Operator`s manual

SECTION 8. PROCESSING AND PROGRAM CONTROL EXAMPLES
8-11
TABLE 8.7-4. Thirty Minute Output From Example
01 110 02 DAY 03 HRMIN 04 M(W1) 05 M(U1) 06 M(V1) 07 M(Tai) 08 M(e1)
09 V(W1) 10 V(U1) 11 V(V1) 12 V(Ta1) 13 V(e1) 14 CV(W1,U1) 15 CV(W1,V1) 16 CV(W1,Ta1)
17 CV(W1,e1) 18 CR(W1,U1) 19 CR(W1,V1) 20 M(W2) 21 M(U2) 22 M(V2) 23 M(Ta2) 24 M(e2)
25 V(W2) 26 V(U2) 27 V(V2) 28 V(Ta2) 29 V(e2) 30 CV(W2,U2) 31 CV(W2,V2) 32 CV(W2,Ta2)
33 CV(W2,e2) 34 CV(U2,V2) 35 CV(U2,Ta2) 36 CV(U2,e2) 37 CV(V2,Ta2) 38 CV(V2,e2)
8.8 FAST FOURIER TRANSFORM
EXAMPLES
8.8.1. EXAMPLE WITHOUT BIN AVERAGING
The 21X was used to generate data
representing two superimposed sine wave
signals, one at 1.25 Hz (amplitude = 1) and the
other at 0.25 Hz (amplitude = 2). The 1024
generated samples simulate a sampling rate of
10 Hz or a 0.1 second scan rate. Figure 8.8-1
shows a plot of the simulated signal. The FFT
was applied to the data and the real and
imaginary, phase and magnitude, and the power
spectra results are shown in Tables 8.8-1, 8.8-2,
and 8.8-3 respectively. A portion of the power
spectra results are illustrated in Figure 8.8-2.
The phase of the cosine wave that describes the
signal at the beginning of the first interval and
the end of the last interval can be determined by
looking at the 21X program (Table 5) that
generated the "original time series data". The
1.25 Hz signal began and ended at 270
degrees. [cos 270 = cos(0 - 90) = sin 0]. The
0.25 Hz signal began at 270 degrees and ended
at 126 degrees. The phases of the 1.25 and
0.25 signals are 270 and 198 respectively (Table
8.8-2).
TIME IN SECONDS
AMPLITUDE
SAMPLED AT 10 Hz
0 10 20 30 40 50 60 70 80 90 100
3
2
1
0
-1
-2
-3
FIGURE 8.8-1. Simulated 1.25 and 0.25 Hz
Signals
FFT ANALYSIS OF 0.25 AND 1.25 Hz
FREQUENCY IN Hz = BINS 0
A
M
P
L
I
T
(
T
h
o
u
0
0.1953 0.3906 0.5859 0.7812 0.976 1.17192 1.36724
1
0
0
0
0
0
0
0
0
0
0
FIGURE 8.8-2. FFT Power Spectra Analysis
of 0.25 and 1.25 Hz Signal