Transform of a Periodic Function

Discussion Problems

Theorem 4.4.3 Transform of a Periodic Function

If f (t) is piecewise continuous on [0, q), of exponential order, and periodic with period T, then +5f(t)6 1

12e^sT

#

₀^Testf(t)dt.

PROOF:

Write the Laplace transform of f as two integrals:

+5f(t)6

#

₀^{Testf(t) dt}

#

_T^{qestf(t) dt.}

When we let t u T, the last integral becomes

#

_T^qestf(t)dt

#

₀^{qes(uT)f(u T)duesT}

#

₀^{qesuf(u) esT+5f(t)6.}

Therefore +{ f (t)}

#

₀^{Testf (t) dt esT+{ f (t)}.}

Solving the equation in the last line for +{ f (t)} proves the theorem.

EXAMPLE 8 Transform of a Periodic Function

Find the Laplace transform of the periodic function shown in FIGURE 4.4.4.

SOLUTION The function E(t) is called a square wave and has period T 2. For 0 t 2, E(t) can be defined by

E(t) e1, 0#t ,1 0, 1#t ,2,

and outside the interval by f (t 2) f (t). Now from Theorem 4.4.3, +5E(t)6 1

12e^2s

#

₀^2estE(t)dt 12¹e^stc

#

₀^{1est1 dt}

#

₁^{2est0 dtd}

1

12 e^2s

12e^s s

1

s(1 e^s). (15)

EXAMPLE 9 A Periodic Impressed Voltage

The differential equation for the current i(t) in a single-loop LR-series circuit is L di

dt Ri E(t). (16)

Determine the current i(t) when i(0) 0 and E(t) is the square-wave function given in Figure 4.4.4.

SOLUTION Using the result in (15) of the preceding example, the Laplace transform of the DE is

LsI(s) RI(s) 1

s(1 e^s)  or  I(s) 1/L

s(sR/L) 1

1e^s. (17) d 1 e^2s (1 e^s)(1 e^s)

FIGURE 4.4.4 Square wave in Example 8

1 3 t

1

2 4

E(t)

To find the inverse Laplace transform of the last function, we first make use of geometric series. With the identification x e^s, s 0, the geometric series

1

1x 12 x x²2 x³ p  becomes  1

1e^s 12e^se^2s2e^3s p.

From 1

s(s R/L) L/R

s 2 L/R

sR/L, we can then rewrite (17) as

FIGURE 4.4.5 Graph of current i(t) in Example 9

2i 1.5 1 0.5

0 t

0 1 2 3 4

4.4.1 Derivatives of Transforms

In Problems 1–8, use Theorem 4.4.1 to evaluate the given Laplace transform.

1. +{te^10t} 2. +{t ³e^t}

3. +{t cos 2t} 4. +{t sinh 3t}

5. +{t sinh t} 6. +{t ² cos t}

7. +{te^2t sin 6t} 8. +{te^3t cos 3t}

In Problems 9–14, use the Laplace transform to solve the given initial-value problem. Use the table of Laplace transforms in Appendix III as needed.

9. y y t sin t, y(0) 0 10. y y te^t sin t, y(0) 0

11. y 9y cos 3t, y(0) 2, y(0) 5 12. y y sin t, y(0) 1, y(0) 1

Exercises

Answers to selected odd-numbered problems begin on page ANS-9.

4.4

4.4 Additional Operational Properties ^| 245 By applying the form of the second translation theorem to each term of both series we obtain

i(t) 1

Ra12 8(t2 1) 8(t2 2)2 8(t2 3) pb 1

Rae^Rt/L 2e^R(t21)/L8(t2 1) e^R(t22)/L8(t22) 2e^R(t23)/L8(t 23) pb or, equivalently,

i(t) 1

R(1 2e^Rt/L) 1 Ra

q

n1(1)ⁿ(12 e^R(t2n)/L)8(t2 n).

To interpret the solution, let us assume for the sake of illustration that R 1, L 1, and 0 t 4. In this case

i(t) 1 e^t (1 e^{t 1}) 8(t 1) (1 e^{(t 2)}) 8(t 2) (1 e^{(t 3)}) 8(t 3);

in other words,

i(t) μ

12e^t, e^t e^(t21),

12e^te^(t21)2 e^(t22), e^t e^(t21)2e^(t22) e^(t23),

 

0#t,1 1#t,2 2#t,3 3#t,4.

(18)

The graph of i(t) for 0 t 4, given in FIGURE 4.4.5, was obtained with the help of a CAS.

I(s) 1 Ra1

s 2 1

s R/Lb(12 e^se^2s2e^3s p) 1

Ra1 s 2 e^s

s e^2s s 2 e^3s

s pb 2 1 Ra 1

s R/L 2 e^s

sR/L e^2s

sR/L 2 e^3s

sR/L pb.

246 ^| CHAPTER 4 The Laplace Transform 13. y 16y f (t), y(0) 0, y(0) 1, where

f(t) ecos4t,

0,  0#t, p t$ p 14. y y f (t), y(0) 1, y(0) 0, where

f(t) e1,

sin t, 0#t, p/2 t$ p/2

In Problems 15 and 16, use a graphing utility to graph the indicated solution.

15. y(t) of Problem 13 for 0 t 2p 16. y(t) of Problem 14 for 0 t 3p

In some instances the Laplace transform can be used to solve linear differential equations with variable monomial

coefficients. In Problems 17 and 18, use Theorem 4.4.1 to reduce the given differential equation to a linear first-order DE in the transformed function Y(s) +{y(t)}. Solve the first-order DE for Y(s) and then find y(t) +¹{Y(s)}.

17. ty y 2t ², y(0) 0

18. 2y ty 2y 10, y(0) y(0) 0

4.4.2 Transforms of Integrals

In Problems 19–22, proceed as in Example 3 and find the convolution f * g of the given functions. After integrating find the Laplace transform of f * g.

19. f(t) 4t, g(t) 3t² 20. f(t) t, g(t)e^t 21. f(t) e^t, g(t) e^{t 22.} f(t) cos2t, g(t)e^t In Problems 23–34, proceed as in Example 4 and find the Laplace transform of f * g using Theorem 4.4.2. Do not evaluate the convolution integral before transforming.

23. +{1 * t ³} 24. +{t ² * te^t} 25. +{e^t * e^t cos t} 26. +{e^2t * sin t}

27. +e

#

₀^{tetdtf 28. +e}

#

₀^{tcost dtf}

29. +e

#

₀^{tetcost dtf 30. +e}

#

₀^{ttsint dtf}

31. +e

#

₀^{ttet2 tdtf 32. +e}

#

₀^{tsintcos(t2 t)dtf}

33. +et

#

₀^{tsint dtf 34. +et}

#

₀^ttetdtf

In Problems 35–38, use (8) to evaluate the given inverse transform.

35. +¹e 1

s(s21)f ^36. +¹e 1 s²(s2 1)f

37. +¹e 1

s³(s2 1)f ^38. +¹e 1 s(s2a)²f 39. The table in Appendix III does not contain an entry for

+¹e 8k³s (s² k²)³f.

(a) Use (4) along with the result in (5) to evaluate this inverse transform. Use a CAS as an aid in evaluating the convo- lution integral.

(b) Reexamine your answer to part (a). Could you have obtained the result in a different manner?

40. Use the Laplace transform and the result of Problem 39 to solve the initial-value problem

y y sin t t sin t, y(0) 0, y(0) 0.

Use a graphing utility to graph the solution.

In Problems 41–50, use the Laplace transform to solve the given integral equation or integrodifferential equation.

41. f(t)

#

₀^{t(t2 t)f(t)dt t}

42. f(t)2t24

#

₀^{tsintf(t2 t)dt}

43. f(t)te^t

#

₀^{ttf(t2 t)dt}

44. f(t)2

#

₀^{tf(t)cos(t2 t)dt 4etsint}

45. f(t)

#

₀^{tf(t)dt 1}

46. f(t) cost

#

₀^{tetf(t2 t)dt}

47. f(t)1t2 8

3

#

₀^{t(t 2t)3f(t)dt}

48. t22f(t)

#

₀^{t(et2 et) f(t2 t)dt}

49. y9(t) 12 sint2

#

₀^{ty(t)dt, y(0) 0}

50. dy

dt 6y(t) 9

#

₀^{ty(t)dt1, y(0)0}

In Problems 51 and 52, solve equation (10) subject to i(0) 0 with L, R, C, and E(t) as given. Use a graphing utility to graph the solution for 0 t 3.

51. L 0.1 h, R 3 , C 0.05 f, E(t) 100 [8(t 1) 8(t 2)]

52. L 0.005 h, R 1 , C 0.02 f, E(t) 100 [t (t 1) 8(t 1)]

53. The Laplace transform +5e^t26 exists, but without finding it solve the initial-value problem

y0 9y 3e^t2, y(0) 0, y9(0) 0.

54. Solve the integral equation

f(t) e^t e^t

#

₀^tetf(t)dt.

4.4.3 Transform of a Periodic Function

In Problems 55–60, use Theorem 4.4.3 to find the Laplace transform of the given periodic function.

55.

FIGURE 4.4.6 Graph for Problem 55 t 1

–1 a

Meander function f(t)

2a 3a 4a

56.

FIGURE 4.4.7 Graph for Problem 56 t 1

Square wave a

f(t)

2a 3a 4a

57.

FIGURE 4.4.8 Graph for Problem 57 Sawtooth function

t a

b f(t)

2b 3b 4b

58.

FIGURE 4.4.9 Graph for Problem 58 Triangular wave

t 1

2

1 3 4

f(t)

59.

FIGURE 4.4.10 Graph for Problem 59 t 1

Full-wave rectification of sin t π 2π 3π 4π f(t)

60.

FIGURE 4.4.11 Graph for Problem 60 t 1

Half-wave rectification of sin t π 2π 3π 4π

f(t)

In Problems 61 and 62, solve equation (16) subject to i(0) 0 with E(t) as given. Use a graphing utility to graph the solution for 0 t 4 in the case when L 1 and R 1.

61. E(t) is the meander function in Problem 55 with amplitude 1 and a 1.

62. E(t) is the sawtooth function in Problem 57 with amplitude 1 and b 1.

In Problems 63 and 64, solve the model for a driven spring/mass system with damping

m d²x

dt² b dx

dt kx f(t), x(0) 0, x9(0) 0, where the driving function f is as specified. Use a graphing utility to graph x(t) for the indicated values of t.

63. m ¹2, b 1, k 5, f is the meander function in Problem 55 with amplitude 10, and a p, 0 t 2p.

64. m 1, b 2, k 1, f is the square wave in Problem 56 with amplitude 5, and a p, 0 t 4p.

Discussion Problems

65. Show how to use the Laplace transform to find the numerical value of the improper integral

#

₀^{qte2tsin4tdt.}

66. In Problem 53 we were able to solve an initial-value problem without knowing the Laplace transform +5e^t26. In this prob- lem you are asked to find the actual transformed function Y(s) +5e^t26 by solving another initial-value problem.

(a) If y e^t2, then show that y is a solution of the initial- value problem

dy

dt 2ty0, y(0) 1.

(b) Find Y(s)+5e^t26 by using the Laplace transform to solve the problem in part (a). [Hint: First find Y(0) by rereading the material on the error function in Section 2.3.

Then in the solution of the resulting linear first-order DE in Y(s) integrate on the interval [0, s]. It also helps to use a dummy variable of integration.]

67. Discuss how Theorem 4.4.1 can be used to find +¹elns23

s1f.

68. Bessel’s differential equation of order n 0 is

ty y ty 0.

We shall see in Section 5.3 that a solution of the initial-value prob- lem ty y t y 0, y(0) 1, y(0) 0 is y J₀(t), called the Bessel function of the first kind of order n 0. Use the procedure outlined in the instructions to Problems 17 and 18 to show that

+5J₀(t)6 1

"s² 1.

[Hint: You may need to use Problem 52 in Exercises 4.2. Also, it is known that J₀(0) 1.]

4.4 Additional Operational Properties ^| 247

248 ^| CHAPTER 4 The Laplace Transform 69. (a) Laguerre’s differential equation

ty (1 t)y ny 0

is known to possess polynomial solutions when n is a nonnegative integer. These solutions are naturally called Laguerre polynomials and are denoted by Ln(t).

Find y Ln(t), for n 0, 1, 2, 3, 4 if it is known that Ln(0) 1.

(b) Show that

+ee^t n!

dⁿ

dtⁿtⁿe^tf Y(s),

where Y(s) +{y} and y Ln(t) is a polynomial solution of the DE in part (a). Conclude that

L_n(t) e^t n!

dⁿ

dtⁿ tⁿe^t, n0, 1, 2, p .