Another 2=A2...  The following is what I think I know about this stuff.
Please tell me if I am off base on any of it. =20

As you say, Barry, there will be a mechanical restoring force and another
one from the feedback.  Consequently, for a given acceleration, there will
be an amount of deflection with just the mechanical system, and a smaller
amount of deflection with the feedback turned on.  If you take the ratio of
these two deflections, it is approximately the loop gain of the closed-loop
system.  And the loop gain is an indication of how well it measures
acceleration.  As you point out, a strong mechanical system (large spring
constant) requires a good feedback system (sensitive position detector,
high gain, and strong feedback coil), because you want the feedback to do
most of the work.

This ratio depends on the frequency of the acceleration signal because the
mechanical response to acceleration depends on the frequency.  And also the
frequency response of the feedback system has a direct bearing on loop gain
vs frequency.  Most feedback systems of this type have a differential term,
a proporational term, and an integral term.  The differential term is the
predominant one due to stability requirments.  This results in a feedback
system with highest gain at higher frequencies, and relatively low gain at
low frequencies. =20

As has been noted before, above the natural period of the mechanical system
its response is proporational to displacement, and below it is
proporational to acceleration.  Since displacement is the double integral
of acceleration, the response to acceleration falls off at 12db/octave
above the natural period.

The combination of the response of the mechanical and feedback systems
determines the loop gain, which determines the accuracy of the feedback
system in measuring acceleration.  The gain of the mechanical system is
flat to acceleration at lower frequencies, but the gain of the feedback
system falls off as frequency goes lower (usually at 6db/octave).  So loop
gain suffers at low frequencies.  And as frequency increases above the
natural period, the gain of the mechanical system falls off at 12db/octave
and the gain of the feedback system increases at 6db/octave.  The
combination falls off at 6db/octave, so loop gain suffers at high
frequencies too.

You can use a mechanical system with a relatively large spring constant
(not much deflection for a given acceleration), but the feedback system has
to be that much better (to keep the loop gain high) at all frequencies of
interest.  Check some numbers and see how it measures up.  I'd be happy to
crunch some numbers if anyone is interested.

By the way, the VBB system normally produces a velocity output, but the
current through the feedback coil is proportional to acceleration, and that
signal can be used as well.

Karl Cunningham
La Mesa, CA.
PSN Station #40
karlc@.......


At 10:40 AM 10/30/1999 -0700, you wrote:
>Larry et al
>2=A2  : If  I  understand it correctly, with a force feedback system,  the
>voltage/current measured is that required to restore the sensor to the
neutral
>position. It would seem that with a system which has a strong mechanical
>restoring force  that the system would be less sensitive ??? What would the
>feedback be used for? I guess if the feedback force is strong relative to=
 the
>mechanical  restoring force you can get the sensor to have any desired
response
>with electronics. eg , if the mechanical restoring force is directed back=
 to
>null before the the desired response time, that the feedback force would
would
>actually be directed away from null ???
>Barry


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