Fall 2001 16.317-5 Linearization Often have a nonlinear set of dynamics given by f(a, u) where a is once gain the state vector, u is the vector of inputs, and f(, is a nonlinear vector function that describes the dynamics Example: simple spring. With a mass at the end of a linear spring (rate k) we have the dynamics but with a "leaf spring,"as is used on car suspensions, we have a nonlinear spring- the more it deflects, the stiffer it gets. Good model now is kla+k 2T which is a cubic sprin Restoring force depends on the deflection a in a nonlinear way. Figure 1: Response to linear k and nonlinear(k1=0, k2= k)springs(code at the end
Fall 2001 16.31 7—5 Linearization • Often have a nonlinear set of dynamics given by x˙ = f(x, u) where x is once gain the state vector, u is the vector of inputs, and f(·, ·) is a nonlinear vector function that describes the dynamics • Example: simple spring. With a mass at the end of a linear spring (rate k) we have the dynamics mx¨ = kx but with a “leaf spring” as is used on car suspensions, we have a nonlinear spring — the more it deflects, the stiffer it gets. Good model now is mx¨ = (k1x + k2x3 ) which is a “cubic spring”. — Restoring force depends on the deflection x in a nonlinear way. 0 2 4 6 8 10 12 −2 −1 0 1 2 X Time Nonlinear Linear 0 2 4 6 8 10 12 −3 −2 −1 0 1 2 3 V Time Nonlinear Linear Figure 1: Response to linear k and nonlinear (k1 = 0, k2 = k) springs (code at the end)
Fall 2001 16.317-6 Typically assume that the system is operating about some nominal state solution a(t)(possibly requires a nominal input u(t)) Then write the actual state as a(t=a(t)+dx(t and the actual inputs as u(t)=u(t)+Su(t) The "8"is included to denote the fact that we expect the varia- tions about the nominal to be small Can then develop the linearized equations by using the Taylor series expansion of f(, about a (t) and a () Recall the vector equation i=f(a, u), each equation of which 正z=f(x,u) can be expanded as 0 dt )=f(x0+6x,b+) f(x0,)+ afi afi Su du where 0f「af;Of ax a and o means that we should evaluate the function at the nominal values of o and 2o The meaning of "small"deviations now clear-the variations in dc and Su must be small enough that we can ignore the higher order terms in the Taylor expansion of f(a, a
Fall 2001 16.31 7—6 • Typically assume that the system is operating about some nominal state solution x0(t) (possibly requires a nominal input u0(t)) — Then write the actual state as x(t) = x0(t) + δx(t) and the actual inputs as u(t) = u0(t) + δu(t) — The “δ” is included to denote the fact that we expect the variations about the nominal to be “small” • Can then develop the linearized equations by using the Taylor series expansion of f(·, ·) about x0(t) and u0(t). • Recall the vector equation x˙ = f(x, u), each equation of which x˙ i = fi(x, u) can be expanded as d dt(x0 i + δxi) = fi(x0 + δx, u0 + δu) ≈ fi(x0 , u0 ) + ∂fi ∂x ¯ ¯ ¯ ¯ ¯ ¯ 0 δx + ∂fi ∂u ¯ ¯ ¯ ¯ ¯ ¯ 0 δu where ∂fi ∂x = ∂fi ∂x1 ··· ∂fi ∂xn and ·|0 means that we should evaluate the function at the nominal values of x0 and u0. • The meaning of “small” deviations now clear — the variations in δx and δu must be small enough that we can ignore the higher order terms in the Taylor expansion of f(x, u)