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September 3, 2003 (^) Start of Presentation
Electrical Circuits I
- This lecture discusses the mathematical modeling
of simple electrical linear circuits.
- When modeling a circuit, one ends up with a set of
implicitly formulated algebraic and differential
equations (DAEs), which in the process of
horizontal and vertical sorting are converted to a
set of explicitly formulated algebraic and
differential equations.
- By eliminating the algebraic variables, it is
possible to convert these DAEs to a state-space
representation.
September 3, 2003 (^) Start of Presentation
Table of Contents
- Components and their models
- The circuit topology and its equations
- An example
- Horizontal sorting
- Vertical sorting
- State-space representation
- Transformation to state-space form
September 3, 2003 (^) Start of Presentation
Linear Circuit Components
- Resistors
- Capacitors
- Inductors
i^ R
va vb u
i^ C
va vb u
i^ L
va vb u
u = va – vb
u = R·i
u = va – vb
i = C· dudt
u = va – vb
u = L· didt
September 3, 2003 (^) Start of Presentation
Linear Circuit Components II
- Voltage sources
- Current sources
- Ground
U 0 = vb – va
U 0 = f(t)
I 0
v^ I a vb u
0 u = vb^ – va
I 0 = f(t)
V 0
V 0 V 0
V 0 = 0
U 0
v^ i a vb
U 0
| (^) +
September 3, 2003 (^) Start of Presentation
Rules for Systems of Equations I
- The component and topology equations contain a
certain degree of redundancy.
- For example, it is possible to eliminate all
potential variables ( vi ) without problems.
- The current node equation for the ground node is
redundant and is not used.
- The mesh equations are only used if the potential
variables are being eliminated. If this is not the
case, they are redundant.
September 3, 2003 (^) Start of Presentation
Rules for Systems of Equations II
- If the potential variables are eliminated, every circuit
component defines two variables: the current ( i )
through the element and die Voltage ( u ) across the
element.
- Consequently, we need two equations to compute
values for these two variables.
- One of the equations is the constituent equation of
the element itself, the other comes from the topology.
September 3, 2003 (^) Start of Presentation
An Example II
Component equations: U 0 = f(t)^ i^ C = C· duC/dt u 1 = R 1 · i 1 uL = L· di (^) L/dt u 2 = R 2 · i (^2)
Node equations: i 0 = i 1 + i (^) L i 1 = i 2 + iC
Mesh equations: U 0 = u 1 + uC uL = u 1 + u 2 uC = u 2
The circuit contains 5 components ⇒ We require 10 equations in 10 unknowns
September 3, 2003 (^) Start of Presentation
Rules for Horizontal Sorting I
- The time t may be assumed as known.
- The state variables (variables, which appear in differentiated form) may be assumed as known.
U 0 = f(t) u 1 = R 1 · i (^1) u 2 = R 2 · i (^2) i (^) C = C· duC/dt uL = L· di (^) L/dt
i 0 = i 1 + i (^) L i 1 = i 2 + iC U 0 = u 1 + uC uC = u 2 uL = u 1 + u 2
U 0 = f(t) u 1 = R 1 · i (^1) u 2 = R 2 · i (^2) i (^) C = C· duC/dt uL = L· di (^) L/dt
i 0 = i 1 + i (^) L i 1 = i 2 + iC U 0 = u 1 + uC uC = u 2 uL = u 1 + u 2
September 3, 2003 (^) Start of Presentation
Rules for Horizontal Sorting IV
- All rules may be used recursively.
U 0 = f(t) u 1 = R 1 · i (^1) u 2 = R 2 · i (^2) i (^) C = C· duC/dt uL = L· di (^) L/dt
i 0 = i 1 + i (^) L i 1 = i 2 + iC U 0 = u 1 + uC uC = u 2 uL = u 1 + u 2
U 0 = f(t) u 1 = R 1 · i (^1) u 2 = R 2 · i (^2) i (^) C = C· duC/dt uL = L· di (^) L/dt
i 0 = i 1 + i (^) L i 1 = i 2 + iC U 0 = u 1 + uC uC = u 2 uL = u 1 + u 2
September 3, 2003 (^) Start of Presentation
U 0 = f(t) u 1 = R 1 · i (^1) u 2 = R 2 · i (^2) i (^) C = C· duC/dt uL = L· di (^) L/dt
i 0 = i 1 + i (^) L i 1 = i 2 + iC U 0 = u 1 + uC uC = u 2 uL = u 1 + u 2
U 0 = f(t) u 1 = R 1 · i (^1) u 2 = R 2 · i (^2) i (^) C = C· duC/dt uL = L· di (^) L/dt
i 0 = i 1 + i (^) L i 1 = i 2 + iC U 0 = u 1 + uC uC = u 2 uL = u 1 + u 2
U 0 = f(t) u 1 = R 1 · i (^1) u 2 = R 2 · i (^2) i (^) C = C· duC/dt uL = L· di (^) L/dt
i 0 = i 1 + i (^) L i 1 = i 2 + iC U 0 = u 1 + uC uC = u 2 uL = u 1 + u 2
⇓
The algorithm is applied, until every equation defines exactly one variable that is solves for.
September 3, 2003 (^) Start of Presentation
Rules for Horizontal Sorting V
- The horizontal sorting can now be performed using symbolic formula manipulation techniques.
U 0 = f(t) u 1 = R 1 · i (^1) u 2 = R 2 · i (^2) i (^) C = C· duC/dt uL = L· di (^) L/dt
i 0 = i 1 + i (^) L i 1 = i 2 + iC U 0 = u 1 + uC uC = u 2 uL = u 1 + u 2
U 0 = f(t) i 1 = u 1 /R 1 i 2 = u 2 /R 2 duC /dt = i (^) C /C di (^) L/dt = uL /L
i 0 = i 1 + i (^) L i (^) C = i 1 - i (^2) u 1 = U 0 - uC u 2 = uC uL = u 1 + u 2
September 3, 2003 (^) Start of Presentation
Rules for Vertical Sorting
- By now, the equations have become assignment statements. They can be sorted vertically, such that no variable is being used before it has been defined.
U 0 = f(t) i 1 = u 1 /R 1 i 2 = u 2 /R 2 duC /dt = i (^) C /C di (^) L/dt = uL /L
i 0 = i 1 + i (^) L i (^) C = i 1 - i (^2) u 1 = U 0 - uC u 2 = uC uL = u 1 + u 2
U 0 = f(t) u 1 = U 0 - uC i 1 = u 1 /R 1 i 0 = i 1 + i (^) L u 2 = uC
i 2 = u 2 /R 2 i (^) C = i 1 - i (^2) uL = u 1 + u 2 duC /dt = i (^) C /C di (^) L/dt = uL /L
September 3, 2003 (^) Start of Presentation
Sorting
- The sorting algorithms are applied just like before.
- The sorting algorithm has already been reduced to a purely mathematical (informational) structure without any remaining knowledge of electrical circuit theory.
- Therefore, the overall modeling task can be reduced to two sub-problems:
1. Mapping of the physical topology to a system of implicitly
formulated DAEs.
2. Conversion of the DAE system into an executable program
structure.
September 3, 2003 (^) Start of Presentation
State-space Representation
- Linear systems:
- Non-linear systems:
d x dt =^ A · x + B · u y = C · x + D · u
; x ( t 0 ) = x 0
d x dt =^ f ( x,u, t ) y = g ( x,u, t )
; (^) x ( t 0 ) = x 0
x ∈ ℜ n u ∈ ℜ m y ∈ ℜ p
x = State vector
u = Input vector
y = Output vector
n = Number of state variables
m = Number of inputs
p = Number of outputs
A ∈ ℜ n^ ×^ n B ∈ ℜ n^ ×^ m C ∈ ℜ p^ ×^ n D ∈ ℜ p^ ×^ m
September 3, 2003 (^) Start of Presentation
Conversion to State-space Form I
U 0 = f(t) u 1 = U 0 - uC i 1 = u 1 /R 1 i 0 = i 1 + i (^) L u 2 = uC
i 2 = u 2 /R 2 i (^) C = i 1 - i (^2) uL = u 1 + u 2 duC /dt = i (^) C /C di (^) L/dt = uL /L
duC /dt = i (^) C /C = (i 1 - i 2 ) /C = i 1 /C - i^2 /C = u 1 /(R 1 · C) – u 2 /(R 2 · C) = (U 0 - uC) /(R 1 · C) – uC /(R 2 · C) di (^) L/dt = uL /L = (u 1 + u 2 ) /L = u 1 /L + u 2 /L = (U 0 - uC) /L + uC /L = U 0 /L
For each equation defining a state derivative, we substitute the variables on the right-hand side by the equations defining them, until the state derivatives depend only on state variables and inputs.
September 3, 2003 (^) Start of Presentation
Conversion to State-space Form II
x 1 = u C
x 2 = iL
u = U 0
y = u C
We let:
x 1 = -
R 1 · C
1
R 2 · C
+^1
[ ] x^1 R 1 · C
. +^1. u
x^. 2 =^1 L .u
y = x 1
