# Looking for simple LC Butterworth Coefficient tables and scaling rules

Discussion in 'Electronic Basics' started by zebedont, Nov 16, 2004.

1. ### zebedontGuest

I have Googled but get a gazillion (exaggerates) hits and it's all a
bit complicated. Perhaps my search terms aren't too good.

This is for low impedance sources so the first component is an
inductor and the load is resistive. I've had a fiddle with LTspice and
remembered something about root(2) and 1/root(2) so for a single LC
section

L1 = root(2)*R/2piF
C1 = 1/root(2)2piFR

And it seems to work. The -3dB point is at the right frequency and the
response is critically damped.

Is that right for Butterworth?

I suppose I don't want to go much above 4th order.

Any help appreciated.

Zebedee

2. ### Kevin AylwardGuest

If you use SuperSpice it will do all of this for your. You use Fantastic
filter dialog and it will design the filter for you. You can then press
the button and it will place the filter schematic on the page so you can
simulate it.

Kevin Aylward

http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

3. ### John LarkinGuest

No, a Butterworth rings a bit for a step input. Critically damped is
closer to Bessel.

It's worth getting a good filter book, like Williams+Taylor, or even
the old classic "Simplified Modern Filter Design" by Geffe. They have
scads of normalized filter tables, response curves, and scaling rules.

Lancaster's Active Filter Cookbook is good to have around, too.

John

4. ### john jardineGuest

It's a second order filter set for a Q of 0.707. It's just a series resonant
tuned circuit with the source and load resistances sized to provide
sufficient 'damping' to knock the Q down to that (optimum) value of
1/root(2). Easy to sort if you just want a quick filter but awkward if you
want to go up to higher roll off rates or different response shapes.
For particular Butterworth, Cheb, Bessel, response shapes it's quite easy
just to set equal source and load resistances and read from a table of
standardised values. If it's of any use, I've posted a .GIF page with some
Butterworth tabled values, onto alt.binaries.schematics.electronics.
(Unequal source and load resistances need an initial fudge factor
calculating ).

By far the easiest filters for calculation are the traditional (but
seemingly forgotten nowadays) 'prototype K' and 'M-derived' filters. These
use 2 simple calcs to make a 3rd order section. Sections then stacked to any
filter order required. Not theoretically perfect but good enough for most
applications.
Of course there are the free filter prog's out there.
regards
john

5. ### zebedontGuest

Thanks for the software. Unfortunately it calculates values assuming
equal source and load impedances. I have a voltage source as the input
to the filter.

I notice that for a second order Butterworth that if I set the source
impedance to a low value and halve the value of the capacitor it gives
me the response I'm looking for.

I tried a similar thing for higher order filters but the damping
doesn't work out properly. Is there some other trick I should be
using?

Zebedee

6. ### zebedontGuest

Thanks, is there another source for this? I don't have access to the
group you mention. Kevins software assumes equal source and load
impedances as well but I am dealing with a voltage source and
resistive load. What is the fudge factor?
I shall go looking for one.
Thanks again

Zebedee

7. ### zebedontGuest

If I pick F=40KHz and R=8R then I get L=45uH and C=350nF, if I use the
(guessed)sums I've given above. In LTspice the AC filter response is
flat to crossover with no peaking and -3dB down at 40KHz.

If I do a transient response on it then the output overshoots and
recovers in about one cycle. That's what I thought critically damped
meant.

Overdamped would rise to the final value without overshoot. I think
I'm getting confused now, maybe critically damped rises to the final
value in the fastest possible time.

I started out by making the impedance of the L and the C equal to the
load resistance at the crossover frequency, 32uH and 500nF, and got a
peaked response.

Then I tried making them twice the load resistance, thinking that (in
ac terms) they appear in parallel to the load. That gave 64uH and
250nF and the response looked overdamped.

Then I made them equal to the load and scaled them by that root(2) and
1/root(2) and the response looked much nicer..... 11.3 ohms, 45uH and
350nF.

So I fiddled the values by root(2) and 1/root(2) and it got better.

Kevins software gives me 45uH and 700nF but uses equal source and load
resistances. If I make the input resistor small then the response
peaks at crossover.

Well, I think I am suitably lost now.

Zebedee

8. ### John LarkinGuest

It's fine to fiddle with parts on a simulator until you get something
you like, as long as the result is realizable with practical parts.
But "Butterworth" and "critically damped" are formal mathematical
definitions, and a filter can't be both. What you have, something that
rings a little and looks flat in frequency response, is probably
intermediate between the classic Butterworth (maximally flat in
frequency response) and Bessel (no overshoot) forms. Nothing wrong
with that.

I often "design" filters, especially digital filters with shiftable
(power-of-2) coefficients, by fiddling until I like what I see. You
can dignify the results by call it a "transitional" filter.

John

9. ### john jardineGuest

I'll email that page and the series-generator that develops those magic
numbers.

10. ### The PhantomGuest

These are the (very nearly) exact values for a 2nd order
Butterworth low-pass with zero ohms source impedance and 8 ohms load.

The Butterworth response is less than critically damped; the poles
are complex. Therefore it has overshoot.
Critically damped means that the poles of the denominator merge at
the negative real axis into a single, second-order pole. The
oscillatory component of the time response just disappears for this
amount of damping. This would be achieved if you keep the same L=45uH
and C=352 nF but reduce R to 5.657 ohms. R^2 = L/(4C) is the
condition for critical damping (the discriminant of the denominator is
zero) for your zero ohm source impedance case.
The critically damped response has no overshoot, so it rises to the
final value in the shortest time without overshoot. The overdamped
case, of course, has no overshoot, but a slower rise.