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 Form24 Linearized analysis of open-loop and clsoed-loop broom balancing25 Stabilization by State-Feedback5 Transfer functionsAssociated with the linear system (input u, output y) governed by the ODEy[n](t) + a1y[n−1](t) + ··· + an−1y[1](t) + any(t)= b0u[n](t) + b1u[n−1](t) + ··· + bn−1u[1](t) + bnu(t)(35)we write “in transfer function form”Y =b0sn+ b1sn−1+ ··· + bn−1s + bnsn+ a1sn−1+ ··· + an−1s + anU (36)The expression in (36) is interpreted to be equivalent to the ODE in (35), justa different way of writing the coefficients. The notation in (36) is suggestive ofmultiplication, and we will see that such an interpretation is indeed useful. ThefunctionG(s) :=b0sn+ b1sn−1+ ··· + bn−1s + bnsn+ a1sn−1+ ··· + an−1s + anis called the transfer function from u to y, and is sometimes denoted Gu→y(s) toindicate this. At this point, the expression in equation (36),Y = Gu→y(s)Uis nothing more than a new notation for the differential equation in (35). Thedifferential equation has a well-defined meaning, and we understand what eachterm represents, and the meaning of the equality sign, =. In the transfer functionexpression, (36), there is no specific meaning to the individual terms, or the equalitysymbol. The expression, as a whole, simply means the differential equation it isassociated with.In this section, we will see that, in fact, we can assign proper equality, and makealgebraic substitutions and manipulations of transfer function expressions, whichwill aid our manipulation of linear differential equations. But all of that requiresproof, and that is the purpose of this section.5.1 Linear Differential Operators (LDOs)Note that in the expression (36), the symbol s plays the role ofddt, and higherpowers of s mean higher order derivatives, ie., skmeansdkdtk. If z is a function oftime, let the notation"b0dndtn+ b1dn−1dtn−1+ ··· + bn−1ddt+ bn#(z) := b0dnzdtn+ b1dn−1zdtn−1+ ···+ bn−1dzdt+ bnzWe will call this type of operation a linear differential operation, or LDO. For thepurposes of this section, we will denote these by capital letters, sayL :=hdndtn+ a1dn−1dtn−1+ ··· + an−1ddt+ aniR :=hb0dndtn+ b1dn−1dtn−1+ ··· + bn−1ddt+ bni56From _Dynamic Systems and Feedback_by Packard, Poola, Horowitz, 2002Copyright 2002, Andrew PackardAll rights reserved.Do not duplicate or redistribute.Using this shorthand notation, we can write the original ODE in (39) asL(y) = R(u)With each LDO, we naturally associate a polynomial. Specifically, ifL :="dndtn+ a1dn−1dtn−1+ ··· + an−1ddt+ an#then pL(s) is defined aspL(s) := sn+ a1sn−1+ ··· + an−1s + anSimilarly, with each polynomial, we associate an LDO – ifq(s) := sm+ b1sm−1+ ··· + bm−1s + bmthen Lqis defined asLq:="dmdtm+ b1dm−1dtm−1+ ··· + bm−1ddt+ bm#Therefore, if a linear system is governed by an ODE of the form L(y) = R(u), thenthe transfer function description is simplyY =pR(s)pL(s)USimilarly, if the transfer function description of a system isV =n(s)d(s)Wthen the ODE description is Ld(v) = Ln(w).5.2 Algebra of Linear differential operationsNote that two successive linear differential operations can be done in either order.For example letL1:="d2dt2+ 5ddt+ 6#andL2:="d3dt3− 2d2dt2+ 3ddt− 4#57Then, on a differentiable signal z, simple calculations givesL1(L2(z)) =hd2dt2+ 5ddt+ 6i³hd3dt3− 2d2dt2+ 3ddt− 4i(z)i=hd2dt2+ 5ddt+ 6i³z[3]− 2¨z + 3 ˙z − 4z´= z[5]− 2z[4]+ 3z[3]− 4z[2]5z[4]− 10z[3]+ 15z[2]− 20z[1]6z[3]− 12z[2]+ 18z[1]− 24z= z[5]+ 3z[4]− z[3]− z[2]− 2z[1]− 24zwhich is the same asL2(L1(z)) =hd3dt3− 2d2dt2+ 3ddt− 4i³hd2dt2+ 5ddt+ 6i(z)i=hd3dt3− 2d2dt2+ 3ddt− 4i³z[2]+ 5 ˙z + 6z´= z[5]+ 5z[4]+ 6z[3]−2z[4]− 10z[3]− 12z[2]z[3]+ 15z[2]+ 18z[1]−4z[2]− 20z[1]− 24z= z[5]+ 3z[4]− z[3]− z[2]− 2z[1]− 24zThis equality is easily associated with the fact that multiplication of polynomialsis a commutative operation, specifically(s2+ 5s + 6) (s3− 2s2+ 3s − 4) = (s3− 2s2+ 3s − 4) (s2+ 5s + 6)= s5+ 3s4− s3− s2− 2s + 24Since composition of two linear differential operations behaves like polynomialmultiplication, we sometimes notate the “product” L1L2to be composition, inother words, given LDOs L1and L2, the LDO L1L2is defined by composition, andon a differentiable signal z, it is[L1L2] (z) := L1(L2(z))We will often use the notation [L1◦ L2] to also denote this composition of LDOs.Similarly, if L1and L2are LDOs, then the sum L1+ L2is an LDO defined by itsoperation on a signal z as [L1+ L2] (z) := L1(z) + L2(z).It is clear that the following manipulations are always true for every differentiablesignal z,L1(L2(z)) + L3(L4(z)) = (L1L2+ L3L4) (z)andL (z1+ z2) = L (z1) + L (z2)and[L1◦ L2] (z) = [L2◦ L1] (z)In terms of LDOs and their associated polynomials, we have the relationshipsp[L1+L2](s) = pL1(s) + pL2(s)p[L1◦L2](s) = pL1(s)pL2(s)In the next several subsections, we derive the LDO representation of an intercon-nection from the LDO representation of the subsystems.585.3 Feedback ConnectionThe most important interconnection we know of is the basic feedback loop. It is alsothe easiest interconnection for which we derive the differential equation governingthe interconnection from the differential equation governing the components.Consider the simple unity-feedback system shown belowS- d - -6ruy−+Assume that system S is described by the LDO L(y) = D(u). The feedbackinterconnection yields u(t) = r(t) − y(t). Eliminate u by substitution, yielding anLDO relationship between r and yL(y) = D(r − y) = D(r) − D(y)This is rearranged to the closed-loop LDO(L + D)(y) = D(r).That’s a pretty simple derivation. Based on the ODE description of the closed-loop, we can immediately write the closed-loop transfer function,Y =pD(s)p[L+D](s)R=pD(s)pL(s) + pD(s)R.Additional manipulation leads to further interpretation. Let G(s) denote the trans-fer function of S, so G =pD(s)pL(s). ThenY =pD(s)pL(s) + pD(s)R=pD(s)pL(s)1 +pD(s)pL(s)R=G(s)1 + G(s)RThis can be
View Full Document
Unlocking...