# over and critically damped systems settling time

I know that for second order systems the settling time(St) equation is:

So my question is, should this same formula be used when the system is over or critically damped? Is it right to use it in that cases?

• Settling time is normally: $T_s\approx \large \frac{4}{\zeta\omega _n}$, and generally it doesn't work for $\zeta > 1$
– Chu
Apr 4, 2017 at 13:48
• So what should I use to find the settling time when ζ>1? Apr 4, 2017 at 13:55
• Solve the time function: unit step response = 0.98 (for 2% settling time and unity gain system).
– Chu
Apr 4, 2017 at 14:03
• Ts is always determined by time at max % error to a step input. where Ts multiplies according to ln (error ratio) e.g. ln(2%) =-3.9 ~4. The damping factor, ζ affect these approximations , and depends on 1 or 2 poles in a 2nd order system. This is different from Rise time which is 63% for Time constant RC=T or typically 10 to 90% for overdamped, but depends on % error again. Apr 5, 2017 at 1:24

TL;DR: NO, you can't use the underdamped settling time formula to find out the settling time of an overdamped system. And you can't use it for a critically damped system either.

## Critically damped case

For the critically damped case ($\zeta=1$), the step response is:

$$v_{out}(t) = H_0 u(t) \lbrack 1 - (1+\omega_0 t) e^{-\omega_0 t} \rbrack$$

If we define the settling time $T_s$ using the same "within 2% of final response" criteria, then:

$$0.02 = (1+\omega_0 T_s) e^{-\omega_0 T_s}\\$$

Solving numerically for $\omega_0 T_s$ (by simply using Excel's solver) we obtain:

$$T_s \approx \frac{5.8335}{\omega_0}$$

## Overdamped case

For the overdamped case ($\zeta>1$), the step response is:

$$v_{out}(t) = H_0 u(t) \left[ 1 - \frac{s_2}{s_2-s_1}e^{s_1 t} - \frac{s_1}{s_1-s_2}e^{s_2 t} \right]$$

where $s_1, s_2$ are the real roots of the transfer function denominator:

$$s_1 = -\zeta \omega_0 + \omega_0 \sqrt{\zeta^2-1} \\ s_2 = -\zeta \omega_0 - \omega_0 \sqrt{\zeta^2-1}$$

For convenience we define:

\begin{align} \Delta &= \frac{s_2-s_1}{2} = - \omega_0 \sqrt{\zeta^2-1} \\ \Sigma &= \frac{s_1+s_2}{2} = - \zeta \omega_0 \\ K &= \frac{\Sigma}{\Delta} = \frac{\zeta}{\sqrt{\zeta^2-1}} \end{align}

So that:

\begin{align} s_1 &= \Sigma-\Delta \\ s_2 &= \Sigma+\Delta \end{align}

If we define the settling time $T_s$ using the same "within 2% of final response" criteria, then:

\begin{align} 0.02 &= \frac{s_2}{s_2-s_1} e^{s_1 T_s} + \frac{s_1}{s_1-s_2} e^{s_2 T_s} = \\ &= \frac{\Sigma + \Delta}{2 \Delta} e^{(\Sigma - \Delta) T_s} - \frac{\Sigma - \Delta}{2 \Delta} e^{(\Sigma + \Delta) T_s} = \\ &= \frac{e^{\Sigma T_s}}{\Delta} \left[ \frac{\Sigma+\Delta}{2} e^{-\Delta T_s} - \frac{\Sigma-\Delta}{2} e^{\Delta T_s} \right] = \\ &= \frac{e^{\Sigma T_s}}{\Delta} \left[ \frac{\Delta}{2} \left( e^{\Delta T_s} + e^{-\Delta T_s} \right) - \frac{\Sigma}{2} \left( e^{\Delta T_s} - e^{-\Delta T_s} \right) \right] = \\ &= \frac{e^{\Sigma T_s}}{\Delta} \left[ \Delta \cosh{(\Delta T_s)} - \Sigma \sinh{(\Delta T_s)} \right] = \\ &= e^{K \Delta T_s} \left[ \cosh{(\Delta T_s)} - K \sinh{(\Delta T_s)} \right] = \\ &= e^{-K |\Delta| T_s} \left[ \cosh{(-|\Delta| T_s)} - K \sinh{(-|\Delta| T_s)} \right] \end{align}

And finally:

$$0.02 = e^{-K |\Delta| T_s} \left[ \cosh{(|\Delta| T_s)} + K \sinh{(|\Delta| T_s)} \right] \\$$

Now that we have rewritten the expression in term of $|\Delta| T_s$ and $K$ (instead of in terms of $s_1$ and $s_2$), we can numerically solve for $|\Delta| T_s$, (by simply using Excel's solver) for any arbitrary given $\zeta>1$.

Example 1: a moderately overdamped system with $\zeta = 1.1$. Thus $K = \frac{1.1}{1.1^2-1} \approx 2.4$, and then solving numerically:

$$T_s \approx \frac{3.172}{|\Delta|} = \frac{3.172}{\omega_0 \sqrt{1.1^2-1}} \approx \frac{6.922}{\omega_0}$$

Example 2: a heavily overdamped system with $\zeta = 5$. Thus $K = \frac{5}{\sqrt{24}} \approx 1.0206$, and then solving numerically:

$$T_s \approx \frac{190.21}{|\Delta|} = \frac{190.21}{\omega_0 \sqrt{24}} \approx \frac{38.827}{\omega_0}$$

There is also an approximation for heavily overdamped ($\zeta \gg 1$) systems based on the dominant pole:

$$v_{out}(t) \approx H_0 u(t) \left[ 1 - e^{s_1 t} \right]$$

If we define the settling time $T_s$ using the same "within 2% of final response" criteria, then:

$$0.02 \approx e^{s_1 T_s}$$

and:

$$T_s \approx \frac{\ln(0.02)}{s_1} = \frac{-\ln(0.02)}{\omega_0 (\zeta-\sqrt{\zeta^2-1})}$$

We can compare this approximation with the exact results that we have derived before.

For $\zeta = 5$:

$$T_s \approx \frac{38.725}{\omega_0}$$

An estimation error just about -0.25%. Quite good indeed.

For $\zeta = 1.1$:

$$T_s \approx \frac{6.096}{\omega_0}$$

An estimation error of approx -12%. Not bad taking into account that $\zeta = 1.1$ is just marginally above the critically damped case!.

## Bonus

We can write a generic settling time expression for $\zeta>1$ as follows

$$T_s = \frac{\psi}{\omega_0}$$

where $\psi$ is a coefficient roughly proportional to the damping factor $\zeta$.

I've numerically calculated the value of $\psi$ for a range of $1<\zeta<9$ using the expression previously derived for settling within 2% of the final value,

$$0.02 = e^{-K |\Delta| T_s} \left[ \cosh{(|\Delta| T_s)} + K \sinh{(|\Delta| T_s)} \right]$$

Then I've calculated (for comparison purposes) 1) the dominant pole approximation, 2) a 3rd order polynomial regression on my numerically calculated dataset, and 3), 4) the relative error due to these two approximations.

Here is an Excel plot with the results: • (jaw hits ground) +1 for epic answer.
– user98663
Apr 11, 2017 at 14:34
• Thank you. I considered calculating and plotting a set of curves representing a range of settling tolerance values (1%, 2%, 5%, 10%...) but then I thought it might be a bit excessive. :D Apr 11, 2017 at 14:41
• Thanks for great answer. As note, I was was solving equation for ζ=10 via secant method, and double got quickly overflown. If you having same problem, just unfold sinh and cosh by definition. I got: 0.5 * ((K + 1) * exp(x * (-K + 1)) + (-K + 1) * exp(x * (-K - 1))) - 0.02, this doesn't overflow :) May 25, 2018 at 14:09
• @EnricBlanco Is there any general solution for finding Ts? I don't want to use Excel or WolframAlpha to calculate it, I know it has two roots by the graph. I've already find it for several hours but failed, I need to automate the calculation Nov 12, 2018 at 13:07
• @Unknown123 I’m afraid there is no closed-form expression, sorry. Nov 12, 2018 at 14:39

The settling time for the underdamped case is well known. I will present solutions for the other two cases (2% definition):

1. Overdamped

The general step response for 2 real and distinct poles $p_1$ and $p_2$ is:

$$y_s(t)=K\left[1 - \frac{p_2}{p_2-p_1}e^{-p_1t} - \frac{p_1}{p_1-p_2}e^{-p_2t}\right]u(t)$$

Doing $p_2=kp_1$, where $k$ is a constant and writing in a normalized form, regardless of the final value $K$:

$$\frac{y_s(t)}{K}=\left[1 - \frac{k}{k-1}e^{-p_1t} + \frac{1}{k-1}e^{-kp_1t}\right]u(t)$$

When $t=t_s$ (settling time), $\frac{y_s(t_s)}{K}$ is equal to 0.98, resulting in:

$$\frac{k}{k-1}e^{-p_1t_s} - \frac{1}{k-1}e^{-kp_1t_s} = 0.02$$

This equation can be solved using numerical methods, for a normalized variable $p_1t_s$. The solution can be simplified if the existence of a dominant pole is admitted, for example $p1$, so that $k \gg 1$. In this case, the second term on left side vanishes fastly and $\frac{k}{k-1}\simeq 1$. Therefore:

$$e^{-p_1t_s} \simeq 0.02$$

Solving for $p_1t_s$:

$$p_1t_s \simeq 3.91$$

or $$t_s \simeq \frac{3.91}{p_1}$$

Using the 5% definition: $t_s \simeq\frac{3}{p_1}$

2. Critically damped

In this case, the normalized response is:

$$y_s(t)= K \left[ 1 - (1 + p_1t)e^{-p_1t}\right]$$

So:

$$\frac{y_s(t)}{K}= 1-\left( 1 + p_1t \right)e^{-p_1t}$$

With a settling time $t_s$ (2% definition):

$$0.02 = (1+p_1t_s)e^{-p_1t_s}$$

This equation can be solved using numerical methods, for a normalized variable $p_1t_s$. With Newton-Raphson I got:

$$p_1t_s \simeq 5.83$$

or $$t_s \simeq \frac{5.83}{p_1}$$

Similarly, using the 5% definition: $t_s \simeq\frac{4.74}{p_1}$

No, you can't use the same formula. The reason being is when you change the poles you also change the settling time. If you solve the equations for a step input and look at the output each equation has different time constants because of the poles of the system. See here:

In the critically damped case, the time constant 1/ω0 is smaller than the slower time constant 2ζ/ω0 of the overdamped case. In consequence, the response is faster. This is the fastest response that contains no overshoot and ringing.