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

<strong>and</strong> ε L,R
p
f
∆x,εp=0
= 0 ∀p.
+
N
3. THE PERTURBED RESONANT FREQUENCY 57
∂f
ε
p=1
L p
∂εL
p
∆x,εL p =0
+
N
∂f
ε
p=1
R p
∂εR
p
∆x,εR p =0
+ ∆x ∂f
∂x
∆x,εp=0
= 0
(4.47)
The first term represents the imperturbed part that is zero. Let us
remember the expression of the element (1, 2) of T N in the perturbed
case
N
i=1
Ti
12
= jωM {sinh x sinh Nx
+
N
ε + p [sinh x sinh (N − 1)x + cosh x cosh (N − 1)x]
p=1
+ cosh x
+ sinh x
N
ε + p cosh (2p − N − 1)x
p=1
N
p=1
ε − p sinh (N − 2p + 1)x
(4.48)
If one deals with the derivative with respect to x of the previous expression,
it is useless to consider more than the first term, because the
others become zero when one substitutes εp = 0 ∀p. In the following,
using x = j π + ∆x, we can simplify
2
sinh x = j cosh ∆x
sinh Nx = cos N π
sinh N∆x
2
where we consider N as an even number, because we want to excite
the π/2 mode. Then, it is valid that
<strong>and</strong> therefore
f(∆x) = −ωM cos N π
cosh ∆x sinh N∆x,
2
∂f
∂x
x=jπ/2
= −ωMN cos Nπ/2. (4.49)