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by: Jada Daniel

RobustControlSystemsI MEM633

Jada Daniel
GPA 3.65


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Class Notes
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This 12 page Class Notes was uploaded by Jada Daniel on Wednesday September 23, 2015. The Class Notes belongs to MEM633 at Drexel University taught by Bor-ChinChang in Fall. Since its upload, it has received 40 views. For similar materials see /class/212375/mem633-drexel-university in Mechanical Engineering at Drexel University.

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Date Created: 09/23/15
MEM633 LeCtUl es 1amp2 A system H is said to be causal if and only if its response to an input does not depend on future Chapter 1 Introduction values of that input 11 Properties of Systems ut yt ya H ut A system H is said to be instantaneous if its output is a function of the input at the present time A system H IS saId to be linear If and only If only HC1u102u2 C1 Hu1 02Hiu2 where 01 and 02 are arbitrary real or complex numbers A system H is said to be dynamic if its output depends on past and present values of the input Asystem H is said to be timeinvariant if and only if HUtr ytr for any ut and any I 12 Representations of Systems 1 Differential Equations ut yt n n1 ad mha d tf l dtn n391oltn1 m b MD m tm m d 2 Transfer Functions a0 yt m1 1W b0ut dt 03 98 A h s 9s Gs 8 Impulse Response 5t 93 ht L N0 its yt ht mixmm 4 Statespace Representation xt state vector Sltt Axt But V0 0X0 13 An Example Choose state variables as Inverted pendulum position system X1 6 x2 6 X3yx4y Then Governing equations State equation 1 X 0 1 0 0 1 0 After linearization X2 0c 0 0 0 X2 B o o o 1 x of X 3 3 y 0 0 0 X 5 14 W P plant ie a system to be controlled y1 system output vector y1 P u d d disturbance input vector u control input vector r reference input vector command 2 controlled output the output to be controlled 2 y1 r y measured output ie a vector which consists of all the variables can be measured Q a controller compensator to be found u Qy Objectives To find a realizable controller Q such that the closedloop system is internally stable and has some desired performance Make Z as small as possible What does it meant by quota small 2quot What kind of disturbances and references we are dealing with Constraints on the control inputs Rebusit stability System remains stable under plant pertur bations Rebust performance 2 remains quotsmallquot under plant perturbations 15 State Eguations x0 Axt But VG 0X0 Choose a new state vector Xt as 1 Xt T Xt where T is a nonsingular matrix Then This transformation is referred as similarity transformation A system can have many statespace descriptions Diagonalization xt Axt But yt CXG 1 To find a nonsingular matrix T such that T AT is a diagonal matrix Suppose we can find a nonsingular matrix T such that 1 I T AT Ad dlagx1x2 xn ie AT TAd Let ei i12n be column vectors of T ie T e1 e2 en then Aei Ki ei i12n It is clear that xi must be an eigenvalue of A and ei a corresponding eigenvector Hence a nonsingular T can be found if and only if A has n linearly independent eigenvectors If M k2 kn are distinct then their corresponding eigenvectors e1 e2 en are linearly independent But converse is not necessarily true 9 Solution of the State Eguation Xt Axt 4 But yt CXG After taking Laplace transform and some simple manipulations we have Xs s1 A1x0 s1 A1BUs The solution Xt is simply the inverse Laplace transform of Xs or At At Xt e x0 e But State transition matrix At 1 1 e L sIA 2 e IAt By CayleyHamilton Theorem At 391 k e Z ocktA k0 where 0510 k012n1 can be solved from M 391 e39 Z ocktt i12n k0 I Transfer functions The transfer function matrix is 1 Hs C 31 A B and the impulse response matrix is At Ht C e B In scalar case the expression for Hs can be written as 1 c 31 A b cAdj sI Ab det sI A bsas bs and as may have some common factors so that we can write bsas brsars where brs ars are relatively prime ie have no common factors except possibly constants The poles and zeros of Hs are defined as the roots of the polynomials ars and brs respectively The definitions of poles and zeros of multivariable systems will be given later 16 Stability BIBO stability External stability l l ar 394 ya Def A system is said to be BIBO stable externally stable if for each M1 lt 00 there exists M2 lt 00 such that ut 3 M1 implies yt 3 M2 Theorem A linear timeinvariant system with impulse response ht is BIBO stable if and only if i hrdr M lt 00 Theorem A linear timeinvariant system with transfer function Hs bsas is BIBO stable if and only if all the poles of Hs lie in the left half of splane ie the real parts of all the poles are negative Remark Suppose bs as are coprime then poles of Hs zeros of as Internal stability Def The linear timeinvariant system 51 Axt But VG CXG is internally stable if the solution xt of Xt Axt withinitialstatex0 tends toward zero as t gt 00 for arbitrary xO Theorem A linear timeinvariant system is internally stable if and only if all the eigenvalues have negative real parts Unstable polezero cancellation Consider a system with transfer function 1 PS a This system is unstable To stabilize it let39s try the compensation technique shown in Fig161 with the compensator s1 V y Fig 161 Cs Then we have the overall transfer function 1 517 1 PltSgtCSgt E m 51 It looks nice but unfortunately this technique will not work In practice it is difficult to ensure exact cancellation because of variations in component values etc Even with perfect cancellation this technique still does not work To see why let39s first set up an analogcomputer simulation as shown in Fig162 Then we have the state equations x1 t x1 t 2 x2t vt x10 x1O x2t x2t vt x20 x20 and the output equation yt X1 0 By solving these equations we have yt x1 t x1O et x2O e39t et e39t vt where denotes convolution We can see that the output yt will grow without bound unless the initial conditions can always be kept zero Now let39s try the following V y Fig163 The analogcomputer simulation of the system is shown in Fig164 Fig 164 Then we have the dynamic equations 561 I X10 vt 3610 xlo x20 2 x10 x20 x20 x20 yt 3610 29620 The solution is t t t yt x1O e x2O e e vt yt looks OK but the system is still internally unstable Feedback Connection Thetransfermatrixfrom ds to ys is 1 Psos 1 391 The feedback system is BIBO stable iff all the 1 poles of 1 PsQs are in LHP ie all the zeros of det 1 PsQs are in the LHP States ace re resentation Q Define then The closedloop system is internally stable iff all the zeros of sI AP BFCQ det BQCP sI AQ are in LHP Theorem Let Ps and Qs be characteristic polynomals of systems P and Q respectively ie Ps det sI AP Qs det sI AQ Then the closedloop system is internally stable iff all the zeros of ltlgtpS QS det 1 F SQSi are in the LHP ie with strictly negative real parts Lemma X W are Invertible then det X Y 1 1 Z W XWIZX YW Lemma Let M mxn matrix N nxm matrix 1m mxm identity matrix n nxn identity matrix Then ImMN InNM Proof of the Theorem Remark Internal stability 3 BIBO stability lt


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