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I've been trying to figure out how to estimate the settling time of a second order system in response to a step input of magnitude 5. The systems transfer function is $$G(s) = \frac{1}{(s+2)(s+4)}$$ and I have already determined the time response with the step input R(s): $$C(s)=R(s)G(s)\qquad \therefore c(t) = \frac{5}{8}+\frac{5}{8}e^{-4t}-\frac{5}{4}e^{-2t}$$ Now I need to estimate the 2% settling time of the response using this information, but I'm not sure how. I know the system is overdamped as ζ>1, so I can't use the normal settling time equation $$T_s =\frac{4}{\zeta\omega_n}$$ I looked into this post: (over and critically damped systems settling time) but the answers only explain long winded ways to get an accurate result. I've already used MATLAB to obtain an exact result of 2.3 seconds, but I need to be able to estimate it without MATLAB.

I was thinking I could try trial and error with different values of t until c(t) is within 2% of the steady state value (which is 0.625) but while this would work I doubt its the correct way to do it, so is anyone able to help me out with a better method?

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3 Answers 3

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For systems with real left-half plane poles, you can usually estimate it by only considering the dominant pole (the pole with the lowest frequency). In your case this would be \$p_d=-2\$. The result gets more accurate as the non-dominant pole (\$p_{nd}\$) moves further away from the dominant pole.

By only considering the dominant pole, you get a rather simple equation:

$$\begin{align} \frac{5}{4}\cdot e^{-2t}&=0.02\cdot \frac{5}{8} \\ t &= -\frac{1}{2}\cdot \ln\left( 0.02\cdot \frac{5}{8}\cdot \frac{4}{5} \right) \approx 2.30258509299s\\ \end{align}$$

The idea is that the non-dominant pole at \$p_{nd}=-4\$ leads to a term \$e^{-4t}\$ which will be damping so quickly that it doesn't affect the overall settling time. The advantage is the simplicity of the equation, and the fact that is actually a pretty common occurrence to have a very dominant pole and far-away non-dominant poles in electronic circuits.

In your specific case, it is possible to analytically calculate the settling time. The time it takes for the time-dependent terms to damp to 2% of the final value can be calculated using (similar to Andy's answer, but using the absolute value):

$$\begin{align} \left| e^{-4t}-2\cdot e^{-2t} \right| &=0.02 \\ &\Updownarrow (y=e^{-2t})\\ y^2-2\cdot y &= \pm 0.02 \\ &\Updownarrow (\text{There are 4 distinct solutions, but I only take the relevant one}) \\ y = e^{-2t} &= 1 - \frac{7}{5\sqrt{2}} \\ &\Updownarrow \\ t &= -\frac{1}{2}\cdot \ln\left(1 - \frac{7}{5\sqrt{2}}\right) \approx 2.30006613189s \end{align}$$

So a factor of 2 for \$p_{nd}/p_d\$ leads to about an error of 0.1% on the calculated settling time when using the dominant-pole approximation. Whether or not this is sufficient I leave to you.

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  • \$\begingroup\$ Thats perfect thankyou! \$\endgroup\$
    – MendelumS
    Jul 22, 2020 at 13:09
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Yes, your inverse Laplace calculation is correct.

The final steady state value will be 5/8 - this is the DC value after a long length of time. So, you are really looking for the rest of the equation to fall in magnitude to 2% of 5/8: -

$$\dfrac{5}{8}e^{-4t} - \dfrac{5}{4}e^{-2t} = \dfrac{5}{8}\cdot \text{0.02}$$

$$=\dfrac{8}{8}e^{-4t} - \dfrac{8}{4}e^{-2t} = \dfrac{8}{8}\cdot \text{0.02}$$

$$= e^{-4t} - 2e^{-2t} = 0.02$$

Does that help?

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  • \$\begingroup\$ I was pretty close then as I was just doing a similar thing to this, but I was using 0.98 instead of 0.02. Why do you use 2% and not 98%? Surely if we want 2% less than the final value then it would be 0.98 x 5/8? \$\endgroup\$
    – MendelumS
    Jul 22, 2020 at 12:05
  • \$\begingroup\$ @Sam your approach and mine are equally valid, it's just I thought one way and you thought the other. \$\endgroup\$
    – Andy aka
    Jul 22, 2020 at 12:10
  • \$\begingroup\$ We would have to use \$-0.02\$ instead of \$+0.02\$ to find the correct settling time (or invert the coefficients of the exponentials, or use the 0.98 @Sam proposed and include the DC term). Probably an oversight, but perhaps the source of the confusion? \$\endgroup\$
    – Sven B
    Jul 22, 2020 at 13:08
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Well, let's solve this in a more general case. We have the following transfer function (assuming real positive value for \$\epsilon\$):

$$\mathcal{H}\left(\text{s}\right):=\frac{\text{Y}\left(\text{s}\right)}{\text{X}\left(\text{s}\right)}=\frac{1}{\left(\text{s}+\epsilon\right)\left(\text{s}+2\epsilon\right)}\tag1$$

When we look at the step-response we are using \$\text{X}\left(\text{s}\right)=\mathcal{L}_t\left[\theta\left(t\right)\right]_{\left(\text{s}\right)}=\frac{1}{\text{s}}\$, so the output is given by:

$$\text{Y}\left(\text{s}\right)=\frac{1}{\text{s}}\cdot\frac{1}{\left(\text{s}+\epsilon\right)\left(\text{s}+2\epsilon\right)}\tag2$$

Using inverse Laplace transform, we find:

$$\text{y}\left(t\right)=\mathcal{L}_\text{s}^{-1}\left[\frac{1}{\text{s}}\cdot\frac{1}{\left(\text{s}+\epsilon\right)\left(\text{s}+2\epsilon\right)}\right]_{\left(t\right)}=\frac{\exp\left(-2\epsilon t\right)\left(\exp\left(\epsilon t\right)-1\right)^2}{2\epsilon^2}\tag3$$

It is not hard to show that when \$t\to\infty\$ (assuming real positive value for \$\epsilon\$), we get:

$$\lim_{t\to\infty}\text{y}\left(t\right)=\frac{1}{2\epsilon^2}\tag4$$

Now, for the settling time, we want to find the time \$t\$ when \$\text{n}\text{%}\$ of the final value is reached:

$$\text{y}\left(t_\text{n}\right)=\frac{\text{n}\text{%}}{100}\cdot\frac{1}{2\epsilon^2}\Longleftrightarrow\space t_\text{n}=\dots\tag5$$

Solving that gives:

$$t_\text{n}=\frac{1}{\epsilon}\cdot\ln\left(\frac{10}{10-\sqrt{\text{n}\text{%}}}\right)\tag6$$

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