Questions and Answers #18 Laplace Transform

Questions and Answers Sheet 18 Laplace Transform Question #1 I want to find the laplace inverse of 𝑠 −3/2 the steps given in the solution manual are as follows:  2  2s 3 2 =2 t  3 I know the first part \frac{2}{\sqrt\pi} is obtained using the gamma function Γ ( ) , but not quite sure how the 2 rest is obtained. Answer: 3 1 The laplace transform of 𝑡 2 is Γ(2) 3 3 1 , so the inverse transform of 𝑠 −2 is 𝑠2 𝑡2 3 2 Γ( ) Question #2 I'm having difficulties calculating a simple Laplace inverse : 𝑆−4 𝑆 2 −2𝑆−11 Answer: When Laplace transforms are rational functions, the typical line of attack is to put it into the form of partial fractions. Note that 𝑆 2 − 2𝑆 − 11 = (𝑆 − 1)2 − 12 has roots 1 ± √12 = 1 ± 2√3 , so: 𝑆−4 𝑆 2 −2𝑆−11 = 𝑆−4 (𝑆−1−2√3)(𝑆−1+2√3) So, we try to write it in the form 𝐴 𝑆−1−2√3 + 𝐵 , with A,B constants. One way of solving for these constants is 𝑆−1+2√3 the cover-up method. We apply this here: 𝐴= 𝐵= 𝑆−4 | 𝑆−1+2√3 𝑆=1+2√(3) 𝑆−4 | 𝑆−1−2√3 𝑆=1−2√(3) = = −3+2√3 4√3 −3−2√3 −4√3 = = 2−√3 4 2+√3 4 Thus, our function's Laplace transform can be written as We now recall that 𝑒 𝑎𝑡 has Laplace transform 1 1 𝑠−𝑎 ( 2−√3 4 𝑆−1+2√3 + 2+√3 𝑆−1−2√3 ) , and this allows us to recover our function as: 𝑓(𝑡) = ((2 − √3)𝑒 (1−2√3)𝑡 + (2 + √3)𝑒 (1+2√3)𝑡 ) 4 1 Question #3 Inverse Laplace Transform without using equations I'm stuck with a question about Inverse Laplace Transform here, but the use of inverse laplace transform equation is forbidden. 𝑌(𝑠) = 𝑒 −𝜋𝑠 𝑠[(𝑠+1)2 +1] Answer: The inverse Laplace transform of −1 1 𝑠 2 +1 is sin 𝑡 . Therefore we have   1 1 ( −1+i )t ( −1−i )t   −t e −e   = e sin ( t ) = 2 2i s + 1 + 1 ( )     ( ) Therefore −1   1 1  e( −1+i )t − 1 e( −1−i )t − 1  1  t −u −    =  0e sin ( u ) du =  2 2i  −1 + i −1 − i   s ( s + 1) + 1    1 1 2 2 = − 𝑒 −𝑡 (sin(𝑡) + cos(𝑡)) The last step is to use the shift theorem {e− s −1 −1 e − as  F ( s ) = f ( t − a )  ( t − a ) and we get 1 1 1 1 } = [ − e − (t − ) (sin(t −  ) + cos(t −  ))]Θ(t −  ) 2 s ( s + 1) + 1 2 2 1 1 = ( + e −t (cos(t ) + sin(t )))Θ(t −  ) 2 2 Question #4 Laplace transform involving step function 𝑓(𝑡) = sin(2𝑡) 𝑒 2𝑡 + 𝑡 ⋅ 𝑢(𝑡 − 4) Answer: sin(2𝑡) F(s) = 𝐿𝑡 [ 𝑒 2𝑡 + 𝑡𝜃(𝑡 − 4)] ∞ (s) sin(2𝑡) ≔ ∫0 𝑒 −s𝑡 ( 𝑒 2𝑡 Where \theta(x) is the Heaviside step function. Now: ∞ F(s) = ∫0 𝑒 −s𝑡 ⋅ ∞ sin(2𝑡) 𝑒 2𝑡 ∞ 𝑑𝑡 + ∫0 𝑡𝑒 −s𝑡 𝜃(𝑡 − 4)𝑑𝑡 = ∞ ∫0 𝑒 −𝑡(2+s) sin(2𝑡) 𝑑𝑡 + ∫4 𝑡𝑒 −𝑠𝑡 𝑑𝑡 + 𝑡𝜃(𝑡 − 4)) 𝑑𝑡 So, we get: 1. When 𝑅[s] > −2 ∞ ∫ 𝑒 −𝑡(2+𝑠) sin(2𝑡) 𝑑𝑡 = 0 𝑠2 2 + 4(2 + 𝑠) 2. When 𝑅[s] > 0 ∞ ∫4 𝑡𝑒 −𝑠𝑡 𝑑𝑡 = 𝑒 −4𝑠 (1+4𝑠) 𝑠2 Question #5 Laplace transform of a exponential. we know that Laplace transform of 𝑥 𝑛 is ℒ[𝑥 𝑛 ] provided n is a positive integer but what is Laplace transform of 𝑎 𝑥 where a is some constant number please help i searched on google so many times and try to solve this by its original defination that is ∞ ℒ[𝑥 𝑛 ] = ∫0 𝑎 𝑥 𝑒 −𝑠𝑥 𝑑𝑥 but how to solve further ( i am using x insted of t here) Answer: ∞ ∞ ∞ 1 ∫0 𝑎 𝑥 𝑒 −𝑠𝑥 𝑑𝑥 = ∫0 𝑒 𝑥 ln 𝑎 𝑒 −𝑠𝑥 𝑑𝑥 = ∫0 𝑒 −(𝑠−ln 𝑎)𝑥 𝑑𝑥 = 𝑠−ln 𝑎 Question #6 Laplace transform of 𝑓(𝑡) = 𝑡𝑒 −𝑡 sin(2𝑡) Answer: Here is an approach. ℒ(sin 2 𝑡) = 2 𝑠 2 +22 = 2 𝑠 2 +4 2 ℒ(𝑒 −𝑡 sin 2 𝑡) = (𝑠+1)2 (te − t sin 2t ) = − Question #7 +4 , using the table. , using frequency shifting. d 2 4( s + 1) ( )= , using frequency differentiation. 2 ds ( s + 1) + 4 (( s + 1) 2 + 4) 2 Laplace transform of 𝑓(𝑡 2 ) Suppose we know the Laplace transform of a function f(t): ∞ 𝐹(𝑠) = ∫0 𝑓(𝑡)𝑒 −𝑠𝑡 𝑑𝑡 Answer: You can change variables 𝑡 2 = 𝑥 ∞ ∞ 𝐻(𝑠) = ∫ 𝑓(𝑡 2 )𝑒 −𝑠𝑡 𝑑𝑡 = ∫ 0 0 𝑓(𝑥) 2 √𝑥 𝑒 −𝑠√𝑥 𝑑𝑥 Now we can express 𝑒 −𝑠√𝑥 /√𝑥 as a Laplace transform, so: =   0  0 f ( x) − s2 /(4t ) − xt e e dt dx 2 t ∞ 𝐹(𝑡) = ∫0 2√𝜋𝑡 𝑒 −𝑠 2 /(4𝑡) 𝑑𝑡 Question #8 Let 𝑥 * (𝑠) = ((𝑠+𝜇 1 ) ̂2 )(𝑠+𝜆1 +𝜆2 ) 1 +𝜇2 )(𝑠+𝜆 ̂2 are just real parameters. be the laplace transform in question, where 𝜇1 , 𝜇2 , 𝜆1 , 𝜆2 , 𝜆 How can i get its inverse by means of convolution, 𝑥(𝑡) = ℒ −1 (𝑥 * (𝑠)) Answer: Let 𝑥 * (𝑠) = ( 1 ) ̂2 )(𝑠 + 𝜆1 + 𝜆2 ) (𝑠 + 𝜇1 + 𝜇2 )(𝑠 + 𝜆 ̂2 are just real parameters. be the laplace transform in question, where 𝜇1 , 𝜇2 , 𝜆1 , 𝜆2 , 𝜆 How can i get its inverse by means of convolution, 𝑥(𝑡) = ℒ −1 (𝑥 * (𝑠)) Question #9 ∞ cos(𝑥𝑡) ∫0 1+𝑡 2 𝑑𝑡 I'm supposed to solve this using Laplace Transformations. Answer: Let f(x) denote the integral, and assume temporarily that 𝑥 > 0. This makes no difference since f(x) is even by definition. Then its Laplace transform Lf(s) defines a continuous function on 𝑠 ∈ (0, ∞) (in fact, it defines an analytic function on 𝑅(𝑠) > 0). Thus we may assume further that 𝑠 ≠ 1 and then rely on the continuity argument. Then ∞ ℒ𝑓(𝑠) = ∫0 𝑓(𝑥)𝑒 −𝑠𝑥 𝑑𝑥 ∞ ∞ cos(𝑥𝑡) = ∫0 ∫0  =   0  = 𝑒 −𝑠𝑥 𝑑𝑡𝑑𝑥 cos( xt ) − sx e dxdt 1+ t2 0 ∞ = ∫0 = 1+𝑡 2 1 𝑠 1+𝑡 2 𝑠 2 +𝑡 2 𝑑𝑡 ∞ 𝑠 1 ∫ (𝑠2 +𝑡 2 1−𝑠 2 0 𝑠 − 𝜋 1 ( − 1) = 1−𝑠 2 2 𝑠 1 1+𝑡 2 ) 𝑑𝑡 𝜋 1 2 1+𝑠 Here, the change of order of integration (∗) is justified by the dominated convergence theorem. Though we proved this for 𝑠 ≠ 1 , it remains valid by continuity argument as mentioned above. Then by the uniqueness of the Laplace transform, we find that 𝜋 𝑓(𝑥) = 𝑒 −𝑥 2 Question #10 𝑠−5 𝐻(𝑠) = (𝑠−3)(𝑠−1) The inverse Laplace transform of H(s) is equal to f*g Answer: Note that we can write 𝐹(𝑠) = 𝑠−5 𝑠−3 =1− 2 𝑠−3 Then, 𝑓(𝑡) = 𝛿(𝑡) − 2𝑒 3𝑡 𝑢(𝑡) and 𝑡 ′ ′ ℎ(𝑡) = ∫0 𝑒 𝑡−𝑡 (𝛿(𝑡 ′ ) − 2𝑒 3𝑡 )𝑑𝑡 ′ = 𝑒 𝑡 − 𝑒 𝑡 (𝑒 2𝑡 − 1) = 2𝑒 𝑡 − 𝑒 3𝑡 Note that partial fraction expansion makes things easier. We simply write 𝐻(𝑠) = 2 𝑠−1 − 1 𝑠−3 Question #11 Trying to find the Laplace transform of cos 𝑡 𝑡 It comes out as infinity, but that doesn't make any sense. Does this mean that this function doesn't have a Laplace transform or is something wrong here? Answer: Well, we are trying to find (solving a more general problem): ℒ𝓉 [ ∞ cos(n⋅𝑡) cos(n⋅𝑡) 𝑡 ] (s) ≔ ∫0 𝑡 ⋅ exp(−s𝑡) d𝑥 Using the 'frequency-domain integration' property of the Laplace transform, we can write: ∞ cos(n⋅𝑡) ℒ𝓉 [ ] 𝑡 (s) = ∫s c𝐿𝑡 [cos(n ⋅ 𝑡)](𝜎)d 𝜎 Using the table of selected Laplace transforms, we find: ℒ𝓉 [cos(n ⋅ 𝑡)](𝜎) = 𝜎 𝜎 2 +n2 So, we get: ℒ𝓉 [ ∞ cos(n⋅𝑡) 𝑡 ] (s) = ∫s 𝜎 𝜎 2 +n2 d𝜎 Let's substitute u ≔ 𝜎 2 + n2 , so we get: t[ 2 +n2 1 cos(n  t ) 1 ](s) =  lim  2 2 t 2 → s + n u 1  lim(ln | 2 → 2 du = 2 2 1  lim[ln | u |]s2 ++nn2 = →  2 2 1 + n2 + n | − ln | s + n |) =  lim | 2 |→  2 → s + n 2 2 2 2 Question #12 Use the Laplace transform to solve the following initial value problem: 𝑦 ′′ + 𝑦 = 2𝑡 with 𝑦(𝜋/4) = 𝜋/2 and 𝑦 ′ (𝜋/4) = 2 − √2 Answer: Laplace-transforming both sides of 𝑦 ′′ + 𝑦 = 2𝑡 𝑠 2 𝑌(𝑠) − 𝑦0 𝑠 − 𝑣0 + 𝑌(𝑠) = 2 𝑠2 Hence 𝑌(𝑠) = 𝑦0 ( 𝑠 𝑠 2 +1 ) + (𝑣0 − 2) ( 1 𝑠 2 +1 )+ Taking the inverse Laplace transform, 2 𝑠2 𝑦(𝑡) = 𝑦0 cos(𝑡) + (𝑣0 − 2) sin(𝑡) + 2𝑡 From the conditions 𝑦(𝜋/4) = 𝜋/2 and 𝑦 ′ (𝜋/4) = 2 − √2 ,we get 𝑦0 + 𝑣0 = 2 𝑦0 − 𝑣0 = 2 Thus 𝑦0 = 2 and 𝑣0 = 0 , and 𝑦(𝑡) = 2 cos(𝑡) − 2 sin(𝑡) + 2𝑡 Question #13 What is the Laplace transform of t cos t into the s domain? Answer: A function of t, f(t), is transformed into a new function of s, F, by a Laplace transformation (s). That is indicated as: ℒ[𝑓(𝑡)] = 𝐹(𝑠) where the actual transform can be acquired from a table of Laplace transforms. For 𝑓(𝑡) = 𝑡 cos(𝑎𝑡): ℒ[𝑓(𝑡)] = ℒ[𝑡 cos(𝑎𝑡)] = 𝑠 2 −𝑎2 2 𝑠 2 +𝑎2 Then, for your function, plug in a=1 to get: ℒ[𝑓(𝑡)] = ℒ[𝑡 cos 𝑡] = = 𝑠 2 –1 𝑠 2 +1 2 𝑠 2 −12 𝑠 2 +12 2