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

obtains
4. THE PERTURBED AXIAL FIELD 61
Ii(ε l p, ε R p ) = (−1) N−i+1
2 {cos [(N − i + 1)∆x]
+ [sin [(N − i)∆x] − tan (∆x) cos [(N − i)∆x]]
− tan (∆x)
−
N
q=i
N
ε + q (−1) (i−q) cos [(2q − N − i)∆x]
q=i
ε − q (−1) (i−q) sin [(2q − N − i)∆x]
N
q=i
ε + q
(4.61)
By noting that εp is a small quantity, we can exp<strong>and</strong> the trigonometric
functions around zero, leading to the following expression of Ii
Ii(ε1, . . . , εN) = (−1) N−i+1
⎧
⎨
N
(N − i + 1)2
2 1 − ε
⎩ ∗ 2 p +
+
−
N
N
ε
p=1
∗ p ε
q=i
+ q
N
N
ε
p=1
∗ p ε
q=i
− q
N 2
p=1
(i−q)
N − i + 1 − (−1)
N
2q − N − i
N
(4.62)
for i = 2s + 1 with s = 0, . . . , N
− 1
2
<strong>and</strong> we note that the first is the imperturbed term, the second term
is independent of the errors positions <strong>and</strong> the last two are dependent,
like for the perturbed resonant frequency.
It is worth noting that the cells current is stationary respect to the
perturbations, because it depends on the square of ε L,R
p .
In the following the term (−1) N−i+1
2 is considered equal to 1, because
it represents the phase relation between accelerating cavities. In fact,
at a certain time, the electric field E(t) in two adjacent accelerating
cavities has a phase difference equal to π, but we are interested to the
field experienced from the particles <strong>and</strong> they see always a positive field,
<strong>and</strong> therefore we consider (−1) N−i+1
2 = 1.
In order to obtain Vci, it is sufficient to multiply the previous currents
by the capacitive impedance of the cavities.
If we come back to the case without errors, the structure we consider
is symmetric <strong>and</strong> terminated on two half cavities, <strong>and</strong> therefore in the
equivalent model, the circuits i = 1 <strong>and</strong> i = N + 1 have a double
capacitance <strong>and</strong> an half inductance. The field in the cavities can be
represented through a bar-plot as in figure 4.4, where the height of