−1 = 0 hC (sI −Unlock A )−1 xofull + hN uo (strial. that is, Download With Free Trial (s −so )hC (sI − A )−1 xo + hN (s )D −1 (s )uo = 0 If so is not a pole of G (s ), then D (so ) is invertible. Thus evaluating at s = so gives hN (so )D −1 (so )uo = 0 Master your semester with Scribd and we have that if s is not a pole of G (s ), then for every ũ & The New York Times hN (s )ũ = 0 o o o o Read Free Foron 30this Days Sign up to vote title   Not useful Cancel anytime. Useful   Special offer forThus students: $4.99/month. hN (sOnly o ) = 0, that is rank N (so ) < p < m, which implies that so is a transmission zero. Conversely, suppose s is a transmission zero that is not a pole of G (s ). Then for a right-   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E Solution 17.9  Solutions Using a coprime right polynomial fraction description G (s ) = N (s )D −1 (s ) = N (s ) adj D (s ) ____________ det D (s ) suppose for some i, j and complex so we have [ N (s ) adj D (s ) ]  o o ij ∞ = Gij (so ) = ___________________ det D (so ) Since the numerator is the magnitude of a polynomial, it is finite for every so , and this implies det D (so ) = is, so is a pole of G (s ). Now suppose s o is such that det D (so ) = 0. By coprimeness of the right polynomial fraction descrip N (s )D −1 (s ), there exist polynomial matrices X (s ) and Y (s ) such that X (s )N (s ) + Y (s )D (s ) = Im for all s. Therefore [ X (s )G (s ) + Y (s ) ] D (s ) = Im for all s, and thus det [ X (s )G (s ) + Y (s ) ] det D (s ) = 1 for all s. This implies that at s = so we must have You're Reading a Preview det [ X (so )G (so ) + Y (so ) ] = ∞ Unlock full access with a free trial. Since the entries of the polynomial matrices X (so ) and Y (so ) are finite, some entry of G (so ) must have infi magnitude. Download With Free Trial Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month. Read Free Foron 30this Days Sign up to vote title   Not useful Cancel anytime. Useful     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition. Khalil - Nonlinear Systems Slides     Save Embed Share Print 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document  CHAPTER 18 Solution 18.2 (a) If x ∈ A (A −1 V ), then clearly x ∈ Im [A ], and there exists y ∈ A −1 V such that x = Ay, which implies Therefore A (A −1 V ) ⊂ V ∩ Im [A ]. Conversely, suppose x ∈ V ∩ Im [A ]. Then x ∈ Im [A ] implies there exi such that x = Ay, and x ∈ V implies y ∈ A −1 V. Thus x ∈ A (A −1 V ), that is, V ∩ Im [A ] ⊂ A (A −1 V ). (b) If x ∈ V +Ker [A ], then we can write x = You're xa + xb ,Reading xa ∈ V , a xPreview b ∈ Ker [A ] −1 and Ax = Axa ∈ A V. Thus x ∈ A −1 (A V ), which ]⊂A V +Ker Unlockgives full access with[A a free trial.(A V ). Conversely, if x there exists y ∈ V such that Ax = Ay, that is, A (x −y ) = 0. Thus writing ∈ A −1 (A V ) ∈ V Free [A ] x = y + (x −yWith +KerTrial Download gives A −1 (A V ) ⊂ V +Ker [A ]. (c) If A V ⊂ W, then using (b) gives A −1 (A V ) = V + Ker [A ] ⊂ A −1 W. Thus V Conversely, V ⊂ A −1 W implies, using (a), Master your semester with Scribd A ⊂ A (A − & The New York Times Therefore A ⊂ . V V W Special offer for students: Only $4.99/month. 1 W ⊂ A −1 W. Read Free Foron 30this Days Sign up to vote title ) = W ∩ Im [A ]   Not useful Cancel anytime. Useful     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload Join RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music  This is the solution manual of Rugh W.J Linear system theory. 2 edition. Khalil - Nonlinear Systems Slides     Save Embed Share Print 1 Download of 106   Table of Contents for Linear Linear System Theory and Linear System Theory, 2/E Solution 18.9   Search document Solutions Clearly C <A | B> = Y if and only if B AB . . . A n −1 B C rank =p and thus the proof involves showing that the rank condition is equivalent to positive definiteness of tf ∫ Ce A (t f −t ) BB T e A T (t f −t ) C T dt 0 This is carried out in Solution 9.11. Solution 18.10 We show equivalence of the negations. First suppose 0 subspace. Then picking a friend F of V we have (A + BF )V ⊂ V Selecting 0 ≠ xo ≠ V ⊂ Ker [C ] is a controlled inva ⊂ Ker [C ] ∈ V, this gives e (A +BF )t xo ∈V , t ≥0 and thus )t You're Reading Ce (A +BF xo = 0a, Preview t≥0 Thus the closed-loop state equation is notUnlock observable, since the zero-input response to x o ≠ 0 is identical to full access with a free trial. zero-input response to the zero initial state. Conversely, suppose the closed-loop state equation is not observable for some F. Then Download With Free Trial n −1 k N = ∩ Ker [C (A + BF ) ] ≠ 0 k =0 Thus 0≠xo ∈ N implies, using the Cayley-Hamilton theorem, Master your semester with = C (A + BF ) x = C (A + BF ) x = . . . 0 = Cx Scribd  Read Free Foron 30this Days Sign up to vote title  inva That is, ( A + BF) x ∈ , which gives (A + BF) ⊂ . Clearly ⊂ Ker [C ], so is a nonzero controlled & The New York Times Useful  Not useful  subspace contained in Ker [C ]. o o N Special offer for students: Only $4.99/month. 2 o N N N o N Cancel anytime.   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Upload Sign In RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1 of 106   Table of Contents for Linear Linear System Theory and  Search document  CHAPTER 19 Solution 19.1 First we show ( W + S) ⊥ = W ⊥ ∩ S⊥ An n × 1 vector x satisfies x ∈ ( W + S) ⊥ if and only if x T (w + s ) = 0 for all w ∈ W and s ∈ S. This is equiv to x T w + x T s = 0 for all w ∈ W and s ∈ S, and by taking first s = 0 and then w = 0 this is equivalent to x for all w ∈ W and x T s = 0 for all s ∈ S. These conditions hold if and only if x ∈ W ⊥ and x ∈ S⊥ , tha x ∈ W ⊥ ∩ S⊥ . You're Reading a Preview Next we show Unlock fullTaccess with a free trial. ( A S) ⊥ = A −1 S⊥ An n × 1 vector x satisfies x ∈ ( A T S) ⊥ if and only if x T y = 0 for all y ∈ A T S, which holds if and on Trial ⊥ as ( Ax )T z =With 0 forFree all z ∈ x T A T z = 0 for all z ∈ S, which is the sameDownload S , which is equivalent to Ax ∈ S , whi equivalent to x ∈ A −1 S⊥ . Finally we prove that ( S⊥ ) ⊥ = S. It is easy to show that S ⊂ ( S⊥ ) ⊥ since x ∈ S implies y T x = 0 fo ⊥ y ∈ S , that is, x T y = 0 for all y ∈ S⊥ , which implies x ∈ ( S⊥ ) ⊥ . To show ( S⊥ ) ⊥ ⊂ S, suppose 0 ≠ x ∈ ( S⊥ ) ⊥ . Then for all y ∈ S⊥ we have x T y = 0. That is, if y T z  T Free Foron 30this Days Sign to vote title y =up all z ∈ S, then x T y = 0. Equivalently, if z T y = 0 for all z ∈ S, then xRead 0. Thus Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month. Ker zT = Ker  T z xT ,  Not useful Cancel anytime. for all z ∈ S Useful    Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload Join RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106 Table of Contents for Linear   Linear System Theory and   Search document Linear System Theory, 2/E Solutions 0 =K k +1 =K V V ∩ A −1 (V k + B ) = V k ∩ A −1 (V k + B ) For k = 0 the claim becomes ( K ⊥ ) ⊥ = K , which is established in Exercise 19.1. So suppose for some noneg integer K we have (W K ) ⊥ = V K . Then, using Exercise 19.1, ⊥ (W K +1 ) ⊥ = W K + A T [ W K ∩ B ⊥ ] A T (W K ∩ B ⊥) ⊥ ∩ A T [ (V K ) ⊥ ∩ B⊥ ] ⊥ = A −1 (V K ) ⊥ = (W K ) ⊥ =VK ∩ But further use of Exercise 19. 1 gives (V K ) ⊥ AT ⊥ ∩ B⊥ ∩ B⊥ ⊥ = A −1 (V K + B) Thus (W K +1 ) ⊥ = V K ∩ A −1 (V K + B) = V K +1 This completes the induction proof, and gives V * = V n = (W n ) ⊥ . You're Reading a Preview Solution 19.4 full access a freeoftrial. We establish the Hint byUnlock induction, for F with a friend V *. For k = 1, k ∩ Free * =Trial * ∩ (A .0 + ( ∩ *) = Σ (A + BF) −Download With j 1 B B V V V B ) j =1 = R1 Assume now that for some positive integer K we have Master your semester with Scribd ( ) − ( ∩ *) = A + BF Σ & The New York Times Then K j 1 B V j =1 R K K −1 ) 30this B on ∩ (Aup ∩For Days = V * Read RFree Sign to vote title  Special offer for students: Only $4.99/month. K +1 Σ (A + BF ) j −1 (B K V *) B V * + (A + BF)  Not useful Cancel anytime. Useful Σ (A + BF) − ( j 1 B V *)     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E Solution 19.7  Solutions The closed-loop state equation . x (t ) = (A + BF )x (t ) + (E + BK )w (t ) +BGv (t ) y (t ) = Cx (t ) is disturbance decoupled if and only if C (sI − A − BF )−1 (E + BK ) = 0 That is, if and only if <A +BFIm [E +BK ]> ⊂ Ker [C ] Thus we want to show that there exist F and K such that (*) holds if and only if Im [E ] ⊂ V * + B, where V maximal controlled invariant subspace contained in Ker [C ] for the plant. First suppose F and K are such that (*) holds. Since <A +BF Im [E +BK ]> is invariant under ( A + is a controlled invariant subspace contained in Ker [C ] for the plant. Then Im [E +BK ] ⊂ <A +BF Im [E +BK ]> ⊂ V * ∈ X there is a v ∈ V * such that (E + BK )x = v. Therefore Ex = v + B (−K x ) You're Reading a Preview which implies Im [E ] ⊂ V * + B. Conversely, suppose Im [E ] ⊂ V * + B, where V * is the maximal controlled invariant subspace conta Unlock access K with a free trial. Im [E +BK ] ⊂ V *. Then we can pi in Ker [C ] for the plant. We first show how to full compute such that That is, for any x friend F of V * and the proof will be finished since we will have Download With Free Trial ]> ⊂ <A +BF V * ⊂ Ker [C ] Im [E +BK If w 1 , . . . , wq is a basis for W, then there exist v 1 , . . . , vq ∈ V * and u 1 , . . . , uq ∈ U such that Ew j = v j + Bu j , j = 1, . . . , q Master your semester with Scribd Let K = −u . . . − u & The New York Times 1 Read Free Foron 30this Days Sign up to vote title q Special offer forThen students: Only $4.99/month. (E + BK ) = Ew BK −1 w1 . . . w q Useful   Not useful Cancel anytime.     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E Since Im [BG 1 ] ⊂ B ∩ R 1 * Solutions ˆ =C P= C 1 1 ˆ C 11 0 0 ˆ =C P= C 2 2 ˆ 0 C 11 0 ⊂ R 1 * and BG 1 = PB̂ 1 we have B̂ 1 = Similarly, Im [BG 2 ] ⊂ B ∩ R 2 * B̂ 11 0 B̂ 13 ⊂ R 2 * gives B̂ 2 = Finally, (A + BF)R i *  0 B̂ 22 ˆ B 23 ⊂ R i *, i = 1, 2, and (A + BF)P = PÂ give Â 11 0 0 0 Â 22 0 Â = Â 31 Â 32 Â 33 You're Reading a Preview That is, with z (t ) = P −1 x (t ), the closed-loop state equation takes the partitioned form . Unlock full access with a free trial. za (t ) = Â 11 za (t ) + B̂ 11 r 1 (t ) . zb (t ) = Â 22 zb (t ) + B̂ 22 r 2 (t ) Download With Free Trial . zc (t ) = Â 31 za (t ) + Â 32 zb (t ) + Â 33 zc (t ) + B̂ 13 r 1 (t ) + B̂ 23 r 2 (t ) ˆ z (t ) y 1 (t ) = C 11 a ˆ z (t ) t) = C Master your semestery (with Scribd & The New York Times 2 Special offer for students: Only $4.99/month. 12 b Read Free Foron 30this Days Sign up to vote title   Not useful Cancel anytime. Useful     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1 of 106   Table of Contents for Linear Linear System Theory and  Search document  CHAPTER 20 Solution 20.1 A sketch shows that v (t ) is a sequence of unit-height rectangular pulses, occurring eve seconds, with the width of the k th pulse given by k/5, k = 0, . . . , 5. This is a piecewise-continuous (actu piecewise-constant) input, and the continuous-time solution formula gives z (t ) = e F (t −to ) t ∫ z (to ) + e F (t −σ) Gv (σ) d σ to Reading a Preview Evaluate this at t = (k +1)T and t o = kT to You're get (k +1)T Unlock trial. FT full access with aFfree (kT+T −σ) z [(k +1)T ] = e z (kT) + ∫ Gv (σ) d σ e kT Let τ = kT+T −σ in the integral, to obtain Download With Free Trial T z [(k +1)T ] = e FT ∫ z (kT) + e F τ Gv (kT +T−τ) d τ 0 Then the special form of v (t ) gives Master your semester with Scribd & The New York Times z [(k +1)T ] = e z (kT) + ∫ − T FT Special offer for students: Only $4.99/month. T u (k )T Read Free Foron 30this Days Sign up to vote title τ τ sgn [u ( e F G dUseful k )] Not useful   Cancel anytime. The integral term is not linear in the input sequence u (k ), so we approximate the integral when  u (k ) is s   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload Join RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download   of 106 Table of Contents for Linear Linear System Theory and   Search document Linear System Theory, 2/E Solutions ũ x̃ = ũ 2 , ỹ = ũ 2 Easy calculation gives the linearized state equation x δ (k +1) = y δ (k ) = −1 0 0 −1 2ũ x δ (k ) + −1 2 u δ (k ) 4ũ x δ (k ) + 2ũ u δ (k ) Since A k = (−1)k I and CB = 0, the zero-state solution formula easily gives y δ (k ) = 2ũ u δ (k ) Thus the zero-state behavior of the linearized state equation is that of a pure gain. Solution 20.10 Φ(k, Computing Φ ( j +q, j ) for the first few values of q ≥ 0 easily leads to the general formul j ): 0 a 1 (k −1)a 2 (k −2)a 1 (k −3)a 2 (k −4) . . . a 1 ( j ) . . . a 2 (k −1)a 1 (k −2)a 2 (k −3)a 1 (k −4) a 2( j) 0 , k − j odd, ≥1 , k − j even, ≥1 You're Reading a Preview a 1 (k −1)a 2 (k −2)a 1 (k −3)a 2 (k −4) . . . a 2 ( j ) 0 Unlock full access with a free trial. 0 a 2 (k −1)a 1 (k −2)a 2 (k −3)a 1 (k −4) . . . a 1 ( j ) Download With Free Trial Solution 20.11 ≥ j +1, j ) = F (k −1)F (k −2) By definition, for k ΦF (k, . . . F ( j +1)F ( j ) Master your semester with (1−k )A (2−k ) . . . A (−1− j )A (− j ) , k ≥ j +1 = A Scribd Read Free Foron 30this Days Sign up to vote title Therefore, for k ≥Times j +1, & The New York  Useful  Not useful T Φ TF (k, j ) = A (− j )A (− j −1) 2, that is, k ≥ j 1, Special offer for students: Only $4.99/month. However, for j 1 ≥ k T T T Cancel anytime. . . . A (−k+ 2)A (−k+1)     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1 of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E Solutions Φ(k, ko ) ≤ Solution 20.16  k −1 1 _____ Φ(k, j ) Φ( j, k ) , k k −k 1 j =k 1 Σ ≥ k 1 +1 ≥ k o +1 Given A (k ) and F we want P (k ) to satisfy F = P −1 (k +1)A (k )P (k ) for all k . Assuming F is invertible and A (k ) is invertible for every k , it is easy to verify that P (k ) = ΦA (k, 0)F −k is the correct choice. Obviously if F = I, then the variable change is P (k ) = ΦA (k, 0). Using this in Exa 20.19, where 1 a (k ) 0 1 A (k ) = gives k −1 P (k ) = ΦA (k, 0) = 1 Σ a (i ) i =0 0 , k ≥1 1 You're Reading a Preview and k −1 trial. Unlock full access with a free P −1 (k +1) = ΦA (0, k +1) = 1 − Σ a (i ) i =0 Download With0Free Trial 1 , k ≥0 Then an easy multiplication verifies the property. Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month. Read Free Foron 30this Days Sign up to vote title   Not useful Cancel anytime. Useful     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1 of 106   Table of Contents for Linear Linear System Theory and  Search document  CHAPTER 21 Solution 21.3 Using z-transforms, (zI − A )−1 = z −1 12 z+ 7 −1 = 1 _________ 2 z +7z+ 12 z +7 1 −12 z and Y (z ) = zc (zI −A )−1 xo + c (zI −A )−1 b U (z ) You're Reading a Preview z z −1 ____ z _________ 1/ 20 _________ − z −19 z −1 1/ 20 + 2 = 2 Unlock full access with a free trial. z +7z+ 12 z −1 z +7z+ 12 =0 Download With Free Trial Therefore the complete solution is y (k ) = 0, k ≥ 0. Solution 21.4 First compute the corresponding discrete-time state equation Master your semester with Scribd (kT) up x ([(k +1)T ] = Fx (kT) + guRead Free Foron 30this Days Sign to vote title & The New York Times y (kT) = hx (kT)  Useful  Not useful Special offer forUsing students: 0, it$4.99/month. is easy to compute A 2 =Only Cancel anytime.     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Upload Sign In Join RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1   of 106 Table of Contents for Linear Linear System Theory and   Search document Linear System Theory, 2/E Solutions x (k ) = (1+r/l )k xo + k −1 Σ (1+r/l ) − − b k j 1 j =0 = (1+r/l )k xo + b (1+r/l )k −1 1−1/(1+r/l )k ____________ 1−1/(1+r/l ) = (1+r/l )k (xo +bl/r ) − bl/r (b) In one year a deposit x o yields x (l ) = (1+r/l )l xo so (1+r/l )l xo − xo _____________ effective interest rate = xo × 100% = [(1+r/l )l − 1] × 100% For r = 0.05, l = 2, the effective interest rate is 5.06%. For r = 0.05, l = 12, the effective interest rate is 5.12% (c) Set 50,000 (−50,000) _______ _________ + You're Reading a0.05 Preview 0.05 0 = x (19) = (1.05)19 xo + and solve to obtain xo = $604,266. Of course this means you have actually won only $654,26 Unlock full access with a free trial. congratulations remain appropriate. Download With Free Trial Solution 21.9 With T = Td / l and v (t ) = v (kT), kT ≤ t ≤ (k +1)T, evaluate the solution formula t z (t ) = e F (t −τ) ∫τ z (τ) + e F (t −σ) Gv (σ−Td ) d σ , t ≥ T Master your semester with Scribd at t = (k +1)T, τ = kT to obtain & The New York Times Special offer for students: Only $4.99/month. Read Free Foron 30this Days Sign up to vote title T ∫ z [(k +1)T ] =e FT z (kT) + e F τ 0  Useful  Not useful d τ G v [(k −Cancel l )T ] anytime.     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload Join RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and   Search document Linear System Theory, 2/E x (k +1) = yˆ (k ) = Solutions 0 .. . 1 .. . . . . . . . 0 0 0 .. . 0 0 0 .. . A 0 . . . B 0 . . . 0 0 . . x (k ) + . 0 0 . . . 1 0 0 1 u (k ) , x (0) = z (0) v (−lT ) . . . v (−2T ) v (−T ) C 0 . . . 0 x (k ) The dimension of the initial state is n +l. The transfer function of this state equation is the same as the tran function of z (k +1) = Az (k ) + Bu (k −l ) y (k ) = Cz (k ) Taking the z -transform, using the right shift property, gives Y (z ) = C (zI −A )−1 Bz −l U (z ) Solution 21.12 Easy calculation shows that for 1 0 0 1 You're Reading , Mba=Preview M a= 0 0 0 0 Ma has a square root, with Unlock full access with a free trial. √Ma = Ma , but M b does not. Download With Free Trial Solution 21.13 By Lemma 21.6, given any k o there is a K -periodic solution of the forced state equation if only if there is an x o satisfying ko +K −1 Φ(ko +K, j +1) f ( j ) [I − Φ(ko +K, ko )]xo = Master your semester with Scribd Σ  Read Free Foron 30this Days Sign up to vote title  Similarly there is Times a K -periodic solution of the unforced state equation ifUseful if there is a z satisfying & The New York Not useful  and only  j =ko Special offer for students: Only $4.99/month. Since there is no [I − Φ(ko +K, ko )]zo = 0 ≠ 0 satisfying (**), it follows that [ I −Φ(k Cancel anytime. o +K k )] is invertible. This implies that for   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1 of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions we have by linear algebra that there exits a nonzero, n × 1 vector p such that [I − Φ(ko +K, ko )]T p = 0 and pT ko +K −1 Σ Φ(ko +K, ∆ j +1) f ( j ) = q ≠ 0 j =ko Now pick any xo . Then it is easy to show that the corresponding solution satisfies p T x (ko +jK ) = p j = 1, 2, . . . . This shows that the solution is unbounded. You're Reading a Preview Unlock full access with a free trial. Download With Free Trial Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month. Read Free Foron 30this Days Sign up to vote title   Not useful Cancel anytime. Useful     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download   of 106 Table of Contents for Linear Linear System Theory and   Search document CHAPTER 22 Solution 22.1 Similar to Solution 6.1. Solution 22.4 If the state equation is uniformly exponentially stable, then there exist γ ≥ 1 and 0 ≤ λ < that Φ(k, j ) ≤ γ λk − j , k ≥ j You're Reading a Preview Equivalently, for every k , Unlock full access withj a free trial. Φ(k +j , k ) ≤ γ λ , j ≥ 0 which implies Download With Free Trial φ j = sup Φ(k +j , k ) ≤ γ λ j k Then φ = lim (γ Master your semester with lim Scribd →∞ →∞ ≤λ & The New York Times j Special offer for students: Only $4.99/month. Now suppose 1/j j 1/j j <1 λ) = λ lim γ1/j j→∞ Read Free Foron 30this Days Sign up to vote title   Not useful Cancel anytime. Useful     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions Similarly, for j > J, Φ(k +j , k ) ≤ sup Φ(k +j , j ) k = φj < (1−ε) j = λ j ≤ γ λj This implies uniform exponential stability. Solution 22.6 For λ = 0 the problem is trivial, so suppose λ ≠ 0 and write k k λk  = k λk = k ( e lnλ ) , k ≥ 0 Let η = −lnλ, so that η > 0 since λ < 1. Then max k λk k≥0 ≤ max t e −η t t≥0 and a simple maximization argument (as in Exercise 6.10) gives 1 ___ You'remax Reading te −η t ≤ a Preview ηe t≥0 Therefore Unlock full access with a free trial. ________ ∆ =β , k ≥ 0 Download −e lnWith λ Free Trial k λk  ≤ 1 To get a decaying exponential bound, write k λk = k ( √λ )k ( √λ )k = 2 β ( √λ )k , k Then Master your semester with Scribd ∞ Σ k λ  & The New York Times k Special offer for students: Only $4.99/month. For j > 1 write k =0 ≥0 Read Free Foron 30this Days Sign up to vote title 2β ≤ _______ 1− √λ   Not useful Cancel anytime. Useful     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1 of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions which is equivalent to ΦA T (−k ) (k, j ) ≤ γ λk − j , k ≥ j which is equivalent to uniform exponential stability of A T (−k ). However for the case of A T (k ), consider the example where A (k ) is 3-periodic with A (0) = 0 2 1/ 2 0 , A (1) = 0 1/ 2 1/ 2 0 , A (2) = 2 0 0 1/ 2 Then ΦA (k ) (3, 0) = 1/ 2 0 0 1/ 2 and it is easy to conclude uniform exponential stability. However ΦA T (k ) (3, 0) = 2 0 0 1/ 8 and it is easy to see that there will be unbounded solutions. You're Reading a Preview Unlock full access with a free trial. Download With Free Trial Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month. Read Free Foron 30this Days Sign up to vote title   Not useful Cancel anytime. Useful     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1 of 106   Table of Contents for Linear Linear System Theory and  Search document  CHAPTER 23 With Q = q I, where q > 0 we compute A T (k )QA (k )−Q to get the sufficient condition uniform exponential stability: Solution 23.1 a 21 (k ), a 22 (k ) ≤ 1− _ν_ , q ν>0 Thus the state equation is uniformly exponentially stable if there exists a constant α < 1 such that for all k ≤α k ),  a 2 (ka)Preview a 1 (Reading You're With Unlock full access with a free trial. Q= q1 0 0 q2 Download With Free Trial where q 1 , q 2 > 0, the sufficient condition for uniform exponential stability becomes existence of a constant such that for all k , q 2 −ν q 1 −ν _____ _____ , a (k ) ≤ (k ) ≤ Master your semester with aScribd  q Free For 30 Days q Read Sign up to vote on this title  These conclusions show uniform exponential stability under weaker one bounded coeffic & The New York Times Useful where Not useful  conditions, can be larger than unity if the other bounded coefficient is suitably small. For example, 2 1 1 2 2 2 Cancel anytime. Special offer for students: Only $4.99/month. sup | a 2 (k ) = α < ∞. Then we can take q 1 = α2 +0.01, q 2 = 1, and ν = 0.01 to conclude uniform expone k   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Upload Sign In RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1 of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions holds if a 21 (k ), a 22 (k ) ≤ α < 1 for all k, but it also holds under weaker conditions. For example suppose th bound is violated only for k = 0, and a 21 (0) > 1 , a 21 (0)a 22 (1) < α Then we can conclude uniform exponential stability. (More sophisticated analyses should be possible . . . Solution 23.6 If the state equation is exponentially stable, then by Theorem 23.7 there is for any symmetr a unique symmetric Q such that A T QA − Q = −M Write M= m1 m2 m2 m3 Q= , q1 q2 q2 q3 and write the discrete-time Lyapunov equation as the vector equation −1 0 1 The condition 0 − 1− a 0 −2 a 20 a0 0 q1 q2 q3 −m 1 −m 2 −m 3 = You're Reading a Preview Unlock full −1access 0 witha 20a free trial. 0 − 1− a 0 a 0 ≠ 0 −2 Free 1 With 0 Download Trial det reduces to the condition a 0 ≠ 0, 1, −2. Assuming this condition we compute Q for M = I, and use the fact Q > 0 since M > 0. The expression −1 0 q Master your semester with Scribd q = 0 −1−a −2 q 1 & The New York Times 1 2 3 0 a 20 a0 0 Special offer forgives students: Only $4.99/month. 2 −1 −1 Read Free Foron 30this Days Sign0 up to vote title 1  −Useful  Not useful Cancel anytime.     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and   Search document Linear System Theory, 2/E Solutions p H A T QAp − p H Qp = −p H Mp That is, ( λ2 −1 )p H Qp = −p H Mp If p H Mp > 0, then λ2 −1 < 0, which gives λ < 1. But suppose p H Mp = 0. Then for k _ k H 2 k 0 = λ p Mp = λ p H Mp λk = p H (A T )k MA k p ≥ 0, = (Re [p ])T (A T )k MA k (Re [p ]) + (Im [p ])T (A T )k MA k (Im [p ]) Since M ≥ 0, this implies 0 = (Re [p ])T (A T )k MA k (Re [p ]) = (Im [p ])T (A T )k MA k (Im [p ]) By hypothesis this implies lim A k (Re [p ]) = lim A k (Im [p ]) = 0 k→∞ k→∞ Therefore lim A k p = lim λk p = 0 k→∞ k→∞ which implies λ < 1. You're Reading a Preview Unlock full access with a free trial. Download With Free Trial Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month. Read Free Foron 30this Days Sign up to vote title   Not useful Cancel anytime. Useful     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload Join RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and   Search document CHAPTER 24 Solution 24.1 Since T A (k )A (k ) = a 22 0 0 a 21 it is clear that /2 (k ) = max [ a 1 (k ), a 2 (k ) ] λ 1max You're Reading a Preview Thus Corollary 24.3 states that the state equation is uniformly stable if there exists a constant Unlock full access with a free trial. k γ such that Π max [ a 1 (i ), a 2 (i ) ] ≤ γ i =j for all k , j with k Download With Free Trial ≥ j. (Note that this condition holds if max [ a 1 (k ), a 2 (k ) ] ≤ 1 for all but a finite number of values of k .) Of course the condition (#) is not necessary. Consider Master your semester with Scribd x (k +1) = & The New York Times 0 1/9 4 0 Free Foron 30this Days to vote title ) up x (Read kSign  Useful  Not useful Cancel anytime. eigenvalues are ± 2 / 3, so the state equation is uniformly stable, but clearly (#) fails. Special offer forThe students: Only $4.99/month.     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload Join RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music  This is the solution manual of Rugh W.J Linear system theory. 2 edition. Khalil - Nonlinear Systems Slides     Save Embed Share Print 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document L in e a r S y s t e m T h e o r y , 2 / E k r (k +1) Π j =ko  S o lu t io n s 1 __________ 1+η( j )ν( j ) k −1 ≤ r (k ) Π j =ko 1 __________ 1+η( j )ν( j ) k Π + ν (k )ψ(k ) j =ko 1 __________ 1+η( j )ν( j ) , k ≥ ko + Iterating this inequality gives r (k ) ≤ k −1 Σ k −1 ν( j )ψ( j ) Π [ 1+η(i )ν(i ) ] , k ≥ k o +1 i = j +1 j =ko and substituting this into (*) yields the result. Solution 24.7 By assumption ΦA (k, j ) ≤ γ for k ≥ j. Treating f ( k, z (k )) as an input, the complete solu formula is z (k ) = ΦA (k, ko )z (ko ) + k −1 Σ Φ (k, j +1) f ( j, z ( j )) , A k ≥ k o +1 j =ko This gives z (k ) ≤ γ z (ko ) + k −1 Σ γ f ( j, z ( j )) j =ko ≤ γ z (ko ) + k −1 Σ γ α z ( j ) , k ≥ k o +1 j j =ko Applying Lemma 24.5, z (k ) ≤ γ z (ko ) exp [ γ k −1 Σα] j j =ko ≤ γ z (ko ) exp [ γ Master your semester with Scribd ≤ γ e z (k ) , & The New York Times This implies uniform stability. γα Special offer for students: Only $4.99/month. For the scalar example o ∞ Σα] j j =ko ≥ ko upFree k Read Foron 30this Days Sign to vote title   Not useful Cancel anytime. Useful     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload Join RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document  CHAPTER 25 Solution 25.1 If M (ko , k f ) is not invertible, then there exists a nonzero, n × 1 vector x a such that 0 = x Ta M (ko , k f )xa k f −1 = x Ta ΦT ( j, ko )C T ( j )C ( j )Φ( j, ko )xa Σ j =ko k f −1 = Σ C ( j )Φ( j, k )x  o a 2 j =ko This implies C ( j )Φ( j, ko )xa = 0 , j = k o , . . . , k f −1 which shows that the nonzero initial state x a yields the same output on the interval as does the zero initial s Therefore the state equation is not observable. On the other hand, for any initial state x o we can write, just as in the proof of Theorem 25.9, Master your semester with Scribd M (k , k )x = O (k , k ) & The New York Times o Special offer for students: Only $4.99/month. If M (k f o T o f y (ko ) . up Read Free Foron 30this Days Sign to vote title . .  Useful  Not useful y (k f −1) Cancel anytime. k ) is invertible, then the initial state is uniquely determined by     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download   of 106 Table of Contents for Linear Linear System Theory and  Search document L in e a r S y s t e m T h e o r y , 2 / E  S o lu t io n s The claim is true if A (k ) is invertible at each k . Let k f = n so that n −1 Φ(n, j +1)b ( j )b T ( j )ΦT (n, j +1) W (0, n ) = Σ j =0 Since Φ(n, j +1) is invertible for j = 0, . . . , n −1, let b (k ) = Φ−1 (n, k +1)ek +1 , k = 0, . . . , n −1 where e k is the k th -column of I n . Then n −1 W (0, n ) = Σe T j +1 e j +1 = In j =0 and the state equation is reachable on [0, n ]. Solution 25.7 Suppose W O (ko , k f ) is invertible. Given a p × 1 vector y f , let − 1 (k , k )y , k = k , . . . , k − 1 u (k ) = B T (k )ΦT (k f , k +1)C T (k f )W O o f f o f and let u (k ) = 0 for other values of k. Then it is easy to show that the zero-state response to this input y y (k f ) = y f . Thus the state equation is output reachable on [ ko , k f ]. Conversely, suppose the state equation is output reachable on [ ko , k f ]. If W O (ko , k f ) is not invertible, there exists a nonzero p × 1 vector y a such that 0 = y Ta WO (ko , k f )ya k f −1 = Σ y C (k )Φ (k , j +1)B ( j )B ( j )Φ (k , j +1)C (k )y T a T f T f T f T f a j =ko k f −1 = Σ y C (k )Φ(k , j +1)B ( j ) T a f f Master your semester with Scribd Therefore & The New York Times y C (k )Φ(k , j +1)B ( j ) = 0 , 2 j =ko Special offer for students: Only $4.99/month. T a f f Read Free Foron 30this Days Sign up to vote title  Not useful . . . Cancel , k f −1anytime. j = k o , Useful But by output reachability, with y f = ya , there exists an input u a (k ) such that     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download   of 106 Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E Solution 25.13  Solutions We will prove that the state equation is reconstructible if and only if C CA . . . z = 0 implies A n z = 0 CA n −1 That is, if and only if the null space of the observability matrix is contained in the null space of A n . First, suppose the state equation is not reconstructible. Then there exist n × 1 vectors xa and xb such xa ≠ xb and C . . . xa = CA n −1 C . . . xb , A n xa ≠ A n xb CA n −1 That is C . . . (x a −x b ) = 0 , A n (xa −xb ) ≠ 0 CA n −1 You're Reading a Preview Thus the condition (*) fails. Now suppose the condition (*) fails Unlock and z isfull such thatwith a free trial. access C . . Download Free 0 and z =With A nTrial z≠0 . CA n −1 Obviously z ≠ 0. Then for x (0) = z the zero-input response is Master your semester with Scribd y (k ) = 0 , k = 0, . . . , n −1  Read Free Foron 30this Days Sign up to vote title  and x (n ) ≠ 0. But the same output sequence is produced by x (0) = 0, and for this initial state x (n ) = 0. Thu & The New York Times Useful  Not useful  cannot determine from the output (+) whether x (n ) = z or x (n ) = 0, which implies the state equation is Special offer forreconstructible. students: Only $4.99/month. Cancel anytime.   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document  CHAPTER 26 Solution 26.2 For the linear state equation x (k +1) = 1 k 1 1 x (k ) + 0 1 u (k ) easy computations give R 2 (k ) = and R 3 (k ) = B (k ) B (k ) Φ(k +1, k )B (k −1) = You're Reading a Preview 0 k 1 1 full access with a free trial. Φ(k +Unlock 1, k )B (k −1) Φ(k +1, k −1)B (k −2) = 0 k 2 k −1 1 1 k From the respective ranks the state equation is 3-step reachable, but Trial not 2-step reachable. Download With Free Solution 26.4 The (n +1)-dimensional state equation A 0 b Master your semester withz (Scribd ) k +1) = z (k ) + Read u (kFree Foron 30this Days to vote title c 0 dSign up & The New York Times Useful  Not useful (k ) y (k ) = 0 × 1 z (k ) − u Special offer for students: Only $4.99/month. has the transfer function 1 n Cancel anytime.     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1 of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions m G (z ) = σl ΣΣG lr l =1 r =1 z ______ (z −λl )r _ Here λ1 , . . . , λm are distinct complex numbers __ such that if λL is complex, then λM = λL for some M . Furtherm the p × m complex matrices satisfy GMr = GLr for r = 1, . . . , σL . From Table 1.10 the corresponding unit p response is m σl k 1−r λ k+ G (k ) = Glr l − l 1 l =1 r =1 ΣΣ Thus we can state that a unit pulse response G (k ) is realizable if and only if (a) there exist positive integers m , σ1 , . . . , σm , distinct complex numbers λ 1 , . . . , λm , and σ 1 + . . . +σm com for all k ≥ 1, and p × m matrices G lr such that (#) holds _ (b) if λ L is complex, then λ M = λL for some M . Furthermore the p × m complex matrices satisfy GMr = r = 1, . . . , σL . Solution 26.8 Suppose the given state equation is minimal and of dimension n. We can write its (str proper, rational) transfer function as . adj( zI −A ) . b _c____________ G (z ) = det (zI −A ) You're Reading a Preview where the polynomial det ( zI −A ) has degree n. If the numerator and denominator polynomials have a com root, then this root can be canceled without changing inverse z transform of G (z ). Therefore, follo Unlock full accessthe with a free trial. Example 26.10, we can write by inspection a dimension-( n −1) realization of the unit pulse response of original state equation. This contradicts the assumed minimality, and the contradiction gives that the polynomials cannot have a common root. Download With Free Trial Now suppose the polynomials det ( zI −A ) and c . adj( zI −A ) . b have no common root, but that the given equation is not minimal. Then there is a minimal realization z (k +1) = Fz (k ) + gu (k ) Master your semester with Scribd Read Free Foron 30this Days Sign up to vote title y (k ) = hz (k ) & The New  Useful  Not useful and we York then have Times Special offer for students: Only $4.99/month. Cancel anytime. . adj (zI −A ) . b h . adj (zI −F ) . g ______________ _c____________ = det (zI F ) det (zI A )     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1 of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions From the latter approach, setting cb = 0, cAb = 1, cA 2 b = 1/ 2, cA 3 b = 1/ 2 easily yields c 1 = 0, c 0 = 1, a 0 = 1/ 4, a 1 = 1/ 2. You're Reading a Preview Unlock full access with a free trial. Download With Free Trial Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month. Read Free Foron 30this Days Sign up to vote title   Not useful Cancel anytime. Useful     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Upload Sign In RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Download Khalil - Nonlinear Systems Slides 1 of 106   Table of Contents for Linear Linear System Theory and  Search document  CHAPTER 27 Solution 27.1 Similar to Solution 12.1. Solution 27.4 Suppose the entry G ij (z ) has one pole at z = 1, that is Gij (z ) = Nij (z ) __________ (z −1)Dij (z ) You're Reading a Preview where all roots of the polynomial D ij (z ) have magnitude less than unity (so D ij (1) ≠ 0), and the polynomial satisfies Nij (1) ≠ 0. Suppose that the m ×Unlock 1 U (zfull ) has all components zero except for U j (z ) = z /(z −1). The access with a free trial. i th -component of the output is given by (z ) Trial z N ijFree ___________ Download With Y (z ) = i (z −1)2 Dij (z ) By partial fraction expansion y i (k ) includes decaying exponential terms, possibly a constant term, and the term (1) _N _____ Master your semester with Scribd k, k≥0  D (1) Read Free Foron 30this Days Sign up to vote title  & The New York Since this term is Times unbounded, every realization of G (z ) fails to be uniform bounded-input, bounded-output sta  Useful  Not useful ij ij Special offer for students: Only $4.99/month. Cancel anytime.   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions µ = sup u (k ) , k≥0 η= ∞ Σ G (k ) k =0 The first constant is finite for a well-defined sequence that goes to zero, and the second is finite by unif bounded-input, bounded-output stability. Then there is a positive integer K 1 such that ∞ ε ε ___ ___ , k ≥ K1 , G (k ) ≤ u (k ) ≤ 2µ 2η k =K 1 Σ Let K = 2K 1 . Then for k ≥ K we have y (k ) ≤ µ K 1 −1 Σ j =0 ≤µ k Σ k ε ___ G (k − j ) G (k − j ) + 2 η k =K 1 q =k −K 1 Σ ε k −K 1 ___ G (q ) G (q ) + 2 η q =0 Σ ε 2η ε ___ η=ε + ≤ µ ___ 2µ Solution 27.8 Similar to Solution 12.12. You're Reading a Preview Unlock full access with a free trial. Download With Free Trial Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month. Read Free Foron 30this Days Sign up to vote title   Not useful Cancel anytime. Useful     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music Join  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and   Search document CHAPTER 28 Solution 28.2 Lemma 16.18 gives that if V 11 and V are invertible, then V −1 = V 11 V 12 V 21 V 22 −1 V −111 +V−111 V 12 V −a1 V 21 V−111 −V−a1 V 21 V−111 = −V−111 V 12 V−a1 −1 V −a1 where Va = V 22 −V 21 V −111 V 12 . From the expression V V −1 = I, written as You're Reading a Preview V 11 V 12 W 11 W 12 =I V 21 full V 22access Wwith 21 W Unlock a 22 free trial. we obtain Download With Free Trial V 11 W 11 + V 12 W 21 = I V 21 W 11 + V 22 W 21 = 0 Under the assumption that V and V are invertible these imply Master your semester with Scribd −V− up W = V − − V− V W , W =Read VFree Foron 30this Days Sign toWvote title & The New Times SolvingYork for W gives  Useful  Not useful 11 22 11 1 11 1 11 12 21 1 22 21 11 Special offer for students: Only $4.99/month. W 11 = (V 11 −V 12 V −1 V 22 −1 21 ) 21 11 Cancel anytime.     Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload Join RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download   of 106 Table of Contents for Linear Linear System Theory and  Search document Linear System Theory, 2/E  Solutions T T K = −Bˆ (Aˆ )n n Σ k T T Aˆ Bˆ Bˆ (Aˆ )k −1 Aˆ n +1 k =0 That is, K = −α B (α A ) T n T n −1 Σ (α A ) (α B)(α B) (α A ) k T (α A )n +1 T k k =0 = −B (A ) T T n n Σ −1 α−2(n −k ) A k BB T (A T )k A n +1 k =0 Solution 28.4 Similar to Solution 13.11. However for the time-invariant case the reachability matrix rank can be used, rather than the eigenvector test, by writing B (A+BK )B (A+BK )2 B .. . = B AB A 2 B .. . I KB KAB+(KB )2 . . . .. . 0 I KB .. . 0 0 I . . . . . . . . . . . . You're Reading a Preview Solution 28.6 Similar to Solution 2.8. Unlock full access with a free trial. Solution 28.8 Download With Free Trial Supposing that the linear state equation is reachable, there exists K such that all eigenvalu A+BK have magnitude less than unity. Therefore ( I −A −BK ) is invertible, and if we suppose A −I B C 0 Master your semester with Scribd  − Free Foron 30this Days Sign up to vote titlem × m matrix is invertible, then C (I −A −BK ) B is invertible from Exercise 28.6. Read Then given any diagonal,  can choose & The New York Times  Useful  Not useful 1 Special offer for students: Only $4.99/month. ˆ N = [C (I −A −BK )−1 B ]−1 Λ Cancel anytime.   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join   Search  Home  Saved 242 views 1  Sign In Upload Join RELATED TITLES 0 Linear System Theory 2 e Sol Uploaded by Shruti Mahadik  Books  Audiobooks  Magazines  News  Documents  Sheet Music  This is the solution manual of Rugh W.J Linear system theory. 2 edition.     Save Embed Share Print Khalil - Nonlinear Systems Slides 1 Download of 106   Table of Contents for Linear Linear System Theory and  Search document  CHAPTER 29 Solution 29.1 The error e b (k ) satisfies eb (k +1) = z (k +1) − Pb (k +1)x (k +1) ˜ (k )C (k )−P (k +1)A (k )]x (k ) + [G˜ (k )−P (k +1)B (k )]u (k ) = F˜ (k )z (k ) + [G b b a b = F˜ (k )z (k ) − F˜ (k )Pb (k )x (k ) You're Reading a Preview = F˜ (k )eb (k ) Therefore e b (k ) → 0 exponentially as k full access with a free trial. →Unlock ∞. Now ∆ e (k ) = x (k ) − xˆ (k ) Download With Free Trial = x (k )−H (k )C (k )x (k )−J (k )z (k ) = −J (k )eb (k ) + [I −H (k )C (k )−J (k )Pb (k )]x (k ) (k )e (k ) = −JScribd Master your semester with  Read Free Foron 30this Days Sign up to vote title k → Therefore if J (k ) is bounded, that is, J (k ) ≤ α < ∞ for all k , then e (k ) → 0 implies e (k ) → 0, as & The New York Times Useful Not useful   x̂ (k ) is an asymptotic estimate of x (k ). b b Special offer for students: Only $4.99/month. Cancel anytime.   Home  Saved  Books  Audiobooks  Magazines  News  Documents  Sheet Music Master your semester with Scribd & The New York Times Special offer for students: Only $4.99/month.  Upload Sign In Read Free For 30 Days Cancel anytime. Join 

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