Fourier
Problem
Problem
Freq4P08
Name_______________________________________
Here is the lab data. Notice the large negative spikes in the oscilloscope
trace.

For the record, you need
to know this about the signal.
-
The signal record length
is 50 milliseconds and the time data in the file goes from .025sec to .075
sec.
-
There are 2000 data points
in the data set.
The FFT of the signal is shown below. (It's a magnitude plot.)
(This in Mathcad, so the smallest index is 0 - not 1 like in Matlab.)

The only thing you see
in the FFT is the spike at DC (frequency = 0). However, enlarging
the scale, you can see that there is something else.

There are spikes at
18 (although that one "slops over" a little bit into the 19th point!) and
at 37 (which may justify calling the first spike at 18.5). Calling
it 18.5 will get us about where we should be. Since the record is
.05 seconds long, we need to multiply 18 by (1/.05) or 20. That gives
us noise at 360 Hz (or 370 for 18.5) and at multiples after that.
Now, here's the data for the second data set (the 9.5 inch situation).
The noise spikes seem to occur a little less frequently.

Now, if we take the
FFT and operate on it as above, we get this plot.

Here the strong components
of this signal (except for the DC term!) seem to occur at 7, 14, 21 and
27 (whoops!). Probably, the fact is that the first one isn't exactly
at 7, but may be a little spread out. Anyhow, the same computations
as before will give frequencies of 140, 280, 420, etc.
To reduce the lowest component by a factor of 10, we would need to have:
-
<
0.1 - at 140 Hz. (The lowest frequency we found was 70 Hz.)
-
That means we have to have:
-
(wRC)2
> 99
-
(wRC) > 9.9
-
RC > 9.9/(2pf)
= 9.9/(2p140) = .011sec (The lower limit
on the time constant of the filter!)