Laplace Transforms and Convolusions: Practice Problems and Solutions, Exams of Mathematics

Download Laplace Transforms and Convolusions: Practice Problems and Solutions and more Exams Mathematics in PDF only on Docsity! EXAM Practice Questions for Exam #3 Math 3350, Spring 2004 April 16, 2004 ANSWERS i Problem 2. Find the Laplace Transform of the following function: f(t) =  0, 0 < t < 1 t2 + 2t, 1 < t < 2 1, 2 < t <∞. Answer : Using the indicator functions of the intervals, we can write f(t) as f(t) = 0I(0,1)(t) + (t2 + 2t)I(1,2)(t) + 1I(2,∞)(t) = (t2 + 2t)I(1,2)(t) + I(2,∞)(t). From our table of indicator functions, we have I(1,2)(t) = u(t− 1)− u(t− 2), I(2,∞)(t) = u(t− 2). Plugging this in, we have f(t) = (t2 + 2t)[u(t− 1)− u(t− 2)] + u(t− 2). Collecting coefficients of the u’s, this is (2.1) f(t) = u(t− 1)[t2 + 2t] + u(t− 2)[1− 2t− t2]. To find the transform of this, we want to use the shifting rule (2.2) L [u(t− a)g(t− a)] = e−asG(s), G(s) = L [g(t)]. Consider the first term in (2.1), u(t − 1)[t2 + 2t]. To get this to match the left-hand side of (2.2), we must have u(t − a) = u(t − 1), so a = 1, and g(t− 1) = t2 + 2t. From this, we have to figure out what g(t) is. To do this, we substitute t + 1 for t, and get g(t) = g(t + 1− 1) = (t + 1)2 + 2(t + 1) = t2 + 2t + 1 + 2t + 2 = t2 + 4t + 3. Then we have G(s) = L [t2 + 4t + 3] = 2 s3 + 4 s2 + 3 s . Applying (2.2) we then get L [u(t− 1)[t2 + 2t]] = e−sG(s) = e−s [ 2 s3 + 4 s2 + 3 s ] . 3 Similarly, consider the second term in (2.1), u(t−2)[1−2t− t2]. Comparing this with (2.2) we have a = 2 and g(t− 2) = 1− 2t− t2. Thus, g(t) = g(t + 2− 2) = 1− 2(t + 2)− (t + 2)2 = 1− 2t− 4− t2 − 4t− 4 = −7− 8t− t2. Then G(s) = L [−7− 8t− t2] = −7 s − 8 s2 − 2 s3 . and using (2.2) we get L [u(t− 2)[1− 2t− t2]] = −e−2s [ 7 s + 8 s2 + 2 s3 ] . Combining these results, we get L [f(t)] = e−s [ 2 s3 + 4 s2 + 3 s ] − e−2s [ 7 s + 8 s2 + 2 s3 ] Problem 3. Find the Inverse Laplace Transform of the following function: F (s) = 1 s + 1 + e−s 1 (s + 2)2 + e−3s 1 s2 + 4 Answer : We know L −1 [ 1 s + 1 ] = e−t.(3.1) L −1 [ 1 (s + 2)2 ] = te−2t.(3.2) L −1 [ 1 s2 + 4 ] = 1 2 L −1 [ 2 s2 + 4 ] = 1 2 sin(2t).(3.3) We use the shifting rule: L −1[e−asF (s)] = u(t− a)f(t− a), f(t) = L −1[F (s)]. Applying this rule, and (3.2) we get L −1 [ e−s 1 (s + 2)2 ] = u(t− 1)(t− 1)e−2(t−1) 4 and from (3.3) we get L −1 [ e−3s 1 s2 + 4 ] = 1 2 u(t− 3) sin(2(t− 3)). Thus, we get L −1[F (s)] = e−t + u(t− 1)(t− 1)e−2(t−1) + 1 2 u(t− 3) sin(2(t− 3)). Problem 4. Solve the following initial value problem, using Laplace Trans- forms: (4.1) y′′ − 2y′ + y = t2u(t− 1) + δ(t− 2), y(0) = 0, y′(0) = 1. Answer : Lets start by finding the Laplace transform of the right-hand side. We know that (4.2) L [δ(t− 2)] = e−2s. To find the transform of the other term, recall the shifting rule (4.3) L [u(t− a)f(t− a)] = e−asF (s), F (s) = L [f(t)]. To match the left-hand side of this with u(t − 1)t2 we must have a = 1 and f(t−1) = t2. To find f(t), substitute t+1 for t in the equation f(t−1) = t2. This gives f(t) = (t+1)2 = t2+2t+1. Thus, F (s) = L [t2+2t+1] = 2/s3+2/s2+1/s. Hence, by the shifting rule, we have (4.4) L [u(t− 1)t2] = e−s [ 2 s3 + 2 s2 + 1 s ] = e−s s2 + 2s + 1 s3 . Now transform both sides of (4.1). The result is s2Y (s)− sy(0)− y′(0)− 2[sY (s)− y(0)] + Y (s) = e−s s 2 + 2s + 1 s3 + e−2s, by (4.2) and (4.4). Putting in the initial conditions and collecting terms gives us (s2 − 2s + 1)Y (s)− 1 = e−s s 2 + 2s + 1 s3 + e−2s. Solving this for Y (s), we get Y (s) = 1 s2 − 2s + 1 + e−s s2 + 2s + 1 s3(s2 − 2s + 1) + e−2s 1 s2 − 2s + 1 = 1 (s− 1)2 + e−s s2 + 2s + 1 s3(s− 1)2 + e−2s 1 (s− 1)2 . 5 so we have (5.3) F (s) = G′(s) = 2s s2 + 1 − 2s s2 + 4 . (this equation defines F (s)). From this we have∫ ∞ s F (σ) dσ = ∫ ∞ s G′(σ) dσ = G(∞)−G(s) = −G(s), (loosely speaking) since G(σ)→ 0 as σ →∞. Thus, we have (5.4) G(s) = − ∫ ∞ s F (σ)dσ. From the inverse version of (5.2), we then have g(t) = L −1[G(s)] = −L −1 [∫ ∞ s F (σ)dσ ] = −f(t) t , where f(t) = L −1[F (s)]. From the table and (5.3), f(t) = 2 sin(t)− 2 sin(2t). Thus, finally, the answer is g(t) = 2 sin(t)− sin(2t) t . You can verify that limt→0+ g(t) exists by L’Hôpital’s Rule. Problem 6. Find the convolution t2 ∗ t directly from the definition. Answer : The definition of convolution is (f ∗ g)(t) = ∫ t 0 f(τ)g(t− τ) dτ. Let f(t) = t2 and g(t) = t. Plugging into the formula above, we have (f ∗ g((t) = ∫ t 0 τ2(t− τ) dτ = ∫ t 0 (tτ2 − τ3) dτ = [ 1 3 tτ3 − 1 4 τ4 ]τ=t τ=0 = 1 3 t4 − 1 4 t4 − [0− 0] = 1 12 t4. 8 Problem 7. Find the convolution cos(2t)∗sin(t) using Laplace transforms. Answer : We want to use the formula (7.1) L [(f ∗ g)(t)] = F (s)G(s). We know, of course, that L [cos(2t)] = s s2 + 4 L [sin(t)] = 1 s2 + 1 . Thus, by (7.1), L [cos(2t) ∗ sin(t)] = s s2 + 4 1 s2 + 1 = s (s2 + 4)(s2 + 1) . All we have to do is take the inverse transform of the right-hand side. The partial fractions decomposition is (by machine) s (s2 + 4)(s2 + 1) = 1 3 s s2 + 1 − 1 3 s s2 + 4 . Thus, we have cos(2t) ∗ sin(t) = L −1 [ s (s2 + 4)(s2 + 1) ] = 1 3 L −1 [ s s2 + 1 ] − 1 3 L −1 [ s s2 + 4 ] = 1 3 cos(t)− 1 3 cos(2t). Problem 8. Express the solution of the following initial value problem using a convolution integral. y′′ − 2y′ + y = r(t), y(0) = 0, y′(0) = 1. Answer : Take the Laplace Transform of both sides of the equation. This gives s2Y (s)− sy(0)− y′(0)− 2[sY (s)− y(0)] + Y (s) = R(s), where R(s) = L [r(t)]. Plugging in the initial conditions and simplifying we get (s2 − 2s + 1)Y (s) = 1 + R(s). 9 Solving this equation for Y (s) gives us Y (s) = 1 s2 − 2s + 1 + R(s) s2 − 2s + 1 = 1 (s− 1)2 + R(s) (s− 1)2 . As we know, L −1 [ 1 (s− 1)2 ] = tet. To take the inverse transform of the other term, we use the formula L −1[F (s)G(s)] = (f ∗ g)(t). Thus, L −1 [ R(s) 1 (s− 1)2 ] = r(t) ∗ tet. From the definition of convolution, r(t) ∗ tet = ∫ t 0 r(τ)(t− τ)et−τ dτ. Putting all this together, the solution of our initial value problem is y(t) = ∫ t 0 (t− τ)et−τr(τ) dτ + tet. 10