Say a AC voltage source, V, provides current to a circuit with a resistor, R, and inductor, L. Then my equations to be solved, in terms of current will be:

$$ V - Ri - L\frac{di}{dt} = 0 $$

I would like to know if there is any electronic component (or an integrated circuit) such that the following operation is performed:

$$ V - Ri - L\frac{di}{dt} - \frac{d^2i}{dt^2}= 0 $$

  • \$\begingroup\$ A differentiator. Usually an op-amp circuit. Basically, the same as well-tuned high pass filter. Think about it...both differentiators and high pass filters produce an output that is larger if the rate of change of the signal is faster. And resistors are basically voltage to current converters so you just use those and feed the voltage into a voltage differentiator. \$\endgroup\$ – DKNguyen Oct 28 '19 at 15:41
  • \$\begingroup\$ @DKNguyen There is a single \$i\$ variable, so I'm wanting to assume that this represents a current that is flowing through an \$R\$ and an \$L\$ and through something with the constant of \$1\$, in series with the indicated time-varying voltage supply. Note that dimensional analysis suggests already that the last term on the right side has the wrong units for summing into the rest. So the constant of \$1\$ must have the required units -- namely, the units of \$\frac{\text{kg}\cdot\text{m}^2}{\text{s}\cdot \text{A}^3}\$. Also, I'm not sure how one might compose that last term with a circuit. \$\endgroup\$ – jonk Oct 28 '19 at 16:15
  • \$\begingroup\$ @jonk For the last term wouldn't you "just" run the output of the first differentiator through a second differentiator? That sounds pretty hairy in practice though with analog. Or am I missing something? \$\endgroup\$ – DKNguyen Oct 28 '19 at 18:09
  • \$\begingroup\$ @DKNguyen It's possible my imagination is more limited than yours and I'll learn from what you suggest. But the first problem with the equation provided is that the units of the last term simply don't pass dimensional analysis. And I don't know of a device that carries the needed units. Do you? (Not that it cannot be cobbled up, somehow. I'm just saying I don't yet see how it can be cobbled up in such a way that the dimensional analysis succeeds and the thing is buildable, too.) But I'm just a hobbyist. So there's that. \$\endgroup\$ – jonk Oct 28 '19 at 18:12
  • \$\begingroup\$ @jonk Are you a hobbiest? I was under the impression you were an RF or analog engineer. One of those voodoo guys. The concern about the dimensional analysis is valid, but I wasn't too concerned without the validity of the OP's equation, only the implementation of whatever it is in hardware. \$\endgroup\$ – DKNguyen Oct 28 '19 at 18:20

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