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### multiport_capaci

Course: MECHANICAL 2.141, Fall 2006
School: MIT
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Word Count: 2070

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COUPLING ENERGY-STORING BETWEEN DOMAINS MULTI-PORT ENERGY STORAGE ELEMENTS Context: examine limitations of some basic model elements. EXAMPLE: open fluid container with deformable walls P=gh h=AV V = Cf P where Cf = A g --fluid capacitor But when squeezed, h (and hence P) may vary with time even though V does not. Seems to imply Cf = Cf(t) i.e., Cf = A(t) g --apparently a &quot;modulated capacitor&quot;...

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COUPLING ENERGY-STORING BETWEEN DOMAINS MULTI-PORT ENERGY STORAGE ELEMENTS Context: examine limitations of some basic model elements. EXAMPLE: open fluid container with deformable walls P=gh h=AV V = Cf P where Cf = A g --fluid capacitor But when squeezed, h (and hence P) may vary with time even though V does not. Seems to imply Cf = Cf(t) i.e., Cf = A(t) g --apparently a "modulated capacitor" PROBLEM! V2 Ep = 2 C f V is constant, therefore no (pressure) work done dV = 0 PdV = 0 --yet (stored) energy may change This would violate the first law (energy conservation) P where did this energy come from? initial stored energy V --a BIG problem! MODULATED ENERGY STORAGE IS PROSCRIBED! SOLUTION Identify another power port to keep track of the work done to change the stored energy F v = dx/dt P Q = dV/dt C introduces a new network element: a multiport capacitor Mathematical foundations: Power variables: Each power port must have properly defined conjugate power and energy variables. Net input power flow is the sum of the products of effort and flow over all ports. P = ei fi i In vector notation: e f e1 2 _ e en P = et f = ft e f1 2 _ f fn Energy variables Energy variables are defined as in the scalar case as time integrals of the flow and effort vectors respectively. generalized displacement q = f dt + qo generalized momentum p = e dt + po MULTI-PORT CAPACITOR A "vectorized" or multivariable generalization of the one-port capacitor. definition A multiport capacitor is defined as an entity for which effort is a single-valued (integrable) function of displacement . e = (q) The vector function () is the capacitor constitutive equation. --a vector field (in the mathematical sense of the word). A multi-port capacitor is sometimes called a C-field or capacitive field. BOND GRAPH NOTATION By convention, power is defined positive into all ports. e1 f1 = dq1/dt e2 e3 f3 = dq3/dt f2 = dq2/dt C An alternative notation: n n denotes the number of ports. More on this multi-bond notation later. STORED ENERGY: determined by integrating the constitutive equation. Ep - Epo = etf dt = etdq = (q)tdq = Ep(q) C potential energy, as it is a function of displacement --a function of as many displacements as there are ports. COUPLING BETWEEN PORTS. Each effort may depend on any or all displacements. ei = i(q1,q2,... qn)all i This coupling between ports is constrained. Mathematically: Energy stored is a scalar function of vector displacement. Stored energy is a scalar potential field. The effort vector is the gradient of this potential field. e = qEp(q) Therefore the constitutive equation, (), must have zero curl. e=0 or =0 In terms of the individual efforts and displacements, ei = E p qi all i ei E p E p e j = = = q j q j qi qi q j q i all i, j. Coupling between ports must be symmetric. The dependence of ei on qj must be identical to the dependence of ej on qi. This is known as Maxwell's reciprocity condition. Later we will see that stability and passivity further constrain the capacitor constitutive equation. EXAMPLE: "CONDENSER" MICROPHONE HEADS UP! There's an error in what follows -- see if you can spot it. The sketch depicts a simple electro-mechanical transducer, a "condenser" microphone. --essentially a moving-plate capacitor. Electrically a capacitor, but capacitance varies with plate separation. Mechanically, electric charge pulls the plates together. DEVICE CAN STORE ENERGY. Energy can change in two ways: mechanical displacement charge displacement --a two-port capacitor F v = dx/dt e i = dq/dt C like a spring in the mechanical domain like a capacitor in the electrical domain Two constitutive equations needed both relate effort to displacement e = e(q, x) F = F(q, x) Assuming electrical linearity: q e = C(x) To find C(x) assume a pair of parallel plates very close together. (i.e., plates are very large compared to their separation) (Fringing effects can be handled in a completely analogous way) C = 0A/x is a permittivity A is plate area One constitutive equation is e= qx A To find the other constitutive equation we could work out the attraction due to the charges on the plates ... AN EASIER WAY: --use the energy equation Ep = Ep(q, x) capacitor effort = gradient of stored energy. (by definition) energy is the same in all domains. compute energy in the electrical domain (easy) gradient with respect to plate separation = force (also easy) Stored electrical energy: 1 1 A Eelectrical = 2 C e2 = 2 x e2 Gradient: F= 1 A E electrical = 2 2 e2 x x Why the minus? --a sign error! ENERGY AND CO-ENERGY There are two ways to integrate the capacitor constitutive equation. --only one of them is energy --the other is co-energy energy: e co-energy energy q _ Ep(q) etdq q2 2C co-energy: is electrical energy _ Ep*(q) qtde 1 2 2 Ce is electrical co-energy THE ERROR WAS TO CONFUSE ENERGY WITH CO-ENERGY Stored electrical energy: Ep = q2x 2A gradient of energy with respect to plate separation: F= E p x = q2 2A Sign error corrected, but ... this equation implies force is independent of plate separation. IS THAT PHYSICALLY REASONABLE? Shouldn't electrostatic attraction weaken as plate separation increases? Would the plates pull together just as hard if they were infinitely far apart? A PARADOX? Cross-check: are the two constitutive equations reciprocal (symmetric)? partial derivatives q F = q A q e = x A As required, the constitutive equations are reciprocal. WHAT'S WRONG? A "PARADOX" RESOLVED: A clue: the electrical constitutive equation e= qx A voltage drop increases with plate separation. for a fixed charge, infinite separation requires infinite voltage. --NOT THE USUAL ARRANGEMENT real devices cannot sustain arbitrarily large voltages. Change "boundary conditions" to input voltage: eA q= x F= 1 eA2 e2A = 2A x 2x2 For fixed voltage, force between plates declines sharply with separation. --much more plausible Mechanically, a spring --albeit a highly nonlinear one. KEY POINT: BOUNDARY CONDITIONS PROFOUNDLY INFLUENCE BEHAVIOR CAUSAL ASSIGNMENT Different "boundary conditions" correspond to different causal assignments. displacement in, effort out on both ports e = e(q, x) = F = F(q, x) = qx q2 A 2A --Integral causality F v = dx/dt Energy function: C e i = dq/dt Ep = Ep(q,x) = q2x 2A Expressing force as a function of voltage and displacement is equivalent to changing the electrical boundary conditions. voltage in, charge out on the electrical port. eA q = e(e, x) = x q2 F = F(e, x) = 2A --differential causality on the electrical port. F v = dx/dt e i = dq/dt C Co-energy function: e2A Ep* = Ep*(e,x) = 2x CO-ENERGY AND LEGENDRE TRANSFORMATIONS Energy and co-energy are related by a Legendre transformation e co-energy energy q The rectangle eq is the sum of energy and co-energy. eq = Ep(q) + Ep*(e) Re-arranging: Ep*(e) = eq Ep(q) This is the negative of a Legendre transformation E p * L{Ep(q)} = Ep(q) - q q = E (e) Commonly used in thermodynamics SINGLE-PORT CAPACITOR: Constitutive equation e = (q) Energy equation Ep = Ep(q) = (q)dq e = Ep(q)/q Co-energy equation Ep*(e) = -1(e)de Legendre transformation L{Ep(q)} e = Ep(q) - eq = -Ep*(e) thus Ep*(e) = eq - Ep(q) Partial differential with respect to e Ep*(e)/e = q Note the positive sign. TWO-PORT CAPACITOR: Constitutive equations e1 = (q1,q2) e2 = (q1,q2) Energy equation Ep = Ep(q1,q2) e1 = Ep/q1 e2 = Ep/q2 Co-energy equations: three possibilities Ep* = Ep*(e1,q2) Ep* = Ep*(q1,e2) Ep* = Ep*(e1,e2) Legendre transformation applied to port 1 Ep*(e1,q2) = e1q1 - Ep(q1,q2) Partial differential with respect to e1 Ep*(e1,q2)/e1 = q1 Partial differential with respect to q2 Ep*(e1,q2)/q2 = - Ep(q1,q2)/q2 Ep*(e1,q2)/q2 = - e2 Note the negative sign. APPLY TO THE CONDENSER MICROPHONE Energy: Ep = q2x 2A Legendre transform: L q2x q2x = Ep(q) eq = eq = Ep*(e) 2A 2A Substitute eA q= x Co-energy: eA2 x eA e2A Ep*(e) = x + e x = 2x 2A This is the "electrical energy" we had used previously. It is actually a co-energy. Thus F= - Note: mechanical force is the negative gradient of electrical co-energy with respect to displacement. --That fixes our sign error. Comment: E p * x = e2A 2x2 In this simple (electrically linear) example, co-energy may as easily be determined without the Legendre transform by substitution for q in the energy equation. REMARKS: Even with the idealizing assumptions above (no electrical saturation, no "fringe effects" in the electrostatic field) the multi-port constitutive equations are profoundly nonlinear e = e(q, x) = F = F(q, x) = qx A e2A 2x2 fundamentally coupled q F e = 0 if q 0 = q x A That is, except when the stored charge is identically zero, --the electrical domain affects the mechanical domain and --the mechanical domain affects the electrical domain. The condenser microphone is not well modeled by one-port energy storage elements in either the mechanical or the electrical domains. Because of inter-domain coupling, this device serves is both a sensor (a microphone) or an actuator (a speaker) --It is commonly used for both purposes. It is an example of an energy-storing transducer. Energy may be stored or removed from either domain Thus energy and power may be transferred between domains. INTRINSIC STABILITY Review the multi-port capacitor definition e = (q) such that Ep - Epo = (q)tdq = Ep(q) That is, q uniquely determines e and hence Ep The converse is not required -- q need not be a well-defined function of e. The constitutive equation(s) may be recovered by differentiation e = qEp(q) In other notation, ei = E p q i all i The constitutive equations may be coupled ei = i(q1,q2,... qn)all i MAXWELL'S RECIPROCITY CONSTRAINT The coupling is constrained such that e=0 In other notation, ei E p E p e j = = = q j q j qi qi q j q i all i, j. Define inverse capacitance 2 e E p C = = q qiq j -1 The inverse capacitance must be a symmetric matrix. STABILITY A physically observable multi-port capacitor must also be intrinsically stable. A further constraint on the constitutive equations. Mathematically: Intrinsically stable if C-1 positive definite --sufficient condition, not necessary EXAMPLE: CONDENSER MICROPHONE (REVISITED) Energy q2x Ep(q, x) = 2A Inverse capacitance 2 q/A e E p 0 = C = = q qiq j q/A x/ -1 Stability determinant C-1 = q 2 A --Unstable for non-zero charge PHYSICALLY REASONABLE -- STABILITY REQUIRES SOMETHING TO OPPOSE THE ATTRACTIVE ELECTROSTATIC FORCE. Include elasticity of the supporting structure Assume elastic linearity (for simplicity) 1 Eelastic = 2 k (x xo)2 Energy (revised) Ep(q, x) = q2x 1 + k (x xo)2 2A 2 2 Constitutive equation q 2A + k (xxo) F e= = e qx A Inverse capacitance (revised) 2 q/A e E p k = C = = q qiq j q/A x/ -1 Stability (revised) determinant C-1 = kx q 2 A A Sufficient condition for stability: k>0 x >0 A k> q2 xA PHYSICAL INTERPRETATION With charge as an input, electrostatic force is independent of displacement. Felectrostatic = F(q, x) = q2 2A Electrostatic force will pull the capacitor plates together until equilibrium is reached. Ftotal = q2 + k (x xo) = 0 2A q2 2Ak xequilibrium = xo This establishes a minimum value for xo if xequilibrium is to be positive. Intuitively, stability about that equilibrium point should only require non-zero mechanical stiffness. Why does the mechanical stiffness have to be any larger? CONSIDER EACH SUFFICIENT CONDITION IN TURN If charge remains constant (q = 0) a displacement from equilibrium of x requires an applied force change of k x k > 0 means that increasing displacement requires increasing applied force --provided charge remains constant If displacement remains constant (x = 0) a displacement from equilibrium of q requires an applied voltage change of x > 0 means that A increasing charge requires increasing applied voltage --provided displacement remains constant x q A HOWEVER If displacement may also change (x 0) q also increases electrostatic force by that decreases displacement by x = q q A q 1 q A k q 1 q q A k A that, in turn, decreases voltage by e = The net voltage increase is x q 1 q q A A k A If increasing charge is to require increasing applied voltage, then q 1 q x > A A k A manipulating: k> q2 means that xA increasing charge requires increasing applied voltage --when both charge and displacement are free to change
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MIT - MECHANICAL - 2.141
NODICITY One of the important ways that physical system behavior differs between domains is the way elements may be connected. Electric circuit elements may be connected in series or in parallel - networks of arbitrary structure may be assembled This no
MIT - MECHANICAL - 2.141
Convection bonds and &quot;pseudo&quot; bondsEven in the simplest case of matter transport, power has two components, one due to the rate of work done, the other due to transported internal energy of the material. &quot;Pseudo&quot; bond graphs depict two distinct bonds. On
MIT - MECHANICAL - 2.141
MULTI-DOMAIN MODELING WHAT'S THE ISSUE? Why not just &quot;write down the equations&quot;? - standard formulations in different domains are often incompatible usually due to incompatible boundary conditions (choice of &quot;inputs&quot;) EXAMPLE: SIMPLE FLUID SYSTEM Scenario
MIT - MECHANICAL - 2.141
CAUSAL ANALYSISThings should be made as simple as possible - but no simpler. Albert Einstein How simple is &quot;as simple as possible&quot;? Causal assignment provides considerable insight.EXAMPLE: AQUARIUM AIR PUMPoscillatory motion in this direction coil leve
MIT - MECHANICAL - 2.141
EXAMPLE: ELECTROMAGNETIC SOLENOID A common electromechanical actuator for linear (translational) motion is a solenoid.Current in the coil sets up a magnetic field that tends to center the movable armature.Electromagnetic Solenoidpage 1 Neville HoganO
MIT - MECHANICAL - 2.141
CO-ENERGY (AGAIN) In the linear case, energy and co-energy are numerically equal. -the value of distinguishing between them may not be obvious. Why bother with co-energy at all? EXAMPLE: SOLENOID WITH MAGNETIC SATURATION. Previous solenoid constitutive eq
MIT - MECHANICAL - 2.141
LINEARIZED ENERGY-STORING TRANSDUCER MODELS Energy transduction in an electro-mechanical solenoid may be modeled by an energy-storing multiport.e= i..ICF . xEnergy transduction in an electric motor may be modeled by a gyrator.e= iGYF . xBut the
MIT - MECHANICAL - 2.141
Massachusetts Institute of Technology Department of Mechanical Engineering2.141 Modeling and Simulation of Dynamic SystemsINTRODUCTIONGOAL OF THE SUBJECT Methods for mathematical modeling of engineering systems Computational approaches are ubiquitous i
MIT - MECHANICAL - 2.141
Junction elements in network models. Classify by number of ports and examine the possible structures that result. Using only one-port elements, no more than two elements can be assembled.Combining two two-ports yields another two-port.At most two one-po
MIT - MECHANICAL - 2.141
EXAMPLE: THERMAL DAMPINGwork in air sealed outletA BICYCLE PUMP WITH THE OUTLET SEALED. When the piston is depressed, a fixed mass of air is compressed. -mechanical work is done. The mechanical work done on the air is converted to heat. -the air tempera
MIT - MECHANICAL - 2.141
NONLINEAR MECHANICAL SYSTEMS CANONICAL TRANSFORMATION S AND NUMERICAL INTEGRATION Jacobi Canonical Transformations A Jacobi canonical transformations yields a Hamiltonian that depends on only one of the conjugate variable sets. Assume dependence on new mo
MIT - MECHANICAL - 2.141
NETWORK MODELS OF TRANSMISSION LINES AND WAVE BEHAVIOR MOTIVATION: Ideal junction elements are power-continuous. Power out = power out instantaneously In reality, power transmission takes finite time. Power out power in Consider a lossless, continuous uni
MIT - MECHANICAL - 2.141
WORK-TO-HEAT TRANSDUCTION IN THERMO-FLUID SYSTEMS ENERGY-BASED MODELING IS BUILT ON THERMODYNAMICS - the fundamental science of physical processes. THERMODYNAMICS IS TO PHYSICAL SYSTEM DYNAMICS WHAT GEOMETRY IS TO MECHANICS. WHY SHOULD WE CARE ABOUT THERM
MIT - MECHANICAL - 2.154
MANEUVERING AND CONTROL OF MARINE VEHICLESMichael S. Triantafyllou Franz S. HoverDepartment of Ocean Engineering Massachusetts Institute of Technology Cambridge, Massachusetts USAManeuvering and Control of Marine Vehicles Latest Revision: November 5, 2
MIT - MECHANICAL - 2.154
111.1KINEMATICS OF MOVING FRAMESRotation of Reference FramesWe denote through a subscript the specific reference system of a vector. Let a vector ex pressed in the inertial frame be denoted as , and in a body-reference frame b . For the x x moment, w
MIT - MECHANICAL - 2.154
2VESSEL INERTIAL DYNAMICSWe consider the rigid body dynamics with a coordinate system affixed on the body. A common frame for ships, submarines, and other marine vehicles has the body-referenced xaxis forward, y-axis to port (left), and z-axis up. This
MIT - MECHANICAL - 2.154
3NONLINEAR COEFFICIENTS IN DETAILThe method of hydrodynamic coefficients is a somewhat blind series expansion of the fluid force in an attempt to provide a framework on which to base experiments and calculations to evaluate these terms. The basic dificu
MIT - MECHANICAL - 2.154
44.1VESSEL DYNAMICS: LINEAR CASESurface Vessel Linear ModelWe rst discuss some of the hydrodynamic parameters which govern a ship maneuvering in the horizontal plane. The b ody x-axis is forward and the y -axis is to port, so positive r has the vessel
MIT - MECHANICAL - 2.154
55.1SIMILITUDEUse of Nondimensional GroupsFor a consistent description of physical processes, we require that all terms in an equation must have the same units. On the basis of physical laws, some quantities are dependent on other, independent quantit
MIT - MECHANICAL - 2.154
6CAPTIVE MEASUREMENTSBefore making the decision to measure hydrodynamic derivatives, a preliminary search of the literature may turn up useful estimates. For example, test results for many hull-forms have already been published, and the basic lifting su
MIT - MECHANICAL - 2.154
7STANDARD MANEUVERING TESTSThis section describes some of the typical maneuvering tests which are performed on full-scale vessels, to assess stability and performance.7.1Dieudonn Spiral e1. Achieve steady speed and direction for one minute. No change
MIT - MECHANICAL - 2.154
88.1STREAMLINED BODIESNominal Drag ForceA symmetric streamlined body at zero angle of attack experiences only a drag force, which has the form 1 (109) FA = - CA Ao U 2 . 2 The drag coefficient CA has both pressure and skin friction components, and hen
MIT - MECHANICAL - 2.154
99.1SLENDER-BODY THEORYIntroductionConsider a slender body with d &lt; L, that is mostly straight. The body could be asymmetric in cross-section, or even flexible, but we require that the lateral variations are small and(Continued on next page)409 SLE
MIT - MECHANICAL - 2.154
1010.1PRACTICAL LIFT CALCULATIONSCharacteristics of Lift-Producing MechanismsAt a small angle of attack, a slender body experiences transverse force due to: helical body vortices, the blunt trailing end, and fins. The helical body vortices are stable
MIT - MECHANICAL - 2.154
11FINS AND LIFTING SURFACESVessels traveling at significant speed typically use rudders, elevators, and other streamlined control surfaces to maneuver. Their utility arises mainly from the high lift forces they can develop, with little drag penalty. Lif
MIT - MECHANICAL - 2.154
1212.1PROPELLERS AND PROPULSIONIntroductionWe discuss in this section the nature of steady and unsteady propulsion. In many marine vessels and vehicles, an engine (diesel or gas turbine, say) or an electric motor drives the(Continued on next page)12
MIT - MECHANICAL - 2.154
13ELECTRIC MOTORSModern underwater vehicles and surface vessels are making increased use of electrical ac tuators, for all range of tasks including weaponry, control surfaces, and main propulsion. This section gives a very brief introduction to the most
MIT - MECHANICAL - 2.154
14TOWING OF VEHICLESVehicles which are towed have some similarities to the vehicles that have been discussed so far. For example, towed vehicles are often streamlined, and usually need good directional stability. Some towed vehicles might have active li
MIT - MECHANICAL - 2.154
15TRANSFER FUNCTIONS &amp; STABILITYThe reader is referred to Laplace Transforms in the section MATH FACTS for preliminary material on the Laplace transform. Partial fractions are presented here, in the context of control systems, as the fundamental link be
MIT - MECHANICAL - 2.154
1616.116.1.1CONTROL FUNDAMENTALSIntroductionPlants, Inputs, and OutputsController design is about creating dynamic systems that behave in useful ways. Many target systems are physical; we employ controllers to steer ships, fly jets, position electri
MIT - MECHANICAL - 2.154
1717.1MODAL ANALYSISIntroductionThe evolution of states in a linear system occurs through independent modes, which can be driven by external inputs, and observed through plant output. This section provides the basis for modal analysis of systems. Thro
MIT - MECHANICAL - 2.154
(Continued on next page)18.2 Roots of Stability Nyquist Criterion87S(s) =e(s) 1 = , r(s) 1 + P (s)C(s)where P (s) represents the plant transfer function, and C(s) the compensator. The closedloop characteristic equation, whose roots are the poles of t
MIT - MECHANICAL - 2.154
1919.1LINEAR QUADRATIC REGULATORIntroductionThe simple form of loopshaping in scalar systems does not extend directly to multivariable (MIMO) plants, which are characterized by transfer matrices instead of transfer functions. The notion of optimality
MIT - MECHANICAL - 2.154
2020.1KALMAN FILTERIntroductionIn the previous section, we derived the linear quadratic regulator as an optimal solution for the full-state feedback control problem. The inherent assumption was that each state was known perfectly. In real applications
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2121.1LOOP TRANSFER RECOVERYIntroductionThe Linear Quadratic Regulator(LQR) and Kalman Filter (KF) provide practical solutions to the full-state feedback and state estimation problems, respectively. If the sensor noise and disturbance properties of th
MIT - MECHANICAL - 2.154
2222.122.1.1APPENDIX 1: MATH FACTSVectorsDefinitionA vector has a dual definition: It is a segment of a a line with direction, or it consists of its projection on a reference system 0xyz, usually orthogonal and right handed. The first form is indepe
MIT - MECHANICAL - 2.154
23 APPENDIX 2: ADDED MASS VIA LAGRANGIAN DYNAMICSThe development of rigid body inertial dynamics presented in a previous section depends on the rates of change of vectors expressed in a moving frame, specifically that of the vehicle. An alternative appro
MIT - MECHANICAL - 2.154
24APPENDIX 3: LQR VIA DYNAMIC PROGRAM MINGThere are at least two conventional derivations for the LQR; we present here one based on dynamic programming, due to R. Bellman. The key observation is best given through a loose example.(Continued on next pag
MIT - MECHANICAL - 2.154
25Further Robustness of the LQRThe most common robustness measures attributed to the LQR are a one-half gain reduction in any input channel, an infinite gain amplification in any input channel, or a phase error of plus or minus sixty degrees in any inpu
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2.160 System Identification, Estimation, and LearningLecture Notes No. 18April 26, 2006 13 Asymptotic Distribution of Parameter Estimates 13.1 Overview^ If convergence is guaranteed, then N * . ^ But, how quickly does the estimate N approach the limit
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2.160 System Identification, Estimation, and LearningLecture Notes No. 19May 1, 200614 Experiment Design14.1 Review of System ID Theories for Experiment Design Key Requirements for System ID ^ Consistent (unbiased) estimate* N 0 ^ Small covariance cov
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2.160 System Identification, Estimation, and LearningLecture Notes No. 20May 3, 200615. Maximum Likelihood15.1 Principle Consider an unknown stochastic processObserved data Unknown Stochastic Processy N = ( y1 , y2 ,, yN )Assume that each observed