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Suppose I have the following step response: $$y_{step}(t)=\frac{kP}{1+kP}(1-e^{\frac{-t}{\tau}})$$ where k is constant and P is the plant. How may I determine the values of k for which there would be no overshoot?

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    \$\begingroup\$ Are there any values of \$k\$ for which there is overshoot? \$\endgroup\$ – Alfred Centauri Mar 13 '14 at 23:14
  • \$\begingroup\$ But how may I check for these values? Given a certain function, how does one normally check for potential overshoots? Do I simply check whether y_step(t) exceeds 1? \$\endgroup\$ – peripatein Mar 14 '14 at 6:49
  • \$\begingroup\$ For a time domain expression, if a function does not overshoot or undershoot but steadily approaches the 'final' value, what does that imply about the (time) derivative of the function? \$\endgroup\$ – Alfred Centauri Mar 14 '14 at 12:27
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The particular step response you show can't overshoot, as k and P are constants. The dynamics of the step response are determined by the exponential function only.

In general, you can check if a linear system will overshoot with a step input by analyzing its frequency response.

The maximum value of the Bode plot of magnitude is related to the overshoot you will have, given a step input, because the step input excites your system at every frequency.

If the maximum of the Bode plot of magnitude of a closed-loop system is 1 (or less, allowing a steady-state error as in the closed-loop system you brought as an example) you have no overshoot.

You could find the topic of critical damping interesting, too.

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  • \$\begingroup\$ P might not be linear. VERY true for a 1st order plant & the rest of this there should not be any overshoot. But a 2nd order plant with a high proportional aspect to it? \$\endgroup\$ – JonRB Mar 14 '14 at 14:34
  • \$\begingroup\$ Actually, P(s)=s^2+a1*s+a0 and Y_step(s)=[kP(s)]/[1+kP(s)]. For what values of k will there be no overshoot? \$\endgroup\$ – peripatein Mar 14 '14 at 18:12
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If there is no overshoot, then, mathematically, it means that the function monotonically increases toward the limiting value at high values of t (which is just the fraction without the exponential factor). This means that its derivative of the function stays positive.

Overshoot means that the function rises and then falls. When it switches from rising to falling, its slope becomes zero very briefly and then negative.

You can analyze the derivative, looking for values of k which make it touch or cross zero.

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  • \$\begingroup\$ Is there any way to check for overshoots without having to use an inverse Laplace (I mean, how else could I determine the derivative wrt time)? My initial expression for y(t) is most likely incorrect. \$\endgroup\$ – peripatein Mar 15 '14 at 7:19
  • \$\begingroup\$ You mean the \$y_\text{step}(t)\$ function isn't necessarily the right one? If it is the actual step function, then it represents the actual time-domain response, and it boils down to, for what values of the parameters does the function not show a "hump". \$\endgroup\$ – Kaz Apr 15 '14 at 22:07

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