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Anti-windup is a concept for feedback controllers with integral terms, e.g. PID, to keep the integral term from „overcharging“ when regulating a large set point error. It basically saturates the integral term to keep the system from overshooting the set point. The classic form of anti-windup, as described above, does not actually ensure satisfaction of ...

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For some systems, the salient criterion is settling time to within some error band. Sometimes you can get faster settling by allowing earlier overshoot. If you need a system to get to within some minimum error of before starting a process, you probably don't care what the system is doing before the process is started, only that it gets there as quickly as ...

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To my knowledge the simplest way to accomplish the transfer function... $$G(s) = \frac{E_o(s)}{E_i(s)} = \frac{0.0364s}{0.0002s + 1} = \frac{KTs}{Ts +1}$$ ...is the following high-pass filter circuit: The transfer function of this circuit is: $$\frac{E_o(s)}{E_i(s)} = \frac{R_2+R_3}{R_3}\frac{R_1 C s}{R_1 C s + 1}$$ So let $R_1C = T = 0.0002$ and $\... 4 To understand why proportional gain won't drive the error point to zero, it is best just to look at the math. Consider the PID loop shown in the image below. The loop algebra in the$sdomain comes out to \begin{align} e(s)&=r(s)-y(s)\\ y(s)&=P\ u(s)\\ u(s)&=\left(k_p+\frac{k_i}{s}+k_ds\right)e(s), \end{align} where I have usedP$... 4 Both the (negated) output and reference are needed to calculate the error. So in the scalar case the reference also gets multiplied by$s\,k_d$, while$T_2(1,1)$only considers the contribution of$r$to$y_1$. In order to get the correct results you should also add the contribution of$s\,r$, which could be expressed as$s\,T_2(1,2)$or$T_2(2,2)$. 4 Although the advertised advantage of a stepper is that it can be driven "open loop" (no position or velocity feedback required for accuracy, instead you just keep count of the steps) this is only true when the speed with which the step signals are being fed to it are below a certain rate threshold. Beyond that threshold, the stepper armature begins to lag ... 4 At the very high level you will need 24V Power supply or a method to generate 24V 24V Motor controller Microcontroller - Arduino is a good place to start There are also prebuild motor controllers that can be programed via computer. These tend to be expensive. I would suggest following web sites similar to the ones listed below. They tend to have blogs, ... 3 Consider a controller$C(s)$for plant$G(s)$and feedback$H(s)$. The closed loop is $$\frac{Y(s)}{U(s)}=\frac{CG}{1+CGH}$$ The ramp is$U(s)=\frac1{s^2}$Thus the error is $$E=Y(s)-\frac{1}{s^2}=(\frac{CG}{1+CGH}-1)\frac1{s^2}$$ According to the final value theorem,$$\lim_{t\to \infty} e(t)=\lim_{s\to 0} sE(s)=\lim_{s\to 0} (\frac{CG-1-CGH}{1+CGH} ... 3 A lead filter implies that the zero has a lower frequency than the pole. While a PD controller with a low-pass-filter does not necessarily imply that order. Also a lead filter (usually) does not have the zero and pole to far apart from each other, meaning that the bode diagram does not get very close to the asymptotes of +1 slope for magnitude and 90° for ... 3 Not really an answer, but I can't comment, so: Does this mean I could in general choose between PID and G-Code? PID and G-code are not two interchangable approaches to control a CNC-machine, they serve different purposes. G-code is basically the programming language to tell the CNC-machine where to drive with the axes, when to do stuff like turn on the ... 3 Step 1: Draw the root locus of the system. Here you can see the two poles of your plant$G(s)$(marked with an x), at$p_1=-9$and$p_2=-1$, the pole of your controller$C(s)$at$p_c = 0$and the zero (marked with an o) at$z_c = -c$(for now just at a random location). The purple squares indicate the poles of the closed loop system for a certain gain$K_p$... 3 I would not call this a cascaded control loop but rather a MIMO control loop. Typically a cascaded control loop has an inner and an outer loop and the output of the outer loop is used as the setpoint for the inner loop. Here you have multiple loops, for some reason running at different update rates, but each with their own setpoint and (presumably) sensors. ... 3 But what if my error is zero and plant works ideally? meaning there is no history of error being anything other than zero. So three terms have been and will be zero. Does that mean the input of my plant is zero? Yes. That means that the desired output of each plant always should have an input equal to zero so that it would be equal to the setpoint? No, ... 3 Even if you can tune a function as per your needs, using proportional control(P) only you can never fully eliminate error from the system. Error is inherent in P control. Using I control along with P, you can fully eliminate the error in the system, but now your system has a specific transient response. D control may be used if you want to control the ... 2 Your system is not reaching the target value. Possibly this is due to static friction in the actuator mechanism, which causes it not to move until the force exceeds some threshold. Assuming you are using a PID controller, the residual error will cause the I error term to accumulate slowly. This will cause the actuator force to increase until it is enough to ... 2 In general, poor performance of PID2 will cause poor performance in PID1, in both disturbance rejection and setpoint tracking. There are some special instances where poor performance of PID2 in the form of high frequency oscillation is attenuated by G1, so the performance of PID1 is not significantly affected. 2 It really depends on the logic required for your system. I don't think there is a hard fast mathematical solution like there is for a standard one-input one-output PID. In many cases a PID may not even be the best logic for your system; especially if the outputs have a complex relationship with the input like a chemical reaction. PID's are attractive ... 2 Quick Answer The$R$in the PI controller equation would best be described as a "weighting factor" that determines the influence of the integral term relative to the proportional term. The larger$R$gets, the more the integrated error affects the output of the controller compared to the instantaneous error. More explanation Often PI controllers ... 2 The A could be acceleration. The acceleration part is then used as feedforward or feedback in addition to the ordinary P, I and D parts. Take for instance a height controller for a quadcopter. For an ordinary PID controller the equation would be$F = K_p \cdot e + K_i \cdot \int e\ dt + K_d \cdot \dot{e}$, where$e = h_{d} - h$(error between desired ... 2 You are correct about the effect of the Proportional and Integral terms in your controller, but don't forget the Derivative part! While the Integral term tends to add energy to a system, the Derivative term subtracts energy by trying to slow down the rate of change of error. It acts like friction or damping in the system. Derivative terms are usually ... 2 This system can be considered "closed loop," since the control input is being determined by some sort of feedback loop (even though the mathematical expression of the feedback loop through the operator is unknown). Also, yes the diagram you've shown is valid. I often work with control systems that have a human operator in the control loop, such as robotic ... 2 There are two answers here. First answer is the theoretical one. It doesn't matter what static power you apply. You could apply any number. The reason is that the plot of y(t) that you end up drawing will be normalized to an input of 1. In other words, compare slides 12 and 13 in this presentation PID control - Simple tuning methods If you apply an ... 2 Alephzero's comment pretty well covers it. "exact spot" need to be quantified. For an elevator, a mm or two is sufficient error. For a phased-array optical telescope, the error must be closer to 100 nm (fraction of visible light wavelength). About the only rule designers apply to elevator algorithms is that the system must be significantly overdamped. ... 2 Ideal Water Tank The closed-loop water level control system looks like Figure. Figure. Closed-loop water level control system Figure 3 can be used as mentioned in comment above : T(s) = 1 / ( A * s ) where Flow = Area * ( dHeight / dTime ) If all parameters set (positively), this system will be stable also. Changing controller parameters will change the ... 2 Your system: You have a PID connected to a servo. So you have a closed system connected to a closed system. A servomotor already makes a control loop, that's what being a servo means. So more than likely the servomotor already has a PID inside it (or some other controller), so you don't really need a PID to connect to the servo. you can just connect the$P_C\$...

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There might be so many methods to perform that. But, the premium method is to use a nonlinear Model Predictive Control (MPC). This is a multi-input/multi-output control method which predicts the future based on the available model. Then it optimizes a series of control inputs to minimize a forecast cost function with respect to the constraints. A nonlinear ...

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So typically there is proportionality and a continuously varying (analogue) output. Correct. This could be an analog voltage or a digital value. In the case of a simple process like a domestic gas boiler, the boiler is either fully on or fully off. Or, say a cooling fan that can only be switched on or off to keep something cold. In these cases, can a ...

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The PID simulation you've built does not care about the units of numbers being passed around your control loop. However, you may want to ensure that the units are correct at different points in the loop to make it easier for you to interpret and debug. To do that, you can simply add gain blocks (no frequency dependence) with the appropriate conversion ...

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Any question that asks "would it make sense" in control systems has the answer "that depends on your plant". In the case of a limited time horizon integrator, you're turning your integrator from something that has infinite gain at DC into (essentially) a low-pass filter. You lose the ability to completely cancel out a constant disturbance. I think that ...

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Ideally, I would imagine there would be sufficient material in the burner and the fan speed would be the only thing to modulate to regulate temperature. If that's the case, then I would have the target temperature as a reference input to the fan loop. Then, what you're left to look at is actual fan speed. As the material in the burner is consumed, the fan ...

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