Page 1Math 217Exam 3Name:ID:Section:This exam has 16 questions:• 15 multiple choice worth 5 points each.• 1 hand graded worth 25 points.Important:• No graphing calculators!• For the multiple choice questions, mark your answer on the answer card.• Show all your work for the written problems. You will be graded on the ease of reading yoursolution.• You are allowed a 3 × 5 note card for the exam.Function Transform Function Transformf0(t) sF (s) − f(0) 11sf00(t) s2F (s) − sf(0) − f0(0) t1s2Rt0f(τ) dτF (s)stnn!sn+1eatf(t) F (s − a) taΓ(a + 1)sa+1u(t − a)f(t − a) e−asF (s)1√πt1√s(f ∗ g)(t) =Zt0f(τ)g(t − τ ) dτ F (s)G(s) cos ktss2+ k2tf(t) −F0(s) sin ktks2+ k21tf(t)R∞sF (σ) dσ cosh ktss2− k2f(t), period p11 − e−psZp0e−stf(t) dt sinh ktks2− k2u(t − a)e−ass12k3(sin kt − kt cos k t)1(s2+ k2)2eat1s − at2ksin kts(s2+ k2)2tneatn!(s − a)n+112k(sin kt + kt cos kt)s2(s2+ k2)2sin(A + B) = sin A cos B + cos A sin Bcos(A + B) = cos A cos B − sin A sin B2 cos A cos B = cos(A − B) + cos(A + B)2 sin A sin B = cos(A − B) − cos(A + B)2 sin A cos B = sin(A − B) + sin(A + B)Page 2Math 217Exam 31. Use partial fractions to decompose the function3s2− 4s + 2(s − 1)(s2− 4s + 4)(a)−1s − 1+s + 2(s − 2)2(b)1s − 1+1s − 2+2s + 1(s − 2)2(c)1s − 1+s + 3s2− 4s + 4(d)1s − 1+2s − 2+6(s − 2)2−→ CORRECT(e)−1s − 1+6s − 2(f)1s − 1+1s2− 4s + 4(g)1s − 1+6(s − 2)2(h)1s − 1+1s − 2+1(s − 2)2(i)−1s − 1+2s − 2+2(s − 2)2(j) None of the aboveSolution: Solve for A, B, C:3s2− 4s + 2(s − 1)(s2− 4s + 4)=3s2− 4s + 2(s − 1)(s − 2)2=As − 1+Bs − 2+C(s − 2)2Page 3Math 217Exam 32. Let f(t) = 3 cos t. The graph of f(t) is the top graph and the function g(t) is on the bottom.-3-2-1 0 1 2 3 0 1 2 3 4 5 6 7 8 9f(t)=3cos(t)-3-2-1 0 1 2 3 0 1 2 3 4 5 6 7 8 9g(t)Which of the following functions is g(t)?(a) u(t)f(t)(b) u(t + 2)f(t + 2)(c) u(t − 2)f(t + 2)(d) u(t + 2)f(t − 2)(e) u(t − 2)f(t − 2) −→ CORRECT(f) u(t)f (3t)(g) u(t + 2)f(3t + 2)(h) u(t − 2)f(3t + 2)(i) u(t + 2)f(3t − 2)(j) u(t − 2)f(3t − 2)(k) None of the abovePage 4Math 217Exam 33. Let f(t) = sin 3t and g(t) = 1. Find the convolution product (f ∗ g)(t).(a) t(b) −t(c)13cos 3t(d) −13cos 3t(e)13(1 − cos 3t) −→ CORRECT(f) −13(1 − cos 3t)(g) cos 3t(h) −cos 3t(i) 1 − cos 3t(j) −1 + cos 3t(k) None of the aboveSolution:(f ∗ g)(t) =Zt0f(τ)g(t − τ ) dτ =Zt0sin 3τ · 1 dτ= −cos 3t3t0=13(1 − cos 3t)4. Letf(t) =Zt0g(t − τ )h(τ ) dτLet F (s) = L{f(t)}, G(s) = L{g(t)} and H(s) = L{h(t)}.Suppose you know G(2) = 8 and H(2) = 3.Find F (2), choose the closest answer.(a) 2(b) 4(c) 4(d) 6(e) 10(f) 16(g) 20(h) 24 −→ CORRECT(i) 40(j) 48(k) 96Solution: By properties of convolutions, we have F (s) = G(s) · H(s).Page 5Math 217Exam 35. Let f(t) = L−1{F (s)} whereF (s) =1s − 4+2(s − 1)2Find f(1), choose the closest answer.(a) 5(b) 57(c) 49(d) 55(e) 57(f) 60 −→ CORRECT f (t) = e4t+ 2tet(g) 64(h) 93(i) 101(j) 143(k) F (s) is not the Laplace transform of a function of exponential order6. Let f(t) = L−1{F (s)} whereF (s) =1s2− 2s + 5Find f(1), choose the closest answer.(a) 0.0(b) 0.2(c) 0.8(d) 1.2 −→ CORRECT f(t) =12etsin 2t(e) 1.9(f) 2.3(g) 2.5(h) 2.7(i) 3.2(j) 3.3(k) F (s) is not the Laplace transform of a function of exponential orderSolution: Complete the square: F (s) =1(s − 1)2+ 4Page 6Math 217Exam 37. Let f(t) = L−1{F (s)} wherelns − 1s + 3Find f(1), choose the closest answer.(a) −15.2(b) −7.6(c) −3.8(d) −2.7 −→ CORRECT(e) −1.9(f) 0.0(g) 1.9(h) 2.7(i) 3.8(j) 7.6(k) F (s) is not the Laplace transform of a function of exponential orderSolution:L−1{F (s)} = −1tL−1{F0(s)} = −1tL−14(s − 1)(s + 3)= −1tL−11s − 1−1s + 3=−et+ e−3tt= −2e−tsinh 2tt8. Let f(t) = L−1{F (s)} whereF (s) =s − 1s + 1Find f(1), choose the closest answer.(a) −20(b) −15(c) −10(d) −5(e) 0(f) 5(g) 10(h) 15(i) 20(j) 25(k) F (s) is not the Laplace transform of a function of exponential order −→ CORRECTSolution: F (s) is a Laplace transform of a function of exp onential order only if lims→∞F (s) = 0.But, lims→∞F (s) = 1.Page 7Math 217Exam 39. Let f(t) = L−1{F (s)} whereF (s) =s2+ 1s(s − 1)2Find f(2), choose the closest answer.(a) 0.0(b) 27.5(c) 28.0(d) 28.5(e) 29.0(f) 29.5(g) 30.0(h) 30.5 −→ CORRECT(i) 31.0(j) 31.5(k) F (s) is not the Laplace transform of a function of exponential orderSolution: There are several approach, partial fractions is perhaps the most straightforward.s2+ 1s(s − 1)2=1s+2(s − 1)2This gives f(t) = 2tet+ 110. LetF (s) = L(2t − 3)2Find F (0.5).(a) 0(b) 2(c) 6(d) 18(e) 22(f) 26(g) 30(h) 34 −→ CORRECT(i) 38(j) 42(k) The given function is not of exponential order and has no Laplace transformSolution: F (s) =9s−12s2+8s3Page 8Math 217Exam 311. LetF (s) = L{t sin 2t}Find F (1.9).(a) 0.00(b) 0.06(c) 0.07(d) 0.08(e) 0.09(f) 0.10(g) 0.11(h) 0.12(i) 0.13 −→ CORRECT(j) 0.14(k) The given function is not of exponential order and has no Laplace transformSolution: F (s) =4s(s2+ 4)212. Let F (s) = L{f(t)} wheref(t) =(sin t if t < π0 if t ≥ πFind F (0.4).(a) 0.0(b) 0.5(c) 0.6(d) 0.7(e) 0.8(f) 0.9(g) 1.0(h) 1.1 −→ CORRECT(i) 1.2(j) 1.5(k) The given function is not of exponential order and has no Laplace transformSolution: F (s) =1 + e−πss2+ 1Page 9Math 217Exam 313. Use the Laplace transform to turn the initial value problem into an algebraic equation in X(s).x00+ 8x0+ 15x = etx(0) = 1, x0(0) = 2(a) (s2+ 8s)X(s) =1s − 1− 15(b) X(s) = (s2+ 8s + 15)(1s − 1+ 2s + 9)(c) (X(s) + 3)(X(s) + 5) =1s − 1(d) s2+ 8s + 15X(s) =1s − 1(e) (s2+ 8s + 15)X(s) = s + 10(f) (s2+ 8s + 15)X(s) = 2s + 9(g) (s2+ 8s + 15)X(s) = 2s + 9 +1s − 1(h) (s2+ 8s + 15)X(s) =1s − 1+ s + 10 −→ CORRECT(i) (s2+ 8s + 15)X(s) =1s − 1(j) (s2+ 8s + 15)X(s) =1s + 1(k) None of the aboveSolution:x00+ 8x0+ 15x = ets2X − sx(0) − x0(0) + 8 [sX − x(0)] + 15X =1s − 1s2X − s −2 + 8 [sX − 1] + 15X =1s − 1(s2+ 8s + 15)X − (s + 10) =1s − 1(s2+ 8s + 15)X =1s − 1+ s + 10Page 10Math 217Exam 314. Use the Laplace transform to turn the initial value problem into an algebraic equation in X(s).Solve for X(s).2x00− 5x0− 12x = cos t x(0) = x0(0) = 0Find the Laplace transform of the solution.(a) X(s) =ss2+ 1(b) X(s) =s(s2+ 1)(s − 4)(2s + 3)−→ CORRECT(c) X(s) =1(s2+
View Full Document
Unlocking...