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My textbook explains how we recognize an underdamped , overdamped or a critically damped system and all their characteristics.

Later on, in another chapter, without going into much detail I'm given the following two transfer functions and the systems' responses. $$G_1(s)=\frac{1}{(s+1)^2}\\ G_2(s)=\frac{s+0,5}{(s+1)^2}$$ enter image description hereenter image description here

It is known and written that these are two critically damped systems. It is then just stated that the overshoot happens because of the s=-0,5 zero of the second system. Should that be a good enough explanation for me ? Because I don't get why it happens.

Also , if I'm asked to adjust the systems so that I have a certain overshoot percentage how would I achieve that ? I'm only aware of a formula for underdamped systems which doesn't stand here.

John Katsantas
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  • I would recommend to expand the denominator \$D(s)\$ and put it under a 2nd-order canonical form \$D(s)=1+\frac{s}{Q\omega_0}+(\frac{s}{\omega_0})^2\$ then identify the terms. Check my comments here to understand how \$Q\$ affects the response (https://electronics.stackexchange.com/questions/316825/conjugate-non-conjugate-and-single-poles-in-an-rlc-circuit-system/316836#316836). You must find the formula linking the overshoot to \$Q\$ (or the damping ratio) in your textbook. Watch-out, it is usually valid without zero(s). – Verbal Kint Jul 17 '17 at 15:48

1 Answers1

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Simply put, the second diagram cannot exist as shown. If an LTI system overshoots at all, there must be an oscillatory component to the decay of the error waveform. Remember, the behavior of the system must be qualitatively identical, regardless of the initial value of the error!

Critically damped is when the system is on the verge of oscillatory decay. There is only one point at which this happens.

On the other hand, minimum settling time to within a certain error tolerance does allow a small amount of underdamping, with oscillation. Perhaps that's the distinction that the book is trying to make.

Dave Tweed
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