3 Short review of ODEs4 Revisiting the Integral Controller5 Transfer functions6 Frequency Responses of Linear Systems7 Saturation and Antiwindup Strategies8 Effect of time delays9 More on ODEs10 Distributions11 DC Motors12 Robustness Margins13 Control of Second-Order System19 Jacobian Linearizations, equilibrium points20 Linear Systems and Time-Invariance21 Matrix Exponential22 Eigenvalues, eigenvectors, stability23 Jordan Form19 Jacobian Linearizations, equilibrium pointsIn modeling systems, we see that nearly all systems are nonlinear, in that the dif-ferential equations governing the evolution of the system’s variables are nonlinear.However, most of the theory we have developed has centered on linear systems.So, a question arises: “In what limited sense can a nonlinear system be viewed asa linear system?” In this section we develop what is called a “Jacobian lineariza-tion of a nonlinear system,” about a specific operating point, called an equilibriumpoint.19.1 Equilibrium PointsConsider a nonlinear differential equation˙x(t) = f(x(t), u(t)) (72)where f is a function mapping Rn× Rm→ Rn. A point ¯x ∈ Rnis called anequilibrium point if there is a specific ¯u ∈ Rm(called the equilibrium input)such thatf (¯x, ¯u) = 0nSuppose ¯x is an equilibrium point (with equilibrium input ¯u). Consider startingthe system (72) from initial condition x(t0) = ¯x, and applying the input u(t) ≡ ¯ufor all t ≥ t0. The resulting solution x(t) satisfiesx(t) = ¯xfor all t ≥ t0. That is why it is called an equilibrium point.19.2 Deviation VariablesSuppose (¯x, ¯u) is an equilibrium point and input. We know that if we start thesystem at x(t0) = ¯x, and apply the constant input u(t) ≡ ¯u, then the state of thesystem will remain fixed at x(t) = ¯x for all t. What happens if we start a littlebit away from ¯x, and we apply a slightly different input from ¯u? Define deviationvariables to measure the difference.δx(t) := x(t) − ¯xδu(t) := u(t) − ¯uIn this way, we are simply relabling where we call 0. Now, the variables x(t) andu(t) are related by the differential equation˙x(t) = f(x(t), u(t))169From _Dynamic Systems and Feedback_by Packard, Poola, Horowitz, 2002Copyright 2002, Andrew PackardAll rights reserved.Do not duplicate or redistribute.Substituting in, using the constant and deviation variables, we get˙δx(t) = f (¯x + δx(t), ¯u + δu(t))This is exact. Now however, let’s do a Taylor expansion of the right hand side,and neglect all higher (higher than 1st) order terms˙δx(t) ≈ f (¯x, ¯u) +∂f∂x¯¯¯¯¯x=¯xu=¯uδx(t) +∂f∂u¯¯¯¯¯x=¯xu=¯uδu(t)But f(¯x, ¯u) = 0, leaving˙δx(t) ≈∂f∂x¯¯¯¯¯x=¯xu=¯uδx(t) +∂f∂u¯¯¯¯¯x=¯xu=¯uδu(t)This differential equation approximately governs (we are neglecting 2nd order andhigher terms) the deviation variables δx(t) and δu(t), as long as they remainsmall. It is a linear, time-invariant, differential equation, since the derivatives ofδxare linear combinations of the δxvariables and the deviation inputs, δu. ThematricesA :=∂f∂x¯¯¯¯¯x=¯xu=¯u∈ Rn×n, B :=∂f∂u¯¯¯¯¯x=¯xu=¯u∈ Rn×m(73)are constant matrices. With the matrices A and B as defined in (73), the linearsystem˙δx(t) = Aδx(t) + Bδu(t)is called the Jacobian Linearization of the original nonlinear system (72), aboutthe equilibrium point (¯x, ¯u). For “small” values of δxand δu, the linear equationapproximately governs the exact relationship between the deviation variables δuand δx.For “small” δu(ie., while u(t) remains close to ¯u), and while δxremains “small”(ie., while x(t) remains close to ¯x), the variables δxand δuare related by thedifferential equation˙δx(t) = Aδx(t) + Bδu(t)In some of the rigid body problems we considered earlier, we treated problemsby making a small-angle approximation, taking θ and its derivatives˙θ and¨θ verysmall, so that certain terms were ignored (˙θ2,¨θ sin θ) and other terms simplified(sin θ ≈ θ, cos θ ≈ 1). In the context of this discussion, the linear models weobtained were, in fact, the Jacobian linearizations around the equilibrium pointθ = 0,˙θ = 0.If we design a controller that effectively controls the deviations δx, then we havedesigned a controller that works well when the system is operating near the equi-librium point (¯x, ¯u). We will cover this idea in greater detail later. This is acommon, and somewhat effective way to deal with nonlinear systems in a linearmanner.17019.3 Tank ExampleConsider a mixing tank, with constant supply temperatures TCand TH. Let theinputs be the two flow rates qC(t) and qH(t). The equations for the tank are˙h(t) =1AT³qC(t) + qH(t) − cDAoq2gh(t)´˙TT(t) =1h(t)AT(qC(t) [TC− TT(t)] + qH(t) [TH− TT(t)])Let the state vector x and input vector u be defined asx(t) :="h(t)TT(t)#, u(t) :="qC(t)qH(t)#f1(x, u) =1AT³u1+ u2− cDAo√2gx1´f2(x, u) =1x1AT(u1[TC− x2] + u2[TH− x2])Intuitively, any height¯h > 0 and any tank temperature¯TTsatisfyingTC≤¯TT≤ THshould be a possible equilibrium point (after specifying the correct values of theequilibrium inputs). In fact, with¯h and¯TTchosen, the equation f(¯x, ¯u) = 0 canbe written as"1 1TC− ¯x2TH− ¯x2#"¯u1¯u2#="cDAo√2g¯x10#The 2 × 2 matrix is invertible if and only if TC6= TH. Hence, as long as TC6= TH,there is a unique equilibrium input for any choice of ¯x. It is given by"¯u1¯u2#=1TH− TC"TH− ¯x2−1¯x2− TC1#"cDAo√2g¯x10#This is simply¯u1=cDAo√2g¯x1(TH− ¯x2)TH− TC, ¯u2=cDAo√2g¯x1(¯x2− TC)TH− TCSince the uirepresent flow rates into the tank, physical considerations restrictthem to be nonegative real numbers. This implies that ¯x1≥ 0 and TC≤¯TT≤ TH.Looking at the differential equation for TT, we see that its rate of change is inverselyrelated to h. Hence, the differential equation model is valid while h(t) > 0, sowe further restrict ¯x1> 0. Under those restrictions, the state ¯x is indeed anequilibrium point, and there is a unique equilibrium input given by the equationsabove.Next we compute the necessary partial derivatives."∂f1∂x1∂f1∂x2∂f2∂x1∂f2∂x2#=−gcDAoAT√2gx10−u1[TC−x2]+u2[TH−x2]x21AT−(u1+u2)x1AT171"∂f1∂u1∂f1∂u2∂f2∂u1∂f2∂u2#="1AT1ATTC−x2x1ATTH−x2x1AT#The linearization requires that the matrices of partial derivatives be evaluated atthe equilibrium points. Let’s pick some realistic
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