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?
E XAMPLE : S OLENOID WITH MAGNETIC SATURATION .
Previous solenoid constitutive equations assumed electromagnetic linearity.
—arbitrarily large magnetic fluxes could be generated.
In reality flux cannot exceed saturation flux.
For sufficiently high currents behavior is strongly nonlinear.
Solenoid & co-energy page 1 © Neville Hogan

M
ODEL THAT PHENOMENON
.
Assume an electrical constitutive equation as follows:
λ (i, x) =
L(x) i λ s
L(x)2i2 + λ s2 where L(x) is position-dependent inductance as before.
For sufficiently small currents the relation is approximately linear. i <<
λ s
L(x)
λ ≈ L(x) i
For sufficiently large currents the flux linkage reaches a limiting value, λ s. i >>
λ s
L(x)
λ ≈ λ s
Solenoid & co-energy page 2 © Neville Hogan

M
ECHANICAL CONSTITUTIVE EQUATION may be found using the stored energy.
—Find the stored energy at a fixed displacement.
—Find force as the gradient with respect to displacement.
That yields the relation between force and flux linkage.
F = F( λ , x)
But flux cannot be specified arbitrarily.
Realistic boundary conditions require current input.
To find the relation between force and current, substitute.
λ = λ (i, x)
Solenoid & co-energy page 3 © Neville Hogan

S
TORED ELECTRICAL ENERGY
(at a fixed displacement)
E = i d λ
Need to invert the relation between flux linkage and current.
—In general, anything but straightforward.
In this case, a little algebra yields the following. i =
λ s
L(x)
λ
λ s2 – λ 2
| λ | ≤ | λ s|
U
SE WITH CAUTION
!
If λ > λ s this expression yields an imaginary number for the current.
Solenoid & co-energy page 4 © Neville Hogan

A
LITTLE CALCULUS
(and some more algebra) yields an expression for energy.
E( λ , x) =
λ s 2
L(x) ⎝
⎜
⎛
1 – 1 – ⎜
⎝
⎛ λ
λ s
⎞
⎟
⎠
2
⎞
⎟
⎠ see Note 1 attached
Partial derivative with respect to displacement:
F
( )
= −
∂ L
∂ x L
λ x s
2
( )
2
⎝
⎜⎜
⎛
⎜
1 − 1 −
⎝
⎜⎜
⎛ λ
λ s
⎠
⎟⎟
⎞
2
⎞
⎟
⎠
⎟⎟
Solenoid & co-energy page 5 © Neville Hogan

Substitute for flux linkage:
F
( )
= −
∂ L
∂ x L
λ x
2 s
( )
2
⎛
⎜
⎜
⎜
⎝
1 − 1 −
⎜
⎜
⎝
⎛
L
( ) i
L
( )
2 i 2 + λ 2
S
⎟
⎟
⎠
⎞
2
⎟
⎟
⎠
⎞
⎟
Still more algebra simplifies this expression:
F
( )
= −
∂ L
∂ x L
λ x s
2
( )
2
⎜
⎜
⎝
⎛
1 −
L
( )
2
λ
S i 2 + λ 2
S
⎟
⎟
⎠
⎞
Solenoid & co-energy page 6 © Neville Hogan

C
OMMENT
:
The assumed electrical constitutive equation
λ =
L
L
( )
( )
2 i i λ
2 s
+ λ s
2 was chosen primarily for pedagogic simplicity
—a more realistic relation may be (far) less tractable
—usually no simple algebraic form exists
Consequently a procedure requiring
(a) inversion
(b) integration with respect to flux linkage and
(c) differentiation with respect to displacement may be impractical.
Solenoid & co-energy page 7 © Neville Hogan

A
N ALTERNATIVE APPROACH
: use co-energy instead of energy.
Total stored energy:
E = i d λ + F dx
Electrical co-energy:
E* = i λ – E = i λ – i d λ – F dx
(a Legendre transformation with respect to current)
Force is the negative gradient of this co-energy.
∂ E *
∂ x
= −
∂ E
∂ x
= − F
Solenoid & co-energy page 8 © Neville Hogan

Electrical co-energy at a fixed displacement:
E* =
⌡
λ di
E*(i, x) =
λ s
L(x) ⎝
⎛ see Note 2 attached
L(x)
2 i
2
+ λ s 2
– λ s
K EY POINT :
Co-energy may be found without inverting the relation between flux linkage and current.
Partial differentiation with respect to displacement:
F
( )
= −
∂ L
∂ x see Note 3 attached
L
λ x s
2
( )
2
⎜
⎜
⎝
⎛
1 −
No further algebra required.
L
( )
2
λ s i 2 + λ s
2
⎟
⎟
⎠
⎞
Solenoid & co-energy page 9 © Neville Hogan

REMARKS
In general, co-energy functions are useful for multiport and/or nonlinear energy storage elements.
Inverting constitutive equations may be avoided.
In the linear case, energy and co-energy are numerically equal. the value of distinguishing between them may not be obvious.
In the nonlinear case, the two are not equal.
Distinguishing between them is important.
In this example, energy is upper-bounded co-energy is not.
In general, energy is conserved co-energy need not be.
Solenoid & co-energy page 10 © Neville Hogan

NOTE 1:
***************************************
Integration to find energy:
E( λ , x) =
λ
⌠
⎮
⌡
λ s
L(x)
0 y dy
λ s2 – y2
| λ | ≤ | λ s| where y is a “dummy variable”. rearrange:
E( λ , x) =
λ
⌠
⎮
⌡
1
L(x)
0 substitution: define y dy
1 –
⎝
⎛
⎜ y
λ s
⎞
⎟
⎠
2
Solenoid & co-energy page 11 © Neville Hogan

sin u = y
λ s
1 –
⎝
⎛
⎜ y
λ s
⎟
⎠
⎞
2 = 1 – sin
2
u = cos
2
u = cos u y = λ s sin u u = sin
-1
⎝
⎛
⎜ y
λ s
⎞
⎟
⎠
– π /2 ≤ u ≤ π /2 dy = λ s cos u du (assuming dx = 0)
E( λ , x) = sin
-1
( λ / λ s)
⎮
1
L(x)
λ s sin u cos u
λ s cos u du
0
E( λ , x) = sin
-1
( λ / λ s)
λ s 2
⎮
L(x) sin u du
0
Solenoid & co-energy page 12 © Neville Hogan

E( λ , x) = sin
-1
( λ / λ s)
λ s 2
⎮
L(x) d(–cos u)
0
E( λ , x) = –
λ s 2
L(x) ⎝
⎛
⎜ cos
⎝
⎛
⎜ sin
-1
⎝
⎛
⎜
λ
λ s
⎞
⎟
⎠
⎟
⎠
⎞
⎟
⎠ cos
⎝
⎛
⎜ sin
-1
⎝
⎜
⎛ λ
λ s
⎟
⎠
⎞
= 1 –
⎝
⎜
⎛ λ
λ s
⎟
⎠
⎞
2
E( λ , x) =
λ s 2
L(x) ⎝
⎜
⎛
1 – 1 – ⎜
⎝
⎛ λ
λ s
⎞
⎟
⎠
2
⎞
⎟
⎠
Does that make sense?
In the limit as λ approaches λ s
E( λ , x) ≈
λ s 2
L(x)
Plausible; implies that only a finite amount of energy may be stored.
In the limit as λ approaches 0, binomial series expansion of the root yields
Solenoid & co-energy page 13 © Neville Hogan

1 –
⎝
⎛
⎜
λ
λ s
⎟
⎠
⎞
2 = 1 –
1
2 ⎝
⎛
⎜
λ
λ s
⎞
⎟
⎠
2 + higher order terms
E( λ , x) ≈
λ s 2
L(x) ⎝
⎛
⎜ 1 – 1 +
1
2 ⎝
⎜
⎛ λ
λ s
⎞
⎟
⎠
2
⎞
⎟
⎠
E( λ , x) ≈
λ
2
2 L(x) as expected for a linear inductor.
Solenoid & co-energy page 14 © Neville Hogan

NOTE 2:
***************************************
Integration to find co-energy: i
E*(i, x) =
⌠
⎮
⌡
0
L(x) i λ s
L(x)2i2 + λ s2
di d
⎛
⎝⎜
L(x)2i2 + λ s2 =
L(x)
2
i di
L(x)2i2 + λ s2
E*(i, x) =
λ s
L(x) ⌡ i
0 d
⎛
⎝⎜
L(x)2i2 + λ s2
E*(i, x) =
λ s
L(x) ⎝⎜
⎛
L(x)2i2 +
Does this make sense?
λ s2 – λ s2
Solenoid & co-energy page 15 © Neville Hogan

In the limit as
L(x) i >> λ s E* λ s i
This is the area of a rectangle of sides i and λ s. Note that co-energy may increase without bound, whereas energy may not.
In the limit as
L(x) i << λ s series expansion of the square root
L(x)
2 i
2
+ λ s 2
= ( λ s 2
)
1/2
+
1
2 (
λ s 2
)
-1/2
L(x)
2 i
2
+ higher order terms
E*(i, x) ≈
λ s
L(x)
⎛
⎜
( λ s 2
)
1/2
+
1
2 (
λ s 2
)
-1/2
L(x)
2 i
2
– ( λ s 2
)
1/2
⎞
⎟
E*(i, x) ≈
1
2 L(x) i
2 as expected for a linear inductor.
Solenoid & co-energy page 16 © Neville Hogan

NOTE 3:
***************************************
Partial differentiation with respect to displacement:
F
( )
= −
∂ E *
∂ x
∂ E *
∂ x
= −
∂ L
∂ x L
λ s
( )
2 ⎝
⎜
⎛
L 2 i 2 + λ 2 s
− λ s ⎠
⎟
⎞
+
L
λ s
( )
⎜
⎜
⎝
⎛
L
L
( ) i 2
2 i 2 + λ s
2
⎟
⎟
⎠
⎞
∂ L
∂ x
∂ E *
∂ x
= −
∂ L
∂ x L
λ x s
( )
2
⎜
⎜
⎝
⎛
L
( )
2 i 2 + λ 2 s
− λ s
L
L
( )
2
( )
2 i 2 + i 2
λ s
2
+ λ 2 s
L
( )
2 i 2
⎟
⎟
⎠
⎞
F
( )
= −
∂ L
∂ x L
λ x s
2
( )
2
⎜
⎜
⎝
⎛
1 −
L
( )
2
λ s i 2 + λ s
2
⎟
⎟
⎠
⎞
Solenoid & co-energy page 17 © Neville Hogan

An expanded form may be easier to understand physically:
F
( )
= −
∂ L
∂ x L
λ x s
2
( )
2
⎜
⎜
⎝
⎛
L
( )
2 i 2 +
L
λ 2 s x
−
( )
2
λ i s
2
L
( )
2 i 2
+ λ s
2
+ λ 2 s
⎟
⎟
⎠
⎞
Solenoid & co-energy page 18 © Neville Hogan