Specifications

60
The decimation is hence a list of 27 indices, k
0
(k
max
)…k
26
(k
max
), of the bins that are
included, and since the list depends on k
max
, it can be tabulated in a 112×27 matrix,
lk
k
,
max
= k
l
(k
max
) (5.7.25)
There will of course always be a l for which
lk
k
,
max
= k
max
, i.e. the bin of the spectral peak is
always included. The same holds for the adjacent bins. As an example, let us assume that the
spectral peak is at a frequency of 0.25 Hz, i.e. k
max
= 50. Then:
k
13
(50) = k
50,13
= 50 (5.7.26a)
The eight adjacent bins are also included:
k
9
(50) = k
50,9
= 46 k
14
(50) = k
50,14
= 51 (5.7.26b)
k
10
(50) = k
50,10
= 47 k
15
(50) = k
50,15
= 52 (5.7.26c)
k
11
(50) = k
50,11
= 48 k
16
(50) = k
50,16
= 53 (5.7.26d)
k
12
(50) = k
50,12
= 49 k
17
(50) = k
50,17
= 54 (5.7.26e)
For the next bins, the step is doubled from 1 to 2:
k
5
(50) = k
50,5
= 38 k
18
(50) = k
50,18
= 56 (5.7.26f)
k
6
(50) = k
50,6
= 40 k
19
(50) = k
50,19
= 58 (5.7.26g)
k
7
(50) = k
50,7
= 42 k
20
(50) = k
50,20
= 60 (5.7.26h)
k
8
(50) = k
50,8
= 44 k
21
(50) = k
50,21
= 62 (5.7.26i)
For the next bins, the step is doubled from 2 to 4:
k
1
(50) = k
50,1
= 22 k
22
(50) = k
50,22
= 66 (5.7.26j)
k
2
(50) = k
50,2
= 26 k
23
(50) = k
50,23
= 70 (5.7.26k)
k
3
(50) = k
50,3
= 30 k
24
(50) = k
50,24
= 74 (5.7.26l)
k
4
(50) = k
50,4
= 34 k
25
(50) = k
50,25
= 78 (5.7.26m)
Finally, the step for the outermost bins is 16:
k
0
(50) = k
50,0
= 6 k
26
(50) = k
50,26
= 94 (5.7.26n)
Thus this 50
th
row of the matrix reads:
f
p
l
k
max
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 … 25 26
0.25 50 6 22 26 30 34 38 40 42 44 46 47 48 49 50 51 52 53 54 56 … 78 94
The complete matrix is found on the next pages, Table 5.7.18.