Biomedical Engineering Reference
In-Depth Information
5.7.2 Proof of Eq. ( 5.94 )
We provide a proof of Eq. ( 5.94 ), which is repeated here for convenience.
K
K
1
2
1
2
T
u k u k
E
1 (
y k
Au k )
ʛ (
y k
Au k )
(
A
,
u
)
k
=
k
=
1
K
K
1
2
1
2
y k ʛ
u k ʓ ¯
=−
y k +
1 ¯
u k .
j
=
1
j
=
The left-hand side of the equation above is changed to:
K
K
1
2
1
2
T
u k u k
E
1 (
y k
Au k )
ʛ (
y k
Au k )
(
A
,
u
)
k
=
k
=
1
K
1
2
y k ʛ
=−
y k
k
=
1
u k u k
K
1
2
u k
A T
y k ʛ
u k
A T
E ( A , u )
ʛ
y k
Au k +
ʛ
Au k +
.
k
=
1
(5.174)
On the right-hand side of Eq. ( 5.174 ), the following relationship:
u k
u k u k
E u u k
E A A T
A
I u k
A T
E
ʛ
Au k +
=
ʛ
+
(
A
,
u
)
E u u k
I u k
¯ A T
ʛ ¯ A
ʨ 1
=
+
M
+
E u u k ʓ
u k
E u tr u k u k ʓ
=
=
tr E u u k u k
tr
u k
+ ʓ 1
=
ʓ
=
( ¯
u k ¯
) ʓ
u k ʓ ¯
= ¯
u k
(5.175)
holds, where constant terms are omitted. In the equation above, we use Eq. ( 5.56 ).
Also, the relationship,
y k ʛ
y k
¯ A T
u k
A T
y k ʛ ¯ A
u k
E
Au k +
ʛ
=
u k + ¯
¯
ʛ
y k
(
A
,
u
)
¯ A T
y k ʛ ¯ A
ʓ 1
u k ʓʓ 1
=
ʓ ¯
u k + ¯
ʛ
y k
u k ʓ ¯
=
2
¯
u k
(5.176)
holds. Substituting Eqs. ( 5.175 ) and ( 5.176 )into( 5.174 ), we get Eq. ( 5.94 ).
 
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