PSN-L Email List Message

Subject: Re: Vertical BB Development
From: Brett Nordgren brett3nt@.............
Date: Mon, 15 Mar 2010 21:34:21 -0400


Hi Matt,

You have just identified one of the problems with=20
the STM-8 feedback loop,  The loop gain reflects=20
the relative ability of the loop to be 'in=20
control'  The numbers you have given indicate=20
that the loop gain might be less than 2 under=20
some conditions, which is true.  At certain=20
frequencies the loop is not able to adequately overpower the spring.

You may be ready to be seriously diving into this=20
if you have=20
Excel.  Try=20
http://bnordgren.org/seismo/loop7.zip    Take a=20
look at the 'macros' tab for a quick idea of how=20
to load data sets so you can load the parameters=20
for the STM-8 and get a good look at what's going=20
on.  In the Inyo FBV we were shooting for a loop=20
gain of over 100 in that frequency region and I=20
think we might have ended up with something around 200.

I ought to mention that I've found that the old=20
XP version of Excel (2002 ?) works much better on=20
these big spreadsheets than 2007.  Here's hoping=20
that 2010 fixes some of the extreme slowness=20
problems I have seen when working with the newer=20
Excel, particularly with another big spring-design workbook.

The documentation pdf  has a section which walks=20
you through the process of doing a sample=20
design.  Several people have gotten quite good at=20
doing feedback seismometer design and analysis=20
with the workbook, but it seemed to typically=20
take two or three weeks for them to get=20
comfortable with the process.  We would now never=20
consider doing a new design or making changes to=20
the loop without first trying it out in loop7.

I also have a number of small spreadsheets that=20
can help out when designing things like inverse filters and such.

Have fun, and let me know whenever you have questions.

Brett

At 07:43 PM 3/15/2010, you wrote:
>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=20
> wrote:
> > Matt,
> >
> > I went through your calculations and now think I understand why they=
 differ
> > 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.......?
> >
> >>        F =3D ma          (Newton=92s Law)
> >>        F =3D F(s)[Y(s) =AD X(s)]   (From  2,3)
> >> So:     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  +  K_p  +  K_i / s  +  s * K_d
> >
> > I think here you are deriving an expression for the gain *around* the=
 loop,
> > 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.  To get the loop gain=
 you
> > 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. =
 In
> > the configuration I described, if Q B >> 1,=20
> which is assured (at all but the
> > highest frequencies) by design, by making Q large enough, the instrument
> > response will closely approximate 1/B.  Interestingly, that means that=
 it
> > doesn't depend on the spring characteristics or on Q (so long as it is=
 high
> > enough), but only on B -- which is the whole point of using feedback. =
 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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> >
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> >
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