CHAPTER LAPLACE TRANSFORM METHODS SECTION 7.1 LAPLACE TRANSFORMS AND INVERSE TRANSFORMS The objectives of this section are especially clear cut They include familiarity with the definition of the Laplace transform L{f(t)} = F(s) that is given in Equation (1) in the textbook, the direct application of this definition to calculate Laplace transforms of simple functions (as in Examples 1–3), and the use of known transforms (those listed in Figure 7.1.2) to find Laplace transforms and inverse transforms (as in Examples 4-6) Perhaps students need to be told explicitly to memorize the transforms that are listed in the short table that appears in Figure 7.1.2 L{t} e st t dt (u st , du s dt ) 1 1 u u ( 1) ue du u e s s s 0 We substitute u = -st in the tabulated integral ue u du eu u 2u C (or, alternatively, integrate by parts) and get L t st t 2t e t dt e s t 0 s s s st L e3t 1 With a = -s and b = the tabulated integral e st e3t 1 dt e e ( s 3) t dt e au e s3 a cos bu b sin bu cos bu du e au C a b2 yields L cos t e st e st ( s cos t sin t ) s cos t dt s 1 s 1 t 0 422 Copyright © 2015 Pearson Education, Inc Section 7.1 423 L sinh t e st et e t dt e ( s 1) t e ( s 1) t dt 1 1 s 1 s s L et e t L sin t e st sin t dt e st 1 cos 2t dt 1 1 s s cos 2t 2sin 2t 1 e st e st 2 s 4 2 s s s t 0 e s dt e st s s 0 L f (t ) L f (t ) L f (t ) 10 L f (t ) 11 L 12 L 3t 5/ 4t 3 13 L t 2e3 t 14 L t 3/ e 10 t 15 L 1 + cosh 5t e st e st e st e s e 2 s dt s s 1 e s se s s2 e st t dt t 3t (1 t )e st 1 e s t dt e st s s s s s s 0 (3/ 2) 3 3/ 3/ s s 2s s 3! 45 24 (7 / 2) 4 7/2 s s 8s / s s s3 (5 / 2) 5/ 5/ s s 10 4s s 10 s s s 25 Copyright © 2015 Pearson Education, Inc 424 LAPLACE TRANSFORMS AND INVERSE TRANSFORMS s s2 s 4 s 4 s 4 16 L sin 2t cos 2t 17 L cos2 2t 18 L sin 3t cos 3t 19 L 1 t 20 Integrating by parts with u = t, dv = e-(s-1)t dt, we get s 11 L 1+cos 4t 2 s s 16 1 L sin 6t 2 s 36 s 36 L 1 3t 3t L tet t 3 e st tet dt 1! 2! 3! 6 3 3 s s s s s s s s te ( s 1) t dt te ( s 1) t st t 1 L t e e dt ( s 1) s 1 s 0 s 21 Integration by parts with u = t and dv = e-stcos 2t dt yields L t cos 2t te st cos 2t dt e st s cos 2t 2sin 2t dt s 4 s L cos 2t L sin 2t s 4 s2 s2 s s s s2 1 s 1 L cosh 6t 1 2 s 36 s 22 L sinh 3t 23 3 1 L 1 L 1 t s 2 s 24 3/ L 1 3/ L 1 s s 25 1 (5 / 2) 2 8t 3/ 1 L 1 / L 1 5/ t 3/ 3 s s s (5 / 2) s 22 t1/ t Copyright © 2015 Pearson Education, Inc Section 7.1 425 26 5 t L 1 e s 5 27 4t 1 L 1 3 L 3e s 4 s 4 28 s 3s 1 1 L 1 3 L L 3cos 2t sin 2t s 4 s 4 s 4 29 s 1 3s 1 L 1 L 3 L sin 3t 3cos 3t 3 s 9 s 9 s 9 30 9 s 9s 1 L 1 L 1 L sinh 2t cosh 2t 2 2 4 s s 4 s 4 31 10s s 1 10 L 1 L 1 + L 10cosh 5t + sinh 5t 25 s s 25 s 25 32 e 3 s L 1 u (t 3) u3 (t ) s 33 [See Example in the textbook.] eikt e ikt 1 1 L sin kt L 2i 2i s ik s ik 2ik k (because i 1) 2i ( s ik )( s ik ) s k2 34 e kt e kt 1 1 2k k L sinh kt L 2 2 sk sk s k s k2 35 Using the given tabulated integral with a = –s and b = k, we find that e st L cos kt e st cos kt dt s cos kt k sin kt 0 s k t 0 e st e0 s lim s cos kt k sin kt 2 ( s k 0) 2 t s k s k s k Copyright © 2015 Pearson Education, Inc 426 36 LAPLACE TRANSFORMS AND INVERSE TRANSFORMS Evidently the function f (t ) sin(et ) is of exponential order because it is bounded; we can simply take c = and M = in Eq (23) of this section in the text However, 2 its derivative f (t ) 2t et cos( et ) is not bounded by any exponential function ect , because et / e ct et 37 ct as t f (t ) = - ua(t) = - u(t - a) so L f (t ) L 1 L ua (t ) e as s 1 (1 e as ) s s For the graph of f, note that f ( a ) u( a ) 38 f (t ) u(t a ) u(t b), so L f (t ) L ua (t ) L ub (t ) e as e bs s 1 e as e bs s s For the graph of f, note that f ( a ) u(0) u( a b) because a b, but f (b) u(b a ) u(0) 39 Use of the geometric series gives L f (t ) 40 L u(t n) n 0 e ns 1 e s e 2 s e 3s s n 0 s 1 (e s ) (e s )2 (e s )3 s 1 s s 1 e s 1 e s Use of the geometric series gives ( 1) n e ns L f (t ) ( 1) L u(t n ) 1 e s e 2 s e 3s s s n 0 n 0 41 n 1 ( e s ) ( e s ) ( e s )3 s 1 s s ( e ) s 1 e s By checking values at sample points, you can verify that g (t ) f (t ) in terms of the square wave function f (t ) of Problem 40 Hence L g (t ) L 2 f (t ) 1 1 1 e s s e s s e s s 1 e s s 1 e s es / es / e s / s/2 s s/2 s / s 1 e e s e e s 2 Copyright © 2015 Pearson Education, Inc e e s/2 s/2 e s / e s / Section 7.1 427 42 s sinh( s / 2) s cosh( s / 2) s Let's refer to (n 1, n ] as an odd interval if the integer n is odd, and even interval if n is even Then our function h (t ) has the value a on odd intervals, the value b on even intervals Now the unit step function f (t ) of Problem 40 has the value on odd intervals, the value on even intervals Hence the function (a b) f (t ) has the value ( a b) on odd intervals, the value on even intervals Finally, the function ( a b) f (t ) b has the value (a b) b a on odd intervals, the value b on even intervals, and hence ( a b) f (t ) b h(t ) Therefore a b b a be s L{h (t )} L{( a b) f (t )} L{b} s(1 e s ) s s(1 e s ) SECTION 7.2 TRANSFORMATION OF INITIAL VALUE PROBLEMS The focus of this section is on the use of transforms of derivatives (Theorem 1) to solve initial value problems (as in Examples and 2) Transforms of integrals (Theorem 2) appear less frequently in practice, and the extension of Theorem at the end of Section 7.2 may be considered entirely optional (except perhaps for electrical engineering students) In Problems 1–10 we give first the transformed differential equation, then the transform X(s) of the solution, and finally the inverse transform x(t) of X(s) [s2 X(s) - 5s] + 4{X(s)} = X ( s) 5s s 5 s 4 s 4 x(t) = L-1{X(s)} = cos 2t [s2 X(s) - 3s - 4] + 9[X(s)] = X ( s) s 3s 4 3 2 s 9 s 9 s 9 x(t) = L-1{X(s)} = cos 3t + (4/3)sin 3t [s2 X(s) - 2] - [sX(s)] - 2[X(s)] = Copyright © 2015 Pearson Education, Inc 428 TRANSFORMATION OF INITIAL VALUE PROBLEMS X ( s) 2 2 1 s s2 ( s 2)( s 1) s s 1 x(t) = (2/3)(e2t - e-t) [s2 X(s) - 2s + 3] + 8[s X(s) - 2] + 15[X(s)] = X ( s) s 13 s 8s 15 s3 s5 x(t) = L-1{X(s)} = (7/2)e-3t - (3/2)e-5t [s2 X(s)] + [X(s)] = 2/(s2 + 4) X ( s) 2 1 2 ( s 1)( s 4) s 1 s x(t) = (2 sin t - sin 2t)/3 [s2 X(s)] + 4[X(s)] = L{cos t} = s/(s2 + 1) X ( s) s s 1 2 ( s 1)( s 4) s 1 s x(t) = L-1 {X(s)} = (cos t - cos 2t)/3 [s2 X(s) - s] + [X(s)] = s/s2 + 9) (s2 + 1)X(s) = s + s/(s2 + 9) = (s3 + 10s)/(s2 + 9) s 10s s s X ( s) 2 ( s 1)( s 9) s 1 s x(t) = (9 cos t - cos 3t)/8 [s2 X(s)] + 9[X(s)] = L{1} = 1/s X ( s) s 1 1 s ( s 9) s s 9 x(t) = L-1 {X(s)} = (1 - cos 3t)/9 s2 X(s) + 4sX(s) + 3X(s) = 1/s 1 1 1 1 X ( s) s( s s 3) s( s 1)( s 3) s s 1 s x(t) = (2 - 3e-t + e-3t)/6 Copyright © 2015 Pearson Education, Inc Section 7.2 429 10 [s2 X(s) - 2] + 3[sX(s)] + 2[X(s)] = L{t} = 1/s2 (s2 + 3s + 2)X(s) = + 1/s2 = (2s2 + 1)/s2 2s2 2s2 1 1 X ( s) 2 3 s s s ( s 3s 2) s ( s 1)( s 2) s 1 s x(t) = L-1 {X(s)} = (-3 + 2t + 12e-t - 9e-2t)/4 11 The transformed equations are sX(s) - = 2X(s) + Y(s) sY(s) + = 6X(s) + 3Y(s) We solve for the Laplace transforms X ( s) s5 s ( s 5) s Y(s) = X ( s ) 2 s 10 s( s 5) s Hence the solution is given by x(t) = 1, 12 y(t) = -2 The transformed equations are s X(s) = X(s) + 2Y(s) s Y(s) = X(s) + 1/(s + 1), which we solve for X ( s) 2 1 3 ( s 2)( s 1) s s 1 ( s 1) Y ( s) s 1 1 1 6 ( s 2)( s 1) s s 1 ( s 1) Hence the solution is x(t) = (2/9)(e2t - e-t - 3t e-t) y(t) = (1/9)(e2t - e-t + 6t e-t) Copyright © 2015 Pearson Education, Inc 430 13 TRANSFORMATION OF INITIAL VALUE PROBLEMS The transformed equations are sX(s) + 2[sY(s) - 1] + X(s) = sX(s) - [sY(s) - 1] + Y(s) = 0, which we solve for the transforms X ( s) Y ( s) 2 1/ 3s s 1/ 3 s 1/ 3s s 1/ s 3s s 1/ s2 / cosh t / 1/ sinh t / x(t) = / sinh t / y(t) = 14 The transformed equations are s2 X(s) + + 2X(s) + 4Y(s) = s2 Y(s) + + X(s) + 2Y(s) = 0, which we solve for s2 1 3 X ( s) 2 s ( s 4) s 4 4 s Y ( s) 1 s2 3 2 8 s s ( s 4) s 4 Hence the solution is x(t) = (1/4)(2t - sin 2t) y(t) = (-1/8)(2t + sin 2t) 15 1/ s2 / Hence the solution is The transformed equations are [s2 X - s] + [sX - 1] + [sY - 1] + 2X - Y = [s2 Y - s] + [sX - 1] + [sY - 1] + 4X - 2Y = 0, which we solve for Copyright © 2015 Pearson Education, Inc Section 7.2 431 X ( s) Y ( s) 12 12 s 3s s3 s3 2 s s 3s s ( s 3/ 2) (3/ 4) s 3s 3s 12 3/2 s 3/ s ( s 3/ 2) ( / 2) ( s 3/ 2) ( / 2) 28 s 15 s3 2s 2s 21 s s s 3s s 3s 3s 28 s 15 21 s s ( s 3/ 2) 3/ 28 3/2 s 3/ 2 21 s s ( s 3/ 2) ( / 2) ( s 3/ 2) ( / 2) Here we've used some fairly heavy-duty partial fractions (Section 7.3) The transforms L e at cos kt sa , ( s a )2 k L e at sin kt k ( s a )2 k from the inside-front-cover table (with a 3/ 2, k / 2) finally yield 16 3t / sin x (t ) e 3t / cos y (t ) 28 9et e 3t / cos 21 3t / 3t / sin The transformed equations are s X(s) - = X(s) + Z(s) s Y(s) = X(s) + Y(s) s Z(s) = -2X(s) - Z(s), which we solve for X ( s) s2 s 1 2 ( s 1)( s 1) s 1 Y ( s) s 1 s 2 ( s 1)( s 1) s 1 s 1 Z ( s) 2 s 2 ( s 1)( s 1) s 1 Copyright © 2015 Pearson Education, Inc 3t / 458 PERIODIC AND PIECEWISE CONTINUOUS INPUT FUNCTIONS 29 1 e as (1 as ) as as e 1 s 1 e 2 as s 1 e 2 as With p = 2/k and f(t) = sin kt for t /k while f(t) = for /k t 2/k, Formula (12) and the integral formula e at a sin bt b cos bt sin bt dt e at C a b2 give L{ f (t )} 1 e 2 s / k e 2 s / k /k e st sin kt dt e s / k k ( k ) e 2 s / k s2 k s / k k 1 e s / k s / k 1 e 1 e s k 30 t / k st s sin kt k cos kt e s2 k t 0 k s k 1 e s / k 2 h(t ) f (t ) g (t ) f (t ) u(t / k ) f (t / k ), so Problem 29 gives H ( s ) F ( s ) e s / k F ( s ) 1 e s / k F ( s ) 1 e s / k e s / k e s / k k k s k e s / k s k e s / k e s / 2k k e s / k e s / k k cosh( s / 2k ) k s coth s / 2k 2 s / k 2 s k e e s k sinh( s / 2k ) s k 2k In Problems 31-42, we first write and transform the appropriate differential equation Then we solve for the transform of the solution, and finally inverse transform to find the desired solution 31 x'' + 4x = - u(t – ) s2X(s) + 4X(s) = e s s e s s 1 X(s) = 1 e s s s 4 s s 4 x(t) = (1/4)[1 - u(t – )][1 - cos 2(t - )] = (1/2)[1 - u(t – )]sin2 t The graph of the position function x(t ) is shown at the top of the next page Copyright © 2015 Pearson Education, Inc Section 7.5 459 x(t) t Π 32 x'' + 5x' + 4x = - u(t – 2) s2X(s) + 5s X(s) + 4X(s) = X(s) = e 2 s s e 2 s = (1 - e-2s)G(s) s s 5s where 3 4e t e 4 t , so g (t ) 12 s s s 12 G ( s) It follows that g (t ) if t 2, x (t ) g (t ) u(t 2) g (t 2) g (t ) g (t 2) if t x(t) 0.2 0.1 t 33 x'' + 9x = [1 - u(t – 2)]sin t Copyright © 2015 Pearson Education, Inc 460 PERIODIC AND PIECEWISE CONTINUOUS INPUT FUNCTIONS X(s) = e 2 s 1 1 e 2 s 2 s 1 s s 1 s 1 1 u(t 2 ) sin t sin 3t The left-hand figure below show the graph of this position function x(t) = x(t) x(t) 0.5 t t 2Π Π 1 2Π 0.1 0.5 34 x'' + x = [1 - u(t – 1)] t u(t 1) f (t 1), where s2 X(s) + X(s) = f (t ) t 1 1 e sG ( s ) e s s s s s It follows that X ( s) e s ( s 1) s ( s 1) s ( s 1) s s 1 s s (1 e s ) e (1 e )G ( s ) e H ( s ) s 1 s s s 1 where g(t) = t - sin t, h(t) = - cos t Hence x(t) = g(t) - u(t – 1)g(t - 1) - u(t – 1)h(t - 1) and so x(t) = t - sin t if t < 1, x(t) = -sin t + sin(t - 1) + cos(t - 1) if t > The right-hand figure above shows the graph of this position function 35 x + 4x + 4x = [1 - u(t – 2)]t = t - u(t – 2)g(t – 2) where g (t ) t (s + 2)2 X(s) = 2 e 2 s s s s Copyright © 2015 Pearson Education, Inc 4Π Section 7.5 461 X ( s) s s 2 2 e 2 s 2s s s 2 2 1 1 1 2 s 1 e s s s s 2 s s s s 2 x(t) = (1/4){-1 + t + (1 + t)e-2t + u(t –2)[1 – t + (3t - 5)e-2(t-2)]} x(t) t 36 100 I ( s ) 1000 e s I ( s) 100 s s s e s I ( s) 1 e s L e 10t s 10 i(t) = e-10t - u(t – 1)e-10(t-1) 37 i'(t) + 104 i (t ) dt = 100[1 - u(t – 2)] I ( s) e 2 s 100 s s 2 s 100 1 e I ( s) 1 e 2 s L sin100t s 10 s I ( s ) 104 i(t) = sin 100t - u(t – 2)sin 100(t - 2) = [1 - u(t – 2)]sin 100t 38 i'(t) + 10000 i (t ) dt = [1 - u(t – )](100 sin 10t) s I(s) + 10000 I(s)/s = 1000(1 - e-s)/(s2 + 100) Copyright © 2015 Pearson Education, Inc 462 PERIODIC AND PIECEWISE CONTINUOUS INPUT FUNCTIONS I ( s ) 1 e s i (t ) s s 1000s 10 1 e s 2 s 100 99 s 10 s 100 s 10000 10 e s L cos10t cos100t 99 10 10 cos10t cos100t u t cos10(t ) cos100(t ) 99 99 10 1 u t cos10t cos100t 99 10 cos10t cos100t if t , i (t ) 99 t if 39 i'(t) + 150 i(t) + 5000 i (t ) dt = 100t[1 - u(t – 1)] sI ( s ) 150 I ( s ) 5000 I ( s) 100 1 100e s s s s s 100 100( s 1) e s s( s 50)( s 100) s( s 50)( s 100) 1 s 98 99 e 50 s s 50 s 100 50 s s 50 s 100 I ( s) i(t) = (1/50)[1 - 2e-50t + e-100t] - (1/50)u(t – 1)[1 + 98e-50(t-1) - 99e-100(t-1)] 40 i'(t) + 100 i(t) + 2500 i (t ) dt = 50t[1 - u(t – 1)] sI ( s ) 100 I ( s ) 2500 I ( s) 50 1 50e s s s s s 50 50( s 1) e s s( s 50) s ( s 50) 1 50 s 50 2450 e 50 s s 50 ( s 50) 50 s s 50 ( s 50) I ( s) i (t ) 1 e 50t 50t e 50t u(t 1) 1 e 50( t 1) 2450t e 50( t 1) 50 50 Copyright © 2015 Pearson Education, Inc Section 7.5 463 41 x'' + 4x = f(t), x(0) = x'(0) = s X ( s) s 4 X ( s) 1 e s (by Example of Section 7.5) s 1 e s ( 1) n e n s s s n 1 (as in Eq (16) of Section 7.5) Now let g (t ) L1 cos 2t 2sin t s ( s 4) Then it follows that n 1 n 1 x (t ) g (t ) 2 ( 1) n un (t ) g (t n ) 2sin t 4 ( 1) n un (t ) sin t Hence 2sin t if 2n t (2n 1) , x (t ) 2sin t if (2n 1) t 2n Consequently the complete solution x (t ) sin t sin t is periodic, so the transient solution is zero The graph of x (t ) : x(t) t 2Π 4Π Copyright © 2015 Pearson Education, Inc 6Π 464 42 IMPULSES AND DELTA FUNCTIONS x'' + 2x' + 10x = f(t), x(0) = x'(0) = As in the solution of Example we find first that s s 10 X ( s ) 10 20 ( 1) n e n s , s s n 1 so X ( s) 10 10( 1) n e n s s ( s s 10) n 1 s ( s s 10) If 10 t g (t ) L 1 e 3cos 3t sin 3t , s ( s 1) then it follows that x (t ) g (t ) 2 ( 1) n un (t ) g (t n ) n 1 The graph of x (t ) : x(t) t 2Π 4Π 6Π SECTION 7.6 IMPULSES AND DELTA FUNCTIONS Among the several ways of introducing delta functions, we consider the physical approach of the first two pages of this section to be the most tangible one for elementary students Whatever the Copyright © 2015 Pearson Education, Inc Section 7.6 465 approach, however, the practical consequences are the same — as described in the discussion associated with equations (11)–(19) in the text That is, in order to solve a differential equation of the form a x(t ) b x(t ) c x (t ) f (t ) where f(t) involves delta functions, we transform the equation using the operational principle L{a(t)} = e-as, then solve for X(s), and finally invert as usual to find the formal solution x(t) Then we show the graph of x(t ) x s2X(s) + 4X(s) = X ( s) x (t ) 1 s 4 2 sin 2t t 2Π Π 3Π 2 s2X(s) + 4X(s) = + e-s X ( s) e s s2 1 sin 2t if t , x (t ) 1 u (t ) sin 2t 2 sin 2t if t x 1 t Π 2Π 3Π Copyright © 2015 Pearson Education, Inc 4Π 4Π 466 IMPULSES AND DELTA FUNCTIONS + e-2s s 2 s e 11 e 2 s X ( s) s ( s 2) ( s 2) s s ( s 2) ( s 2) s2X(s) + 4sX(s) + 4X(s) = x (t ) 1 e 2 t 2t e 2 t u(t 2)(t 2)e 2( t 2) x 0.5 0.25 4 t s2 2 2 s s s ( s 1) [s2X(s) - 1] + 2sX(s) + X(s) = + X ( s) 2s2 s ( s 1) x(t) = -2 + t + 2e-t + 3te-t x 1.5 0.5 Copyright © 2015 Pearson Education, Inc t Section 7.6 467 (s2 + 2s + 2)X(s) = 2e-s X ( s) 2e s ( s 1) if t , x(t) = 2u(t – ) e ( t ) sin(t - ) = ( t ) sin t if t 2 e x 0.5 t Π s2X(s) + 9X(s) = e-3s + X ( s) x (t ) s s 9 2Π 3Π s s 9 e 3 s s2 1 1 t sin 3t u(t 3 )sin 3(t 3 ) t sin 3t u(t 3 ) sin 3t 6 x -2 Copyright © 2015 Pearson Education, Inc t 468 IMPULSES AND DELTA FUNCTIONS [s2X(s) - 2] + 4sX(s) + 5X(s) = e-s + e-2s X ( s) e s e 2 s ( s 2) x(t) = 2e-2t sin t + u(t)e-2(t-) sin(t - ) + u2(t)e-2(t-2) sin(t - 2) = [2 - e2 u(t – ) + e4 u(t – 2)] e–2t sin t x 0.3 t Π 2Π 3Π [s2X(s) - 2s - 2] + 2[sX(s) - 2] + X(s) = - e-2s X ( s) s e 2 s e 2 s ( s 1) s ( s 1) ( s 1) x(t) = (2 + 5t)e-t - u(t – 2)(t - 2)e-(t-2) x t Copyright © 2015 Pearson Education, Inc Section 7.6 469 s2X(s) + 4X(s) = F(s) X(s) = x (t ) 10 11 t (sin 2u ) f (t u ) du X(s) = F ( s) ( s 3) x (t ) t ue 3u f (t u ) du (s2 + 6s + 8)X(s) = F(s) x (t ) F ( s) ( s 3) t e 3u (sinh u ) f (t u ) du s2X(s) + 4sX(s) + 8X(s) = F(s) X(s) = x (t ) 13 s2X(s) + 6s X(s) + 9X(s) = F(s) X(s) = 12 F ( s) s 4 (a) F ( s) ( s 2) t e 2 u (sin 2u ) f (t u ) du mx(t ) = (p/ )[u0(t) - u(t)] ms2X(s) = (p/ )[1/s - e-s /s] (b) mX(s) = (p/ )[(1 - e-s)/s3] mx(t) = (p/2)[t2 - u(t)(t - )2] If t > then mx(t) = (p/2)[t2 - (t2 - 2t + 2)] = (p/2)(2t - 2) Hence mx(t) pt as (c) mv = (mx) = (pt) = p Copyright © 2015 Pearson Education, Inc 470 IMPULSES AND DELTA FUNCTIONS 14 sX(s) = e-as; X(s) = e-as /s; 15 Each of the two given initial value problems transforms to x(t) = u(t - a) (ms2 + k)X(s) = mv0 = p0 16 Each of the two given initial value problems transforms to (as2 + bs + c)X(s) = F(s) + av0 17 i' + 100i = 1(t) - 2(t), i(0) = (b) e s e 2 s I(s) s 100 I ( s) i(t) = u1(t)e-100(t-1) - u2(t)e-100(t-2) 18 i''(t) + 100 i(t) = 10 (t) - 10 (t - ) (b) (s2 + 100) I(s) = 10 - 10 e-s 10 10e s s 100 s 100 I ( s) i(t) = sin 10t - u(t)sin 10(t - ) sin10t if t , = [1 - u(t – )]sin 10t = if t 19 s 100 I ( s ) 10 ( 1) n e n s /10 n 0 I ( s) i (t ) 10 ( 1) n e n s /10 n 0 s 100 (1) e n n s /10 n 0 ( 1)n un /10 (t ) sin10(t n /10) n 0 10 s 100 u(t n /10)sin10t n 0 because sin(10t - n) = (-1)n sin 10t Hence i(t) = (n + 1)sin 10t if n /10 < t < (n + 1)/10 Copyright © 2015 Pearson Education, Inc Section 7.6 471 20 s 100 I ( s) 10 (1)n e n s / n 0 I ( s) i (t ) 10 ( 1) n e n s / n 0 s 100 (1) e n n s / n 0 ( 1)n un / (t )sin10(t n / 5) n 0 10 s 100 ( 1) u(t n / 5)sin10t n n 0 Hence i(t) = sin 10t + (–1)1sin 10t + + (-1)n sin 10t if n / t (n 1) / 5, n Thus i (t ) sin10t in this interval if n is even, but is zero in this interval if n is odd 21 s 60s 1000 I ( s ) 10 ( 1) n e n s /10 n 0 I ( s) 10 ( 1) n e n s /10 n 0 s 60s 1000 n 0 (1) e n n s /10 10 ( s 30) 100 i(t) = (-1)n un/10(t) g(t - n/10) where g(t) = e-30t sin 10t, and so g(t - n /10) = exp[-30(t - n /10)] sin 10(t - n /10) = e3n e-30t(-1)n sin 10t Therefore n 3n 30t i t u t e e sin10t 10 n 0 If n /10 < t < (n + 1) /10 then it follows that 3 i(t) = (1 + e + + e 3n )e -30t e3 n 1 30t sin 10t = e sin10t e3 The graph of i (t ) is shown at the top of the next page Copyright © 2015 Pearson Education, Inc 472 IMPULSES AND DELTA FUNCTIONS i(t) 0.1 Π 2Π 3Π 4Π 5 5 t Π 0.1 22 ( s 1) X ( s ) e 2 n s n 0 X ( s) x (t ) e 2 n s n 0 s u n 0 n (t )sin t 2n u(t 2n ) sin t n 0 Hence x(t) = (n + 1)sin t if 2n < t < 2(n + 1) The graph of x (t ) : x(t) t 2Π 4Π 6Π 8Π Copyright © 2015 Pearson Education, Inc 10 Π ... permit going further in this chapter, Sections 7. 1? ?7. 3 provide a self-contained introduction to Laplace transforms that suffices for the most common elementary applications 24 24 , so L{t4 e t}... students Whatever the Copyright © 2015 Pearson Education, Inc Section 7. 6 465 approach, however, the practical consequences are the same — as described in the discussion associated with equations... 2sin x C sin x dx x cos x x sin x cos x C Copyright © 2015 Pearson Education, Inc 446 DERIVATIVES, INTEGRALS, AND PRODUCTS OF TRANSFORMS from #40 and #41 inside the back cover of