I am studying control systems, and in my studies I have encountered the topic of control scheme of the form

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and I have seen thaht this control scheme is really useful since it is possible to add a lag or lead compensator outside the closed loop, so it won't give stability problems.

At this point I was curious to know a little more and in particular I would like to know if it is possible to use a prefilter which is a delay. In such a way it would be possible to avoid the problem of stability given by a system with delay.

So, to explain better, I would have for example:


and so I could design a stable closed loop without delay, and then filter the reference with a delay, so use:


instead of using a closed loop with delay, which gives stability problems.

Do you think it would be a good idea? Or it would be impossible or conceptually wrong?


1 Answer 1


1 - It can be a good idea to put a filter $C_r(s)$ in order to smooth the reference. For example, if you have a speed regulator, and the reference jumps from 100 km/h to 200 km/h, having a low-pass filter will smooth the actual reference fed to the closed-loop. It could prevent excitation of unmodeled dynamics, actuator wind-up, etc. That being said if $C_r(s)$ is a simple delay, it has no purposes as far as I can see.

2 - $C_r(s)$ only affects the reference fed to the closed-loop. It will not affect stability of the closed-loop like you mentioned. It also has no effect on disturbance rejection or errors due to modelization (such as unmodeled dynamics).

  • $\begingroup$ Thanks for answering. Suppose I have a system with a time delay. If for example I want to plot the step response at t=1, the if the system has a delay in it, it will introduce some problems for the stability of the system. But if I consider a prefilter as a delay, the step response will be plotted at t=1, but there won't be stability problems. I don't know if it makes sense or I am just confusing things.I ask beacuse it looks like the same thing, but in the case of a prefilter I don't have to worry about stability. What do you think? Thanks again. $\endgroup$
    – J.D.
    Feb 9, 2020 at 15:00
  • 1
    $\begingroup$ There won't be any stability problems but it still has no value in real life $\endgroup$
    – Ben
    Feb 9, 2020 at 15:59

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