The Laplace Transform - Iit.edu
The Laplace Transform Let f be a function. Its Laplace transform (function) is denotedby the corresponding capitol letter F. Another notation is L(f ). Input to the given function f is denoted by t; input to its Laplacetransform F is denoted by s. By default, the domain of the function f f(t) is the set of all nonnegative real numbers. The domain of its Laplace transformdepends on f and can vary from a function to a function.1
Definition of the Laplace Transform The Laplace transform F F(s) of a function f f(t) is defined byL(f )(s) F (s) tsef (t) dt.0 The integral is evaluated with respect to t,hence once the limits are substituted, what isleft are in terms of s.2
Example: Find the Laplace transform of the constant functionf (t) 1, 0 t .Solution:F (s) tse0 limb limb limb limb f (t) dt tse(1) dt0b tse0 ts be s0dtprovided s 0.e0e bs s s1e bs s s 3
At this stage we need to recall a limit from Cal 1: xeHence,Thus, 0 0e bs limb s if x .if x if s 0.if s 01F (s) , s 0.sIn this case the domain of the transform is the set ofall positive real numbers.4
Table of Transformsf (t) 1, t 0F (s) 1s , s 0f (t) t , t 0F (s) sn 1 ,f (t) eat , t 0F (s) s 01s a ,s anf (t) sin(kt), t 0f (t) cos(kt), t 0f (t) sinh(kt), t 0f (t) cosh(kt), t 0n!F (s) ks2 k2F (s) ss2 k2F (s) F (s) ks2 k2 ,ss2 k2 ,s k s k 5
The Laplace Transform is LinearIf a is a constant and f and g are functions, thenL(af ) aL(f )L(f g) L(f ) L(g)(1)(2)For example, by the above property (1) 3605!55L(3t ) 3L(t ) 3 6 6 , s 0.ssAs an another example, by property (2)s1 2, s 5.L(e cos(3t)) L(e ) L(cos(3t)) s 5 s 95t5t6
An example where both (1) and (2) are used, 7! 1 , s 0.L(3t 8) L(3t ) L(8) 3L(t ) 8L(1) 3 8 8ss777The Laplace transform of the product of two functionsL(f g) L(f )L(g).As an example, we determineL(3 e )6t 2 L(3 e )(3 e ) L(9 6e e6t6t6t12t)L(9) L(6e6t L(e12t )6t12t9L(1) 6L(e ) L(e )619. s s 6 s 12The respective domains of the above three transformsare s 0, s 6, and s 12; equivalently, s 12.7
The Inverse TransformLea f be a function and L(f ) F be its Laplacetransform. Then, by definition, f is the inversetransform of F. This is denoted by 1L(F ) f.As an example, from the Laplace Transforms Table,we see that6.L(sin(6t)) 2s 36Written in the inverse transform notation 1L 6s2 36 sin(6t).8
Recall that L(tn ) n!sn 1. Hence, for example, L 1 s7!8 t7 .Here, by examining the power 5L 1 s11 .8swe saw that n 7.Now considerHere n 1 11. Hence n 10.Now, we need to make the numerator to be 10!. 1L 5s11 1 5Ls11 5 1 10!L 10!s115 10t . 10! 19
More Examples of Inverse TransformsConsider L 1s k2, 1 2L 7s 15s2 2 . The form of the denominator,is that of the Laplace transforms of sin/cos.7s 15s2 2 7s15 1 L L2s 2s2 2 s1 1 1 15L 7L2s 2s2 2 215 1 7 cos( 2t) L2 2s2 15 7 cos( 2t) sin( 2t)2 110
Partial FractionsConsider the rational expressions3s 53s 5 s2 3s 10(s 5)(s 2)The denominator is factored, and the degree of thenumerator is at least one less that that of thedenominator, in fact, it is exactly one less than thedegree of the denominator.We can, therefore, put the rational expression inpartial fractions. This means for constants A and B,11
we have the decomposition3s 5AB .(s 5)(s 2)s 5 s 2To determine A and B, first clear the denominators:AB3s 5 (s 5)(s 2) (s 5)(s 2) (s 5)(s 2). (s 5)(s 2)(s 5)(s 2) Thus we have the polynomial equality:3s 5 A(s 2) B(s 5) (A B)s 2A 5B.By comparing the coefficients of s and constantcoefficients, we get two equations in A and B.A B 32A 5B 512
We can solve for A and B by using Cramer’s rule 3det5 A 1det2 1 31det2 5 5 , and B .11 1det 52 5Now by the definition of the determinant, adetcHence,bd ad cb.120, and B .A 7713
We can now determine the inverse transform 1L 3s 5(s 5)(s 2) BA L s 5 s 2 11 1 1 BL ALs 5s 220 5t 1 2te e . 77 1This could also have been directly determined byusing a formula from your Table of LaplaceTransforms from the text.This inverse transform will be used in slide #19 tosolve an IVP.14
Partial Fractions: More ExamplesPut3s 4(s 2)(s2 7)in partial fractions.Since s2 7 is a quadratic, when it is put in partialfractions, its numerator must be the general polynomialof degree one.3s 4(s 2)(s2 7) As 2 Bs Ds2 73s 4ABs D 222 (s 2)(s 7) (s 7)(s 2)(s 7) 2 (s 2) 2 s 7s 2 (s 2)(s 7) Hence, we have the equality of polynomials:3s 4 A(s2 7) (Bs D)(s 2) (A B)s2 ( 2B D)s 7A 2DComparing the coefficients of s2, s, and constant coefficients,A B 0, 2B D 3, and 7A 2D 4.15
From the first equation, we get that B -A. Sub in thesecond, to getThen,2A D 3(1)7A 2D 4(2) 2 33 1detdet7 44 2 10 13 , and D .A 11 112 12 1detdet7 27 2Hence,101310, B A , D .A 11111116
Transforms of DerivativesGiven a function y y(t), the transform of itsderivative y can be expressed in terms of theLaplace transform of y: L(y ) sL(y) y(0).The corresponding formula for y can be obtainedby replacing y by y (equation 1 below).L(y ) sL(y ) y (0)(1) s(sL(y) y(0)) y (0)Hence,(2) s2 L(y) sy(0) y (0).(3)L(y ) s L(y) sy(0) y (0). 2 17
Solving IVP with Laplace Transforms 2ty 5y e, y(0) 3.Example 1: Solve the IVPSolution: Taking the Laplace transform of both sides,L(y 5y) L( e 2t )1 L(y ) 5L(y) s 21sL(y) y(0) 5L(y) s 21(s 5)L(y) 3 s 213s 5(s 5)L(y) 3 s 2s 23s 5L(y) (s 5)(s 2)(1)(2)(3)(4)(5)(6)18
Hence, by the definition of the inverse transform,y L 1 3s 5(s 5)(s 2) 20 5t 1 2t e e .77The above inverse transform was found in slide # 14.Example 2: Solve the IVP:y 7y 10e2t , y(0) 0, y (0) 3.Solution: Taking the Laplace transform of both sides,L(y ) 7L(y) 10L(e ) 2ts2 L(y) sy(0) y (0) 7L(y) 010s 2319
Whence,(s 7)L(y) 3 10s 2(s 7)L(y) 3 10s 222Then, y L 1 L(y) 3(s 2) 10s 23s 4(s 2)(s2 7)3s 4(s 2)(s2 7) Now using the.partial fraction decomposition in slide # 15 and 16,y y 10 111 L10 2t11 e 1s 21011 10 111 L cos( 7t) ss2 713 11 7 13 111 L 1s2 7 .sin( 7t).20
Lemma: Let L(y(t)) Y (s). Then L(eat y(t)) Y (s a).Example 1 L(cos(kt)) ss2 k2 .To find L(eat cos(kt))we replace s by s-a. Hence,L(e cos(kt)) ats a(s a)2 k2 .The inverse version is also useful: 1L s a(s a)2 k2 eat cos(kt).Notice matching s-a on the numerator and denominator.The corresponding sine version is 1L k(s a)2 k2 e sin(kt).at21
Find: L 14s 1s2 10s 34 .Here the denominator does not factor over the reals.Hence complete the square.s 10s 34 s 10s 25 25 34 (s 5)2 9. this s must now be4s 14s 1 1 1Ls2 10s 34 L(s 5)2 9made into (s 5). 1 4(s 5) 19 1 4(s 5) 20 1 L L2(s 5)2 9(s 5) 922 4L 4ecos(3t) (s 5)(s 5)2 9 19L 3 1 4e 5t cos(3t) 19L3(s 5)2 9 1 5t19 5t3 e 11(s 5)2 9 sin(3t).22
Lemma:Let L(f (t)) F (s). Then L(tf (t)) F (s).atNowL(e) Example 1:s a. 11atHence, L(te ) s a (s a)2 , s a. 1at teIn other words, L 1 (s a).2 s 2 1).Example 2: Find f (t) L (lns 3 s 2Solution: Now L(f (t)) ln. Then by the Lemma,s 3 s 2 L(tf (t)) ln (ln(s 2) ln(s 3))s 31s a ,Next Slide23
1111 ) .L(tf (t)) (s 2 s 3s 3 s 2Hence,tf (t) L 111 1) L ()(s 3s 2 e eHence,3t 2t3t 2te ef (t) t., t 0.24
Unit Function and Piece-wise Defined FunctionsLet a 0. The Heaviside unit function U(t–a) is defined by 1U (t a) 0if t a.if 0 t a1aThe unit function can be used to express piecewisefunctions.25
Example 1: Let f be the piecewise defined functionf (t) 46if 0 t 8.if t 8Now consider the function 4 (6 4)U (t 8).If 0 t 8, then U(t–8) 0. ThenIf t 8, then U(t–8) 1.4 (6 4)U (t 8) 4.Then4 (6 4)U (t 8) 4 6 4 6.Thus, we see thatf (t) 4 2U (t 8).26
Example 2: Consider the piecewise defined function tf (t) t2 3tif 0 t 2if 2 t 6 .if 6 tWe can express f in terms of unit functions.f (t) t (t t)U (t 2) (t t )U (t 6).232Notice how the coefficients of the unit functions arerelated to the outputs by the piece-wise definedfunction.27
Truncating a Functiony f(t)y g(t)aThe graph of g has been obtained by truncating that of f.g(t) 0f (t)if 0 t a.if t ag(t) f (t)U (t a).28
Truncating a Functiony f(t)aby g(t)y gabThe graph of g has been obtained by truncating that of f. 0g(t) f (t) 0if 0 t aif a t b .if b tg(t) f (t)U (t a) f (t)U (t b).29
Translating and Truncating a Functiony f(t) y f(t-c)cay gy g(t)ba bThe graph of g has been obtained translating thegraph of f by c units to the right and thentruncating it. 0g(t) f (t c) 0if 0 t aif a t b .if b t30
y f(a) 0y(t) f (t a)U (t a) f (t a)if 0 t a.if a taProposition 1 Let a 0 and L(f (t)) F (s). ThenL(f (t a)U (t a)) e as F (s).Example 1: Find L(sin(t 2)U (t 2)).Solution: Here a is 2 and f (t 2) sin(t 2). Weneed f(t) to determine F(s). We can get it from theformula for f(t–2) by replacing t by t 2.f (t 2 2) sin(t 2 2), i.e., f (t) sin(t). Hence,1 2s 2sL(sin(t 2)U (t 2)) F (s)e s2 4 e .31
Example 2: Determine L(t U (t 2)).Solution: Recall the formula2L(f (t a)U (t a)) e as F (s), where F (s) L(f ).2In this case, a 2, and f (t a) f (t 2) t .to obtain F(s), we first need f(t). In order todo thatreplace t in the formula for f(t–2) by t 2.f (t 2) 2tf (t 2 2) (t 2) t 4t 4f (t) t2 4t 4F (s) L(t2 ) 4L(t) 4L(1)244 2 , s 0.3sss2L(tU (t 2)) Thus 2s32 4s2 2 4se 2s , s 0.32
πLsin(Example 3: Determine2 t)U (t 3) .Solution: Comparing U (t 3) with U (t a), we get a 3.Hence, f (t 3) sin( π2 t). Now to obtain f (t), replace tby t 3 in the formula for f (t 3). Then,f (t 3 3) πsin( 2 (t 3)) πsin( 2 t π2 3)).Now an elementary trig identity statessin(3 π2 θ) cos(θ).Whence, f (t) cos( π2 t). Thus F (s) 2 s π2 .s 4L πsin( 2 t)U (t 3) e 3ssπ22s 4.33
Let 1L(F (s)) f (t).ThenL 1 (e as F (s)) U (t a)f (t a). s 1 cos(2t).Example 1: Recall L2s 4Hence 1L s 7ses2 4 cos(2(t 7))U (t 7).The presence of e 7s caused two changes to cos(2t) :the input t was replaced by t 7 and then cos(2(t 7))was multiplied by U (t 7).34
Example 2: Determine L 1 7se(s 5)2 4 .First consider the inverse without the factor e 7s . 1LThus, 1L 7s1(s 5)2 4e(s 5)2 4 e sin(2t).5tReplace t by t–7. 5(t 7) esin(2(t 7)) U (t 7).35
ConvolutionsLet f and g be functions. The convolution of f withg is defined by t f (τ )g(t τ ) dτ.f g(t) 0Thus in a convolution integral, in general, you will see aτ factor (the t in the output by f replaced by τ ), and at τ factor (the t in the output by g replaced by t τ ).sin 3t e5t t0sin(3τ ) e dτ. 5(t τ )the τ factor the t τ factor36
Example 1: Let f (t) t2 and g(t) 2t 3. Find f g.Solution:f g(t) t0 t0f (t)g(t τ ) dτ t0τ 2 (2(t τ ) 3) dτ.τ (2t 3 τ ) dτ (2t 3)2 (2t 3) 5t43 t . 12 3 tτ30 4 tτ40 t0 (2t 2τ dτ t33) 3 t03τ dτt4437
Example 2:Express the following integral as a convolution. 0tτ cos(t τ ) dτ.3‘τ ’ factorReplace τ by t toget the first factorof the convolution 0t‘t τ ’ factorReplace t τ by t toget the second factorof the convolutionτ cos(t τ ) dτ t cos(t)3338
Example 3: t0e2(t τ ) τ 3 dτ t3 e2t .Example 4:In the following example ‘t τ ’ factor is missing.That is because the second factor of the convolution t 3was a constant.3τ dτ t 1. t τ t τ t t 2τ ttdτ 0 e edτ e e .Example 5: 0 e0We see that the convolution e e is not the constantfunction 1. Here is an alternative view of the same integral t t0et 2τ dτ t0t t 2τt t 2τeedτ e 0 edτ et (e 2t 1).However, in general f (g h) (f g) (f h).39
Laplace Transform of ConvolutionsThe Laplace transform of the product of two functionsis not equal to the product of the two transforms:L(f g) L(f )L(g).The convolution behaves far better:L(f g) L(f )L(g).Example 1: Without evaluating the integral, findL t 2(t τ ) 3eτ0 dτ .By Example 3 of the previous slideL t 2(t τ ) 3eτ0 dτ L t3 e2t L(t3 )L(e2t ) 3!s4 (s 2) ,s 2.40
Example 2: L Lt t τe0 sin(t τ ) dτt t τee0sin(t τ ) dτ t t τ L e 0 e sin(t τ ) dτ L et (et sin t) ttL(e sint) L(e)L(sin t) Now11s 1 s2 1 ,s 1.The effect of multiplying this input to the Laplacettransform by e is to replace the s in the output bys–1. Hence L e (e sin t) tt11(s 1) 1 (s 1)2 1 1(s 2)((s 1)2 1) ,s 2.41
Example 3: Find LSolution:L t0 t sin(τ ) dτ . ttt sin(τ ) dτ L t 0 sin(τ ) dτ0 L (t(sin t 1))Recall that L(tf (t)) F (s), where L(f (t)) F (s).Now, L ((sin t 1)) L(sin t)L(1) 1 1s2 1 s ,Hence, L (t(sin t 1)) , s 0 1s(s2 1) 3s2 1 s2 (s2 1)2 ,s 0.s 0.42
Integro-differential EquationsExample: Solve: f (t) t tsin(τ)f(t τ)dτ.input to f0Solution: Notice that f (t) t sin t f (t). Now use theLaplace transform to convert the convolution productto regular products.L(f (t)) L(t) L(sin t f (t)) Hence, (1 1s2 1 )L(f (t)) 1s2s2 1 1( s2 1 )L(f (t)) 1s2Thus, L(f (t)) s2 1s4 1s2 i.e.,1s4 .1s2 1s2 1 L(f (t).s2( s2 1 )L(f (t)) 1s2Hence, f (t) L 1 s12 L 1 s14 L 1 s12 1 1 3!3! Ls4 t t36.43
ySolve: (t) 1 sin t ty(τ)dτ,y(0) 0.Take Laplace0 t transforms to get L(y (t)) L(1) L(sin t) L 0 y(τ ) dτ ,sL(y(t)) y(0) 1s 1s2 1 L(y(t) 1),sL(y(t)) 1s 1s2 1 L(y(t))L(1),sL(y(t)) 1s 1s2 1L(y(t))s 1s 1s2 1 ,(s 1s )L(y(t)) 1s 1s2 1 ,(s2 1)L(y(t))s1s 1s2 1 ,sL(y(t)) L(y(t)),sNext Slide44
L(y(t)) s1(s2 1) sL(y(t)) 1(s2 1)y(t) L 1 s1,22(s 1) s 1s,22(s 1)1(s2 1)y(t) sin t 1 L1tsint2 s(s2 1)2 ,45
Using the Dirac’s Delta FunctionSolve the IVPy 4y δ(t 2), y(0) 6.L(y ) 4L(y) L(δ(t 2)) 2sesL(y) y(0) 4L(y) s 2s 2see 6 (s 4)L(y) y(0) sse 2s6e 2se 2s6 L(y) (s 4) s(s 4)(s 4)4s4(s 4)y(t) 6e 4te1 U (t 2) 4 4(t 2)4U (t 2).46
Systems of Equations and Laplace TransformLet x x(t) and y y(t) be functions of t. Supposex 3x y x x y 1(1)t e , x(0) 0, y(0) 0.(2)By taking the Laplace Transforms, we will solve for x andy by converting the equations into two in L(x) and L(y).From (1)From (2)L(x ) 3L(x) L(y ) L(1)L(x ) L(x) L(y ) L(et ). From (3) sL(x) x(0) 3L(x) sL(y) y(0)From (4)(3)(4) sL(x) x(0) L(x) sL(y) y(0) 1s1.s 147
Using x(0) 0 and y(0) 0, and collecting like terms1(s 3)L(x) sL(y) s1(s 1)L(x) sL(y) s 1Using Cramer’s rule, s 3detdets 1 , L(y) L(x) s 3 ss 3 sdetdets 1 ss 1 sL(x) 1s1s 1sss1 s 1s2 3s s2 s, L(y) 1s1s 1s 3s 1s2 s 1 s3s s2 s48
L(x) 1 ss 14s, L(y) s 3s 1 s 1s4s1s 3s 11 , L(y) L(x) 4s 4(s 1)4s(s 1)4s2Hence, x(t) 14 14 et . In L(y) put the first fraction inpartial fractions and distribute the s2 in the second 311to get L(y) 4s s 1 4s 4s12 . Hence,L(y) 1s 1s 1 14s2 .Hence, y 1 e t14 t.The partial fraction calculations:s 34s(s 1) 4A 4B 1As Bs 1and s 3 4A(s 1) 4Bs. 4A 3.Hence,Solve for A and B.49
Periodic FunctionsThis part is repeatedf(t)0tf(2T t)f(T t)T2TT t2T t3TThe graph is made by repeating a beginning part. Thesmallest T 0 such that f(t T) f(t) for all t is calledperiod of the function. The Laplace transform of f is1L(f )(s) 1 e T s T tsef (t) dt.050
Example: Find the Laplace Transform of the periodicfunction f whose graph is given below.12a-14a6aThe period of f is 2a and, i.e., f(t) f(t 2a), for all t andf (t) 1 1if 0 t a.if a t 2aApplying the formula for the transform of a periodic function, we have1L(f (t))(s) 1 e 2as 2a tsef (t) dt,051
1 1 e 2as1 1 e 2as a tsef (t) dt 0 tse01 1 e 2as1 1 e 2as 2a tse f (t) dtaa dt 2a tsea ts a ts 2aee s 0 s a dt1eee s s s s as 2as as 11 as 2as(1 2e e)(1 e 2as ) s52
11 as as(1 e )(1 e ) as ass (1 e )(1 e )2x as1 (1 e as ),s 0 s (1 e as )Recall)e (e e )(1 e(e e ) x x tanh(x) x x x(e e )e (e e )(1 e 2x )x x xWhence,L(f )(s) x xastanh( 2 )s 2x, s 0.53
The Inverse Transform Lea f be a function and be its Laplace transform. Then, by definition, f is the inverse transform of F. This is denoted by L(f) F L 1(F) f. As an example, from the Laplace Transforms Table, we
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Laplace Transform in Engineering Analysis Laplace transforms is a mathematical operation that is used to “transform” a variable (such as x, or y, or z, or t)to a parameter (s)- transform ONE variable at time. Mathematically, it can be expressed as: L f t e st f t dt F s t 0 (5.1) In a layman’s term, Laplace transform is used
Introduction to Laplace transform methods Page 1 A Short Introduction to Laplace Transform Methods (tbco, 3/16/2017) 1. Laplace transforms a) Definition: Given: ( ) Process: ℒ( ) ( ) 0 ̂( ) Result: ̂( ), a function of the “Laplace tr
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