where C f = A ρ g fluid capacitor But when squeezed, h (and hence P) may vary with time even though V does not. Seems to imply C f = C f (t)

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

2 PROBLEM! E p = V 2 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!

3 SOLUTION Identify another power port to keep track of the work done to change the stored energy F v = dx/dt introduces a new network element: a multiport capacitor C P Q = dv/dt 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 = e i f i i In vector notation: e _ e 1 e 2 en P = et f = ft e f _ f 1 f 2 fn

4 Energy variables Energy variables are defined as in the scalar case as time integrals of the flow and effort vectors respectively. generalized displacement = f dt + o 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 = Φ() The vector function Φ( ) is the capacitor constitutive euation. a vector field (in the mathematical sense of the word). A multi-port capacitor is sometimes called a C-field or capacitive field.

5 BOND GRAPH NOTATION By convention, power is defined positive into all ports. e 1 e 3 f 1 = d 1 /dt C f 3 = d 3 /dt e 2 f 2 = d 2 /dt An alternative notation: n denotes the number of ports. More on this multi-bond notation later. n C STORED ENERGY: determined by integrating the constitutive euation. E p - E po = etf dt = etd = Φ()td = E p () potential energy, as it is a function of displacement a function of as many displacements as there are ports.

6 COUPLING BETWEEN PORTS. Each effort may depend on any or all displacements. e i = Φ i ( 1, 2,... n ) 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 = E p () Therefore the constitutive euation, Φ( ), must have zero curl. or e = 0 Φ = 0

7 In terms of the individual efforts and displacements, i p i E e = all i i j j p i i p j j i e E E e = = = all i, j. Coupling between ports must be symmetric. The dependence of ei on j must be identical to the dependence of ej on i. This is known as Maxwell's reciprocity condition. Later we will see that stability and passivity further constrain the capacitor constitutive euation.

8 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.

9 DEVICE CAN STORE ENERGY. Energy can change in two ways: mechanical displacement charge displacement a two-port capacitor F v = dx/dt C e i = d/dt like a spring in the mechanical domain like a capacitor in the electrical domain Two constitutive euations needed both relate effort to displacement e = e(, x) F = F(, x)

10 Assuming electrical linearity: 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 = ε 0 A/x ε ο is a permittivity A is plate area One constitutive euation is e = x εa To find the other constitutive euation we could work out the attraction due to the charges on the plates...

11 AN EASIER WAY: use the energy euation E p = E p (, 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)

12 Stored electrical energy: E electrical = 1 2 C e2 = 1 2 Gradient: εa x e2 F = E electrical x = 1 2 εa x2 e 2 Why the minus? a sign error!

13 ENERGY AND CO-ENERGY There are two ways to integrate the capacitor constitutive euation. only one of them is energy the other is co-energy e co-energy energy energy: E p () _ etd 2 2 C co-energy: is electrical energy E p *() _ tde 1 2 C e2 is electrical co-energy

14 THE ERROR WAS TO CONFUSE ENERGY WITH CO-ENERGY Stored electrical energy: E p = 2x 2εA gradient of energy with respect to plate separation: F = E p x = 2 2εA Sign error corrected, but... this euation implies force is independent of plate separation.

15 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 euations reciprocal (symmetric)? partial derivatives F e x = εa = εa As reuired, the constitutive euations are reciprocal. WHAT S WRONG?

16 A PARADOX RESOLVED: A clue: the electrical constitutive euation e = x εa voltage drop increases with plate separation. for a fixed charge, infinite separation reuires infinite voltage. NOT THE USUAL ARRANGEMENT real devices cannot sustain arbitrarily large voltages.

17 Change boundary conditions to input voltage: = eεa x 1 F = 2εA eεa x 2 e2εa = 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

18 CAUSAL ASSIGNMENT Different boundary conditions correspond to different causal assignments. displacement in, effort out on both ports e = e(, x) = x εa F = F(, x) = 2 2εA Integral causality Energy function: E p = E p (,x) = 2x 2εA F v = dx/dt C e i = d/dt

19 Expressing force as a function of voltage and displacement is euivalent to changing the electrical boundary conditions. voltage in, charge out on the electrical port. = e(e, x) = eεa x F = F(e, x) = 2 2εA differential causality on the electrical port. Co-energy function: E p * = E p *(e,x) = e 2εA 2x F v = dx/dt C e i = d/dt

20 CO-ENERGY AND LEGENDRE TRANSFORMATIONS Energy and co-energy are related by a Legendre transformation e co-energy energy The rectangle e is the sum of energy and co-energy. e = E p () + E p *(e) Re-arranging: E p *(e) = e E p () This is the negative of a Legendre transformation L{ E p ()} = E p () - E p = E * (e) Commonly used in thermodynamics

21 SINGLE-PORT CAPACITOR: Constitutive euation e = Φ() Energy euation E p = E p () = Φ()d e = E p ()/ Co-energy euation E p *(e) = Φ-1(e)de Legendre transformation thus L{ E p ()} e = E p () - e = -E p *(e) E p *(e) = e - E p () Partial differential with respect to e E p *(e)/e = Note the positive sign.

22 TWO-PORT CAPACITOR: Constitutive euations e1 = Φ(1,2) e2 = Φ(1,2) Energy euation E p = E p (1,2) e1 = E p /1 e2 = E p /2 Co-energy euations: three possibilities E p * = E p *(e1,2) E p * = E p *(1,e2) E p * = E p *(e1,e2)

23 Legendre transformation applied to port 1 E p *(e1,2) = e11 - E p (1,2) Partial differential with respect to e1 E p *(e1,2)/e1 = 1 Partial differential with respect to 2 E p *(e1,2)/2 = - E p (1,2)/2 E p *(e1,2)/2 = - e2 Note the negative sign.

24 APPLY TO THE CONDENSER MICROPHONE Energy: E p = 2x 2εA Legendre transform: L 2x 2εA Substitute = E p() e = 2x 2εA e = E p * (e) = eεa x Co-energy: E p *(e) = eεa x 2 x 2εA + e eεa x = e 2εA 2x

25 This is the electrical energy we had used previously. It is actually a co-energy. Thus F = Note: E p * = e 2εA x 2x2 mechanical force is the negative gradient of electrical co-energy with respect to displacement. That fixes our sign error. Comment: In this simple (electrically linear) example, co-energy may as easily be determined without the Legendre transform by substitution for in the energy euation.

26 REMARKS: Even with the idealizing assumptions above (no electrical saturation, no fringe effects in the electrostatic field) the multi-port constitutive euations are profoundly nonlinear e = e(, x) = x εa F = F(, x) = e 2εA 2x2 fundamentally coupled F = e x = εa 0 if 0

27 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. or 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) 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.

28 INTRINSIC STABILITY Review the multi-port capacitor definition e = Φ() such that E p - E po = Φ()td = E p () That is, uniuely determines e and hence E p The converse is not reuired need not be a well-defined function of e. The constitutive euation(s) may be recovered by differentiation e = E p () In other notation, Ep e i = all i i The constitutive euations may be coupled e i = Φ i ( 1, 2,... n ) all i

29 MAXWELL S RECIPROCITY CONSTRAINT The coupling is constrained such that e = 0 In other notation, i j j p i i p j j i e E E e = = = all i, j. Define inverse capacitance = = j i p 2-1 E C e The inverse capacitance must be a symmetric matrix.

30 STABILITY A physically observable multi-port capacitor must also be intrinsically stable. A further constraint on the constitutive euations. Mathematically: Intrinsically stable if C-1 positive definite sufficient condition, not necessary EXAMPLE: CONDENSER MICROPHONE (REVISITED) Energy E p (, x) = 2x 2εA Inverse capacitance C -1 Stability 2 e Ep = = i j = 0 /εa /εa x/εα determinant C-1 = εa 2 Unstable for non-zero charge

31 PHYSICALLY REASONABLE STABILITY REQUIRES SOMETHING TO OPPOSE THE ATTRACTIVE ELECTROSTATIC FORCE. Include elasticity of the supporting structure Assume elastic linearity (for simplicity) E elastic = 1 2 k (x x o)2 Energy (revised) E p (, x) = 2x 2εA k (x x o)2 Constitutive euation F e = 2 2εA + k (x x o) e = x εa Inverse capacitance (revised) C -1 2 e Ep = = i j = k /εa /εa x/εα

32 Stability (revised) determinant C-1 = kx εa εa 2 Sufficient condition for stability: k > 0 x εa > 0 k > 2 xεa

33 PHYSICAL INTERPRETATION With charge as an input, electrostatic force is independent of displacement. F electrostatic = F(, x) = 2 2εA Electrostatic force will pull the capacitor plates together until euilibrium is reached. F total = 2 2εA + k (x x o) = 0 x euilibrium = x o 2 2εAk This establishes a minimum value for x o if x euilibrium is to be positive. Intuitively, stability about that euilibrium point should only reuire non-zero mechanical stiffness. Why does the mechanical stiffness have to be any larger?

34 CONSIDER EACH SUFFICIENT CONDITION IN TURN If charge remains constant ( = 0) a displacement from euilibrium of x reuires an applied force change of k x k > 0 means that increasing displacement reuires increasing applied force provided charge remains constant If displacement remains constant ( x = 0) a displacement from euilibrium of reuires an applied voltage change of x εa x εa > 0 means that increasing charge reuires increasing applied voltage provided displacement remains constant

35 HOWEVER If displacement may also change ( x 0) also increases electrostatic force by εa that decreases displacement by x = 1 εa k that, in turn, decreases voltage by e = 1 εa k εa The net voltage increase is x εa 1 εa k εa If increasing charge is to reuire increasing applied voltage, then x εa > 1 εa k εa manipulating: k > 2 xεa means that increasing charge reuires increasing applied voltage when both charge and displacement are free to change

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