1
\$\begingroup\$

Is a low pass RC circuit the same thing as a first order Buterworth filter?

I was just curious because i googled Buterworth filter and they always show the filter with an inductor in the circuit.

Consider the transfer function for a low pass RC circuit:

$$H(s) = \frac{1}{1+RCs}$$

and the magnitude response for a first order butter worth filter:

$$|H(j \omega)|= H(j\omega)~H(-j\omega)=\frac{1}{\sqrt{1+(\frac{\omega}{\omega_c})^2}}$$

The same thing or different animals?

\$\endgroup\$
1
  • \$\begingroup\$ Relative to @LvW's comments below, have you tried a "1st order Bessel"? Or Chebyshev? \$\endgroup\$ Jan 30 '20 at 13:39
3
\$\begingroup\$

A first order filter has only one free component, so only one degree of freedom, its 3dB bandwidth. You have an arbitrary choice of one component, and then the other is set by the RC time constant needed.

It's only when you get to second order filters (or higher), you get an extra degree of freedom (or more), so now you can define a shape as well as a bandwidth. It's now meaningful to discuss whether it's a Butterworth, Cheby, Bessel or whatever shape.

Whereas a synthesis program for an Nth order filter of any shape can generally set N equal to 1, the resulting filter is trivially identical to the basic first order RC.

When you make the substitution s=jw, and compare like with like, then yes, a first order Butterworth describes the same response as an RC.

It's more useful to say that the Butterworth response reduces to a low pass RC response when it's first order, rather than a low pass RC response is a first order Butterworth.

\$\endgroup\$
8
  • \$\begingroup\$ "A first order filter has only one free component" can you please elaborate? I mean, when I make an first order circuit with a resistor and a capacitors I have 2 elements. If I place them on a breadboard, both are free, when I solder them, both aren't free... \$\endgroup\$
    – Huisman
    Jan 30 '20 at 8:39
  • 1
    \$\begingroup\$ @Huisman A 1st order low pass filter can be completely understood analytically from \$\frac1{1+s}\$. But focusing only on the characteristic equation only one degree of freedom is added by \$\frac1{1+\tau\,s}\$ which allows setting the roll off. That's all you get. In 2nd order, you have a new added term, either \$Q\$ or else \$\zeta\$, which sets and entirely new and separately interesting behavior. This extends beyond 2nd order, though the descriptions becomes more difficult to wrap your mind around, which is why effort is given to breaking down higher order into lower order combinations. \$\endgroup\$
    – jonk
    Jan 30 '20 at 10:36
  • \$\begingroup\$ @jonk I know, I think I wouldn't have used the word "component" but "variable". Component is used for electrical components, esp on a SE.EE site... \$\endgroup\$
    – Huisman
    Jan 30 '20 at 10:54
  • \$\begingroup\$ @Huisman Oh. I see. You were quibbling with that word. Still, the idea of a degree of freedom stands. There is only one degree of freedom in the 1st degree characteristic equation. \$\endgroup\$
    – jonk
    Jan 30 '20 at 10:59
  • \$\begingroup\$ +1 for the effort making the answer nonambiguous \$\endgroup\$
    – Huisman
    Jan 30 '20 at 11:12
1
\$\begingroup\$

Is a low pass RC circuit the same thing as a first order Butterworth filter?

Yes. First order RC passive filters are always Butterworth filters. The term in the transfer function that controls the damping factor (shape), doesn't exist in first order passive filters.

Edit:

Yes, same shape. First order RC passive filters always have a Butterworth shaped response, although since they cannot have any other shape, they are usually not referred to as Butterworth filters. The term in the transfer function that controls the damping factor (shape), doesn't exist in first order passive filters.

/Edit

First order passive filter schematics are often not shown in filter articles because they are trivial. Second (and higher) order passive filters are often made using inductors, they are more interesting, so articles will discuss them.

\$\endgroup\$
10
  • 1
    \$\begingroup\$ No - I think the answer is not correct. The BUTTERWORTH response is available for a filter order of n=2 or higher. A second-order BUTTERWORTH function has a pole-Q of Qp=1/SQRT(2). A 1st-order lowpass has a Q-value of 0.5 only. For 1st-order filter responses we cannot discriminate between different responses... \$\endgroup\$
    – LvW
    Jan 30 '20 at 10:35
  • 1
    \$\begingroup\$ @LvW See the table with n=1 at this wiki. \$\endgroup\$
    – jonk
    Jan 30 '20 at 10:56
  • 1
    \$\begingroup\$ @jonk, at first - I do not blindly trust wiki - and secondly, they have listed Butterworth polynominals ....and, formally, they have included also a first-order expression (just to complete the table). OK - why not? But you certainly will agree that I also could claim that this is a 1st-order Bessel-Thomson or 1st-order Chebyshev expression. Buth this would be crazy (in my eyes). Therefore, it is misleading to call this "Butterworth" response because this implies that there would be other 1st-order possible responses. But this is not the case. \$\endgroup\$
    – LvW
    Jan 30 '20 at 11:08
  • \$\begingroup\$ @LvW I'm only saying that all 1st order filters can be analyzed by that case. The only change possible is the cutoff and that is an irrelevant detail for analytical purposes. Butterworth filters have a definition, "flattest amplitude." I'm just saying that the OP is allowed to call it Butterworth. Also, while I won't debate your assertion, it is still better to just see it that way as it remains consistent in shape as the higher order filters are later introduced. Just my opinion. \$\endgroup\$
    – jonk
    Jan 30 '20 at 11:18
  • \$\begingroup\$ @LvW - good point, edited to make it more clear. \$\endgroup\$
    – Mattman944
    Jan 30 '20 at 13:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.