Higher Engineering Mathematics - BS Grewal

Chapter 62 Properties of Laplace transforms 62.1 The Laplace transform of eat f (t) (ii) L{eat sin ωt} From Chapter 61, the definition of the Laplace transform ω of f (t ) is: Since L{sin ωt } = s2 + ω2 from (iv) of Table 61.1, page 584. then L{eat sin ωt} = (s − ω + ω2 from equa- a)2 ∞ tion (2) (provided s > a). (1) L{ f (t )} = e−st f (t ) dt (iii) L{eat cosh ωt} 0 s ∞ Since L{cosh ωt } = s2 − ω2 from (ix) of Table 61.1, page 584. Thus L{eat f (t )} = e−st (eat f (t )) dt 0 ∞ (2) then L{eat cosh ωt} = s− a from equa- − a)2 − = e−(s−a) f (t ) dt (s ω2 0 (where a is a real constant) tion (2) (provided s > a). Hence the substitution of (s − a) for s in the transform A summary of Laplace transforms of the form eat f (t ) is shown in Table 62.1. shown in equation (1) corresponds to the multiplication of the original function f (t ) by eat . This is known as a Table 62.1 Laplace transforms of the form shift theorem. eat f (t ) 62.2 Laplace transforms of the form Function eat f (t ) Laplace transform eat f(t) (a is a real constant) L{eat f (t )} (i) eat t n From equation (2), Laplace transforms of the form (ii) eat sin ωt n! eat f (t ) may be deduced. For example: (iii) eat cos ωt (s − a)n+1 (iv) eat sinh ωt (i) L{eat tn} (v) eat cosh ωt ω (s − a)2 + ω2 Since L{t n} = s n! from (viii) of Table 61.1, s−a n+1 (s − a)2 + ω2 page 584. ω (s − a)2 − ω2 then L{eat tn} = n! from equation (2) (s − a)n+1 s−a above (provided s > a). (s − a)2 − ω2

588 Higher Engineering Mathematics Problem 1. Determine (a) L{2t 4e3t } 10 10 (b) L{4e3t cos 5t }. = (s + 3)2 − 22 = s2 + 6s+9 − 4 (a) From (i) of Table 62.1, 10 = s2 + 6s + 5 L{2t 4e3t } = 2L{t 4e3t } = 2 4! (b) L{2e3t (4 cos 2t − 5 sin 2t )} (s − 3)4+1 2(4)(3)(2) 48 = 8L{e3t cos 2t } − 10L{e3t sin 2t } (s − 3)5 (s − 3)5 = = 8(s − 3) 10(2) − 3)2 + 22 − 3)2 + = (s − (s 22 (b) From (iii) of Table 62.1, L{4e3t cos 5t } = 4L{e3t cos 5t } from (iii) and (ii) of Table 62.1 =4 s −3 = 8(s − 3) − 10(2) = 8s − 44 (s − 3)2 + 52 (s − 3)2 + 22 s2 − 6s + 13 = 4(s − 3) Problem 4. Show that − 6s +9+ s2 25 3e− 1 x sin2 48 2 1)(4s2 + 4(s − 3) L x = (2s + 4s + 17) = s2 − 6s +34 Since cos 2x = 1 −2 sin2 x, sin2 x = 1 (1 − cos 2x). Problem 2. Determine (a) L{e−2t sin 3t } 2 (b) L{3eθ cosh 4θ}. Hence, L 3e− 1 x sin2 x 2 (a) From (ii) of Table 62.1, L{e−2t }= 3 = 3 =L 3e− 1 x 1 (1 − cos 2x ) (−2))2 +2)2 2 sin 3t 2 (s − + 32 (s + 9 33 = 3L e− 1 x − 3L e− 1 x cos 2x = s2 + 4s + 4 + 9 = s2 + 4s + 13 2 2 22 ⎛ ⎞⎛ − −1 ⎞ 2 (b) From (v) of Table 62.1, ⎝⎜⎜⎜ s ⎟⎟⎠⎟ − −1 +22 L{3eθ 4θ } = 3L{eθ 4θ } = 3(s − 1) = 3 ⎝⎜⎜ 1 ⎟⎟⎠ − 3 2 − 1)2 − 42 2 s −1 2 cosh cosh (s − s 2 2 = 3(s − 1) = 3(s − 1) from (iii) of Table 61.1 (page 584) and (iii) s2 −2s +1−16 s2 − 2s −15 of Table 62.1 above, Problem 3. Determine the Laplace transforms of 3 s+1 (a) 5e−3t sinh 2t (b) 2e3t (4 cos 2t − 5 sin 2t ). 2 =3− 2 (a) From (iv) of Table 62.1, 2 s+1 2 2 s+1 + 22 2 L{5e−3t sinh 2t } = 5L{e−3t sinh 2t } 3 6s + 3 2 =− = 5 (s − (−3))2 − 22 2s + 1 4 s2 + s + 1 + 4 4