lecture notes, me 340 fall 2013 64Lesson 12 - First-order form, equilibrium points, and linearizationExample 72. Consider the two-degree-of-freedom mechanical system gov-erned by the equations of motionm1¨x1+ c1˙x1+ k1x1+ K(x1− x2)+c(˙x1−˙x2)= f1(367)m2¨x2+ c2˙x2+ k2x2+ K(x2− x1)+c(˙x2−˙x1)= f2(368)and shown in the figure below.Figure 51: A two-degree-of-freedommechanism.We may rewrite this as a system of first-order differential equations byintroducing the new variables y1(t) !˙x1(t) and y2(t) !˙x2(t), in whichcase= f1(369)= f2(370)Let X1(s)=L(x1(#))(s),X2(s)=L(x2(#))(s),Y1(s)=L(y1(#))(s),and Y2(s)=L(y2(#))(s). The corresponding Laplace domain representa-tion becomesY1= sX1− x1(0) (371)Y2= sX2− x2(0) (372)and= F1(373)= F2(374)where F1(s)=L( f1(#))(s),F2(s)=L( f2(#))(s).Let Z =!X1X2Y1Y2"T. We may write the system of first-order equations in matrix form as(sI − A)· Z = z(0)+B (375)!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!lecture notes, me 340 fall 2013 65whereA =00100001−−−−(376)andB =00F1/m1F2/m2(377)It follows thatZ = ·(z(0)+B) (378)The inverse Laplace transformL−1(Z(#))(t) (379)may now be obtained from the convolution of(380)and(381)as described in the Lesson 5 lectures notes. Note that the entries of the ma-trix (sI − A)−1are rational functions in s with poles equal to the values ofs for which det(sI − A)=0, i.e., the eigenvalues of the matrix A.Recall that the inverse Laplace transform of1(s − γ1)(s −γ2)(382)equals!eγ1#∗eγ2#"(t)= (383)provided that γ1#= γ2and!eγ#∗eγ#"(t)= (384)if γ1= γ2= γ. Thus, if γ is an eigenvalue of A, then the entries ofL−1!(#I − A)−1"(t) (385)will include exponential terms of the form p(t)eγtfor some polyno-mial38p(t). Eigenvalues in the right half of the complex plane will38The polynomial results from repeatedeigenvalues. Show this by computingthe convolution (eγ#∗#neγ#)(t).result in exponentially growing functions of time, whereas eigen-values in the left half of the complex plane will result in exponen-tially decaying functions of time. Eigenvalues on the imaginary axis!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!lecture notes, me 340 fall 2013 66yield oscillatory terms that neither grow nor decay, unless they arerepeated eigenvalues, in which case there will be terms that growpolynomially in time.Example 73. For f1= f2= 0, the configurationx1= x2= y1= y2= 0(386)in the mechanical system in the previous example is an equilibrium, sincethe system remains in this configuration in the absence of any forcing. Theequilibrium configuration is asymptotically stable if any small perturba-tion results in a free response that decays back to the equilibrium as t → ∞.By the analysis in the example, it follows that the equilibrium is asymptot-ically stable if and only if all eigenvalues of A lie in the left half of the com-plex plane. In contrast, if any eigenvalue lies in the right half of the complexplane, then there exist free responses, for initial conditions arbitrarily closeto the equilibrium, that blow up as t → ∞.Lagrange’s equations describe the dynamics of a mechanical sys-tem using one (or more) differential equations involving generalizedcoordinates. Since these equations are typically not linear39, methods39We say that they are nonlinear!particular to linear systems, such as transfer functions, convolutions,unit-impulse responses, etc, no longer apply. We need additionalideas to make this connection.An ordinary differential equation (ODE) is any equation of a functionx(t) with the formg(x(t),˙x(t),...,x(n)(t), u(t)) = 0(387)where x(t) is an unknown function, and u(t) is a (known) forcingfunction.40If an ODE has the special form40Either or both of x(t) and u(t) may bevector functions rather than scalars.g(x(t),˙x(t),...,x(n)(t), u(t)) = + u(t)(388)then it is a linear ODE41; otherwise, we call this a nonlinear ODE.41When u(t)=0, the solutions to thelinear differential equation satisfy theprinciple of superposition; if x(t) andy(t) are solutions, then αx(t)+βy(t) isalso a solution for any scalars α, βA complete theory for linear ODEs exists; nonlinear ODEs aremuch harder to deal with in general42. Two common strategies for42For nonlinear ODEs, additionalassumptions are required even toguarantee that a solution exists!working with nonlinear ODEs – linearization or numerical simula-tion – are both simplified by first converting to first-order form43.A43This is also called state space formdifferential equation in first order form is expressed asdzdt(t)= f (z(t), t) (389)in which z is a vector called the state, and f is the vector field.Example 74. Consider a pendulum of mass m on a massless rod of length!, rotating around a hinge. The pendulum is connected via a spring with!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!lecture notes, me 340 fall 2013 67stiffness k and relaxed length !0to a massless slider, which moves along ahorizontal rod a distance ! above the pendulum hinge such that the spring isalways vertical. Suppose that motion takes place in a vertical gravitationalfield of strength g and that a viscous force with coefficient c and a horizontalforce F(t) act on the pendulum mass.Figure 52: A single-degree-of-freedompendulum mechanism.Let θ denote the counterclockwise angle of rotation of the pendulumrelative to the horizontal. For notational convenience, letα =mg + k!0k!(390)such that α = 2 if the force of gravity equals the force in the spring whenθ = −π/2. The kinetic and potential energies are then given byT = , V = (391)Moreover, the contribution to the generalized force corresponding to thegeneralized coordinate θ from the nonconservative forces equalsFnc= (392)Lagrange’s equations of the second kind then becomem!2¨θ −k! cos θ(! −! sin θ − !0)+mg! cos θ + c!2˙θ + F(t)! sin θ = 0(393)We can rewrite the equation of motion as a first-order system by lettingz1! θ and z2!˙θ, in which case the corresponding vector field is given byf (z, t)=)z2−*kc1(!s1−! + !0)+mgc1+ c !z2− F(t)s1+/m!,(394)where c1= cos z1and s1= sin z1. When F(t)=0, the vector field dependsonly on z and not explicitly on the value of t. When this happens, we saythat the system is autonomous.An equilibrium of an autonomous system with vector field f (z) isany point zesuch that f (ze)=0; if z(0)=ze, then the solution to thesystem of differential equations is given by z(t)=zefor all