45 10. Operators and the exponential response formula 10.1. Operators. Operators are to functions as functions are to num-bers. An operator takes a function, does something to it, and returns this modified function. There are lots of examples of operators around: —The shift-by-a operator (where a is a number) takes as input a func-tion f(t) and gives as output the function f(t−a). This operator shifts graphs to the right by a units. —The multiply-by-h(t) operator (where h(t) is a function) multiplies by h(t): it takes as input the function f(t) and gives as output the function h(t)f(t). You can go on to invent many other operators. In this course the most important operator is: —The differentiation operator, which carries a function f(t) to its de-rivative f →(t). The differentiation operator is usually denoted by the letter D; so Df(t) is the function f →(t). D carries f to f →; for example, Dt3 = 3t2 . Warning: you can’t take this equation and substitute t = 2 to get D8 = 12. The only way to interpret “8” in “D8” is as a constant function, which of course has derivative zero: D8 = 0. The point is that in order to know the function Df(t) at a particular value of t, say t = a, you need to know more than just f(a); you need to know how f(t) is changing near a as well. This is characteristic of operators; in general you have xto expect to need to know the whole function f(t) in order to evaluate an operator on it. The identity operator takes an input function f(t) and returns the same function, f(t); it does nothing, but it still gets a symbol, I. Operators can be added and multiplied by numbers or more generally by functions. Thus tD+4I is the operator sending f(t) to tf →(t)+4f(t). The single most important thing associated with the concept of op-erators is that they can be composed with each other. I can hand a function off from one operator to another, each taking the output from the previous and modifying it further. For example, D2 differentiates twice: it is the second-derivative operator, sending f(t) to f →→(t). We have been studying ODEs of the form mx¨ + bx˙ + kx = q(t). The left hand side is the effect of an operator on the function x(t), namely, the operator mD2 + bD + kI. This operator describes the system (composed for example of a mass, dashpot, and spring).46 We’ll often denote an operator by a single capital letter, such as L. If L = mD2 + bD + kI, for example, then our favorite ODE, mx¨ + bx˙ + kx = q can be written simply as Lx = q. At this point m, b, and k could be functions of t. Note well: the operator does NOT take the signal as input and return the system response, but rather the reverse: Lx = q, the operator takes the response and returns the signal. In a sense the system is better modeled by the “inverse” of the operator L. In rough terms, solving the ODE Lx = q amounts to inverting the operator L. Here are some definitions. A differential operator is one which is algebraically composed of D’s and multiplication by functions. The order of a differential operator is the highest derivative appearing in it. mD2 +bD +kI is an example of a second order differential operator. This example has another important feature: it is linear. An opera-tor L is linear if L(cf) = cLf and L(f + g) = Lf + Lg. 10.2. LTI operators and exponential signals. We will study al-most exclusively linear differential operators. They are the operators of the form L = an(t)Dn + an−1(t)Dn−1 + + a0(t)I. ··· The functions a0, . . . , an are the coefficients of L. In this course we focus on the case in which the coefficients are constant; each ak is thus a number, and we can form the characteristic polynomial of the operator, p(s) = ans n + an−1s n−1 + + a0.··· The operator is Linear and Time Invariant: an LTI operator. The original operator is obtained from its characteristic polynomial by for-mally replacing the indeterminate s here with the differentiation oper-ator D, so we may write L = p(D). The characteristic polynomial completely determines the operator, and many properties of the operator are conveniently described in terms of its characteristic polynomial.47 Here is a first example of the power of the operator notation. Let r be any constant. (You might as well get used to thinking of it as a possibly complex constant.) Then Dert = re rt . (A fancy expression for this is to say that r is an eigenvalue of the operator D, with corresponding eigenfunction ert.) Iterating this we find that Dk e rt = r k e rt . We can put these equations together, for varying k, and evaluate a general LTI operator p(D) = anDn + an−1Dn−1 + + a0I··· on ert . The operator Dk pulls rk out as a factor, and when you add them all up you get the value of the polynomial p(s) at s = r: (1) p(D)e rt = p(r)e rt . It is crucial here that the operator be time invariant: If the coefficients ak are not constant, then they don’t just pull outside the differentiation; you need to use the product rule instead, and the formulas become more complicated—see Section 12. Multiplying (1) by a/p(r) we find the important Exponential Response Formula: A solution to (2) p(D)x = ae rt is given by the rt e(3) xp = a p(r) provided only that p(r) = 0. ∈The Exponential Response Formula ties together many different parts of this course. Since the most important signals are exponential, and the most important differential operators are LTI operators, this single formula solves most of the ODEs you are likely to face in your future. The function xp given by (3) is the only solution to (2) which is a multiple of an exponential function. If r has the misfortune to be a root of p(s), so that p(r) = 0, then the formula (3) would give a zero in the denominator. The conclusion is that there are no solutions which are multiples of exponential functions. This is a “resonance” situation. In this case we can still find an explicit solution; see Section 12 for this.48 Example 10.2.1. Let’s solve (4) 2¨x + ˙x + x = 1 + 2e t . This is an inhomogeneous linear equation, so the general solution as the form xp + xh, where xp is any particular solution and xh is the general homogeneous solution. The characteristic polynomial is p(s) = 2s2 + s + 1, with
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