# Taking the inverse laplace transform gives y t 2 e 2

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Taking the inverse Laplace transform gives y ( t ) = 2 e - 2 t cos 3 t + 5 3 e - 2 t sin 3 t + 1 3 e - 2( t - π ) sin 3( t - π ) U ( t - π ) . Since sin 3( t - π ) = - sin 3 t , we obtain the answer y ( t ) = 2 e - 2 t cos 3 t + 5 3 e - 2 t sin 3 t - 1 3 e - 2( t - π ) sin 3 t U ( t - π ) . 9. Take the Laplace transform on both sides of the differential equation. L { y ( t ) } + L { y ( t ) } = L δ ( t - π 2 ) + 2 L { δ ( t - π ) } Since y (0) = 3 and y (0) = - 1, we get ( s 2 Y ( s ) - 3 s + 1) + Y ( s ) = e - π 2 s + 2 e - πs . Solving for Y ( s ) gives Y ( s ) = 3 s - 1 s 2 + 1 + e - π 2 s + 2 e - πs s 2 + 1 = 3 s s 2 + 1 - 1 s 2 + 1 + e - π 2 s 1 s 2 + 1 + 2 e - πs 1 s 2 + 1 . Taking the inverse Laplace transform gives y ( t ) = 3 cos t - sin t + sin( t - π 2 ) U ( t - π 2 ) + 2 sin( t - π ) U ( t - π ) . Since sin( t - π 2 ) = - cos t and sin( t - π ) = - sin t, we obtain the answer y ( t ) = 3 cos t - sin t - cos t U ( t - π 2 ) - 2 sin t U ( t - π ) .
32 CHAPTER 2. SOLUTIONS 10. We will use the formula L { f ( t ) } = 1 1 - e - sT T 0 e - st f ( t ) dt. Since f ( t ) = 1 , 0 t < a 0 , a t < 2 a and T = 2 a , we have L { f ( t ) } = 1 1 - e - 2 as 2 a 0 e - st f ( t ) dt = 1 1 - e - 2 as a 0 e - st f ( t ) dt + 2 a a e - st f ( t ) dt = 1 1 - e - 2 as a 0 e - st (1) dt + 2 a a e - st (0) dt = 1 1 - e - 2 as a 0 e - st dt = 1 1 - e - 2 as 1 - e - as s = 1 (1 + e - as )(1 - e - as ) 1 - e - as s = 1 s (1 + e - as ) . 11. Since V ( t ) = 2 - 2 U ( t - 1), the differential equation of the LR circuit is i ( t ) + 4 i ( t ) = 2 - 2 U ( t - 1) with initial conditions i (0) = 0. Take the Laplace transform on both sides of the DE to get sI ( s ) + 4 I ( s ) = 2 s - 2 e - s s . Solving for I ( s ) gives I ( s ) = 2 s ( s + 4) - 2 e - s s ( s + 4) . Let G ( s ) = 2 s ( s + 4) . Using partial fractions, we obtain G ( s ) = 1 2 1 s - 1 s + 4 = g ( t ) = 1 2 ( 1 - e - 4 t ) . Since I ( s ) = G ( s ) - e - s G ( s ) , we can take the inverse Laplace transform to get the current.
2.4. STEP FUNCTIONS AND IMPULSES 33 (a) i ( t ) = 1 2 ( 1 - e - 4 t ) - 1 2 1 - e - 4( t - 1) U ( t - 1) (b) We can express i ( t ) as follows. i ( t ) = 1 2 ( 1 - e - 4 t ) , 0 t < 1 1 2 ( e - 4( t - 1) - e - 4 t ) , t 1 For t = 1 . 5, we get i (1 . 5) = 1 2 e - 4(1 . 5 - 1) - e - 4(1 . 5) 0 . 0664 . At t = 1 . 5 seconds, the current is about 0 . 0664 A. (c) For t 1, we have i ( t ) = 1 2 e - 4( t - 1) - e - 4 t . Therefore, lim t →∞ i ( t ) = 0 . (d) The graph of i ( t ) is the following. 0.6 0.2 0.0 t 2.0 1.8 1.6 1.4 1.2 1.0 0.5 0.8 0.4 0.3 0.1 0.4 0.2 0.0 Observe that the current i ( t ) is continuous even though the voltage is discontinuous at t = 1. 12. We have δ ε ( t - a ) = 1 ε U ( t - a ) - 1 ε U ( t - ( a + ε )) .
34 CHAPTER 2. SOLUTIONS (a) Taking Laplace transform, we get L { δ ε ( t - a ) } = 1 ε e - as s - 1 ε e - ( a + ε ) s s = e - as s 1 - e - εs ε . (b) We will use l’Hˆ opital’s rule to compute the limit. lim ε 0 L { δ ε ( t - a ) } = lim ε 0 e - as s 1 - e - εs ε = e - as s lim ε 0 1 - e - εs ε = e - as s lim ε 0 se - εs 1 = e - as s ( s ) = e - as 2.5 Convolution 1. We will use the convolution theorem. L { f ( t ) * g ( t ) } = F ( s ) G ( s ) (a) L 1 * t 4 = L { 1 } L t 4 = 1 s 4! s 5 = 24 s 6 (b) Observe that t 0 cos θ dθ = 1 * cos t . L t 0 cos θ dθ = L { 1 * cos t } = L { 1 } L { cos t } = 1 s s s 2 + 1 = 1 s 2 + 1 (c) L { e t * sin t } = L { e t } L { sin t } = 1 s - 1 1 s 2 + 1 = 1 ( s - 1)( s 2 + 1)
2.5. CONVOLUTION 35 (d) Observe that t 0 cos θ sin( t - θ ) = cos t * sin t . L t 0 cos θ sin( t - θ ) = L { cos t * sin t } = s s 2 + 1 1 s 2 + 1 = s ( s 2 + 1) 2 2. We will use the convolution theorem. L - 1 { F ( s ) G ( s ) } = f ( t ) * g ( t ) = t 0 f ( θ ) g ( t - θ ) (a) Let F ( s ) = 1 s and G ( s ) = 1 s - 1 , then f ( t ) = L - 1 { F ( s ) } = 1 and g ( t ) = L - 1 { G ( s ) } = e t .
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