Fast Fourier Transform - Signal Processing of Random Physiological Signals - page 153

13.6.3 Third-Stage Calculations

G

X ( 0 )

G

X ( 1 )

G

X ( 2 )

G

X ( 3 )

G

X ( 4 )

G

X ( 5 )

G

X ( 6 )

G

X ( 7 )

0

X (0)

=

G 1 +

8 G 5 =

(0 + × 0) + (1 + × 0)(0 + × 0) = 0 + × 0

W j j j j

W 8 G 6 =

X (1) = G 2 + (2 + j × 0) + (0 . 707 + j × 0 . 707)(1 . 414 − j × 1 . 414)

= 4 + j × 0

W 8 G 7 =

X (2) = G 3 + (0 + j × 0) + (0 + j × 1)(0 + j × 0) = 0 + j × 0

W 8 G 8

X (3) = G 4 + = (2 + j × 0) + ( − 0 . 707 + j × 0 . 707)(1 . 414 + j × 1 . 414)

= 0 + j × 0

0

X (4) = G 1 − W G 5 = (0 + j × 0) − (1 + j × 0)(0 + j × 0) = 0 + j × 0

8

W 8 G 6 X (5) = G 2 − = (2 + j × 0) − (0 . 707 + j × 0 . 707)(1 . 414 − j × 1 . 414)

= 0 + j × 0

W 8 G 7 X (6) = G 3 − = (0 + j × 0) − (0 + j × 1)(0 + j × 0) = 0 + j × 0

W 8 G 8 X (7) = G 4 − = (2 + j × 0) − ( − 0 . 707 + j × 0 . 707)(1 . 414 + j × 1 . 414)

0
Each of the eight samples for x ( nt ) is given by
x (0)

= 4 + j ×

= 1 + j × 0 x (4) =− 1 + j × 0

x (1) = 0 . 707 + j × 0 x (5) =− 0 . 707 + j × 0

x (2) = 0 + j × 0 x (6) = 0 + j × 0

x (3) =− 0 . 707 + j × 0 x (7) = 0 . 707 + j × 0