Resonances in Repetitive Systems
283
From eq. (11.7), we have
n−1
m=0
sin (x + my) =
n−1
m=0
cos
x + my −
π
2
=
cos [x + (n − 1) y/2 − π/2] sin (ny/2)
sin (y/2)
=
sin [x + (n − 1) y/2] sin (ny/2)
sin (y/2)
.
As a result, we obtain
n−1
m=1
sin (iμ) =
n−1
m=1
sin [μ + (m − 1) μ] =
n−2
m=0
sin (μ + mμ)
=
sin (nμ/2) sin [(n − 1) μ/2]
sin (μ/2)
,
n−1
m=1
sin [(2n − m) μ] =
n−1
m=1
sin [(2n − 1) μ − (m − 1) μ]
=
n−2
m=0
sin [(2n − 1) μ − mμ] =
sin (3nμ/2) sin [(n − 1) μ/2]
sin (μ/2)
,
n−1
m=1
sin [(2n − 3m) μ] =
n−1
m=1
sin [(2n − 3) μ − 3 (m − 1) μ]
=
n−2
m=0
sin [(2n − 3) μ − 3mμ] =
sin (nμ/2) sin [3 (n − 1) μ/2]
sin (3μ/2)
.
The position and angle after n turns are
x n
a n
=
x 0 cos (nμ) + a 0 sin (nμ)
− x 0 sin (nμ) + a 0 cos (nμ)
+ k s β
3
2
0
[ x 0 cos (nμ) + a 0 sin (nμ)]
2
+
1
2
k s β
3
2
x
2
0 + a
2
0
sin [(n − 1) μ/2]
sin (μ/2)
sin (nμ/2)
cos (nμ/2)
+
1
4
k s β
3
2
x
2
0 − a
2
0
sin [(n − 1) μ/2]
sin (μ/2)
sin (3nμ/2)
cos (3nμ/2)
+
1
4
k s β
3
2
x
2
0 − a
2
0
sin [3 (n − 1) μ/2]
sin (3μ/2)
− sin (nμ/2)
cos (nμ/2)
+
1
2
k s β
3
2
x 0 a 0
sin [(n − 1) μ/2]
sin (μ/2)
− cos (3nμ/2)
sin (3nμ/2)
+
1
2
k s β
3
2
x 0 a 0
sin [3 (n − 1) μ/2]
sin (3μ/2)
cos (nμ/2)
sin (nμ/2)
,