Why is it said that larger bandwidth leads to better command following , better disturbance rejection and speedy response , but the practical bandwidth being limited by external noise?
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$\begingroup$ The statement isn't always true, if "better" means something practical. Why would you want a driverless car whose the steering could respond to a million inputs per second? A hundred inputs per second (with a bandwidth 10,000 times smaller) would still outperform most human drivers. Even 10 inputs per second would probably be good enough. $\endgroup$– alephzeroCommented Nov 14, 2018 at 13:15
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2$\begingroup$ Would Electrical Engineering or Signal Processing be a better home for this question? $\endgroup$– Qmechanic ♦Commented Nov 14, 2018 at 17:22
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$\begingroup$ Boe, do you have any experience with process control? If you do, I can probably give you an answer of reasonable length. If not, you may need to study a process control book first. $\endgroup$– David WhiteCommented Nov 14, 2018 at 17:45
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2 Answers
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This isn't just limited to control systems -- any kind of linear system behaves this way. Think about it this way. The output of a system that has a larger bandwidth in general responds faster to changes in the input. Imagine you have an input signal that exhibits an abrupt change at some time $t$. Well, abrupt changes in time-domain generally correspond to high-frequency components in the Fourier domain (you need higher frequencies to be able to construct that abrupt change out of a sum of sinusoids). So if the system had a really small bandwidth, these higher frequency components would get filtered out, and the output of the system wouldn't be as abrupt. So in general, the higher your system's bandwidth, the more responsive it is to fast changes in the input, and the lower the bandwidth, the more sluggish will the system be.
However, although you have a more responsive system, you also get a lot more noise in the output. E.g., imagine if the input was contaminated white noise (as is often the case) with spectral density $N$. The noise power in the output of the system is $$P_{n} = \int_{-\infty}^\infty df ~ N \vert H(f)\vert^2 = N \int_{-\infty}^\infty df ~ \vert H(f)\vert^2,$$ where $H(f)$ is the frequency response of the system. With everything else being constant, it's easy to see that the integral on the right hand side increases with larger bandwidths (i.e. a larger support for the $H(f)$ function). So a larger bandwidth basically means you're filtering out less of the noise, which might make your system unsuitable in practice, depending on the application.
answered Jun 17, 2022 at 21:42
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Re. "Why is it said that larger bandwidth leads to better command following , better disturbance rejection and speedy response , but the practical bandwidth being limited by external noise?"
I think this refers to a closed loop feedback control system.
See:
https://www.electronics-tutorials.ws/systems/feedback-systems.html
https://en.wikibooks.org/wiki/Control_Systems/Feedback_Loops#Feedback
and the following best basic tutorial:
https://www.tutorialspoint.com/control_systems/control_systems_feedback.htm
When the output of a system is fed back and subtracted from the input (the command) and the difference (the error) is used to drive the input of the controlled portion of the system, the result is a 'closed loop feedback control system'--which knows what the output is doing and uses its value to adjust the command. It will have greater bandwidth than the system without the feedback loop, greater noise rejection, and better ability to follow the input command because of its wider, flatter bandwidth. {see the forgoing links--the theory is involved}
It will tend to reject noise that is injected within the loop but will be susceptible to noise injected into the input because of its greater bandwidth.
An open loop system (with no feedback) with larger bandwidth would not have these advantages and in fact because of the greater bandwidth would be more susceptible to disturbances and noise. Because the frequency response is not as flat as a closed loop system it would not follow the input as well.
answered Nov 14, 2018 at 17:01
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$\begingroup$ user45664, the control system doesn't know what the output is doing. For feedback control, the output signal is subtracted from the setpoint to give an error signal. For a single-input/single-output (SISO) control algorithm, the error signal is multiplied by a controller gain, and usually a bit of integral function is used, to drive the error signal towards zero (e.g., get the process to the setpoint). $\endgroup$ Commented Nov 14, 2018 at 17:49
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$\begingroup$ A bit of derivative can also be used--a classical PID controller. But I still say the closed loop control system knows what the output is doing, when compared with an open loop control system. $\endgroup$ Commented Nov 14, 2018 at 18:25
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$\begingroup$ Derivative is tricky because it magnifies noise and leads to a jittery control response. And note - when you take the delta (setpoint - feedback) to look at the error signal, you only know what the delta is ... you don't know what the output is at that point. In other words, the input to the standard closed loop PID controller is the error signal, not the output signal. $\endgroup$ Commented Nov 14, 2018 at 19:51
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$\begingroup$ A state variable controller has access to, for example, position, velocity, and acceleration, each of which it individually controls. $\endgroup$ Commented Nov 14, 2018 at 20:17
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$\begingroup$ user45664, let's take this to chat $\endgroup$ Commented Nov 14, 2018 at 20:59