# How does the second order time constant affect circuit behavior?

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)

• +1 nice question, good explanation, decent formatting. Sep 6, 2019 at 14:07
• 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/… Sep 6, 2019 at 19:48

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.

• 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? Sep 6, 2019 at 14:37
• 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, $\tau_s=\sqrt{\frac{1}{s}\cdot\frac{1}{s}}=\frac{1}{s}$ Sep 6, 2019 at 14:48
• 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. Sep 6, 2019 at 14:49
• 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. Sep 6, 2019 at 14:50
• 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. Sep 6, 2019 at 15:37