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 0 0 -4 5 -9 0 -1 35 -1 80 1 T 1 1 T 2 Module ... 10 0 -1 0 -2 0 -3 0 0 -4 5 -9 0 -1 35 -1 80 1 T 1 1 T 2 Module (dB)Phase (degree) Figure 1. 2...
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... 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...
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... modern perspective”, Proceed- ings of the IEEE, vol. 69, no. 11 , p. 13 80 14 19, 19 81. [KWA 91] K WAKERNAAK H., SIVAN R., Modern Signals and Systems, Prentice-Hall, 19 91. [LAR 75] DE LARMINAT P., ... Prentice-Hall, 19 75. [PAP 71] P APOULIS A., Probabilités, variables aléatoires et processus stochastiques, Dunod, Paris, 19 71. 15 0 Analysis and Control of Linear Sy...
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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 1 0000 1 2 2 1 0 10 00 0. .00 0 10 0 01 0...
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Analysis and Control of Linear Systems - Chapter 12 pptx
... plane, of the direct sensitivity functions d σ and complementary sensitivity functions c σ : )()()()()( )()()( 11 111 1 11 1 −−−−−− −−− +∆ ∆ = qRqBqqqSqA qqSqA d σ [12 .13 ] )()()()()( )()( 11 111 1 11 1 −−−−−− −−− +∆ = qRqBqqqSqA qRqBq c σ ... −−− ∆=− + 11 1 ()() ()() ()() r Sq ut Rq yt T q wt [12 .25] with, based on relations [12 .17 ] and [12 .19 ]: − −−−−−−− − =∆+ 1...
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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 0 2 212 2 11 1 11 12 00 0 21 2 21 00 0 0 (): : 00 ee ee ses ss es ses ss es ee WW WW WWW WW WW WWW ... matrix. Methodology of the State Approach Control 409 with: () () () () () () () () [] () [] . ,, , , 21 1 2 212 11 1 2 212 2 1 2 212 1 21 1 22 21 1 22 21 22...
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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 ... T q T 1 1 − − = ∆ δ and operator )1( )1( 1 −+= − ∆ qqw D will be used sometimes. Discrete-Time Systems 99 In addition, if we note by 1d p and by 2d p the...
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Analysis and Control of Linear Systems - Chapter 4 ppt
... EH 011 *1* 11 11 1 1 {0}, , iin AA AAA A [4 .18 ] +− − == ⇒ == 011 *1* 22 222 2 ,, iin HE 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 l ij being then equal to 0): 11 0 Analysis and Control of Linear Systems The object of this chapter is to describe cert...
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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...
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Analysis and Control of Linear Systems - Chapter 7 docx
... K 2 N 20 1+ 1 2 n 2 N 20 [7.22] Equation [7 .15 ] thus becomes: Q e0 + Q 0 − K 1 N 10 1+ 1 2 n 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: Q 0 − n 1 S 1 = A 1 dn 1 dt [7.25] Based ... development of the square root leads to: Q s1 = K 1 N 10 1+ 1 2 n 1 N 10 [7. 21] 222 Analysis and Contro...
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