M427K Handout: Applications of the Laplace Transform: ImpulseFunctions and the Convolution IntegralSalman ButtMarch 28, 2006Impulse FunctionsAn impulse function is a function that applies a very large amount of force for a short amountof time. The classic example is the Dirac delta function, which is constructed as follows. Definethe function δτ(t) as:The impulse of δτis merely the integralNow observe the following:The Dirac delta function can be thought of as the “idealized” version of this function. That is, theDirac delta funtion δ is defined to be a function that satisfies the propertiesIt is also referred to as the unit impulse function. Note that this is not an elementary function tobe found in a calculus textbook. It does not satisfy the theorems we have on the existence of theLaplace transform of a function, but we can in fact formally define its Laplace transform:Using elementary calculus, we find in fact that for t0> 0And for t0= 0, we take the limit of the ab ove equation as t0→ 0 to findWe can also compute the integral of the delta function with any continuous function f over theentire real line:1Z∞∞δ(t − t0)f(t)dt = limτ →0Z∞∞dτ(t − t0)f(t)dtLet’s see an example of a differential equation with the Dirac delta function as a forcing function:2y00+ y0+ 2y = δ(t − 5), y(0) = 0, y0(0) = 0The Convolution IntegralRecall that I never stated the Laplace transform and inverse Laplace transform are multiplica-tive. That is, never did I say thatL {fg} (s) = L { f } (s)L {g} (s) L−1{F G} (t) = L−1{F } (t)L−1{G} (t)The above statement is NOT true. What is true is that the inverse Laplace transform of aproduct of two functions is the convolution of the inverse Laplace transforms of the two functions.That isL−1{F G} (t) = (f ∗ g)(t) (1)where f = L−1{F } (t), g = L−1{G} (t) and f ∗ g is the convolution of f and g, defined to bethe integral2Here are some important facts about the convolution of functions:1.2.3.4.Note, though, that some properties of standard multiplication do not hold true for convolution. Inparticular f ∗ 1 6= f nor is f ∗ f nonnegative [see text]. The proof of Equation (1) can be found onpp. 347-8 of your text (as the proof to the text’s Theorem 6.6.1). I will not go over that, but willinstead discuss examples of how to use this fact to find the inverse Laplace transforms of thornyfunction.Let’s look at an example:y00+ 4y = g(t), y(0) = 3, y0(0) = −1This example should help you see how the convolution integral can be very helpful in solvingdifferential equations in general. To wit, suppose we have the initial value problemay00+ by0+ cy = g(t), y(0) = y0, y0(0) = y00where a, b, c, y0, y00are real constants. Applying the Laplace transform we have the algebraic equa-tionSolving for Y (s) we find3If we were to apply the Laplace transform to the left- and right-hand sides, we would findwhere φ(t) = L−1{Φ} (t), ψ(t) = L−1{Ψ} (t). So we only need to find φ, ψ. But φ is easy tocompute: it is merely the solution to the homogeneous differential equationay00+ by0+ cy = 0, y(0) = y0, y0(0) = y00We know how to solve this (using the table), so we are set here. Now obse rve that ψ(t) is thesolution to the initial value problemay00+ by0+ cy = g(t), y(0) = 0, y0(0) = 0To find ψ, we write Ψ = G(s)H(s) where H(s) = (as2+ bs + c)−1. This function H is called thetransfer function. Now to find ψ, we use what we know about convolution integrals to getψ(t) = L−1{G(s)H(s)} (t) = (g ∗ h)(t) = (h ∗ g)(t)where g = L−1{G} (t), h = L−1{H} (t) are elementary functions to compute. Note that thefunction h(t) is called the impulse response. Adding φ to ψ, we get our solution y. Observe thatthis is precisely what we did in the above