Resonances in Repetitive Systems
285
where
z 1x = A 11
x
2
0 cos
2 μ +
x 0 a 0 sin (2μ) + a
2
0 sin
2 μ
+ A 12
−
1
2
x
2
0 sin (2μ) +
x 0 a 0 cos (2μ) +
1
2
a
2
0 sin (2μ)
+ A 22
x
2
0 sin
2 μ −
x 0 a 0 sin (2μ) + a
2
0 cos
2 μ
,
z 1a = B 11
x
2
0 cos
2 μ +
x 0 a 0 sin (2μ) + a
2
0 sin
2 μ
+ B 12
−
1
2
x
2
0 sin (2μ) +
x 0 a 0 cos (2μ) +
1
2
a
2
0 sin (2μ)
+ B 22
x
2
0 sin
2 μ −
x 0 a 0 sin (2μ) + a
2
0 cos
2 μ
,
z 2x = −
A 11
x
2
0 + A 12
x 0 a 0 + A 22 a
2
0
cos μ−
B 11
x
2
0 + B 12
x 0 a 0 + B 22 a
2
0
sin μ,
z 2a =
A 11
x
2
0 + A 12
x 0 a 0 + A 22 a
2
0
sin μ−
B 11
x
2
0 + B 12
x 0 a 0 + B 22 a
2
0
cos μ,
z 3x = 0,
z 3a = k s β
3
2
x
2
0 cos
2 μ +
x 0 a 0 sin (2μ) + a
2
0 sin
2 μ
.
Since the one turn map in the new coordinates is a rotation up to the second
order, we obtain the following equations
A 11 cos
2 μ −
1
2
A 12 sin (2μ) + A 22 sin
2 μ − A 11 cos μ − B 11 sin μ = 0,
A 11 sin (2μ) + A 12 cos (2μ) − A 22 sin (2μ) − A 12 cos μ − B 12 sin μ = 0,
A 11 sin
2 μ +
1
2
A 12 sin (2μ) + A 22 cos
2 μ − A 22 cos μ − B 22 sin μ = 0,
B 11 cos
2 μ −
1
2
B 12 sin(2μ) + B 22 sin
2 μ + A 11 sin μ − B 11 cos μ = −k s β
3
2 cos
2 μ,
B 11 sin(2μ)+B 12 cos(2μ)−B 22 sin(2μ)+A 12 sin μ−B 12 cos μ = −k s β
3
2 sin(2μ),
B 11 sin
2 μ +
1
2
B 12 sin(2μ) + B 22 cos
2 μ + A 22 sin μ − B 22 cos μ = −k s β
3
2 sin
2 μ.
After tedious but straightforward algebraic and trigonometric manipulations,
the solution is obtained, which is
A 11 = −k s β
3
2
cos (μ/2) cos μ
2 sin (3μ/2)
, A 12 = 0, A 22 = −k s β
3
2
cos
3 (μ/2)
sin (3μ/2)
,
B 11 = −
1
2
k s β
3
2 , B 12 = k s β
3
2
cos (μ/2) cos μ
sin (3μ/2)
, B 22 = 0.
As a result, the transformation defines a coordinate system in which the motion is a rotation up to the second order. In other words, we have
x
2
+ a
2
= ,