Fourier Analysis Scaling

[Bob Schultz]

I have used canned FFT Instruments for years, usually to look for
known frequencies and a few harmonics, which makes the job easier.
Now I anticipate an application for the Fourier tool in Excel.

I looked at search results from this group and its sub groups, and
picked out this example.

The "input time" series:
7.3
4
5
5
4
5
6.5
5
4
5
4
4
4
4
5
6
Returned a reply (from Dana) of absolute values:
4.8875
0.286260483
0.126397985
0.195660461
0.154166667
0.231847721
0.110946058
0.165927123
0.443882023
0.10145626
0.108473983
0.251175644
0.179166667
Which look reasonable, DC near the average amplitude, then mostly
noise.
When I run the data set, Excel returns much bigger values:
77.8000
4.0994
6.1274
3.7501
2.3324
5.7044
2.1760
3.0251
1.8000
3.0251
2.1760
5.7044
2.3324
I read somewhere the data needs to be scaled for number of samples,
time intervals, or something, so I looked at the ratio of my results
to the example:
15.918
14.320
48.477
19.166
15.129
24.604
19.613
18.232
4.055
29.817
20.060
22.711
13.018
Certainly not a constant scaling factor!

I wonder what I am missing? I tried some square waves, sine waves,
and sums of sinewaves with amplitudes <= "1". I even created a
"control" panel with entries in seconds, Hz, "volts", etc. to generate
data. The frequencies and harmonics resolve about as expected, but
the amplitudes are way big, exactly 32 for a "zero leakage" result
from a 64 point input 1 second sample of 14 Hz, 1 V, AND 18 Hz 0.5 V.

Thanks Dana for your contributions to the groups

Bob Schultz

 

[Dana DeLouis]

When I run the data set, Excel returns much bigger values:

77.8000
4.0994
6.1274

Hi Bob. If you don't get a better answer, here are some thoughts.
The equation that Excel's fft uses is more for the field of Signal
processing, than Data Analysis. Which is ok for certain things. (My
favorite is using it to multiply two very large numbers.. :>) )
However, you appear to want to analyze the data.

For what I think you are doing, just divide Excel's fft output by 16, the
number of data points. This is the easiest way to convert Excel's output.
The average of the data given would therefore be 77.8 / 16, or 4.8625.

This number is "off" a little from your example (4.8875) because that
example had 24 data points.

Remember, since you only have real data, there are 8 frequencies in your
signal. You did not list all of Excel's output, but you should notice that
the bottom half should be the complex conjugate of the upper half. Since
you already took the Abs values, the relative magnitudes are the same above
your 1.8 value as below. Here's a small section...

2.1760
3.0251
1.8000
3.0251
2.1760

Anyway, hope this helps a little. :>)

 

[Bob Schultz]

Thank you Dana

I did not say much about my "application". I make two REALLY BIG
LC(R) circuits with equal component values, less manufacturing
tolerances. I charge the two Cs to the same voltage, discharge one
through a SCR-Diode switch, let the zero crossing of that capacitor
voltage trigger discharge of the second, and etc. Except for delays
due to triggering on zero crossings, which helps correct phase toward
90 degrees, the capacitor voltages are exponentially decaying
sinusoids. On an XY plot, the voltages form a tight spiral with a
tick at one or two axis crossings (the switching points). Observing
the tick size is pretty subjective. I am hoping to measure the
harmonic series that result from XT and YT data for quality monitoring
purposes.

I did note the original example (noisy DC) data had 24 points and
assumed you returned Excel results from the first 16, since 2^N=24
does not solve for integer N. And yes, many of the ratios I got from
Excel are close to 16, but they go up over 48. Might you be using a
windowing function? Sorry for returning 13 values and not 8, but on
second glance my second 8 are not symmetric with the first 8 either.
But values 2-8 are symmetric to values 10-16! Too weird?

Bob Schultz