# Constrain inputs for the gain function of LQR Controller?

Posted 5 months ago
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 Hi there, I'm working on a small project, whereby I'm controlling an inverted pendulum via Flywheel and a bldc motor. I have a NonlinearStateSpacemodel with a single input u[t] and single output theta[t] (the angle of the pendulum body) . I have completely simulated the system, and created a control scheme that seems to work well, even made a nice little animation to visualize. (please excuse the framerate)Now comes to the point of programming my micrcontroller, and running/verifying the experiment. However, I've come to notice the control equation ends up producing a solution that can could calculate thousands of ampres...However the motor itself can only take 2...Is there a way within Mathematica and the control system to constrain inputs for the gain function directly using builtin functions? Or must this be done via a semi-smooth newtonian solver or something "hand made"I can gladly post code should a person want to play with my system, or more info is required.Thanks for the help!
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Posted 5 months ago
 Mor,LQR regulators let you put weights on the states and the inputs. It seems that your scaling is wrong and you are weighing the states far more than the inputs. Post your code if you want me to be more specific. The problem is that you need to constrain the inputs more. As to your general application: Here is an example of the inverted pendulum solved in Wolfram SystemModeler with an LQG Regulator. They use SystemModeler (WSM) to load the regulator into an Arduino using ModelPlug. Another resource: At the 2018 Wolfram Conference, I demonstrated how to take a State Space Model from WSM and move it to MMA (you already have yours in MMA) and design an LQR controller. Also, in a different talk the Wolfram developer showed new functionality with MMA 12 that allows you to load your regulator designed in MMA into an embedded processor (so soon you will not even have to program the processor yourself!). I believe that both talks (certainly mine!) will be repeated live in this free Wolfram U Webinar Series November 28. (The Webinar will also be posted after the live event)I believe that webinar will have multiple talks that are relevant to what you are doing. In the meantime, please post your code and I can comment on it.Regards,Neil
 Hi Neil!Ahh thanks for the infos with SysModeler, I'll most definitely have a look at it! and I signed up for the webinar ;) Look forward to seeing your presentation. Mathematica being able to program embedded processors will be a godsend. I'll wait to upgrade then :)Ahh Since I couldn't figure out how to post the nonlinear state space model in the code, I'll post the entire thing up until the gains. params = {m1 -> 0.2928, m2 -> 0.1102, l -> 0.087596, k -> 0.09, c1 -> 0.0089, c2 -> 0.00001, I1 -> (885.758 10^-6 + 0.087596^2*0.354), I2 -> (339.10 10^-6 + 130 10^-6 + 0.09^2 0.1102), \[Tau] -> 33.5 10^-3, g -> 9.81 }; rparams = {\[Alpha] -> (I1 + I2 + l^2 m1 + k^2 m2), \[Beta] -> g (l m1 + k m2)} ; Here are the system vectors to build the Lagrangian. rx = l Sin[\[Theta][t]]; ry = -l Cos[\[Theta][t]]; sx = k Sin[\[Theta][t]]; sy = -k Cos[\[Theta][t]]; \[Tau]\[Theta] = I2 \[Omega]''[t]; \[Tau]\[Omega] = -\[Tau] u[t] + I2 \[Theta]''[t]; \[Eta]\[Omega] = c1 \[Omega]'[t]; \[Eta]\[Theta] = c2 \[Theta]'[t]; fx = -f[t] l Sin[\[Theta][t]]; fy = f[t] l Cos[\[Theta][t]]; T1 = 1/2 (m1 (D[rx, t]^2 + D[ry, t]^2) + I1 \[Theta]'[t]^2) (*pendulum body*); T2 = 1/2 (m2 (D[sx, t]^2 + D[sy, t]^2) + I2 \[Theta]'[t]^2 + I2 \[Omega]'[t]^2)(*flywheel*); V1 = m1 g ry (*pendulum body*); V2 = m2 g sy (*fly wheel*); Make the odes eqn1 = FullSimplify[D[D[L, \[Theta]'[t]], t] - D[L, \[Theta][t]]] eqn2 = FullSimplify[D[D[L, \[Omega]'[t]], t] - D[L, \[Omega][t]]] Set params, ICS, solve and plot deqns = {eqn1 == -\[Tau]\[Theta] - \[Eta]\[Theta] + fx + fy , eqn2 == -\[Tau]\[Omega] - \[Eta]\[Omega]} ics = {\[Theta][0] == \[Pi] - 5 \[Pi]/180, \[Theta]'[0] == 0, \[Omega][0] == 0, \[Omega]'[0] == 0}; sol = NDSolve[{{deqns /. rparams, ics} /. params /. {u[t] -> 0, f[t] -> 0}}, {\[Theta][t], \[Omega][t]}, {t, 0, 60}]; pen1 = {l Sin[\[Theta][t]], -l Cos[\[Theta][t]]} /. sol; A plot of just the homogenous side. Falls like expected, the down black arrow is the g force Heres the StateSpaceModel: model = StateSpaceModel[ deqns, {{\[Theta][t], \[Pi]}, {\[Theta]'[t], 0}, {\[Omega][t], 0}, {\[Omega]'[t], 0}}, {u[t], f[t]}, {\[Theta][t]}, t] /. rparams /. params And now the LQR, Q and R matrices : q = DiagonalMatrix[{1, 100, 100, 0}]; r = 20 {{1}}; gains = Join[ LQRegulatorGains[{model, 1}, {q, r}], {ConstantArray[0, 4]}] controlmodel = SystemsModelStateFeedbackConnect[model, gains]; {phi, phidot, omega, omegadot} = StateResponse[{controlmodel, {7 \[Pi]/180, 0, 0, 0}}, {0, bumpy[t]}, {t, 40}]; And the calculated result:  86.6683 (-\[Pi] + \[Theta][t]) + 2.23607 \[Omega][t] + 13.1289 Derivative[1][\[Theta]][t] + 1.44953 Derivative[1][\[Omega]][t] After this I went and did a nonlinear system: nlmodel = NonlinearStateSpaceModel[ deqns, {{\[Theta][t], \[Pi]}, {\[Theta]'[t], 0}, {\[Omega][t], 0}, {\[Omega]'[t], 0}}, {u[t], f[t]}, {\[Theta][t]}, t] nlmodeln = nlmodel /. rparams /. params then a new LQR calculation and plotting the StateResponse gainsn = Join[ LQRegulatorGains[{nlmodeln, 1}, {q, r}], {ConstantArray[0, 4]}] controlmodeln = SystemsModelStateFeedbackConnect[nlmodeln, gains]; {phin, phidotn, omegan, omegadotn} = StateResponse[{controlmodeln, {\[Pi], 0, 0, 0}}, {0, bump[t]}, {t, 40}] Plot[{phin, omegan}, {t, 0, 11}, PlotRange -> All, ImageSize -> Large, PlotTheme -> "Scientific", GridLines -> Automatic, PlotLabels -> {\[Theta], \[Omega], \[FormalW]}, FrameLabel -> {t, A, "Function"}, RotateLabel -> False, LabelStyle -> ( FontSize -> 14)] So, As you can see, it's a pretty nice stabilized system. (well, I think so, haha!) Though I couldn't figure out how to plot the input (the control force u[t]) Once I actually programmed this and saw the system trying to react and saw controlforce skyrocket...Well, we're here where we are.Just to explain some functions and params. theta is the pendulum body omega the flywheel angular velf[t] is the outside disturbance which is described later as "bump[t]" bump[t_] := 0.2 E^(-10 (t - 3)^2) + 0.4 E^(-10 (t - 3.5)^2) - 1/2 E^(-10 (t - 9)^2) and u[t] the control input. At this point, the BLDC motor I have, which is a maxon EC45 Flat, has a maximum RPM of 6200, and an input of (+/-) 2amps, hence the need to restrict the input/control signal to +/- 2.Thanks for the help/ infos, really appreciate it!