How can I rewrite a transfer function in terms of resonance frequency \$\omega_0\$ and damping factor Q? Referred to as "standard form" in the university materials.

I'm still at it, trying to understand LCL filters, and found a gap in the university material. They always let us calculate the transfer function, then the standard form was given, so we just had to fill in the blanks and use the given function to draw a Bode plot. Now that I have a real circuit, I'm stuck. The university book only contains this section on the matter

Circuit analysis book

Nilsson & Riedel has a section devoted to Bode diagrams in the appendix. It says all you need to do is divide away the poles and zeros and factor the result. Poles and zeroes seems to refer to the coefficients of the highest exponents in the numerator and denominator.

None of this is very revealing to me. Say I have the following transfer function. This is indeed in the general form, but how on earth do you factorise that? Getting rid of the poles and zeros is not very helpful either.


simulate this circuit – Schematic created using CircuitLab

$$ H(j\omega)=\frac{j\omega C_fR_f+1}{j\omega(L_1+L_2) + (j\omega)^2C_fR_f(L_1+L_2)+ (j\omega)^3 L_1L_2C_f}\\ H(j\omega)=\frac{C_fR_f}{L_1L_2C_4}\frac{j\omega+\frac{1}{C_fR_f}}{j\omega\frac{L_1+L_2}{L_1L_2C_4} + (j\omega)^2\frac{R_f(L_1+L_2)}{L_1L_2}+ (j\omega)^3 }\\ H(j\omega)=\frac{\omega C_fR_f-j}{\omega(L_1+L_2) + j\omega^2C_fR_f(L_1+L_2)+ j^2\omega^3 L_1L_2C_f}\\ $$

I put that in Wolfram Alpha, and it gave the following roots for the denominator. Besides being humongous, I don't feel they bring me much closer to a solution. first root second root


The factorization finally clicked, and I came up with the following for the undamped case: $$ \begin{align} H(j\omega)&=\frac{1}{(j\omega-0)((L_1+L_2) + (j\omega)^2L_1L_2C_4)} \\ j\omega&=\frac{\pm j \sqrt{4L_1L_2C_4(L_1+L_2)}}{2L_1L_2C_4} \\ H(j\omega)&=\frac{1}{(j\omega-0)(j\omega-j\frac{ \sqrt{4L_1L_2C_4(L_1+L_2)}}{2L_1L_2C_4})(j\omega+j\frac{ \sqrt{4L_1L_2C_4(L_1+L_2)}}{2L_1L_2C_4})} \\ &=\frac{1}{(j\omega)(\frac{L_1+L_2}{L_1L_2C_4}+(j\omega)^2)} \\ &=\frac{\frac{L_1L_2C_4}{L_1+L_2}}{(j\omega)(1+(j\omega)^2\frac{L_1L_2C_4}{L_1+L_2})} \end{align} $$ Putting this in standard form gives $$ \begin{align} H(j\omega)&=\frac{1}{(j\frac{\omega}{\omega_0})(1+j\frac{\omega}{\omega_1 Q}+(j\frac{\omega}{\omega_1})^2)} \\ Q&=0 \\ \omega_0&=1 \\ \omega_1&=\frac{L_1+L_2}{L_1L_2C_4} \end{align} $$

I hope that's not terribly wrong.

  • \$\begingroup\$ Wouldn't it have been useful to divide everything by j noting that 1/j = -j? This gets you a 2nd order formula in the denominator I believe. \$\endgroup\$
    – Andy aka
    Dec 25, 2015 at 13:05
  • \$\begingroup\$ From the 'general form of a transfer function' divide through by \$a_n\$, then factorise and/or truncate appropriately. It's more convenient to start with the TF in Laplace form and do the \$s\rightarrow j\omega\$ transformation further down the page. \$\endgroup\$
    – Chu
    Dec 25, 2015 at 13:23
  • \$\begingroup\$ Dividing by j did not do anything useful for omega, unless i misunderstand you. I aded the result to the question. \$\endgroup\$
    – Pepijn
    Dec 25, 2015 at 13:28
  • \$\begingroup\$ Expressing the roots in symbolic form is not useful. You need to work with the numerical values of the coefficients. \$\endgroup\$
    – Chu
    Dec 25, 2015 at 14:27
  • \$\begingroup\$ With the worked out examples from university the symbolic form always came out to be something like jw/w0 where w0 would be something like R/L*sqrt(1/RC) or whatever. I was hoping to find a similar equation for the resonance frequency fro the LCL filter. \$\endgroup\$
    – Pepijn
    Dec 25, 2015 at 17:58

2 Answers 2


To get to the standard form, you factorize the nominator and denominator polynomials. Then your polynomials will be of the the form \$K_{1}(s - z_1)(s - z_2)\cdots (s - z_n)\$ and \$K_2(s - p_1)(s - p_2)\cdots (s - p_n)\$. Then identify any complex conjugate pairs among the \$z_k\$ and multiply them out. If, for example, \$z_1 = z_2^*\$, then $$ (s - z_1)(s - z_2) = s^2 - 2 \mathrm{Re}(z_1) s + |z_1|^2 = |z_1|^2\left(1 - \frac{2\mathrm{Re}(z_1)}{|z_1|^2}s + \frac{s^2}{|z_1|^2}\right). $$ Now identify $$ \begin{align} \omega_1^2 &= |z_1|^2\\ 1/Q_1 &= -\frac{2 \mathrm{Re}(z_1)}{|z_1|} \end{align} $$ and you get the prescribed form of the second order term. For the remaining roots \$z_k\$, which will be real, extract the factors as $$ (s - z_k) = -z_k (1 - \frac{s}{z_k}). $$ and identify \$\omega_k = -z_k\$.

Repeat for the denominator roots \$p_k\$, and gather the constants to the front to get the factor \$K\$. The roots you got from Wolfram Alpha are, up to the factors of \$i\$ that connect \$s\$ to \$\omega\$, exactly the \$p_k\$. Sometimes they do indeed end up somewhat hairy, but often it's possible to simplify them by identifying common factors (such as paralleled resistors, products RC that always appear together etc).

Finally, if the polynomial has root \$0\$ with multiplicity \$k\$, these will be factors of the form $$ \left(\frac{s}{\omega_m}\right)^k, $$ which you can bring to the front. The factors \$\omega_m\$ are now ambigous, as you can in principle include any of them in \$K\$, but often in practice there's some meaningful choice. For example, if you're designing a filter with a certain passband, you take \$K\$ to be the passband gain (and phase), and take the remaining part to be \$\omega_m^k\$.

The roots \$z_k\$ of the nominator are called the zeroes of the transfer function, as those are the complex values of \$s\$ where the transfer function is indeed the value zero. The roots \$p_k\$ of the denominator are the poles, since those are the values of \$s\$ where the transfer function diverges, which indeed looks like pole sticking out of the \$s\$ -plane if you plot it.

Note that factorizing a polynomial (over the complex numbers) requires finding its roots. For a second order polynomial, the quadratic formula gives you the answer immediately. For third and fourth order polynomials there's the cubic and quartic formulas. The cubic formula is already quite long, and the quartic formula is about a full page in small print, so it's often not useful in practice. For orders higher than five, there is no general formula, although special cases can often be solved.

In addition to using the general formulas, the circuit topology often provides considerable simplifications. For example, in the case of two second order sections separated by a buffer, you can analyze the two sections separately using the quadratic, and the standard form of the combined transfer function is directly the product of the standard forms of the individual sections. The same applies for any number of sections separated by buffers, which is one of the main reasons that high order filter are usually designed as series of second order sections.

If, in the end, you cannot find explicitly the roots, or they're too complicated to use, you can still learn about the your circuit by studying the discriminants, which tells you about potential complex conjugate or real roots. In your specific case (assuming you roots are correct, I didn't check), the discriminant is the term inside the square roots, $$ \Delta = C_f L_2 R_f^2 + C_f L_1 R_f^2 - 4 L_1 L_2. $$ If this is negative, you have a complex conjugate pair of roots leading to a second order term, and it's positive, you get two real roots. You can divide by \$L_2\$ and \$C_f\$ to get the expression $$ \tilde{\Delta} \triangleq R_f^2\left(1 + \frac{L_1}{L_2}\right) - 4 \frac{L_1}{C_f}, $$ which has the same sign as the discriminant. From here you see, for example, that if \$C_f\$ is small enough, or \$R_f\$ is small enough, you get a complex conjugate pair.

  • \$\begingroup\$ I had to leave in the middle of the writing the answer, sorry for any sign errors there. The idea should be there, though. More details later, if necessary. \$\endgroup\$
    – Timo
    Dec 25, 2015 at 14:52
  • 1
    \$\begingroup\$ Okay, so the reason they gave the standard form in university is that factorising by hand is nigh impossible? Once you have the factors, the rest seems manageable. For the circuits in university, the damping and resonance always came out to nice simple equations. For the LCL filter, not so much it seems. I'd be grateful for any more details, but this is already quite helpful. \$\endgroup\$
    – Pepijn
    Dec 25, 2015 at 18:10
  • \$\begingroup\$ @Pepijn: I added a bit more detail, and changed the link to factorization, as the Wikipedia article was indeed about a more general case than needed here. For low-order filters (yours is third order, but there is a pole at \$s = 0\$ which factorizes immediately, so the problem is really only second order), the factorization can be done analytically, by hand or with a computer algebra system such as Mathematica, Maple, or indeed Wolfram Alpha. \$\endgroup\$
    – Timo
    Dec 27, 2015 at 11:17
  • \$\begingroup\$ The realisation that I really only have a second order problem finally clicked in my head, and I was able to make progress. I updated the question with my results. Thanks! \$\endgroup\$
    – Pepijn
    Jan 4, 2016 at 20:19

The transfer function of this circuit can be determined in a few lines without writing a single equation. Use the Fast Analytical Circuits Techniques or FACTs to get there. First analyze the circuit at \$s = 0\$, in dc: open caps and short the inductors. You have \$ H_0=L_2/(L_1+L_2) \$. Then, determine the time constants of that circuit. To do so, reduce the excitation to 0 V or replace \$ V_{in} \$ by a short circuit. You see that \$ L_{1} \$ comes in \$||\$ with \$ L_{2} \$ (\$ L_{eq} \$): this is a degenerate case and the network loses one order. This is a 2nd-order circuit despite the presence of three energy-storing elements. From that circuit, determine the resistance seen from \$C_f\$ temporary removed from the circuit while \$ L_{1} \$ and \$ L_{2} \$ are replaced by a short circuit: you see \$ R_f \$. The first time constant is \$ \tau_{1}= R_fC_f\$. Then, do the same from \$ L_{eq} \$ terminals (\$ C_f \$ is in its dc state and removed) and you see an infinite resistance: \$ \tau_2= L_1/R_{inf}=0\$. You have the first \$ b_1 \$ coefficient: \$ b_1=R_fC_f \$. For the 2nd-order term, determine the resistance driving \$ L_{eq} \$ while \$ C_f \$ is replaced by a short circuit: you see \$ R_f \$ and \$ \tau_{12}= L_f/R_f\$. The second coefficient \$ b_2 \$ is simply \$ \tau_{1}\tau_{12}=R_fC_fL_{eq}/R_f = C_f(L_1||L_2)\$. This is it, you have the denominator \$D(s) = 1 + sR_fC_f + s²C_f(L_1||L_2)\$. Now the zero. What combination of impedance in that circuit prevents \$V_{in}\$ to produce a response \$V_{out}=0\$ at the zero frequency? Well, if the series combination of \$R_f\$ and \$1/sC_f\$ becomes a transformed short circuit, the response is nulled. This is our zero: \$\omega_z = 1/R_fC_f\$. The complete expression is then:

$$ H(s)=\frac{L_2}{L_1+L_2}\frac{1+sR_fC_f}{1+sR_fC_f + s²C_f(L_1||L_2)} $$


$$ H(s) = H_0\frac{1+s/\omega_z}{1 + \frac{s}{Q\omega_0} + (\frac{s}{\omega_0})^2} $$


\$H_0=L_2/(L_1+L_2), \omega_z = 1/R_fC_f, \omega_0 = 1/\sqrt{L_{eq}C_f}, Q = \frac{1}{R_f}\sqrt{L_{eq}{C_f}}\$

Please note that, in theory, for \$s = 0\$, the input resistance of that circuit is a short circuit :-| The real circuit would thus include ohmic losses for \$L_2\$ (\$r_{L2}\$) and \$L_1\$ (\$r_{L1}\$). In this case, the dc gain would become \$r_{L2}/(r_{L1}+r_{L2})\$, the denominator would become a third order and a new zero would appear (\$\omega_{z2} = r_{L2}/L_2\$).

If you want to know more about FACTs, please look at this PPT


You cannot beat the facts in terms of execution speed and simplicity of the result.

  • \$\begingroup\$ The linked pdf was an interesting read and one I saved for future reference. Good share! \$\endgroup\$
    – efox29
    Mar 15, 2019 at 4:02

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