I am reading about the 2nd order transfer function of a 2nd order system (like the mass-spring-damper system). I am constantly seeing the following form as the standard one:

\begin{equation} H(s) = \frac{\omega_0^2}{s^2 + 2 \zeta \omega_0 s + \omega_0^2} \end{equation}

But my questions are:

  1. Why there are not zeros in this standard form? From what I am reading the general 2nd order transfer function can have zeros but in that form there are none and it is said to be the standard one.

  2. Why there is no gain? I hardly found out sources that even mentioned the gain. Their standard form was:

\begin{equation} H(s) = \frac{K \cdot \omega_0^2}{s^2 + 2 \zeta \omega_0 s + \omega_0^2} \end{equation}

  1. From what I understand the general form, the one that should be called standard, should be derived from the general form of the 2nd order differential equation:

\begin{equation} a_{2}\frac{d^{2}y(t)}{dt^{2}}+a_{1}\frac{dy(t)}{dt}+a_{0}y(t)=b_{m}\frac{d^{m}x(t)}{dt^{m}}+b_{m-1}\frac{d^{m-1}x(t)}{dt^{n-1}}+...+b_{0}x(t) \end{equation}

But from what I see it is derived from the not so general form:

\begin{equation} \frac{d^{2}y(t)}{dt^{2}}+a_{1}\frac{dy(t)}{dt}+a_{0}y(t)=x(t) \end{equation}

Why is that?

I have already read those questions (1),(2) and although they are not very similar I am mentioning them in case anyone see them as duplicates.

  • \$\begingroup\$ Like you mentioned about shock absorber, it has a resonance frequency and damping factor (omega, xi). All real systems behave in similar manner. Laplace is an engineering approach to solve diff. equations, therefore the forms are made for real systems, not just pure math. \$\endgroup\$ Commented Nov 15, 2015 at 7:32
  • \$\begingroup\$ Possible duplicate of Standard form of 2nd order transfer function (Laplace transform)? - I believe the first link in the question to be a duplicate and if not then my answer would be exactly the same for that as it would be for this question because the notch filter has a zero (not at the origin) and the high pass has a zero at the origin. The low passs doesn't have a zero. \$\endgroup\$
    – Andy aka
    Commented Nov 15, 2015 at 10:36
  • 1
    \$\begingroup\$ A zero is when you have an "s+k" in the numerator, do you understand that. If so what don't you understand about it when I said in my comment above "the notch filter has a zero (not at the origin) and the high pass has a zero at the origin"? \$\endgroup\$
    – Andy aka
    Commented Nov 15, 2015 at 17:01
  • 1
    \$\begingroup\$ The standard model contains zeroes (numerator) when the filter is a notch or high pass. An s on the top line means a zero at jw=0 and for a notch it means a zero at the notch frequency. This: (1-\$\dfrac{s^2}{\omega_N^2}\$) defines the zero for a notch filter i.e. when s = \$\omega_N\$. \$\endgroup\$
    – Andy aka
    Commented Nov 15, 2015 at 17:16
  • 1
    \$\begingroup\$ People stop voting it as a duplicate! I have not only added from the start a sentence pointing out that I have read the "duplicate" question but I don't even ask the same thing. Also I have explained why my question is different in so many comments. Don't vote if you haven't read the whole post. \$\endgroup\$
    – Adam
    Commented Nov 16, 2015 at 17:41

5 Answers 5


Start with a general system. Assume that all poles are distinct to keep the analysis simpler.

$$\frac{K \left(s+z_1\right) \left(s+z_2\right) \ldots \left(s+z_m\right)}{\left(s+p_1\right) \left(s+p_2\right) \ldots \left(s+p_n\right)}$$

Assuming there are \$q\$ real poles and \$r\$ pairs of complex conjugate poles, the partial fraction expansion yields the following:

$$\sum _{j=1}^q \frac{a_j}{s+p_j}+\sum _{k=1}^r \frac{b_k \left(s+\zeta _k \omega _k\right)+c_k \omega _k \sqrt{1-\zeta _k^2}}{s^2+2 \zeta _k \omega _k s+\omega _k^2}$$

The inverse Laplace transform gives the unit impulse response.

$$\sum _{j=1}^q a_j e^{-p_j t}+\sum _{k=1}^r b_k e^{-\zeta _k \omega _k t} \cos \left(\omega _k \sqrt{1-\zeta _k^2} t\right)+\sum _{k=1}^r c_k e^{-\zeta _k \omega _k t} \sin \left(\omega _k \sqrt{1-\zeta _k^2} t\right)$$

The gain \$K\$ and the zeros \$z_i\$ contribute to the coefficients \$a_j\$, \$b_k\$, and \$c_k\$. These determine the contribution of the simpler terms to the total response. The simpler terms come from the first- and second-order systems, which are the building blocks of the larger system. So the second order system to consider is:

$$\frac{1}{s^2+2 \zeta _k \omega _k s+\omega _k^2}$$

However, it would be more uniform to consider the following:

$$\frac{\omega _k^2}{s^2+2 \zeta _k \omega _k s+\omega _k^2}$$

(This has the effect of scaling the \$b_k\$ and \$c_k\$ coefficients.) Why is this more uniform? Because now all the second-order systems have unit DC gains and their relative importance is given by the \$b_k\$ and \$c_k\$ coefficients (which capture the effect of the zeros and gain in the system).

Now to give specific answers to your questions:

  1. There are no zeros in the standard form, because the effect of the zeros are captured by the constants \$a_j\$, \$b_k\$ and \$c_k\$. These constants determine the relative importance of the simpler first-and second order systems in the overall response of the system.
  2. Same as above.
  3. The general form is much more general than just a second-order system. Even then it can be decomposed into simpler first- and second-order systems.
  • \$\begingroup\$ Thanks for the accept. It was a nontrivial task to get all those equations in! But if I didn't it would all be handwaving. \$\endgroup\$ Commented Nov 16, 2015 at 17:14
  1. because whatever is being modeled turns out to be a 2nd-order low-pass filter.

  2. because the gain at DC is evidently 1 (or 0 dB).

  3. same answer as 1.

  1. A zero would would complicate the dynamics and detract from conceptualisation/ROT, as would a higher order denominator. Unfortunately, not all systems can be adequately modelled by a standard 2nd order TF.

  2. Some standard forms include a gain term. A gain term does not affect the shape of the transient response - just the magnitude and steady-state value

  3. The 2nd order inhomogeneous ODE defines or approximates many fundamental engineering systems


You are right, the general second-order transfer function is a biquadratic function H(s)=N(s)/D(s) with

N(s)=Ao+A1s+A2s^2 and D(s)=1+B1s+B2s^2

(Remark: Because of a division by a suitable factor it is always possible to have "1" in the denominator D)

This function can be realized using Biquad-blocks (using opamps). However, in most cases some special forms are used only, for example:

  • Lowpass: A1=A2=0 with DC gain Ao
  • Bandpass: Ao=A2=0 with midband gain A1/B1
  • Highpass: Ao=A1=0 with max. gain A2/B2.
  • Bandstop: A1=0
  • Allpass: Ao=1, A1=-B1, A2=B2.

EDIT: After introducing the pole frequency wp and the pole quality factor Qp into the transfer function, we arrive at: H(s)=N(s)/D(s) with

N(s)=wp^2(Ao+A1s+A2s^2) and D(s)=s^2+swp/Qp+wp^2

Note (your terminology): Qp=1/2*theta and wo=wp.


Here are some generalized equations for 2nd order filters: -

enter image description here

Note that the formula in the question is the generalized form for a low pass filter and note also that for the equation to form a "zero", "s" has to be present in the numerator. It isn't present in the low pass filter but an "s" is present is: -

  • Highpass
  • Bandpass
  • Notch

For high pass and band pass, the zero is when jw = 0 but for a notch filter, the zero occurs when s = \$\omega_N\$.

  • \$\begingroup\$ To complete Andy`s contribution: The symbol ωn is the so-called "pole (angular) frequency" which is the length of the pointer from the origin to the pole location. \$\endgroup\$
    – LvW
    Commented Nov 16, 2015 at 18:28

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