Theory, Design and Tests on a Prototype Module of a Compact ...

58 4. CIRCUIT MODEL
For the derivative with respect to ε L,R
p , we have to consider all the
terms, <strong>and</strong> by considering the following trigonometric identities
sinh (N − 2p + 1)x = j(−1) p cos N π
cosh (N − 2p + 1)∆x
2
cosh x = j sinh ∆x
sinh (N − 1)x = j sin (N − 1) π
cosh (N − 1)∆x
2
cosh Nx = cos N π
cosh N∆x
2
cosh (N − 1)x = j sin (N − 1) π
(4.50)
sinh (N − 1)∆x
2
the expression (4.48) becomes
N
Ti = −ωM cos N π
N
cosh ∆x sinh N∆x + j sin (N − 1)π
2 2
i=1
12
· [− cosh ∆x cosh (N − 1)∆x − sinh ∆x sinh (N − 1)∆x]
N
− ε
p=1
+
p sinh ∆x cosh (2p − N − 1)(j π
+ ∆x)
2
N
−j
p=1
ε − p cosh ∆x(−1) p cos N π
cosh (N − 2p + 1)∆x
2
= f(∆x, ε L p , ε R p )
p=1
ε + p
(4.51)
In the derivative with respect to ε L,R
p , the only term different from zero
comes from i = p. For example, the term in ε L p is
∂f
∂εL = −j
p
ωM
2 ·
sin (N − 1) π
[cosh ∆x cosh (N − 1)∆x + sinh ∆x sinh (N − 1)∆x]
2
− 2j sinh ∆x cosh (2p − N − 1)(j π
+ ∆x)
2
cosh ∆x(−1) p cos N π
cosh (N − 2p + 1)∆x
2
And finally, the resonance equation is
jωM cos N π
N
ε
2
+ p + jN∆x + (−1) p
N
ε −
p = 0
p=1
<strong>and</strong> therefore the unknown is
∆x = j 1
N
L εp + ε
N
R p
2
p=1
+ (−1) p εR p − ε L p
2
p=1
= j
N
p=1 ε∗ p
N
(4.52)