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CALTECH CDS 101 - Homework Set 4

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CALIFORNIA INSTITUTE OF TECHNOLOGYControl and Dynamical SystemsCDS 101/110R. M. MurrayFall 2002Homework Set #4 Issued: 21 Oct 02Due: 28 Oct 02Note: In the upper left hand corner of the first page of your homework set, please put theclass you are taking (CDS 101, CDS 110) and the number of hours that you spent on thishomework set (including reading).All students should complete the following problems:1. For each of the following linear systems, determine whether the equilibrium point is asymptoticallystable and, if it is, plot the step response and Bode plot for the system. If there are multiple inputs oroutputs, plot the response for each pair of inputs and outputs.(a) Coupled mass spring system. Consider the coupled mass spring system we saw in class, which hasa damper on only one of the masses:u(t) = sin ωtm mkkbq1q2kThe equations of motion are given byddtq1q2˙q1˙q2=0 0 1 00 0 0 1−2k/m k/m 0 0k/m −2k/m 0 −b/mx +000k/muy =·1 0 0 00 1 0 0¸xUse m = 250, k = 50, b = 10 for the parameter values. Note that this system does not diagonalizein the same way as the version given in homework set #2 (due to the asymmetric damper).(b) Bridged Tee Circuit. Consider the following electrical circuit, with input viand output vo:1 2 3~vivoR1R2C1C2ddt·vc1vc2¸="−1C1³1R1+1R2´−1C1R2−1C2R2−1C2R2#·vc1vc2¸+"1C1³1R1+1R2´1C2R2#uy =£0 1¤·vc1vc2¸+ viwhere vc1and vc2are the voltages across the two capacitors. Assume that R1= 100Ω, R2= 100Ωand C1= C2= 1µF.v1.1, 26 Oct 022. (MATLAB/SIMULINK) Consider the inverted pendulum on a cart, as show in the figure below:xfmθM(M + m)¨x + ml cos θ¨θ = −b ˙x + ml sin θ˙θ2+ F(J + ml2)¨θ + ml cos θ¨x = −mgl sin θM = 0.5 kg m = 0.2 kgb = 0.1 N/m/sec l = 0.3 mJ = 0.006 kg m2This system has been modeled in SIMULINK in the file hw4cartpend.mdl, available from the courseweb page.The linearization of the system was given in class:ddtxθ˙x˙θ=0 0 1 00 0 0 10m2gl2J(M + m) + Mml2−(J + ml2)bJ(M + m) + Mml200mgl(M + m)J(M + m) + Mml2−mlbJ(M + m) + Mml20x +00J + ml2J(M + m) + Mml2mlJ(M + m) + Mml2uy =£1 0 0 0¤xNote: in the SIMULINK model, the output has been set to include all of the states (y = x). You willneed this for part (c) below.(a) Use the MATLAB linmod command to numerically compute the linearization of the originalnonlinear system at the equilibrium point (x, θ, ˙x,˙θ) = (0, π, 0, 0). Compare the eigenvalues of theanalytical linearization to those of the one you obtained with linmod and verify they agree.(b) We can design a stabilizing control law for this system using “state feedback”, which is a controllaw of the form u = −Kx (we will learn about this more next week). The closed loop systemunder state feedback has the form˙x = (A − BK)x.Show that the following state feedback stabilizes the linearization of the inverted pendulum on acart: K = (−1, 18.7, −1.7, 3.5).(c) Now build a simulation for the closed loop, nonlinear system in SIMULINK. Use the file hw4cartpend.mdlfor the nonlinear equations of motion in it (you should look in the file and try to understand howit works). Simulate several different initial conditions and show that the controller locally asymp-totically stabilizes x0. Include plots of a representative simulation for an initial condition that isin the region of attraction of the controller and one that is outside the region of attraction. (Hint:remember that the equilibrium point that we linearized about was not zero. You will need toaccount for this in your controller by implementing u = −K(x − x0) for the nonlinear system.)(d) Optional. Use MATLAB to write an animation of your results and post them on the web.2Only CDS 110a/ChE 105 students need to complete the following additional problems:3. Consider the following discrete time systemz[k + 1] = Az[k] + Bu[k]y[k + 1] = Cz[k + 1].In this problem we will derive the stability conditions, step response, and “convolution equation” fordiscrete time systems. (Don’t worry, they are much easier than the ODE versions.)(a) Assume that the matrix A has a full basis of eigenvectors {vi}, so that any initial condition canbe written as a linear combination of these eigenvectors. Let λibe the associated eigenvectorsand show the discrete time system is asymptotically stable if and only if |λi| < 1.(b) Derive a formula for the transient response to an initial condition x0(the analog of eATx0forcontinuous time systems). Your answer should be in the form y[k] = CΦ[k]x0where Φ[k] dependson the matrix A.(c) Derive a formula for the output response of the system to a general input uk. Your result shouldbe expressed as a sum involving terms of the from Φ[k − j] (similar to the terms eA(t−τ )forcontinuous time systems).Note: you can find the answer to this in many books on linear control systems, including those onreserve in the library. You are encouraged to look at them, but make sure you understand the answeryou write down and include enough detail for the TAs to follow your derivation.4. Consider the motion of a small model aircraft powered by a vectored thrust engine, as shown below.net thrust(x, y)θf2f1adjustable flapsLet (x, y, θ) denote the position and orientation of the center of mass of the fan. We assume that theforces acting on the fan consist of a force f1perpendicular to the axis of the fan acting at a distance rand a force f2parallel to the axis of the fan. Let m be the mass of the fan, J the moment of inertia,γ the gravitational constant, and D the damping coefficient. Then the equations of motion for the fanare given by:m¨x = f1cos θ − f2sin θ − d ˙xm¨y = f1sin θ + f2cos θ − mγ − d ˙yJ¨θ = rf1.3It is convenient to redefine the inputs so that the origin is an equilibrium point of the system with zeroinput. If we let u1= f1and u2= f2− mg then the equations becomem¨x = −mg sin θ − d ˙x + u1cos θ − u2sin θm¨y = mg(cos θ − 1) − d ˙y + u1sin θ + u2cos θJ¨θ = ru1.(1)These equations are referred to as the planar ducted fan equations (this is an experiment that we usefor CDS 111).Use the following values for the parameters of the system:γ = 0.52 m/sec2m = 4.25 kgr = 26 cm J = 0.0475 kg m2d = 0.1 kg/secThe reason that gravity γ is not 9.8 m/sec2is because of the presence of a


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