Splitting an error signal into DC and High Freq.

Look up "cross-over networks". ...Jim Thompson

You mean like an LC thing? (I'd have to set the Q.) I was thinking I could use an active filter. (S-V) and take the HP and LP outputs. But don't I get more phase shift that way?

George H.

No. Run the math.

You can easily make cross-overs with active filters. ...Jim Thompson

What, you want me to work for a living? :^)

Did I say the output is for a control loop?

A two pole filter has to have 180 degree phase shift across the frequency band. RC has only 90 degree's. I don't know if that matters for a control loop.. and what happens in the cross-over region.

I did this ~10 years ago. I could go look in my notebook and (perhaps) find it. It's easier just to do it again from scratch.

Is there anything wrong with my RC approach? (I've got it wired up, I'll test it tomorrow.)

I may very well have used an active filter last time.

George H.

It'll need buffering. But no instrument amps. Equal C, equal R.

Vin o--------------o-------------. | | .-. | | | --- | | --- '-' | | | o-------------------------------o LF out | | | o-----------------o HF out | | | .-. --- | | --- | | | '-' | | VVV VVV

Right I could have done that too. (I wanted an knob to change the x-over freq.... no dual ganged pots.) And this way I'm pretty much guaranteed that HF and LF will have the same time constant.

I screwed up the phase on the instrument amp the first time.

The sum of HF + LF was equal to the input. That's reassuring.

When I tired the same with a State Variable filter, the LP + HP summed to zero at the X-over. (I never knew that before.) I guess I'd have to add the BP too? (My 'scope can't sum three signals.)

(BP = band pass, HP = high pass, LP = low pass) ('cause Rick will ask.)

George H.

LP: w^2 / (s^2 + 2*zeta*w*s + w^2) HP: s^2 / (s^2 + 2*zeta*w*s + w^2)

LP + HP = (s^2 + w^2) / (etc.), which is a plain ol' notch filter.

Absent adjustability I belive my way is better (and they do sum to the original signal).

But yes, if you need to tweak things, an instrumentation amp is probably better than a dual pot.

I assume you mean me? Thanks for the consideration, but in this context those abbreviations were clear to me. I don't understand what you mean about the LP + HP summing to zero though. I assume you mean they sum to be the same as the input?

Right, it made sense afterwards. Hey, is this analogy correct? So in an SV filter with HP, BP and LP output, can I think it as a series LRC circuit, With HP the signal across an ideal inductor, BP that across the resistor, and LP across an ideal capacitor.

That seems right.

George H.

No, the sum is zero.. no signal. at "resonance" the LP and HP amplitudes are equal, and the phase difference is 180 deg.

As Tim said I already knew this, 'cause you can make a notch by adding LP and HP... but for me knowing (book-wise) and doing are two different things.

George H.

Ok, I didn't get the resonance part.

Resonance is where the BP output is maximum.

GH

I'm saying that I didn't catch the use of the term crossover to refer to the point of equal attenuation of the two outputs. I was thinking of a physical crossover circuit.

At first flush yes -- I'd have to work it out to be sure.

Ditto for a current-driven parallel RLC with the signal being the currents involved.