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MAE 3401 – Modeling and Simulation
SECTION 4: MATHEMATICAL MODELING
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K. Webb MAE 3401
In the last section of notes, we saw how to create a bond graph model from a physical system model. The next step in the modeling process is the creation of a mathematical model
Introduction2
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Mathematical Modeling – Introduction
You’re already familiar with some techniques for creating mathematical models for physical systems
For example:
First, create a free‐body diagram:
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Mathematical Modeling – Introduction
Next, apply Newton’s 2nd law
rearranging:(1)
This is a mathematical model A second‐order, linear, constant‐coefficient, ordinary differential equation
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Reduction to a System of 1st‐Order ODE’s
Can reduce this 2nd‐order ODE to a system of two 1st‐order ODE’s
We know that(2)
and(3)
Using (2) and (3), rewrite (1), the original ODE
where (4)
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Reduction to a System of 1st‐Order ODE’s
Equations (4) is a system of first‐order ODE’s that is equivalent to (1)
Rearranging (4):
(5)
These equations can be put into matrix form :
(6)
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Reduction to a System of 1st‐Order ODE’s
Let’s say we want to consider the displacement of the mass as the output of the system
We can add an output equation to the mathematical model
(7)
We can rewrite (7) in a matrix form similar to (6):
(8)
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Mathematical Model
Together, (6) and (8) comprise the mathematical model for our mechanical system:
(9)
Note that , , , , and are all functions of time The is dropped to simplify the notation The convention used here is to only include the for inputs, e.g.
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State‐Space Representation
The system model of (9) is the state‐space representation of the system, or the state‐variable equations for the system
Can be expressed in generic form as
(10)where
: the state vector : derivative of the state : vector of inputs : vector of outputs
: system matrix : input matrix : output matrix : feed‐through matrix
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MIMO vs. SISO Systems
(10)
Note that the state‐space model (10) allows for vectors of inputs and outputs, and
Multi‐input, multi‐output (MIMO) systems and will be vectors
Single‐input, single‐output (SISO) systems and will be scalars
In this course, we’ll mostly focus on SISO systems For now, we’ll assume the more general MIMO case
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System State and State Variables
The vector is the state vector Elements of are the state variables of the system
The state of the system is a complete description of the current condition of the system From our energy‐based perspective, the state describes all of the energy in a system, i.e. where it is stored, at a given point in time
The state variables are a (not the) minimum set of system variables required to completely describe the state of a system
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State Variables are Not Unique
The state vector, i.e. the choice of state variables, for a system is not unique In this example, we have chosen displacement and velocity as the state variables, i.e.
Could have chosen other quantities – later, we will State variables need not even have direct physical significance
Different state‐space representations for the same system are related by similarity transforms Beyond the scope of this class
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The Feed‐Through Matrix
(10)
is the feed‐through or feed‐forward matrix Very often zero for physical systems, as in our example
Non‐zero implies that the input affects the output instantaneously There exists a direct feed‐through path from the input to the output
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State‐Space Vector and Matrix Dimensions
(10)
Assume the state space model of (10) represents an ‐order, ‐input, ‐output MIMO system
The state vector is an column vector
The system has inputs, so the input vector is an column vector
There are outputs, so the output vector is a column vector
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State‐Space Vector and Matrix Dimensions
(10)
If is , then its derivative, , is also
The product must have the same dimensions as , The system matrix, , is a square matrix
The product must also be The vector of inputs, , is , so is
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State‐Space Vector and Matrix Dimensions
(10)
The vector of outputs, , is The product must also have dimension
is , so must be
The product must also have the same dimension as , The vector of inputs, , is , so is
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State‐Space Vector and Matrix Dimensions
For an ‐input, ‐output, MIMO system:
Term Dimension
1
1
1
1
Term Dimension
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State‐Space Vector and Matrix Dimensions
For SISO system, u and y, as well as , are scalars:
Term Dimension
1 1
1 1
1
1
Term Dimension
1
1
1 1
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State‐Space Model Explained
Remember, our reason for modeling a system is to enable the analysis of its dynamic behavior
Basic idea of the state space model: If the current state of a system is known, and the current and future values of the inputs are known, then the trajectory of the system (i.e. the time‐evolution of its state variables) can be determined
Don’t need explicit knowledge of the history of the system or its inputs – no past information All history is accounted for in the current value of the state
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State‐Space Model – Physical Significance
Consider the physical meaning of the state‐space system model
The time derivative of a system’s state variables can be expressed as a linear combination of the current state variables and the current inputs
The outputs of a system can be expressed as a linear combination of the current state and the current inputs
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State‐Space Model – Utility
Again, our goal is to analyze a system’s time‐domain behavior – the time‐evolution of its state variables
Knowledge of the current state variables, as well as the current rate of change of those state variables, allows us to do this
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Where We’re Going
In the previous example, we derived the state‐space model for a mechanical system by applying Newton’s 2nd law For an electrical system we could have applied Kirchhoff’s and Ohm’s laws
Can always derive a mathematical model by applying domain‐specific laws to the physical model
Our approach will be to derive state equations from bond‐graph system models
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State Equations from Bond‐Graph Models
Bond graphs are energy‐basedmodels Our choice of state variables will be those that describe the storage of energy within a system at a given instant in time
State variables will be energy variables of the independent energy‐storage elements in a system Displacements of capacitors Momenta of inertias
Only independent ’s and ’s State variables represent a minimum set of system variables needed to completely describe the state
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State Equation Derivation24
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Deriving State Equations from Bond Graphs
Start with the same mechanical system model:
Two independent energy‐storage elements State variables will be the energy variables associated with these two elements:
The computational bond graph:
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State Equation Derivation – State Variables
State equation will be of the form:
In general, state variables will be momenta and displacements Their derivatives will be efforts and flows, respectively For this example:
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State Equation Derivation – Preparation
Annotate the computational bond graph with state variable derivatives Efforts on the independent Inertias and the flows on the independent Capacitors
Apply constitutive laws to annotate the other power variables on the ’s and ’s
Annotate the known source power variables and indicate as functions of time
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State Equation Derivation – Procedure
Objective: derive a set of equations, each expressing a state variable derivative as a linear combination of state variables and inputs Determine the and matrices
First, choose a state variable and write its derivative as an effort or flow:
(1)
Next, use the causality assigned to the bond graph to work from (1) to a state equation Express as a linear combination of states and inputs Will ultimately relate an effort or flow to a state variable by applying a constitutive relationship for an energy‐storage element
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State Equation Derivation
is an effort on a 1‐jct Caused by , , and , and is known, so
(2)
Relate to using the const. law for the resistor
(3)
is the flow on a 1‐jct, set by , related to s.v. by the const. law for the inertia
(4)
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State Equation Derivation – Procedure
Still need to eliminate related to state variable through
constitutive law for the capacitor
(7)
Substituting (4) into (3)
(5)
And substituting (5) back into (2)
(6)
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State Equation Derivation
Substituting (7) into (6) yields the first of two state equations
(8)
Next, follow a similar procedure for
(9)
is the flow on a 1‐jct, set by , related to state variable by the const. law for the inertia
(10)
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State Equation Derivation
Substituting (10) into (9) yields the second of two state equations
(11)
Combine (8) and (11) into the state‐variable model for our system in matrix form
010 (12)
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State Equation Derivation
Can now replace the computational bond graph parameters in (12) with physical system parameters
010 (13)
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State Equation Derivation – Output Equation
Can also define an output equation as part of our state‐space model
Suppose we want to consider the velocity of the mass as our output Constitutive relation relates an inertia’s flow to its momentum:
(14)
The output equation would be:
1/ 0 (15)
Equations (13) and (15) comprise the complete state‐space system model
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State Equation Derivation – Output Equation
Perhaps, instead, we want to consider the displacement of the mass as our output Same as spring displacement – a state variable
State‐space model, including output equation, becomes:
10
10
(16)
0 1
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State Equation Derivation – Causality
In this example, assignment of causality yielded the simplest result: All energy‐storage elements ended up in integral causality – all were independent
No resistors had their causality arbitrarily assigned
Lack of derivative causality and/or algebraic loops (resistor fields) results in straightforward state equation derivation Unfortunately, the inverse is also true
Next, we’ll look at two more examples without derivative causality or algebraic loops
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State Equation Derivation – Example 1
Consider the mechanical example from Section 3
Four independent energy‐storage elements Fourth‐order system Four state variables:
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State Equation Derivation – Example 1
Annotate the bond graph: State variable derivatives Efforts on independent inertias Flows on independent capacitors
Use constitutive laws and state variables to express: Flows on independent inertias Efforts on independent capacitors
Known source quantities
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State Equation Derivation – Example 1
Choose a state variable derivative and express it as an effort or a flow
(1)
Known source efforts can remain Need to eliminated Effort on a 0‐jct, set by
(2)
Substituting (2) into (1) yields the first of four state equations
(3)
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State Equation Derivation – Example 1
Move on to the next state variable
(4)
and are both flows on 1‐jct’s set by and , respectively
(5)
(6)
Substituting (6) and (5) into (4) yields the second state equation
(7)
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State Equation Derivation – Example 1
Move on to
(8)
which gives the third state equation
(9)
Finally, derive the equation for
(10)
is the effort on a 0‐jct, set by
(11)
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State Equation Derivation – Example 1
is related to state variable
(12)
can be related to using the constitutive law for resistor
(13)
And, is the flow on a 1‐jct, set by
(14)
Substituting (11), (12), and (14) into (10) yields the final state equation
(15)
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State Equation Derivation – Example 1
Combine the state equations into matrix form
0 0 0
0 0
0 0 0
0
1 1 00 0 00 0 00 0 1
(16)
Can rewrite, substituting in physical parameters and are the displacements of springs and , respectively
0 0 00 0
0 0 0
0
1 1 00 0 00 0 00 0 1
(17)
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State Equation Derivation – Example 1
Let the position of each mass to be our outputs Two outputs
Displacement of is
(18)
Displacement of spring is
(19)
so
(20)
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State Equation Derivation – Example 1
Combine (20) and (18) into our output equation Multiple outputs, so will be a matrix
Complete state‐space model, including output equation:
0 0 00 0
0 0 0
0
1 1 00 0 00 0 00 0 1
(21)
0 1 1 00 0 1 0
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State Equation Derivation – Example 2
A slightly modified version of the electrical circuit from Section 3:
The computational bond graph for this circuit:
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State Equation Derivation – Example 2
Three independent energy‐storage elements Third order
State variables:
Annotate the computational bond graph
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State Equation Derivation – Example 2
Begin with equation for
(1)
(2)
is the effort on a 0‐jct, set by the effort on
(3)
Substituting (2) and (3) into (1) gives the first of three state equations
(4)
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State Equation Derivation – Example 2
Next, move on to
(5)
is set by
(6)
The transformer modulus relates to , which is the flow on a 1‐jct, set by
(7)
is algebraically related to and
(8)
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State Equation Derivation – Example 2
The transformer relates back to , which is set by
(9)
Substituting (9) into (8) gives
(10)
Equation (10) can be substituted into (7)
(11)
Using (11) and (6) in (5) gives us our second state equation
(12)
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State Equation Derivation – Example 2
Finally, derive the equation for
(13)
is the flow on a 1‐jct, which is set by
(14)
Substituting (10) into (14)
(15)
Substituting (15) in (13) gives us our third state equation
(16)
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State Equation Derivation – Example 2
Combine the state equations in matrix form
0
0
100
(17)
Replacing computational bond graph parameters with physical parameters
0
0
100
(18)
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State Equation Derivation – Example 2
Choosing the voltage across as our output, the complete state‐space system representation is
10
1 1 1
01 1
100
(19)
0 0 1/
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State Equation Derivation – Example 2
Instead let the voltage across be the system output That is, the effort associated with Effort is the time derivative of momentum, so
(20)
The output equation can be extracted from (19)
0 (21)
Note that, in this case, the feed‐through term, , is non‐zero
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Algebraic Loops or Resistor Fields55
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Algebraic Loops – Example 1
Consider the following electrical circuit
Causality assignment is completed by arbitrarily assigning the causality of resistor (or ) System contains an algebraic loop (resistor field)
Presence of the algebraic loop will complicate the state equation derivation a bit
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Algebraic Loops – Example 1
Second‐order system State variables are:
Begin deriving equations as usual
(1)
(2)
(3)
(4)
has reentered the formulation, and we’re back where we started in (1) An algebraic loop
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Algebraic Loops – Procedure
1. The output of the resistor whose causality was arbitrarily assigned – in this case, though would work equally well – is the auxiliary variable
2. Derive an expression relating the auxiliary variable to the state variables, inputs, and to itself
3. Proceed with the state equation derivation as usual, but leave the auxiliary variable in the formulation along with state variables and inputs
4. Substitute the result from step 2 into the result from step 3
One auxiliary variable for each algebraic loop present Multiple loops require solution of a system of equations
Apply this procedure first, whenever causality assignment involves an arbitrary assignment of resistor causality
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Algebraic Loops – Example 1
Follow causality to derive an expression for auxiliary variable
(5)
(6)
(7)
is the aux. variable, so it can remain in the expression Substituting (7) into (6) into (5)
(8)
(9)
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Algebraic Loops – Example 1
Solve (9) for
(10)
Going back to (1), we had
(1)
Substituting in (10) yields the first state equation
(11)
Now, whenever appears in the formulation, substitute in the expression in (10)
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Algebraic Loops – Example 1
Moving on to
(12)
We already have an expression for in (6) and (7)
(13)
Substituting in (10) to eliminate 1 1 1
Rearranging gives the second state equation
(14)
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Algebraic Loops – Example 1
Assembling (11) and (14) in matrix form gives our state variable system model
(15)
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Algebraic Loops – Example 1
Substitute in physical parameters and define an output equation for the voltage across the capacitor,
(16)
0 1/
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Algebraic Loops – Example 2
Next, consider a mechanical system
Causality assignment is completed by arbitrarily assigning the causality of resistor (or )
A very similar bond graph to the electrical circuit in the previous example
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Algebraic Loops – Example 2
A second‐order system with state variables:
An algebraic loop is present, so we’ll immediately go to the procedure outlined in the previous example
Auxiliary variable is Express in terms of state variables, inputs, and itself
(1)
(2)
is the auxiliary variable, so it can remain in the expression
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Algebraic Loops – Example 2
Substitute (2) into (1)
(3)
Then solve for
(4)
(5)
Now, Proceed with the state equation derivation Whenever the auxiliary variable, , appears in the formulation, it
will be replaced with (5)
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Algebraic Loops – Example 2
Start with
2 (6)
Substituting (5) into (6) gives the first state equation
(7)
Moving on to
(8)
(9)
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Algebraic Loops – Example 2
Substitute (9) into (8)
(10)
Substituting (5) in for gives the second state equation
(11)
(12)
In matrix form:
01 (13)
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Algebraic Loops – Example 2
Note that the origin of the algebraic loop in this example was a modeling assumption The connection point between the spring and dampers was considered massless Instead we could account for the mass of this junction
Now, there are no arbitrary causality assignments and no algebraic loops
State equation derivation will be greatly simplified
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Algebraic Loops – Example 2
System is now third‐order, due to the additional independent energy‐storage element
State equation, after replacing physical parameters:
0 0
0010
(14)
Looks very different from the original second order model, but if ≪ , their behaviors are nearly identical
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Derivative Causality71
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Derivative Causality – Example
Consider the mechanical system from Section 3
The computational bond graph:
Two independent energy‐storage elements Second‐order State variables are:
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Derivative Causality – Example
is in derivative Causality Not independent Does not contribute a state Its energy variable (would be a for an ‐element) is algebraically related to the state variables
Annotate the bond graph Include the energy variable annotation for the dependent inertia
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Derivative Causality – Example
is not a state variable State equation derivation requires first determining the algebraic relationship between and the state variables, and
When or enters the formulation, substitute in this relationship or its derivative
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Derivative Causality – Procedure
1. For the dependent energy‐storage element, apply the constitutive law ‘backwards’ ‐ i.e. express the energy variable as a function of a power variable
a. Inertia: express momentum as a function of flowb. Capacitor: express displacement as a function of effort
2. Follow causality to relate that power variable to the state variables and inputs
3. Substitute the expression from step 2 into that from step 1
4. When the energy variable (or its derivative) enters the formulation, substitute in the expression from step 3
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Derivative Causality – Example
Apply the constitutive law for ‘backwards’
Express as a function of
(1)
Follow causality to express in terms of state variables and inputs
(2)
Substituting (2) into (1)
(3)
Now proceed with derivation, using (3) when needed
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Derivative Causality – Example
Begin state equation derivation with
(4)
relates to
(5)
(6)
Substituting (6) into (5)
(7)
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Derivative Causality – Example
A term has entered the formulation Differentiate (3)
(8)
Substitute (8) into (7)
(9)
Solve (9) for
/ (10)
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Derivative Causality – Example
Rearranging (10) gives the first of two state equations:
(11)
Next, move on to
(12)
The second state equation:
(9)
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Derivative Causality – Example
The state‐space system model:
10
0
With physical parameters:
10
0
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Derivative Causality – Example
Derivative causality in this case resulted from a modeling decision The lever was assumed to be rigid Adding some compliance to the lever arm eliminates derivative causality (see Section 3 notes) Increases system model to fourth‐order Equation derivation simplified at the cost of model complexity
In general, derive an expression for the energy variable of each energy‐storage element in derivative causality Multiple elements in derivative‐causality will require solution of a system of equations