Cap-Offset Block Filter Returns A Fractional Derivative Of Order 0 < nu < 1

Feb 12, 2011 3 Replies

formatting link



See the part on the fractional derivative of a constant.



This confirms that the cap filter is outputing something proportional to a fractional derivative with an order between 0 and 1.



A big cap is near 0 and a small cap is near 1.



There is no integration step. It's a one step process after all.


Bret Cahill


If you are trying to do electronics I suggest you get some real capacitors and see how they behave. A signal genernator, some C's, R's and L's, a 'scope. You could really learn something.

George H.

Cap-Offset Block Filter Returns A Fractional Derivative Of Order 0 < nu < 1?

Blither-Cahill pounds keyboard imbecile factoid word arrangement syndrome fail 'tard!

Things will definitely get a little messier using real components but the point here is even using ideal caps you aren't getting the original signal back. What you get is a fractional derivative of order nu close to zero which is close to the original signal (minus offset, of course).

Same with inductors except nu is between 0 and -1.

What's interesting is that fractional derivatives are as difficult to take as either integer or fractional integrals which makes sense since integration and differentiation are considered the same operation here except for the sign of nu.

As noted on a math site, the only reason taking counting number derivatives is easy is because a lot of terms just happen to drop out.

Fractional derivatives need to be mentioned in undergraduate calculus courses because counting number derivatives are deceptively easy.

I first thought about fractional derivatives playing around with caps on SPICE. I didn't know it dated back to Leibniz but I was 100%% sure I wasn't the first.

Ain't no low hanging fruit on Math Tree. Them mathematicians done stripped Math Tree of bark, branches, roots . . .

Bret Cahill

Join the Discussion

Have something to add? Share your thoughts — no account required.

Didn't find your answer?

Ask the community — no account required