In this 2nd order, low-pass filter circuit


simulate this circuit – Schematic created using CircuitLab

I am interested in the transfer function

$$ H(s) = V_L (s) / V_g (s) $$

which is (hoping that I didn't make mistakes)

$$H(s) = V_g \displaystyle \frac{R_L}{s^2 L C R_L + s(C R_g R_L + L) + R_L + R_g}$$

Which could be the benefits of having

$$ R_L = \sqrt{L/C} $$


I have some notes referring to this condition as to a "matching condition" and this recalls some transmission lines concepts, but I don't know how the transfer function could be simplified by applying that condition.

Even with \$ R_g = R_L = \sqrt{L/C} \$ it can be written as

$$H(s) = V_g \displaystyle \frac{1}{s^2 L C R_L + s(C R_L + L / R_L) + 2}$$

$$H(s) = V_g \displaystyle \frac{1}{s^2 LC + 2s\sqrt{LC} + 2}$$

but, again, I don't see anything useful.

Does the complex conjugate poles have a particular position? Or what else?


Your selection \$R_g=R_L=\sqrt{L/C}\$ gives a conjugate-complex pole pair with a quality factor \$Q_p=0.707\$. Hence, this dimensioning gives you a passive second-order lowpass with Butterworth response (maximally flat). More than that, it can been shown that for each passive ladder structure (and your circuit is the simplest form of a ladder) the sensitivity to parts tolerances is at its theoretical minimum for \$R_g=R_L\$ (matched input and output terminations).

"Does the complex conjugate poles have a particular position?"

Yes - the position of the pole pair has an extraordinary property: The poles are located in the complex \$s\$-plane with a (negative) real part which is identical to the imaginary part (Re=Img). Hence, there is an angle of \$45^{\circ}\$ between a line pointing to the pole and the real axis.

UPDATE: The classical second-order function is \$H(s)=N(s)/D(s)\$. For a lowpass we have \$N(s)=A_o\$ (gain at \$\omega=0\$) and \$D(s)=[1+s/(\omega_pQ_p)+(s/\omega_p)^2]\$.

For finding the maximum of \$H(s)\$ we have to write down the magnitude of the complex function \$H(s=j\omega)\$. As a next step we find the first derivation (differential quotient) and set it to zero. So we find the frequency \$\omega\$, max where \$|H(j\omega)|\$ has its maximum - and inserting this frequency \$\omega\$, max into the expression for the magnitude \$|H(j\omega)|\$ we find the VALUE of the maximum, which is: $$|H,\text{max}|=\frac{A_oQ_p}{\sqrt{1-1/(2Q_p)^2}}$$

From these expressions we can derive that we have \$\omega\$,max=0 and \$|H,\text{max}|=A_o\$ for the special case \$Q_p=\sqrt{0.5}=0.7071\$. This allows the following interpretation:

For \$Q_p=0.7071\$ there is no amplitude peaking and the maximum is reached at \$\omega=0\$. More than that, the magnitude response for this 2nd-order filter has a "maximum flat" characteristic (Butterworth response).

  • \$\begingroup\$ chosen just for the completeness of answer. Thank you. Can you suggest any link where is provided an analytic proof of your statements? In particular, the fact that the Qp factor of 0.707 originates a response with no overshoot. \$\endgroup\$
    – BowPark
    Sep 6 '15 at 15:52
  • \$\begingroup\$ OK - I will add some explanations (regarding Qp=0.7071) in my answer (update). The proof of the statement regarding sensitivity minimization (not easy to understand) can be found in textbooks dealing in detail with filter applications. If it is important for you I will try to find a corresponding ref. \$\endgroup\$
    – LvW
    Sep 6 '15 at 16:56
  • \$\begingroup\$ Ok, no, don't worry for the second proof, but if you can post something about the first (regarding Qp) I will be very grateful to you. \$\endgroup\$
    – BowPark
    Sep 6 '15 at 16:59

I think your on the right track - poles it is!

In filtering, you generally want some region with as little loss as possible and other region with as much loss as possible. In your bode plot each low pass pole will create a knee in the curve where the slope drops by 10 dB / decade. Therefore for a multiple order filter, having all your poles at essentially the same frequency will give you the steepest knee separating your regions of passband and stopband.

Note that your denominator can now would only have a +1 (not +2) if you compared the output of the source (passed the source impedance, Rg) to the final load output. With an end of +1, your denominator can be factored into (s*sqrt(LC)+1)^2, and voila you have two poles at exactly s=1/sqrt(LC)

  • \$\begingroup\$ Roll-off is 20dB/dec for 1st order factors. \$\endgroup\$
    – Chu
    Sep 5 '15 at 6:35

The damping coefficient, \$\zeta=\dfrac{1}{\sqrt{2}}\$ , which is the lowest possible \$\zeta\$ value that does not produce an amplitude resonance peak; i.e. it gives the sharpest corner in the amplitude frequency response with unity amplitude ratio at resonance.


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