98 19. Convolution 19.1. Superposition of infinitesimals: the convolution integral. The system response of an LTI system to a general signal can be re-constructed explicitly from the unit impulse response. To see how this works, start with an LTI system represented by a linear differential operator L with constant coefficients. The system response to a signal f(t) is the solution to Lx = f(t), subject to some specified initial conditions. To make things uniform it is common to specify “rest” initial conditions: x(t) = 0 for t < 0. We will approach this general problem by decomposing the signal into small packets. This means we partition time into intervals of length say �t: t0 = 0, t1 = �t, t2 = 2�t, and generally tk = k�t. The kth signal packet is the null signal (i.e. has value zero) except between t = tk and t = tk+1, where it coincides with f(t). Write fk(t) for the kth packet. Then f (t) is the sum of the fk(t)’s. Now by superposition the system response (with rest initial condi-tions) to f(t) is the sum of the system responses to the fk(t)’s sepa-rately. The next step is to estimate the system response to a single packet, say fk(t). Since fk(t) is concentrated entirely in a small neighborhood of tk, it is well approximated as a rate by a multiple of the delta function concentrated at tk, β(t − tk). The multiple should be chosen so that the cumulative totals match up; that is, it should be the integral under the graph of fk(t), which is itself well approximated by f(tk)�t. Thus we replace fk(t) by f(tk) �t β(t − tk).· · The system response to this signal, a multiple of a shift of the unit impulse, is the same multiple of the same shift of the weight function (= unit impulse response): f(tk) �t w(t − tk).· · By superposition, adding up these packet responses over the packets which occur before the given time t gives the system response to the signal f(t) at time t. As �t � 0 this sum approximates an integral taken over time between time zero and time t. Since the symbol t is already in use, we need to use a different symbol for the variable in the integral; let’s use the Greek equivalent of t, δ (“tau”). The tk’s get99 replaced by δ in the integral, and �t by dδ: � t (1) x(t) = f(δ)w(t − δ) dδ 0 This is a really wonderful formula. Edwards and Penney call it “Duhamel’s principle,” but they seem somewhat isolated in this. Per-haps a better name would be the “superposition integral,” since it is no more and no less than an integral expression of the principle of superposition. It is commonly called the convolution integral. It describes the solution to a general LTI equation Lx = f (t) subject to rest initial conditions, in terms of the unit impulse response w(t). Note that in evaluating this integral δ is always less than t, so we never encounter the part of w(t) where it is zero. 19.2. Example: the build up of a pollutant in a lake. Every good formula deserves a particularly illuminating example, and perhaps the following will serve for the convolution integral. It is illustrated by the Mathlet Convolution: Accumulation. We have a lake, and a pollutant is being dumped into it, at a certain variable rate f (t). This pollutant degrades over time, exponentially. If the lake begins at time zero with no pollutant, how much is in the lake at time t > 0? The exponential decay is described as follows. If a quantity p of pollutant is dropped into the lake at time δ, then at a later time t it will have been reduced in amount to pe−a(t−π ). The number a is the decay constant, and t − δ is the time elapsed. We apply this formula to the small drip of pollutant added between time δ and time δ + �δ. The quantity is p = f(δ)�δ (remember, f(t) is a rate; to get a quantity you must multiply by time), so at time t the this drip has been reduced to the quantity e −a(t−π )f(δ)�δ (assuming t > δ; if t < δ , this particular drip contributed zero). Now we add them up, starting at the initial time δ = 0, and get the convo-lution integral (1), which here is � t (2) x(t) = f(δ)e −a(t−π ) dδ. 0 We found our way straight to the convolution integral, without ever mentioning differential equations. But we can also solve this problem by setting up a differential equation for x(t). The amount of this chem-ical in the lake at time t+�t is the amount at time t, minus the fraction100 that decayed, plus the amount newly added: x(t + �t) = x(t) − ax(t)�t + f(t)�t Forming the limit as �t � 0, we obtain (3) x˙ + ax = f(t), x(0) = 0. We conclude that (2) gives us the solution with rest initial conditions. An interesting case occurs if a = 0. Then the pollutant doesn’t decay at all, and so it just builds up in the lake. At time t the total amount in the lake is just the total amount dumped in up to that time, namely � t f(δ) dδ , 0 which is consistent with (2). 19.3. Convolution as a “product”. The integral (1) is called the convolution of w(t) and f(t), and written using an asterisk: � t (4) w(t) f(t) = w(t − δ)f(δ) dδ, t > 0.↑ 0 We have now fulfilled the promise we made at the beginning of Sec-tion 18: we can explicitly describe the system response, with rest initial conditions, to any input signal, if we know the system response to just one input signal, the unit impulse: Theorem. The solution to an LTI equation Lx = f (t), of any order, with rest initial conditions, is given by x(t) = w(t) f(t),↑ where w(t) is the unit impulse response. If L is an LTI differential operator, we should thus be able to recon-struct its characteristic polynomial p(s) (so that L = p(D)) from its unit impulse response. This is one of the things the Laplace transform does for us; in fact, the Laplace transform of w(t) is the reciprocal of p(s): see Section 21. The expression (4) can be interpreted formally by a process known as “flip and drag.” It is illustrated in the Mathlet Convolution: Flip and Drag.MIT OpenCourseWare http://ocw.mit.edu 18.03 Differential Equations���� Spring 2010 For information about citing these materials or our Terms of Use, visit:
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