Cap-Offset Block Filter Is Actually Taking a Derivative & Then Integrating

And a piece of wire does that, too.

John

But the offset block or cap filter is _not_ really a one step process.

It's actually a 2 step process that just looks like one step.

Bret Cahill

It's continuous, no steps.

The voltage transfer function of a capacitor depends on what it's loaded with.

in----------C-----+-------out | | | R | | | gnd

is the standard AC-coupling thing, a single-pole RC highpass.

Gain is 0 at DC, approaching unity at high frequencies. The corner radian frequency, where the gain is 0.707, is at w=1/(R*C)

Really, you should take an EE101 course if you're interested in this stuff. Get it right.

John

But is the derivative then antiderivative POV is valid?

Bret Cahill

This is a philosophical question.

If a lot of large caps are in series the voltages seem to approach the derivative as you go down the circuit.

If a lot of small caps are in parallel what do the currents indocate?

Not that I can see. The math doesn't work.

John

A big inductor can integrate. In series after a small cap you recover the original signal minus offset.

Bret Cahill

An inductor or capacitor does not "integrate" or "differentiate". Only mathematicians do that!

Those operations approximate the relationship between voltage and current in those devices which actually are not inductors or capacitors in reality. "Real" inductors and capacitors only approximate ideal ones. And even worse, you don't understand that even these ideal components do not act as you suggest. The actual setup requires a resistance that is used to convert current to voltage. You think you are being cool, but you really don't have a clue.

. . .

Everyone can agree than a small enough cap will put out something proportional to the derivative of the original signal while a large enough inductor will put out something proportional to the integral.

The question is, can one circuit made up of just capacitors and resistors take a derivative in one part of the circuit and then reintegrate back to the original signal minus offset somewhere else?

The derivative and the anti-derivative back to the original signal should be voltages. Currents or voltages from currents are not allowed.

If it is possible to get that far it _might_ be possible to argue that there are actually two processes going on in the single cap offset block filter.

Bret Cahill

Nope, it depends on whether it is a series or parallel circuit.

Do you have a clue what the derivative and integral of a sine wave is?

You haven't a clue what you are talking about.

Read and understand:

This is EE101 stuff and what happens is fully explained in the link.

ver

A lot of large caps in series will eventually approach the derivative. This may be all that is necessary.

Bret Cahill

Gibbering nonsense; see:

Scroll down to "For capacitors in series".

Consider the reverse as 2 step process -- small inductor integrates then takes the derivative.

The offset remains just as would be predicted from the above POV.

This isn't a formal proof but it's almost 100% certain someone has treated the subject before, much more comprehensively as well as formally.

Bret Cahill

A lot of ideal math concepts come from non ideal physical situations. Newton wouldn't have bothered with calculus w/o an interest in mechanics.

It might be better to say that the output of a cap is somewhere between the original function (minus DC offset) and the derivative or the output of an inductor is always somewhere between the original function and its integral.

This is juicier fare than what was in the OP.

In math taking a derivative is generally considered a discrete event. Like getting pregnant either you take a derivative or integral or you do not.

With capacitors and inductors you have a continuum.

Has this idea ever been approached in math?

Bret Cahill

ver

Not long after they got the idea for whole number derivatives and integrals:

Nonsense. Not even worth discussing.

John

this integrating and differentiating with capacitors only works (and only approximately) when the output voltage is much lower than the input voltage.

once you're in the pass band of the filter you're getting a different sort of effect.