Time constant of second order system

Question

I have a second order system with open-loop transfer function $G(s)={a\over(s+b)(s+c)}$, $a,b,c$ are already known. This system has unity feedback.

I am interested in the time consant of this system, but I need to know should I calculate based on the open-loop or closed-loop transfer function?

The second question is how to calculate the time consant of a second order system? On this webpage (Second Order Systems), it says a second order system may be the combination of two first order systems. If the time constant for the two first order system is $\tau_1$ and $\tau_2$, the time constant for the second order system is $\tau^2=\tau_1 \tau_2$. Can I compute according to this equation?

Answer 1

Time constant for a system should be calculated based on the closed loop transfer function.

For the open loop transfer function you are considering, the closed loop transfer function is given by: $$TF = \frac{G(s)}{1 + G(s)H(s)}$$

For a unity feedback: $H(s) = 1$ \begin{align} \therefore TF &= \frac{\frac{a}{(s+b)(s+c)}}{1 + \frac{a}{(s+b)(s+c)}}\\ &= \frac{a}{(s+b)(s+c) + a} \\ &= \frac{a}{s^2 +(b+c)s + (a + bc)} \end{align} Write the above equation in the form: $$TF = \frac{a}{s^2 + 2\zeta\omega_n s + \omega_n^2}$$ where: $2\zeta\omega_n = (b+c)$ and $\omega_n^2 = (a + bc)$.

The time constant is given by $T = \frac{1}{\zeta\omega_n}$. You would get this same value when you break the second-order system into two first order systems and then find their corresponding time constants. And finally, use the formula that you have stated.

Answer 2

The equation you give, $\tau^2 = \tau_1\tau_2$ is on the web page you linked to.

However you need to understand that $\tau_1$ and $\tau_2$ are first order time constants but $\tau$ is a second order time constant. The first and second order constants do not have exactly the same physical interpretation, and they are not even in the same physical units!