I learned that the transfer function of a second order circuit can be expressed in the following standard way: $$ \frac{K}{\tau_s^2 s^2 + 2 \zeta \tau_s s + 1} $$ Where:

  • \$K\$ is the gain
  • \$\zeta\$ is the damping factor
  • \$\tau_s\$ is the second order time constant

In trying to understand the meaning of \$\tau_s\$, I found this link that states:

The second order process time constant is the speed that the output response reaches a new steady state condition.

This statement made me have some doubts about how to interpret of this constant.

My questions:

  • Considering, for example, the two transfer functions below:

$$ \begin{array}{c|c} H_1(s)=\dfrac{5}{2s^2+3s+1} \qquad & \qquad H_2(s)=\dfrac{5}{8s^2+6s+1} \end{array} $$

Both have \$0\$ as the steady state condition. Since \$\tau_2 >\tau_1\$, can we conclude that the output of the circuit represented by \$H_2(s)\$ will reach \$0\$ faster than the one represented by \$H_1(s)\$? Can we conclude that the transient response vanishes faster in \$H_2(s)\$, because it has a bigger time constant?

  • If we are dealing with an overdamped second order circuit that is a combination of two first order circuits, such as the following

$$ \left(\frac{K}{\tau_{p1}\,s + 1}\right) \left(\frac{1}{\tau_{p2}\,s + 1}\right) = \frac{K}{\tau_{p1}\tau_{p2}\,s^2 + \left(\tau_{p1}+\tau_{p2}\right)s + 1} $$          we can say that \$\tau_{p1}\cdot\tau_{p2} = \tau_s^2 \implies \tau_s=\sqrt{\tau_{p1}\cdot\tau_{p2}}\$

If \$T_1\$ is the time constant of one of the cascaded first order circuit \$\left(T_1=-\dfrac{1}{\tau_{p1}}\right)\$ and \$T_2\$ of the other \$\left(T_2=-\dfrac{1}{\tau_{p2}}\right)\$, we have:

$$ \tau_s=\sqrt{\frac{T_1+T_2}{T_1\cdot T_2}} $$ So, if one defines \$T_s\triangleq \dfrac{1}{\tau_s}\$, will this value have any meaning, like \$T_1\$ and \$T_2\$ have for a first order circuit (for example, the time to decrease/reach certain value)?

A correction about the expressions in the second question (thanks to @TimWescott):

The poles are \$p_1=-\dfrac{1}{\tau_{p1}}\$ and \$p_2=-\dfrac{1}{\tau_{p2}}\$. Thus, we have: $$ \begin{array}{c|c|c} T_1=-\dfrac{1}{p_1}=\tau_{p1} \qquad & \qquad T_2=-\dfrac{1}{p_2}=\tau_{p2} \qquad & \qquad \tau_s=\displaystyle\sqrt{T_1\cdot T_2} \end{array} $$

Therefore, what I meant was \$T_s\triangleq \tau_s\$. (So, I could have used \$\tau_{p1}\$, \$\tau_{p2}\$ and \$\tau_s\$ directly)

  • 3
    \$\begingroup\$ +1 nice question, good explanation, decent formatting. \$\endgroup\$
    – Ariser
    Sep 6, 2019 at 14:07
  • 1
    \$\begingroup\$ If you want to learn how to combine time constants to form transfer functions, you are ready the fast analytical circuits techniques or FACTs. Have a look at this seminar which is a smooth introduction on the subject: cbasso.pagesperso-orange.fr/Downloads/PPTs/… \$\endgroup\$ Sep 6, 2019 at 19:48

1 Answer 1


Both have 0 as the steady state condition.

You are confusing transfer functions for signals. A transfer function describes a system's behavior; it is a characteristic of some other thing. A signal is more or less a function that evolves in time.

\$H_1\$ and \$H_2\$ both have a DC gain of five, so if the systems they represent are excited by a signal that reaches a steady-state condition, their outputs will each reach a steady-state condition five times the value the input settles to.

If \$T_1\$ is the time constant of one of the cascaded first order circuit (\$T_1 = −\frac{1}{\tau_{p1}}\$).

Time constant implies a variable with units of seconds. Your \$T_1\$ has units of 1/seconds.

will this value have any meaning, ... for a first order circuit (for example, the time to decrease/reach certain value)?

Not particularly. Google "dominant pole". Basically, in a heavily overdamped system that doesn't have significant pole-zero cancellation, the slowest pole wins, and eventually dominates the response.

  • \$\begingroup\$ So, if i excite both circuits with a step function \$1\,H(t)\$, the output will reach a steady-state. In this case, will \$H_2(s)\$ reach it first because of its time constant? \$\endgroup\$ Sep 6, 2019 at 14:37
  • \$\begingroup\$ Another thing: isn't the unit of \$\tau_{p1}\,1/s\$, so for \$T_1\$, we have \$1/(1/s)=s\$? This make sense to me because \$\tau_s=\sqrt{\tau_{p1}\cdot\tau_{p2}}\$ and it is "a speed". Then, \$\tau_s=\sqrt{\frac{1}{s}\cdot\frac{1}{s}}=\frac{1}{s}\$ \$\endgroup\$ Sep 6, 2019 at 14:48
  • \$\begingroup\$ The \$s\$ in \$H(s)\$ is not the unit seconds. It's a variable, and its units are \$\mathrm{s}^-1\$ (Note the Laplacian \$s\$ is denoted as italic, while seconds \$\mathrm{s}\$ is normal. Don't blame me, I wasn't born yet when it was established). I don't recall hearing of a committee from the early 1900's convening to settle on the Most Confusing Variable Name, but they did. \$\endgroup\$
    – TimWescott
    Sep 6, 2019 at 14:49
  • \$\begingroup\$ The time constant of \$H_2\$ is roughly 2.8 seconds, and it is underdamped (if I'm doing my math-in-the-head correctly). The dominant pole time constant of \$H_1\$ is 2 seconds -- so I would expect \$H_1\$ to settle quicker. \$\endgroup\$
    – TimWescott
    Sep 6, 2019 at 14:50
  • \$\begingroup\$ I know that the \$s\$ in \$H(s)\$ is the complex frequency. But you're right about the units. I was thinking in terms of poles, what I meant to say was \$T_1=-\frac{1}{p_1}\$. Now I realized that \$p_1=-\frac{1}{\tau_{p1}}\$, so \$T_1=\tau_{p1}\$ and \$\tau_s\$ has time unit and not frequency/speed. \$\endgroup\$ Sep 6, 2019 at 15:37

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