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I would like to know how to set minimum system bandwidth according to the following requirement :

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I have a system regulating an output. In nominal operation, the system has to regulate its output to \$S_{out1}\$. Then when an event occurs, the system change its reference (input) in order to change its output. The new output is \$S_{out2}\$.

The time required for changing from \$S_{out1}\$ to \$S_{out2}\$ is \$t_{out1->out2}\$. (Suppose the system takes no time for detecting the event).

My idea was to think about a sinewave and at the derivative around 0 where the slope of the sinewave is maximal in order to extract the frequency of the sinewave and so the bandwidth

\$f(t) = Asin(2*Pi*f*t)\$

And then,

\$f = \frac{slope}{2*Pi*A}\$

But it depends on the amplitude of the sinewave and it is not a parameter that I really know. So it makes me think that it is a bad approximation.

Jess
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  • But you don't expect your system to handle any arbitrary input amplitude, do you? You must have a maximum input amplitude specification, use it. – Designalog Apr 01 '23 at 14:06

1 Answers1

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In general, a critically damped system has the fastest settling time - an underdamped system will overshoot and thereby take longer to settle, while an overdamped system will only reach the target value asymptotically. As a result, the minimum system bandwidth for your desired settling time is equal to the bandwidth of a critically damped resonator that settles to your desired accuracy (i.e. 1%) within the given time.

See this answer for details on how to calculate the settling time from a given resonant frequency, and vice-versa.

As a general approximation, you can equate the resonant frequency of a critically damped low-pass to its bandwidth with reasonable accuracy.

Also, if you choose your bandwidth as f=1/Ts (resonant frequency period equal to settling time), you will find that a 2nd order critically damped system will have settled to within 1.36% of its final value after the settling time, according to the formula in the answer I linked.

This is all independent of the absolute amplitude of the signals involved since the system in question is linear.

Jonathan S.
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  • Thank you for your answer. If the system is not a second order system does it still work ? – Jess Apr 01 '23 at 14:34
  • Yes, it will still work - a critically damped system has the fastest settling time for any order, but you will have to calculate the impulse and frequency response yourself. The results won't vary much, though. – Jonathan S. Apr 01 '23 at 14:35
  • You mean that a larger order system can be approximated to a 2nd order system ? By frequency response you mean the bode plot ? Thank you :) – Jess Apr 01 '23 at 14:55
  • You don't usually encounter 4th order (or higher) systems - and if you do, you'd try very hard to reduce the order because systems higher than 2nd order are extremely difficult to stabilize due to the large phase shift. And I mean that the difference in the "frequency response to settling time" relationship of systems of varying order is at most a small constant factor. The Bode plot is a way to visualize the frequency response. – Jonathan S. Apr 01 '23 at 15:26
  • If the phase shift occurs at high frequencies, and I just need a bandwidth at lower frequencies, it seems possible to stabilize a system whatever its order. It more depends on the position of the different phase shift rather than the order of the system, isn't it ? Thank you :) – Jess Apr 02 '23 at 13:03
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    @Jess You're right about that. What I meant is that a closed-loop system should behave like a 2nd order (or lower) low-pass over the frequency band of interest (where it has loop gain greater than 1), so that it is possible to stabilize it. In other words: The system should have a phase shift of less than 180° up to the frequency where it goes down to unity gain. As a result, you can consider most *stable* closed-loop systems to be at most 2nd order with sufficient accuracy, even if they're not *technically* 2nd order. – Jonathan S. Apr 02 '23 at 13:30
  • OK that is very interesting ! It means that the formula given for the settling time for a 2nd order system seems to be a good approximation if we can consider a high order system working as a second order system in the range of frequencies where the system bahaves like a 2nd order system :D I do not know if it is really possible to think like this :D – Jess Apr 02 '23 at 13:41