Campbell Scientific AVW1 and AVW4 Interfaces for Vibrating Wire Sensors - Manual (Page 33)

C-1
APPENDIX C. THEORY AND ADDITIONAL DETAILS
C.1 SWEPT FREQUENCY THEORY
Example: f2 = 31 hundred Hz
f1 = 24 hundred Hz
X = (30*f1*f2)/(f2-f1)
= 3189
where f1 and a f2 are the starting and ending
frequencies in hundreds of Hz respectively.
One clock cycle (CC) occurs every 813.8 ns or
at a rate of 1.2288 mHz.
Rule #1: 65535 > X > 256
the 256 constraint is somehow due to an 8 bit
constraint. The 65535 constraint is some limit
where the swept frequency can no longer be
done in exactly 15 ms.
The minimum increment in frequency is 1 clock
cycle.
How many times do we have to change the half
period by 1 clock cycle to cover the frequency
range?
1/(F1 * 2) = 1/(2400 * 2) =
.208333 ms = half period
1/(F2 * 2) = 1/(3100 * 2) =
.161290 ms = half period
Change in half period =
.208333 - .161290 = .047043 ms
# of clock cycle increments to cover the
.047043 ms half period range is:
= (.047043 ms/.0008138 ms) =
57.81 freq. increments
How much time is there between frequency
increments to cover the frequency range in 15
ms?
time = 15 ms/58 frequency increments =
.2586 ms/freq. incr.
C.2 ADDITIONAL THEORY ON
MULTIPLIER AND OFFSET
The result (X) of Instruction #28 is:
X = 1/((t ms)2) =
1,000,000/((t s)2
where t is the period in milliseconds. Since
frequency (f) is the inverse of period, this can
also be expressed as:
X = (f kHz)2 =
[(f Hz)2]/1,000,000