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

56 4. CIRCUIT MODEL
matrix of the whole chain is
Ttot = ˙ T N + ˙ T N−1
N
ε + p +
p=1
cosh x
sinh x
N
ε + p [♥ + ♦] +
p=1
N
p=1
ε − p ♣ (4.46)
where the symbols represent the following matrices
⎛
sinh (N − 1)x
⎞
jωM sinh x cosh (N − 1)x
♥ = ⎝
⎠
cosh (N−1)x
sinh (N − 1)x
jωM sinh x
⎛
⎞
− sinh (2p − N − 1)x jωM sinh x cosh (2p − N − 1)x
♦ = ⎝
⎠
− cosh (2p−N−1)x
jωM sinh x
⎛
cosh (N − 2p + 1)x
sinh (2p − N − 1)x
⎞
jωM sinh x sinh (N − 2p + 1)x
♣ = ⎝
⎠
− cosh (N − 2p + 1)x
sinh (N−2p+1)x
− jωM sinh x
It is apparent that the matrix (4.46) has an imperturbed term, a perturbed
one not depending on the errors positions along the chain <strong>and</strong>
two terms that depend instead.
3. The perturbed resonant frequency
The resonant frequencies of the whole chain can be obtained from
the zeros of the element of place (1, 2) in the transmission matrix,
representing the chain impedance seen from one side when the opposite
side is short-circuited.
We mainly refer to the π/2 mode <strong>and</strong> want to analyze the effect
of the errors on the resonant frequency that goes from ω0 to
ω = ω0 + ∆ω. We can represent this situation by a function: ∆ω =
F (δω01, δω02, . . . , δω0N). We obtain the expression for the element (1, 2)
from the formula (4.46) which is
N
i=1
Ti
12
= h(ω, δω01, . . . , δω0N) = g(ω0 + ∆ω, ε L p , ε R p )
<strong>and</strong> then the resonant frequency can be seen as an implicit function
g(ω0+∆ω, ε L p , ε R p ) = 0, the solution of which is the perturbed frequency
we are looking for. In that case we deal with
g(ω0 + ∆ω, ε L p , ε R p ) = f(j π
2 + ∆x, εL p , ε R p ) = 0
It could be convenient to calculate ∆x which is the error of x, rather
than directly calculate ∆ω. Let us exp<strong>and</strong>, up to the first order, the
function f around the imperturbed value ∆x = 0, namely x = x0 = j π
2