Sorry for my English in terms of control theory - up to now I used only German phrases in that field.

There is a controlled process of order 2 with

$$F_p(s)=\frac{1}{(1+s)^2} $$

which shall be controlled in "standard-form" by the controller of type I

$$F_c(s)=\frac{1}{sT} $$

The total transfer function from setpoint to output is $$F(s)=\frac{F_p F_c}{1+F_p F_c} = \frac{1}{sT(1+s)^2+1}$$

How do I get the time-function at the output, when a step-function is applied at the input?

I cannot manage to get the complex roots of the denominator:

$$sT(1+s)^2+1 = K(s-s_1)(s-s_2)(s-s_3)$$

The original question was:

How to choose T, so that the system is right on the edge of a swing-over (aperiodic limiting case) behavior.

I have no Idea, how to solve this analytically.

By using Wolfram alpha

enter image description here

I found roots, but there are two are complex conjugated roots and one real one. The analytic representation is quite awful, but, anyway, there are complex solutions. Each complex solution gives rise to a damped oscillation, so how can it be, that T can be chosen at all to yield no overswing (see image of inverse Laplace)? BTW: By simulation of the closed loop I'm convinced, that there IS a limit for T to yield aperiodic behavior.

enter image description here

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    \$\begingroup\$ Partial fractions allows you to break down the formula into smaller easier bits to solve. \$\endgroup\$
    – Andy aka
    Commented Oct 21, 2021 at 10:32
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    \$\begingroup\$ What did you mean with "right on the edge of a swing-over (aperiodic limiting case) behavior"? That it should be similar to a 2nd order, critically damped (e.g. not under damped, not over damped)? Like a Bessel filter? \$\endgroup\$ Commented Oct 21, 2021 at 10:51
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    \$\begingroup\$ I mean critically damped. I found the solution and posted it here as an answer. My English expressions for control engineering are poor... \$\endgroup\$
    – MichaelW
    Commented Oct 21, 2021 at 11:00
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    \$\begingroup\$ @MichaelW Now it makes sense. I'm also not a native English speaker, so the chances that I might get the meaning wrong are increased... \$\endgroup\$ Commented Oct 21, 2021 at 11:27
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    \$\begingroup\$ Yes, when critically damped there is an overlapping dual pole. As more damping occurs, the poles separate along the sigma axis and remain 0 on the jw axis. \$\endgroup\$
    – Andy aka
    Commented Oct 21, 2021 at 12:33

1 Answer 1


To answer my own question: I think the solution is as follows (it took me incredible long to find it out, although the problem seems quite simple):

Assuming to be at the edge of just having no complex solution, there must be one double root and one single root:

So the ansatz is:

$$Ts(1+s)^2+1 = T(1+s_1)^2(1+s_2)$$

This gives by coefficient comparison

$$s_1=1/3, s_2=4/3, T=27/4$$

So the transfer function for just having no overswing is

$$F(s) = \frac{1}{\frac{27}{4}s(1+s)^2+1}$$

and the I-control is

$$F_c(s) = \frac{4}{27}\cdot \frac{1}{s}$$

The simulation is excatly as expected:

The parameter is

$$K_I = 1/T$$

showing the behavior at the boundary. The limit for T is 27/4, a slightly smaller value gives rise to overswing.

enter image description here

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    \$\begingroup\$ If this is the answer you're looking for then you can select it with the check mark. \$\endgroup\$ Commented Oct 21, 2021 at 18:46
  • \$\begingroup\$ Is my own answer. It cannot be checked by yourself. \$\endgroup\$
    – MichaelW
    Commented Oct 22, 2021 at 6:07
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    \$\begingroup\$ I didn't mean the voting (the up or down arrows), I meant the green check mark right beneath them. You are well entitled to select your own answer. \$\endgroup\$ Commented Oct 22, 2021 at 8:39

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