55 12. Resonance and the exponential shift law 12.1. Exponential shift. The calculation (10.1) (1) p(D)e rt = p(r)e rt extends to a formula for the effect of the operator p(D) on a product of the form ertu, where u is a general function. This is useful in solving p(D)x = f(t) when the input signal is of the form f(t) = ertq(t). The formula arises from the product rule for differentiation, which can be written in terms of operators as D(vu) = v Du + (Dv)u. If we take v = ert this becomes D(e rt u) = e rtDu + re rt u = e rt(Du + ru) . Using the notation I for the identity operator, we can write this as (2) D(e rt u) = e rt(D + rI)u. If we apply D to this equation again, D2(e rt u) = D(e rt(D + rI)u) = e rt(D + rI)2 u , where in the second step we have applied (2) with u replaced by (D + rI)u. This generalizes to Dk(e rt u) = e rt(D + rI)k u. The final step is to take a linear combination of Dk’s, to form a general LTI operator p(D). The result is the Exponential Shift Law: (3) p(D)(ertu) = ertp(D + rI)u The effect is that we have pulled the exponential outside the differential operator, at the expense of changing the operator in a specified way. 12.2. Product signals. We can exploit this effect to solve equations of the form p(D)x = e rt q(t) , by a version of the method of variation of parameter: write x = ertu, apply p(D), use (3) to pull the exponential out to the left of the op-erator, and then cancel the exponential from both sides. The result is p(D + rI)u = q(t) ,56 a new LTI ODE for the function u, one from which the exponential factor has been eliminated. Example 12.2.1. Find a particular solution to x¨ + ˙x + x = t2e3t . With p(s) = s2 + s + 1 and x = e3tu, we have x¨ + ˙x + x = p(D)x = p(D)(e 3t u) = e 3t p(D + 3I)u . Set this equal to t2e3t and cancel the exponential, to find p(D + 3I)u = t2 or u˙ + 3u = t2 . This is a good target for the method of undetermined coefficients (Section 11). The first step is to compute p(s + 3) = (s + 3)2 + (s + 3) + 1 = s 2 + 7s + 13 , so we have u¨ + 7 ˙u + 13u = t2 . There is a solution of the form up = at2 + bt + c, and we find it is up = (1/13)t2 − (14/132)t + (85/133) . Thus a particular solution for the original problem is xp = e 3t((1/13)t2 − (14/132)t + (85/133)) . Example 12.2.2. Find a particular solution to ˙x + x = te−t sin t. The signal is the imaginary part of te(−1+i)t, so, following the method of Section 10, we consider the ODE z˙ + z = te(−1+i)t . If we can find a solution zp for this, then xp = Im zp will be a solution to the original problem. We will look for z of the form e(−1+i)tu. The Exponential Shift Law (3) with p(s) = s + 1 gives z˙ + z = (D + I)(e(−1+i)t u) = e(−1+i)t((D + (−1 + i)I) + I)u = e(−1+i)t(D + iI)u. When we set this equal to the right hand side we can cancel the expo-nential: (D + iI)u = t or u˙ + iu = t. While this is now an ODE with complex coefficients, it’s easy to solve by the method of undetermined coefficients: there is a solution of the form up = at + b. Computing the coefficients, up = −it + 1; so zp = e(−1+i)t(−it + 1). Finally, extract the imaginary part to obtain xp: zp = e −t(cos t + i sin t)(−it + 1)57 has imaginary part xp = e −t(−t cos t + sin t). 12.3. Resonance. We have noted that the Exponential Response For-mula for a solution to p(D)x = ert fails when p(r) = 0. For example, For example, suppose we have x˙ + x = e−t . The Exponential Response Formula proposes a solution xp = e−t/p(−1), but p(−1) = 0 so this fails. There is no solution of the form cert . This situation is called resonance, because the signal is tuned to a natural mode of the system. Here is a way to solve p(D)x = ert when this happens. The ERF came from the calculation p(D)e rt = p(r)e rt , which is valid whether or not p(r) = 0. We will take this expression and differentiate it with respect to r, keeping t constant. The result, using the product rule and the fact that partial derivatives commute, is p(D)tert = p →(r)e rt + p(r)tert If p(r) = 0 this simplifies to (4) p(D)tert = p →(r)e rt . Now if p→(r) = 0 we can divide through by it and see: ∈The Resonant Exponential Response Formula: If p(r) = 0 then a solution to p(D)x = aert is given by tert (5) xp = a p→(r) provided that p→(r) = 0. ∈In our example above, p(s) = s + 1 and r = 1, so p→(r) = 1 and xp = te−t is a solution. This example exhibits a characteristic feature of resonance: the solu-tions grow faster than you might expect. The characteristic polynomial leads you to expect a solution of the order of e−t . In fact the solution is t times this. It still decays to zero as t grows, but not as fast as e−t does. Example 12.3.1. Suppose we have a harmonic oscillator represented by x¨ + �n2x, or by the operator D2 + �n2I = p(D), and drive it by the58 signal a cos(�t). This ODE is the real part of z¨ + �2 z = ae i�t ,nso the Exponential Response Formula gives us the periodic solution i�ntezp = a . p(i�) This is fine unless � = �n, in which case p(i�n) = (i�n)2 + �n 2 = 0; so the amplitude of the proposed sinusoidal response should be infinite. The fact is that there is no periodic system response; the system is in resonance with the signal. To circumvent this problem, let’s apply the Resonance Exponential Response Formula: since p(s) = s2 + �n2 , p→(s) = 2s and p→(i�n) = 2i�0, so tei�nt zp = a . 2i�n The real part is a xp = t sin(�nt) . 2�n The general solution is thus a x = t sin(�nt) + b cos(�nt − π) . 2�n In words, all solutions oscillate with pseudoperiod 2ν/�n, and grow in amplitude like at/(2�n). When �n is large—high frequency—this rate of growth is small. 12.4. Higher order resonance. It may happen that both p(r) = 0 and p→(r) = 0. The general picture is this: Suppose that k is such that p(j)(r) = 0 for j < k and …
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