{ sin(t), t [0, sketch the graph of this f(t) = L 1 {F(p)}.

Transcription

1 EM Solved roblems Lalace & Fourier transform c Habala 3 EM Solved roblems Lalace & Fourier transform Find the Lalace transform of the following functions: ft t sint; ft e 3t cos3t; 3 ft e 3s ds; { sint, t,; ft t + t 3; 5 ft 6 ft sint, elsewhere; Find the inverse Lalace transform of the following functions: 8e 3 7 F ++ ; 8 F ++5 ; t 9 F e e + e, sketch the grah of this ft L {F Use the Lalace transform to solve the following roblems: y y x + e x, y 3 {, t,; ÿ + y y +, ẏ +, elsewhere, y t + ys ds 6, y + { sint, t, 3 y y ; y, elsewhere, { t, t,; ẋ x x+, elsewhere, 5 Use the Lalace transform to find the general solution of y + y 8e t sint { t, t,; 6 Consider the function ft, t, Find its Fourier series, its sine Fourier series, and its cosine Fourier series For each series determine its sum

2 EM Solved roblems Lalace & Fourier transform c Habala 3 Solutions We first use the rule for L{t ft to get rid of t: L{t sint d d L{sint ] + + We first use the rule for L{e at ft to get rid of the exonential: L{e 3t cos3t L{cos3t It is also ossible to first get rid of 3t, since it is everywhere in the function We use the choice ft e t cost, then f3t is the given function Using the aroriate formula we get rid of 3t, then we continue as above get rid of exonential, transform cosine When doing two oerations, one has to be careful about the roer order of grammar rules: L{e 3t cos3t 3 L{e t cost /3 3 L{cost /3 3 / /3 3 /3+ 3 We use the rule for integrals: { t L e 3s ds L{e3t / We have to use the shift rule There are two ways to do it: a We change the given function t+ to feature the exression t 3 just like the jum function, then use the shift rule: L{t + Ht 3 L{t 3] + 5 Ht 3 e 3 L{t + 5 Ht e b It is also ossible to use the shift rule and think of it as a substitution, for t 3 one uts, say, s, so t s + 3: L{t + Ht 3 e 3 L{s + 3] + Hs e 3 L{s + 5 Hs e There are two ossibilities First, one can use the definition: L{ft fte t dt sinte t dt This integral is standard two integrations by arts, then solving an equation, but it is a long calculation It might be easier to exress the given function as a roduct of sine and a discrete unit signal on the interval,]: ft sint Ht Ht ] sintht sintht Thus see the revious roblem L{ft L{sintHt sintht L{sintHt L{sintHt + e L{sins + Hs + e L{ sinshs + + e + e There is no rule for the absolute value One ossibility is to use the definition, but integrating absolute value is a roblem, such an integral would have to be slit deending on whether the sine is ositive or negative It will be therefore easier to think of f as a function that is eriodic with T, since indeed, when you draw the grah, you find out that the function f is actually just one hill of the sine reeated over and over In fact, the first eriod, f T, is exactly the function from art d By the theorem on eriodic functions and the revious art, L{ft L{f T e e + + e 7 To find the inverse, it is enough to decomose the fraction into artial fractions: L { + L { ++ L { L { + L { cost + sint e t

3 EM Solved roblems Lalace & Fourier transform c Habala 3 8 First we will use the aroriate rule to get rid of the exonential The inverse Lalace of the remaining fraction is done first by comleting a square, then by using the shift rule for the inverse Lalace and finally using the dictionary: L { 8e L { t 3 Ht 3 L { t 3 Ht 3 e t L { 8 t 3 + Ht 3 e t sint t 3 Ht 3 e 3 t sin t 3 {, t,3; Ht 3 e 3 t sin t 3, t 3 When rewriting the answer as a slit function, we used the fact that Ht 3 for t < 3 and Ht 3 for t 3 Note how t 3 was substituted back: It also became a art of the exonential, due to the order in which the rules were used It is ossible to also first handle the shift, then the rest: L { 8e 3++3 e t L { 8e 3+3 e t L { e 3 e 3 L { 8e e t e 3 L { e e 3 t L { + t 3 Ht 3 + e 3 t sint t 3 Ht 3 e 3 t sin t 3 Ht 3 Now the substitution of t 3 alied only to the sine art itself, because e 3 t was already done, it was not a art of the inverse Lalace when the get rid of exonential rule was alied 9 We just have to use the rule for getting rid of exonential and then the dictionary: L { e e + e L { L { e + t L { + t Ht t t + e t t Ht y t t ] + e t Ht t + e t t Ht, t < ; t, t,; e t, t x We aly the Lalace transform to both sides of the equation, denoting Y L{y: L{y y L{x + e x L{y L{y L{x + L{y y + ] L{y + Y 3 Y + Y Therefore Y To find yt we use the inverse Lalace transform First we need to relace the rational functions in Y with artial fractions:

4 EM Solved roblems Lalace & Fourier transform c Habala 3 Thus Y + + Consequently + yt L { + + L { + L { + L { e t + e t + e t L { e t + e t + e t t, t Since the equation and the solution exist on the whole real line, it is very likely that the solution actually works on IR This would have to be checked by substituting y into the equation and indeed it works We aly the Lalace transform to both sides It might be a good idea to first figure out the Lalace transform on the right We can do it by definition, since this should be easy: L{ft e t dt ] e t e Alternative: L{ft L{Ht Ht e L{Ht e The left hand-side transforms denoting Y L{y like this: L{ÿ + y Y ] + Y Y + Therefore the equation becomes Y e Y e + Thus yt L { e + The inverse Lalace of the first two terms is easy, for the other two we use the artial fractions decomosition: Y e e + Thus yt L { e + sint + cost L { + + t Ht + sint + cost cost t Ht + sint + cost cost Ht + sint + cost + cost Ht { + sint + cost, t,; sint, t We again surrise surrise! use the Lalace transform denoting Y L{y: L { t y + ys ds L{6 L{y + L { t ys ds 6L{ Y + Y 6 Y + Y 6 Thus Y and therefore Y Consequently yt L { { { + 3L + + L + 3sint + cost, t Again, the solution robably works for all t IR 3 How do we transform the right-hand side denoted by ft? By definition: L{ft / sinte t dt This is not an easy integral We try another way and write ft sint Ht H ] t Then

5 EM Solved roblems Lalace & Fourier transform c Habala 3 L{ft L{sintHt L { sinth t + e L { sin s + Hs + e L{cossHs Thus the whole equation transforms as Y Y + e + Y Partial fractions decomosition yields Thus Consqeuently yt L { , + e e Y e e 9 et cost sint L { t / H t 9 et cost sint e t cos t + sin t H t 9 et cost sint e t sint cost H t { 9 et cost sint, t, ; 9 e e t, t We denote the function on the right by ft and first calculate its Lalace transform One ossibility is the definition, using integration by arts: ] L{f te t dt t e t e t dt e e + + Alternatively, one can write ft t Ht t Ht and use the grammar: L{f L{t L{tHt e L{s + Hs e + Either way, alying the Lalace transform to both sides of the given equations and denoting L{x X we get X + ] X e e X e + e + Partial fractions decomosition yields + + and We substitute into X and get X e Thus using the inverse Lalace transform and a bit of grammar xt t L { t Ht t t Ht { t, t,];, elsewhere 5 Since we want a general solution, we need to introduce arameters, namely we need two, a and b On the other hand, we need to know two initial conditions to be able to aly LT to y Thus we decide to set y + a, y + b Transforming the equation and using L{e t sint L{sint leads to 8 Y a b] + Y + Y a We aly artial fractions: A+B + + C+D +5 Thus yt L { a + + L { a+ + + b+ + b b A, B, C, D a + cost + b + sint e t L { + + e t L { + Acost + B sint e t cost + e t sint, t We check that this solution in fact works on IR For solutions using classical aroach see Solved roblems ODE of order

6 EM Solved roblems Lalace & Fourier transform c Habala 3 6 a The Fourier series: T, ω T a T ft dt t dt + dt 3 ; a k T ft coskωt dt t cost dt + cost dt ] ] t sint sint dt + sint sin + ] cost + sin sin k cos cos + k k b k T ft sinkωt dt t sint dt + sint dt ] ] t cost + cost dt + cost cos cos cos + k + Thus f 3 + ] sint k sin sin k k k k cost sint] On the left: Periodic extension of f On the right: The sum of this series by the Jordan criterion b The sine Fourier series: T, ω T a a n ; b k T ft sinkωt dt t sin k t dt + sin k t dt t cos k t] + cos k t dt + cos k t] cos k + sin k t] cos cos k cos k + k sin k Thus f k sin k k sin k] sin k t k cos k k sin k k On the left: Odd eriodic extension of f On the right: The sum of this series by the Jordan criterion b The cosine Fourier series: T, ω T b n ; a T ft dt t dt + dt 3 ; a k T ft coskωt dt t cos k t dt + cos k t dt t sin k t] sin k t dt + sin k cos k t] + sin k t] sin sin k

7 EM Solved roblems Lalace & Fourier transform c Habala 3 sin k + k cos k cos sin k k cos k Thus f 3 + k k cos k ] cos k t Even eriodic extension of f, it is also the sum of this series by the Jordan criterion