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# CSUN ME 501A - Introduction to Laplace Transforms

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1Introduction to Laplace TransformsIntroduction to Laplace TransformsLarry CarettoMechanical Engineering 501ABSeminar in Engineering AnalysisOctober 19, 20042Outline• Review last class and homework• Definitions of Laplace transforms • Getting a transform by integration• Finding transforms (and inverse transforms) from tables• Linearity of transforms• Applications to differential equations3Review last week• Power series solution of differential equations and Frobenius method– Indicial equation roots– Second solution depends on nature of indicial equation roots– Use to find solution to Bessel’s Equation• Bessel solutions: y = A Jν(x) + B Yν(x)• Find Bessel Functions in tables or compute in Excel and Matlab4Tonight’s Homework• Page 205, problem 5, power series method solution yieldsL+−−+−−+−−+−+=400130012001010)(241415)(667)(223)(xxaaxxaaxxaaxxaay)2)(1(2)1(312++−+=++mmaamammm• First few terms are• Note that we get a0= y(x0) and a1= y’(x0)5Tonight’s Homework II• Page 216, problem 9, define z = x + 2 to get from of Frobenius method y’’ + y’/z -y/z2= 0 • Page 232, problem 7, Use y = ux½and z = kx2/2 to show that y’’ + k2x2y = 0 reduces to Bessel’s equation.–Get x½u’’ + 2(1/2)x-½u’ + k2x2 ux½= 0– Problem is getting z as independent variable6Tonight’s Homework III• General rule for transforming derivatives with one independent variabledxduufanduufwithdzudfdxdzdxudf=== )()()()(dzdukzdzdudxdzdxdu21)2(==⎥⎦⎤⎢⎣⎡=⎟⎠⎞⎜⎝⎛=⎟⎠⎞⎜⎝⎛=dzdukzdzddxdzdxdudzddxdzdxdudxddxud2122)2(• Second derivative• First derivative is relatively easy27Tonight’s Homework IV• Full steps for second derivativedzdukdzudkzdzduzkdzudkzkzdzdukzdzdkzdzdukzdzddxdzdxdudzddxdzdxdudxddxud+=⎥⎥⎦⎤⎢⎢⎣⎡+=⎥⎦⎤⎢⎣⎡=⎥⎦⎤⎢⎣⎡=⎟⎠⎞⎜⎝⎛=⎟⎠⎞⎜⎝⎛=−2221212221212121212222)2()2()2()2()2()2(8Series Solution Problem• Page 205 problem 5, get power series solution to y’’ – 3y’ + 2y = 0.0)-(2)-(3)-()1(00010020=+−−∑∑∑∞=∞=−∞=−nnnnnnnnnxxaxxnaxxann∑∑∑∞=+∞=−∞=−++=−++=−002220020)-()2)(1()-()1(00)-()1(mmmnnnnnnxxammxxannxxann9Series Solution Problem II• Rearrange y’ sum∑∑∑∞=+∞=−∞=−+=+=001110010)-()1()-(0)-(mmmnnnnnnxxamxxnaxxna• Reassemble individual revised sums0)-(2)-()1(3)-()2)(1(00001002=++−++∑∑∑∞=∞=+∞=+mmmmmmmmmxxaxxamxxamm• Get common sum[]0)-(2)1(3)2)(1(0012=++−++∑∞=++mmmmmxxaamammMust equal zero10Laplace Transform Definition∫∞−==ℑ0)()()]([ sFdttfetfst• Transforms from a function of time, f(t), to a function in a complex space, F(s), where s is a complex variable• The transform of a function, is written as F(s) = L[f(t)] where L denotes the Laplace transform (use with equation editor)• Laplace transform defined as the following integralℑ11Inverse Laplace Transforms• Although not used in practice, the inverse transform formally defined as integral in complex plane∫∞+∞−−−=ℑ=iistdsesFsFtfσσ)()]([)(1• In this integral σ is selected so that the integral exists • This is entire part of the complex plane to the left of the real value, σ∫∞−0)( dttfetσ12Laplace Transform Practice• We can find simple transforms, F(s) by integrating the Laplace transform defi-nition, e.g., f(t) = e-atwith Re(s + a) > 0• In practice, we use tables of transform pairs to find F(s) for given f(t) and to find the inverse, f(t) for given F(s)• Examples on next chart∫∞∞+−−−−+=+−==ℑ00)(1][asasedteeetasatstat313Simple Laplace Transformss/(s2- ω2)cosh ωtω/(s2- ω2)sinh ωtAdditional transforms in Table 5.9, pp 297-299 of Kreyszigs/(s2+ ω2)cos ωtω/(s2+ ω2)sin ωteatcos ωt1/(s – a)eatΓ(x+1)/sx+1txeatsin ωtn!/sn+1tnF(s)f(t)F(s)f(t)22)(ωω+− as22)()(ω+−−asas14Laplace Transforms are Linear• This gives two practical results that can be combined for table use– L [f1(t) + f2(t)] = L [f1(t)] + L [f2(t)] – L [af(t)] = a L [f(t)] (a and b constants)– L [af1(t) + bf2(t)] = aL [f1(t)] + bL [f2(t)] • Example: what is L[3cos(2t) – 4e-5t]?• From linearity: 3L[cos(2t)] – 4L[e-5t]• Find transforms of cos(ωt) and eatfrom tables15Linear Property Continued• Last chart showed that L[3cos(2t) –4e-5t] = 3L[cos(2t)] – 4L[e-5t]• Transform table gives22][cos1][ωω+=ℑ+=ℑ−sstandaseat5443]4[4]2[cos3]42cos3[255+−+=ℑ−ℑ=−ℑ−−sssetettt:Conclusion16First Shifting Theorem• Importance: used to get transforms of functions multiplied by exponentials• Derivation: from definition of transform; replace s by s – a • Simple example: what is L[e2tcos3t]– Could get this from table as (s – 2) / [(s – 2)2–32], but use as example of shifting theorem∫∫∞−−∞−−===ℑ0)(0)()()()]([ asFdttfedttfeetfetasatstat17Shifting Theorem I Application• Use first shifting theorem to obtain L[e2tcos3t]• For f(t) = cos 3t, Fcos3t(s) = s/(s3+ 9)•For L[e2tcos3t], theorem says we use F(s) for cos 3t, replacing s by s – 2 ∫∫∞−−∞−−===ℑ0)(0)()()()]([ asFdttfedttfeetfetasatstat()922)2(]3cos[23cos2+−−=−=ℑsssFtett18Inverse by Shifting Theorem I• What is f(t) if F(s) = s/(s2– 3s + 2)?• Try to get denominator in form of (s – a)2•For (s –a)2= s2– 2as – a2to match first two terms we must have 2a = 3 or a = 1.5• But (s – 1.5)2= s2– 3s + 2.25 does not match denominator of s2– 3s + 2• But, (s – 1.5)2–0.52= s2– 3s + 2.25 –0.25 = s2– 3s + 2 does match denominator. This gives419Slide 2 on Inverse by Theorem I• F(s) = s/[(s – 1.5)2–0.52] gives s – 1.5 in denominator, but not in numerator• Get s – 1.5 in numerator as follows• Now have consistent use of s = 1.5• Find transforms with s in place of s – 1.52222225.0)5.1(5.15.0)5.1(5.15.0)5.1( −−+−−−=−− sssss221221][sinh][coshβββββ−=ℑ−=ℑ−−stsst20Slide 3 on Inverse by Theorem I• From last chart and shifting theorem2222225.0)5.1(5.15.0)5.1(5.15.0)5.1( −−+−−−=−− sssss⎥⎦⎤⎢⎣⎡−ℑ=⎥⎦⎤⎢⎣⎡−ℑ=−−221221sinhcoshβββββstsst)()]([asFtfeat−=ℑ()()5.05.15.05.15.05.0sinh5.05.15.0cosh2215.12215.1⎭⎬⎫⎩⎨⎧⎥⎦⎤⎢⎣⎡−−ℑ=⎥⎦⎤⎢⎣⎡−−ℑ=−−stesstett21Result of Inverse by Theorem I• Our original F(s) was s/(s2–3s + 2)• We showed that this was equal to the following• Taking the

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