Analysis and Control of Linear Systems - Chapter 1 pptx
Analysis and Control of Linear Systems - Chapter 1 pptx
... Figures 1. 23, 1. 24, 1. 25 and 1. 26 illustrate the diagrams presenting the aspect of the frequency response for a second order system with different values of ξ. 10 0 -1 0 -2 0 -3 00 -4 5 -9 0 -1 35 -1 80 1 T 1 1T2Module ... 10 0 -1 0 -2 0 -3 00 -4 5 -9 0 -1 35 -1 80 1 T 1 1T2Module (dB)Phase (degree) Figure 1. 23. Bode diagram of a second order system with ξ ≥ 1 20 Analysis and Control of Linear Systems Table 1. 1 sums up the features of a ... of Linear Systems Figure 1. 19. Unit-step response of the second order system ξ ≥ 1 1= ξ: critical state. The roots of the denominator of the transfer function are real and merged, and...
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Analysis and Control of Linear Systems - Chapter 0 pptx
... systems analysis and control of linear systems/ edited by Philippe de Larminat. p. cm. ISBN -1 3 : 97 8 -1 -9 0520 9-3 5-4 ISBN -1 0 : 1- 9 0520 9-3 5-5 1. Linear control systems. 2. Automatic control. I. Larminat, ... 11 .5.4. Examples 363 11 .6. Conclusion 370 11 .7. Bibliography 370 xii Analysis and Control of Linear Systems Chapter 12 . Predictive Control 373 Patrick BOUCHER and Didier DUMUR 12 .1. ... spaces, linear applications 11 0 4 .1. 2. Invariant sub-spaces 11 1 4 .1. 3. Polynomials, polynomial matrices 11 3 4 .1. 4. Smith form, companion form, Jordan form 11 4 4 .1. 5. Notes and references 11 5 4.2....
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Analysis and Control of Linear Systems - Chapter 5 pptx
... modern perspective”, Proceed-ings of the IEEE, vol. 69, no. 11 , p. 13 80 14 19, 19 81. [KWA 91] KWAKERNAAK H., SIVAN R., Modern Signals and Systems, Prentice-Hall, 19 91. [LAR 75]DE LARMINAT P., ... Prentice-Hall, 19 75.[PAP 71] PAPOULIS A., Probabilités, variables aléatoires et processus stochastiques, Dunod,Paris, 19 71. 15 0 Analysis and Control of Linear Systems where 1 0,N 1 is the ... transform of theconvolution integral is the product of transforms:(y 1 ⊗ y2)=y 1 y2[5.9] 15 2 Analysis and Control of Linear Systems Figure 5 .1. Typical power spectrum of an MA (left)or...
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Analysis and Control of Linear Systems - Chapter 11 pptx
... on [11 .14 ]: ∂≤ ∂∂ − 1 Xmax{V,B(PU)} [11 .16 ] If P ≠ U, then: ∂−≥∂ 11 B(P U) B [11 .17 ] and since by construction: ∂<∂ 1 VB [11 .18 ] ∂∂ − ≥∂ 11 max{ V, B (P U)} B [11 .19 ] 340 Analysis and ... ⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎣⎡−−+−−+−−+=⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣⎡−−−−−=−−−−−− 11 11 111 10000 1 22 1 0 10 000..000 10 0 01 0 00AnnnnnnnnnnnnBρρσττσρρσττσρρσττσσσσσσ####%%####%%%####%%% [11 .59] 11 .4.3. Management of saturations 11 .4.3 .1. Method ... ∂<∂ 1 VB [11 .13 ] by replacing in [11 .12 ] we obtain: =− − 1 XVB(PU) [11 .14 ] the solution for [11 .3] with minimum degree in X will be obtained for P = U or: =⎫⎬=+⎭ 01 XVYY AU [11 .15 ]...
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Analysis and Control of Linear Systems - Chapter 12 pptx
... plane, of the direct sensitivity functions dσ and complementary sensitivity functions cσ: )()()()()()()()( 11 111 1 11 1−−−−−−−−−+∆∆=qRqBqqqSqAqqSqAdσ [12 .13 ] )()()()()()()( 11 111 1 11 1−−−−−−−−−+∆=qRqBqqqSqAqRqBqcσ ... −−−∆=− + 11 1()() ()() ()()rSq ut Rq yt T q wt [12 .25] with, based on relations [12 .17 ] and [12 .19 ]: −−−−−−−−−=∆+ 1 111 111 1 1 ()() (()()() ()())()rrPqTq Aq q Sq q Bq RqAq [12 .26] ... )()()()( 11 1twqBqtyqArrr−−−= [12 .17 ] where:)()()( 11 1 −−−= qPqBqBr. 376 Analysis and Control of Linear Systems 12 .2. Generalized predictive control (GPC) 12 .2 .1. Formulation of the control law The objective of...
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Analysis and Control of Linear Systems - Chapter 13 pptx
... ⎡⎤⎢⎥⎢⎥⎡⎤⎢⎥⎢⎥⎢⎥==⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎢⎥⎢⎥⎢⎥⎣⎦00 0 0 11 2 12 00 0 0 11 1 11 12 11 112 00 0 02 212 2 11 1 11 1200 0 21 2 21 00 00(): :00eeeeses ss esses ss eseeWWWWWWW WW WWWWW ... matrix. Methodology of the State Approach Control 409 with: () ()() ()()()() ()[]()[].,,,, 21 12 212 11 12 212 2 1 2 212 1 21 122 21 122 21 22 1 222 1 22 1 2222 1 222DDDDIDDDCDDIDCDDDIDCCDDDIBDDDDIBBBDDDDIBACDDIBCDDIBCDDDIBAAKKbfKKKKbfKKKKbfKKKKKKKKKKbf−−−−−−−−−−+=−−+=⎥⎥⎦⎤⎢⎢⎣⎡−−+=⎥⎥⎦⎤⎢⎢⎣⎡−+−−−+= ... section 13 .6.2). Under the dual assumptions of those stated previously, we obtain: 1 212 1 211 2)()()(−+Σ−== DDDBCDsKTTTKIS2H [13 .14 ] with: 1 2 1 21 21 21 2 21 1 1 10 ( positive semi-defined)()()()0TT...
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Analysis and Control of Linear Systems - Chapter 3 pps
... 2002. 94 Analysis and Control of Linear Systems NOTE 3.3.– the class of rational systems that can be described by [3 .16 ] or [3 .18 ] is a sub-class of DLTI systems. To be certain of this, let ... TqT 1 1−−=∆δ and operator )1( )1( 1 −+=−∆qqw D will be used sometimes. Discrete-Time Systems 99 In addition, if we note by 1dp and by 2dp the poles and 1dz the zero of )(qH ... and the mutual operator q 1 are natural and hence very dispersed. Starting, successively, from representations [3 .11 ], [3 .12 ] and [3 .15 ], we obtain expressions [3 .16 ], [3 .17 ] and [3 .18 ] of...
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Analysis and Control of Linear Systems - Chapter 4 ppt
... EH 011 *1* 11 11 1 1 {0}, ,iinAA AAA A [4 .18 ] +− −== ⇒ == 011 *1* 22 222 2,,iinHE HEAXA A AA A [4 .19 ] The regularity of the beam [pE-H] can be translated: ⊕=** 12 , AAXi.e. +=** 12 ... eigenvalues of A appear (certain powers lij being then equal to 0): 11 0 Analysis and Control of Linear Systems The object of this chapter is to describe certain structural properties of linear systems ... ⎥⎥⎦⎤⎢⎢⎣⎡−++=⎥⎦⎤⎢⎣⎡−+⎥⎦⎤⎢⎣⎡+⎥⎦⎤⎢⎣⎡ 1 1 21 10 11 01 21 00 10 22pppppp [4 .11 ] A polynomial matrix is called unimodular if it is square, reversible and polynomial reverse. A square polynomial matrix is unimodular if and...
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Analysis and Control of Linear Systems - Chapter 6 ppsx
... the condition of detectability by a careful choice of matrix Q. 16 2 Analysis and Control of Linear Systems case of multi -control systems. As indicated in Chapter 2, the equations of state can ... stronger control. 17 6 Analysis and Control of Linear Systems Figure 6.6. Stabilization by quadratic optimization 16 6 Analysis and Control of Linear Systems Figure 6.4. Structure of the observer ... [6 .14 ] We can verify that this system is controllable: 2 11 10 rank ) (rank =−=⎟⎠⎞⎜⎝⎛ABB [6 .15 ] We obtain, with= 12 ()Kkk : λλλλλ−−− = =++ +++2 21 12 1 det ( ( )) (1 ) 1 IABK...
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Analysis and Control of Linear Systems - Chapter 7 docx
... K2N20 1+ 1 2n2N20[7.22]Equation [7 .15 ] thus becomes:Qe0+ Q0− K 1 N 10 1+ 1 2n 1 N 10 = A 1 dN 1 dt= A 1 dn 1 dt[7.23]If:S 1 =2√N 10 K 1 [7.24]then:Q0−n 1 S 1 = A 1 dn 1 dt[7.25]Based ... development of the square root leads to:Qs1= K 1 N 10 1+ 1 2n 1 N 10 [7. 21] 222 Analysis and Control of Linear Systems iteration, unstable), interrupt the simulation and the calculation of sensitivity ... 1. 38 9 .18 e 1 + 2.23e 1 i −4.72e 1 Standard deviation 5.2e 1 5.2e 1 1 .19 e−2 5.0e−2Parameter b0b 1 a 1 a2a3Value −8.93e 1 1.23 1. 36 2.50e−2 4.22e 1 Standard deviation 2.4e 1 2.4e 1 4.9e−2...
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