[sdiy] Re: Leapfrog

[Magnus Danielson]

From: "jhaible" <jhaible at debitel.net>
Subject: Re: [sdiy] Re: Leapfrog
Date: Wed, 7 Jan 2004 23:13:17 +0100
Message-ID: <001001c3d56b$768bdb80$9c76b9d9 at debitel.net>

Jürgen,

> > Also, making a "sweeped sine" simulation isn't everything. It is actually
> a
> > rather poor method for characterizing certain aspects. A "smooth" responce
> is
> > not necessarilly a sign of good characteristics, but an "unsmooth"
> responce is
> > a good indication that it probably will not be that good.
> 
> Well yes of course, that's the general problem with Spice. You don't see any
> instability with AC analysis. Typically, if you increase a filter resonance
> until the real circuit breaks into self-oscillation, in Spice you see the
> peaks getting higher, and suddenly getting smaller again. So you have to
> start from the "safe" area, and increase your critical parameter in tiny
> steps to find the threshold to self oscillation by seeing the peak getting
> smaller again.

Well, while what you say is true, that was not at all what I had in mind. I was
refering to my normal point that a "smooth" (or "flat") amplitude responce of a
linear circuit (simulated, measured or calculated) does not hold as evidence
for sounding good. It is however a good indication that it _could_ sound good.
I still assume stability as such, so that is not the issue.

The issue is about having poles and zeros messing about at and around a certain
area of the splane, an area including the jw-axis stretch for
audio-frequencies, but also a fair bit off this axis. When poles and zeroes end
up within this area we start to consider them real problems, but sometimes we
are unable to put a name to the problem. An incorrectly placed zero can sound
as a "hollowness" for certain signals and an incorrectly places pole can sound
as a excessive emphasis for certain signals.

Thus, the amplitude responce (when having excessive deviations) can give you a
hint that this is going on, but since you can have various compensations by
other defects, it may also not be able to tell you about the problem. So, when
the tools are lacking (if they are lacking, it may be in the operator not
"asking the right questions") the ear has to take up and listen to it with some
set of reference material (music, test-CDs or whatever).

> But this wasn't about stability - it was about this circuit being an
> integrator function. Have you seen my last mail?

Yes, but at the time I was too tired to make a good responce.

> I've analysed the circuit for opamp gain = infinite (i.e. V at +in = V at -in) and
> it's indeed a pole at zero. Any real (finite) gain will be an approximation,
> of course. But good enough above 1Hz for a TL071.

Hmm... let's see now.

If V1 is our input signal, V2 is the unearthed end of the capacitor hooked to
the positive input of the op-amp, V3 is the negative input of the op-amp and
V4 is the output of the op-amp.

Then, V2 can be expressed in terms of V1 and V4 as:

        R2//(1/sC)           R1//(1/sC)
V2 = --------------- V1 + --------------- V4
     R1 + R2//(1/sC)      R2 + R1//(1/sC)

(This is comes from analysis where you shorten out all voltage sources except
the one you are describing the expression for, and the report the relation
between the sourced voltage to the target voltage, then you do this for all
sources into your target voltage adding each contribution... a neat trick)

Then, we have the feedback term of the op-amp:

        R4
V3 = ------- V4
     R3 + R4

Then, assuming an ideal op-amp, i.e. assuming infinit gain, we can assume that
V4 becomes such that V2 and V3 is equal. So, let's just set them equal and
twist the formulas to get V4... so:

V2 = V3

giving

   R4           R2//(1/sC)           R1//(1/sC)
------- V4 = --------------- V1 + --------------- V4
R3 + R4      R1 + R2//(1/sC)      R2 + R1//(1/sC)

Now, we simplify this by assuming that R1=R2=R3=R4=R (meaning Ro in your
schematic):

  R          R//(1/sC)          R//(1/sC)         R      1
----- V4 = ------------- V1 + ------------- V4 = -- V4 = - V4
R + R      R + R//(1/sC)      R + R//(1/sC)      2R      2

since we want the V4/V1 form we re-arrange it some:

 1     R//(1/sC)           R//(1/sC)
(- - -------------) V4 = ------------- V1
 2   R + R//(1/sC)       R + R//(1/sC)

         R//(1/sC)
       -------------
V4     R + R//(1/sC)              2[R//(1/sC)]          2[R//(1/sC)]
-- = ----------------- = ---------------------------- = -------------
V1   1     R//(1/sC)     R + R//(1/sC) - 2[R//(1/sC)]   R - R//(1/sC)
     - - -------------
     2   R + R//(1/sC)

Now is the time to expand the R//(1/sC) term, which we do separately:

              R/sC        R
R//(1/sC) = -------- = -------
            R + 1/sC   1 + sRC

thus giving

          R             1
       -------       -------
V4     1 + sRC       1 + sRC          1         1
-- = ----------- = ----------- = ----------- = ---
V1          R             1      1 + sRC - 1   sRC
     R - -------   1 - -------
         1 + sRC       1 + sRC

Ah! There we have it! Yes, it DOES become an integrator. Now I am convinced!
OK? ;O)

Naturually could R3 and R4 be of another value than R, as long as R3 = R4.
Also, R1 and R2 can take on different values as seems fit as long as R1 = R2.

Also, since it does act like a true integrator, it has the instability of an
integrator and in this case meaning the DC instability (since it is oscillating
at 0 Hz with it's pole ON the jw-axis). Yes, the feedback loops of the ladder
will put this instability to rest. So, you're fine!

> Whether the whole thing is stable for, say, 1% resistors, if the negative
> resistance of the NIC slightly dominates the positive resistor(s), is a
> different question. I _expect_ it to be stable, because of the tight leapfrog
> feedback loops, and the well-damped first and last stages, but we'll only
> know when we really build it. Maybe one should build a simple Biquad filter
> from one inverting damped, and one non-inverting undamped integrator, to see
> what happens.

Possibly. OK, so life now contains a non-inverting one-op-amp integrator.
Great!

Now I want a one-op-amp integrator having both a positive and negative input.

Using alternating positive and negative integrators I could also make the
ortonormal ladder filters at a cost of one op-amp per pole. Or? Need to check
the details. I think the logic is the same as in the normal ladder-filters, but
I need to check the details.

<short dwelling with pen and paper>

Yeap! It checks out. One has to be carefull thought, since sometimes along the
ladder it is only the coefficients that change sign, at other times you need
to change sign of the of the state variables... or maybe that only happends in
the loaded end? That's for some more thought. I will drop an email on
ortonormal ladder filter design at a later time.

Anyway, thanks for the contribution Jürgen!
Now ladder filter doesn't seems as expensive as before, so that is an
improvement in the right direction!

Cheers,
Magnus