analog circuit to compute sine of voltage

Just for fun.....

Is this a one off project ??

If its to re-store or re-build a pendulum clock, it would be kind of cool that you use tubes for the control of the motor.

This would make a cool SteamPunk project.

donald

I sort of like the suggestion to use a sine wave oscillator and a sample-and-hold. If you use a divide-by-two counter and a 4046 to phase lock a square wave to half the sine wave frequency, you can put your control voltage in to shift the phase of the square wave over more than a whole cycle of the sine, with ~0.1% linearity. Sample the sine wave using the edge, and clean up the glitches with an RC lowpass.

For motor control, that should be lots fast enough.

Cheers,

Phil Hobbs

From Last October...

Easiest way to make an adjustable phase shifter without amplitude variation is push-pull drive....

+E o | \ / Variable R \ | o----> Output | | | --- C --- | | | o -E

( -E is 180° out-of-phase from +E )

I first used this in 1960 to adjust the phase on a two-phase smear camera (mirror rotator) motor doing ~20,000 RPM (at MIT's Building

20).

...Jim Thompson

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|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
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Right, but the OP wants something like an AD639, which is more or less independent of frequency.

Cheers,

Phil Hobbs

I think I've been making this unnecessarily complicated. Instead of a linear potentiometer attached to the shaft, fed into a sin function circuit, I could simply get a sin potentiometer (I just found out there are such things, eg.

As for the control loop, I think the torque due to gravity of a rigid unbalanced body acting as a pendulum is solely dependent on the sin of the position angle (the mass, rotational inertia, and radius of the center of gravity being fixed). And if the torque produced by a D.C. motor is reasonably proportional to the current, I'd just have to arrange things so the the sin potentiometer proportionally controls the current drawn by the motor.

But this thing's not monotonic- he wants +/- pi range, not the rather easier +/- pi/2. I guess you could still do it, but it might be easier to put some analog switches and comparators in there and just implement the 0~pi/2 curve, just as you'd likely do it digitally.

Best regards, Spehro Pefhany

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The restore force on a pendulum is only approximately sinusoidal, so you might be able to use a crude approximation of a sine. A simple resistors-and-diodes attenuator with the right breakpoints can handle it for B*V in the range of (-1, +1) though it will fail, of course, at (-pi/2, +pi/2).

Pendulums rarely swing more than +/- PI/4. It would be a strange one indeed where the weight goes above the center of rotation.

If I needed to go a full circle, I think I could still do it woth the R-R op-amp trick. It would take one more op-amp to make a signal that does this:

! A ! B ! C ! ...*........ ..*.*....... .*...*...... *.....*...*. .......*.*.. ........*... ! 0V

In A and C the extra op-amp is clipping. This extra op-amp would be inverting so that when combined with the input by resistors, you get the needed shape.

you mean a simple RC circuit? (resistor and capacitor network) for a charged time constant?

"

Google triangle to sine conversions - diode wave shaping. Same technique.

Diode wave shaping will get you -90 to +90 degrees and no more.

Only digital will get the range you ask for.

Not true. If you use the PLL method I suggested earlier, you can get as many cycles range as you like by increasing the division ratio. (And if you want to quibble about the digital divider, you can multiply the sine frequency instead of dividing the square wave.)

Cheers,

Phil Hobbs

If you have a triangular wave generator and run it's output through an integrator, you'll have a sine wave. A comparator which compares the triangular wave with the input signal can be used as a control for a sample & hold switch.

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Not quite. Integeral(x) = 1/2 x^2 (plus C ;-) ). Alternate proof: the derivative of a sine wave is not a triangle wave, so the integral of a triangle wave isn't a sine wave.

It would look mighty close though. The O.P. didn't spec error at all.

Tim

-- Deep Fryer: A very philosophical monk. Website @

Just got it, thanks, great schematic...

Incorrect. An integrator will not ermove all of those harmonics...

Synchro?

Bob

Synchro? I didn't read much into the problem..(Was tired.)

I later realized the OP wanted the analog version of a pocket calculator not an oscillator.

Speaking of... Can't recall who..somebody posted using multipliers.. Considering the Taylor series: sin(x) = x - (x^3)/3! + (x^5)/5! - (x^7)/7! + ...

Taps off a chain of multipliers???

x-----+----------+--------+-------+---------+ | | | | | x >multi1--->multi2--->mulit3---multi4---multi5 ..and so on | | | | -9.5dB -15dB | | | +--------> sum/sub amp

Goofed again.....forgot the factorial. attenuate multi2 by -15.6dB attenuate multi4 by -41.6dB

No matter...the whole idea still sucks and a uC solution is neater.

D from BC British Columbia Canada.