Hi Brett

I think we're both right. Draw a feedback loop with 1/s^2 in the
forward path, and k/m + s*lambda/m in the feedback portion and you'll
get the "Quadratic Polynomial/Spring Mass" expression you have in your
derivation. (lambda is viscous damping). Then around that loop put the
electrical feedback. By a few diagram manipulations you get that the
electrical feedback is operating in parallel with the mechanical
feedback.

STM's spring constant is 4.9 N/m, but the electrical proportional
feedback is ~ (380000 V/m)*(13 N/A)/(561000 V/A) =3D 8.8 N/m. So the
mechanical portion is still significant.


Matt

On Mon, Mar 15, 2010 at 10:08 AM, Brett Nordgren  w=
rote:
> Matt,
>
> I went through your calculations and now think I understand why they diff=
er
> from the MathCad results.
>
> At 05:55 PM 3/14/2010, you wrote:
>>
>> Thanks everybody for all their comments. It should take me a while yet
>> to parse all that information.
>> The Inyo looks like it might fit inside a pressure cooker? That might
>> help to isolate it from barometric pressure variation.
>> **********
>> I=92ll try to expand on how I derived the transfer function:
>>
>> 1.) Neither the ground nor the mass is stationary from the perspective
>> of an inertial frame.
>
> Correct.
>
>> 2.) The only forces that can act on the mass are from the spring and
>> from the feedback transducer.
>
> And the force resisting the acceleration of the mass as it is forced to
> follow ground motion. The spring force (variation) is designed to be
> negligible.
>
>> 3.) Both the spring force and the feedback transducer force depend
>> only on the distance between the ground and mass (and derivative and
>> integral of that distance).
>>
>> Those statements gave me this equation of motion for the mass:
>> X(s) is the mass position from an inertial frame.
>> Y(s) is the mass position from the intertial frame.
>
> One is ground position.......?
>
>> =A0 =A0 =A0 =A0F =3D ma =A0 =A0 =A0 =A0 =A0(Newton=92s Law)
>> =A0 =A0 =A0 =A0F =3D F(s)[Y(s) =AD X(s)] =A0 (From =A02,3)
>> So: =A0 =A0 M * s ^ 2 * X(s) =3D F(s) [Y(s) =AD X(s)]
>>
>> Then it=92s just algebra to get the transfer function.
>>
>> Now F(s) =3D K_m =A0+ =A0K_p =A0+ =A0K_i / s =A0+ =A0s * K_d
>
> I think here you are deriving an expression for the gain *around* the loo=
p,
> the 'loop gain' , which is an important concept, but it is not the
> instrument response.
>
> What needs to be considered is that the instrument's output is taken
> following Q, the 'forward' portion of the loop. =A0To get the loop gain y=
ou
> multiplied Q by G*(1/R_p + C + 1/(T*R_i)), the 'feedback' portion of the
> loop, which for simplicity we can call B. So the loop gain is just Q B. =
=A0In
> the configuration I described, if Q B >> 1, which is assured (at all but =
the
> highest frequencies) by design, by making Q large enough, the instrument
> response will closely approximate 1/B. =A0Interestingly, that means that =
it
> doesn't depend on the spring characteristics or on Q (so long as it is hi=
gh
> enough), but only on B -- which is the whole point of using feedback. =A0=
The
> 'feedback.pdf' reference is a simple explanation of how that works.
>
>> Where K_m is the mechanical spring constant, K_p, K_i, and K_d are the
>> constants of the PID controller. For example, K_p =3D Q*G/R_p where Q is
>> the position sensor sensitivity in V/m, G is N/m, and R_p is the
>> proportional feedback resistor. Likewise, K_d=3D G*Q*C, and K_i=3DQ*G/(T
>> *R_i) where T is the integrator time constant.
>
> Regards,
> Brett
>
>
>
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