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ME 130 Applied Engineering AnalysisChapter 5Review of Laplace Transform and Its Applicationsin Mechanical Engineering Analysis(CONDENSED VERSION)Tai-Ran Hsu, ProfessorDepartment of Mechanical EngineeringSan Jose State UniversitySan Jose, California, USA2018 Version1

Chapter Learning Objectives To learn the application of Laplace transform in engineering analysis To learn the required conditions to transform variable or variables in functions byLaplace transform To learn the use of available Laplace transform table for transformation of functionsand the inverse transformation To learn to use partial fraction and convolution methods in inverse Laplacetransforms To learn the Laplace transform for ordinary derivatives and partial derivatives ofdifferent orders To learn how to use Laplace transform method to solve ordinary differentialequations2

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: (5.1)L f t e st f t dt F st 0 where F(s) expression of Laplace transform of function f(t) involving parameter s In a layman’s term, Laplace transform is used to “transform” a variable in a functioninto a parameter So, after the transformation that variable is no longer a variable anymore,but should be treated as a “parameter”, i.e a “constant under a specific condition” This “specific condition” for Laplace transform is: Laplace transform can only be used to transform variables that cover a range from“zero (0)” to infinity, ( ), for instance: 0 t Any variable that does not vary within this range cannot be transformed usingLaplace transform Because time variable t is the most common variable that varies from (0 to ), functions withvariable t are commonly transformed by Laplace transform Lapalce transform is a valuable “tool” in solving: Differential equations for example: electronic circuit equations, and In “feedback control” for example, in stability and control of aircraft systems3

Laplace transform of simple functions:with 0 t :For f(t) t2L f (t ) e0 st t dt e2 st 2t 2 2t 2 2 3 s 2s s 0 2 F (s)3s(a)For f(t) eat with a constant and 0 t :L f t 0 e st e at dt e s a t dt 011e s a s as a0(b)For f(t) Cosωt withω constant and 0 t :L Cos t e st Cos t dt 0 st es sCostSint 22s s 2 20(c)Appendix 1 of the printed notes provides a Table of Laplace transforms of simple functionsFor example, L[f(t)] of polynomial t2 in Equation (a) is Case 3 with n 3 in the Table,exponential function eat in Equation (b) in Case 7, andtrigonometric function Cosωt in Equation (c) in Case 184

Properties of Laplace TransformLaplace transform of functions by integration:is not always easy.Lt f t 0e st f t dt F s (5.1)Laplace transform (LT) Table in Appendix 1 is useful, but does not always have therequired answer for the specific functions Following properties will be useful in finding the Laplace transform for specific functions:1. Linear operators:L[a f(t) b g(t)] a L[f(t)] b L[g(t)]where a, b constant coefficientsExample 5.4:Find Laplace transform of function: f(t) 4t2 – 3Cos t 5e-t with 0 t :By using the linear operator, we may break up the transform into three individualtransformations:L(4t2 – 3Cos t 5e-t) 4L[t2] – 3L[Cos t] 5L[e-t] F(s)Case 3 with n 3HenceCase 18 with ω 1F (s) Case 7 with a -1 from the LT Table83s5 s3 s2 1 s 15

Properties of Laplace Transform – Cont’d2. Shifting property:If the Laplace transform of a function, f(t) is L[f(t)] F(s) by integration or fromthe Laplace Transform (LT) Table, then the Laplace transform of G(t) eatf(t) can beobtained by the following relationship:L[G(t)] L[eatf(t)] F(s-a)(5.6)where a in the above formulation is the shifting factor, i.e. the parameter s inThe transformed function f(t) has been shifted to (s-a)Example 5.5:Perform the Laplace transform on function: F(t) e2t Sin(at), where a constantWe may use the Laplace transform integral to get the solution, or we could get the solutionby using the LT Table with the shifting property:aSince we can find L[ f (t )] L[ Sin at ] 2(Case 17)s a2We may use the shifting property to get the Laplace transform ofF(t) e2t Sin(at), by“shifting the parameter s by 2, orL[ F (t )] L[e 2t Sin at ] a( s 2) 2 a 26

3. Change of scale property:If we know L[f(t)] F(s) either from the LT Table, or by integral, we may find theLaplace transform of function f(at) by the following expression:1 s F a a where a scale factor for the changeL[ f (at )] (5.7)Example 5.6:Perform the Laplace transform of function F(t) Sin3t.Since we know the Laplace transform of f(t) Sint from the LP Table as:L[ f (t )] L[ Sint ] 1 F (s)s2 1We may find the Laplace transform of F(t) using the Change scale property to be:L[ Sin 3t ] 113 3 s 2s2 9 1 3 7

Inverse Laplace TransformWe define the Laplace transform of a function f(t) to be:Laplace transformFrom hereLt f t There are times we need to do:Lt f t to here 00to theree st f t dt F s e st f t dt F s From thereInverse Laplace transform3 Ways to inverse Laplace transform: Use LP Table by looking at F(s) in right column for corresponding f(t) in middle column- chance of success is not very good Use partial fraction method for F(s) rational function (i.e. fraction functionsinvolving polynomials), and The convolution theorem involving integrations8

The Partial Fraction Method for Inverse Laplace Transform The expression of F(s) to be inversed should be in partial fractions as:F (s) P( s)Q( s)where polynomial P(s) is at least one order less than the order of polynomial Q(s) “Break” up the above rational function into summation of “simple fractions” with whichthe corresponding inverse Laplace transforms can be obtained from the LT Tables:F ( s) AnA1A2P( s) . Q( s) s a1 s a 2s an(5.8)where A1,. A2, .An, and a1, a2, .an are constants to be determined bycomparing coefficients of terms on both sides of the equality:AnP( s)A1A2 . Q( s) s a1 s a2s an The inverse Laplace transform of F(s) P(s)/Q(s) becomes:Inverse LT of a fraction function The sum of the Inverse LT of individual fractions 9of the function by partial fractions

Example 5.7:Perform the inverse Laplace transform of the function:F (s) 3s 7P s 2Q s s 2 s 3Solution:We may express F(s) in the following partial fraction form:F (s) ABA s 1 B s 3 3s 73s 7 s 3 s 1 s 2 2 s 3 ( s 3)( s 1) s 3 s 1where A and B are constant coefficientsAfter expanding the above rational function and equating the terms in numerator:3s 7 A(s 1) B(s-3) (A B)s (A – 3B)We may solve for A and B from the simultaneous equations:A 4 and B -1A B 3and A – 3B 7 yieldThus we have:3s 741 s 2 2s 3 s 3 s 1The required Laplace transform is: 3s 7 1 1 1 1 3t t L 1 4LL 4e e s 3 s 1 ( s 3)( s 1) 10

Example 5.8:Perform the inverse Laplace transform:Solution: P s 3s 1 1 L 1 F ( s ) L 1 L s 3 s 2 s 1 Q s We may break up F(s) in the above expression in the form:The polynomial in numeratoris always one order lessthat in the denominator3s 1P s ABs C Q s ( s 1)( s 2 1) s 1s2 1By following the same procedure, we have coefficients A 2, B -2 and C 1, or:3s 122s1 ( s 1)( s 2 1) s 1s2 1 s2 1We will thus have the inversed Laplace transform function f(t) to be:3s 11 1 2 1 2 s 1 tf (t ) L 1 F ( s ) L 1 3LLL 2e 2 Cos t Sin t 222 s 1 s 1 s s s 1 s 1 11

Inverse Laplace Transform by Convolution Theorem (P.151) This method involves the use of integration of expressions involving LT parameter s - F(s) There is no restriction on the form of the expression of s – they can be rational functions,or trigonometric functions or exponential functions The convolution theorem works in the following way for inverse Laplace transform:If we know the following:L-1[F(s)] f(t) and L-1[G(s)] g(t), with stF ( s) e st f (t ) dt and G ( s) 0 e g (t ) dt0most likely from the LP TableThen the desired inverse Laplace transformed: Q(s) F(s) G(s) can be obtained bythe following integrals:L Q( s) L F ( s)G ( s) 1 1 tf ( ) g (t ) d 0(5.9)ORL Q( s) L F ( s)G ( s) 1 1 t0f (t ) g ( ) d (5.10)12

Example 5.9:Find the inverse of a Laplace transformed function with:Q( s ) ss2 a2 2Solution:We may express F(s) in the following expression:Q( s) ss2 a2 2 s1 s2 a2 s2 a2It is our choice to select F(s) and G(s) from the above expression for the integrals inEquation (5.9) or (5.10).s1 andGsF s 2 Let us choose:s a2s2 a2From the LT Table, we have the following:L 1 F s Cos at f t and L 1 G s Sin at g t aThe inverse of Q(s) F(s)G(s) is obtained by Equation (5.9) as: sL 1 s 2 a 2 Sin a(t )t Sin att Cosad 02a2a One will get the same result by using another convolution integral in Equation (5.10), orusing partial fraction method in Equation (5.8)13

Example 5.11:Use convolution theorem to find the inverse Laplace transform: Q ( s ) 1( s 1)( s 2 4)We may express Q(s) in the following form:Q( s) 111 ( s 1)( s 2 4) s 1 s 2 4(a)We choose F(s) and G(s) as:F (s) 1or e t f t s 1andG (s) 11orSin 2t g t s2 42Let us use Equation (5.10) for the inverse of Q(s) in Equation (a):q (t ) t0f (t ) g ( ) d t01 1 e ( t ) Sin 2 d e t2 2 t0e Sin 2 d After the integration , we get the inverse of Laplace transform Q(s) to be:q (t ) 1 t11 t e Sin 2 2 Cos 2 1e Sint Cost e22 2552 1 2 0 10 t14

Laplace Transform of Derivatives (P.153) We have learned the Laplace transform of function f(t) by:Lt f t 0e st f t dt F s We realize the derivative of function f(t):f ' (t ) (5.1)df (t )is also a FUNCTIONdtSo, there should be a possible way to perform the Laplace transform of the derivativesof functions, as long as its variable varies from zero to infinity. Laplace transform of derivatives is necessary steps in solving DEs using Laplace transform By following the mathematical expression for Laplace transform of functions shown inEquation (5.1), we have:L f ' (t ) 0e st f ' (t ) dt Let: 0 df (t ) e st dt dt 0udv uv(5.11) 0 vdu0The above integration in Equation (5.11) can be performed by “Integration-by-parts:”If we let:u e stdu seand stdt df t dv dt 15v f(t)

By substituting the above ‘u’, “du”, “dv” and “v” into the following:L f ' (t ) 0e st f ' (t ) dt df (t ) dte st dt 0 0udv uv(5.11) 0 vdu0We will have:L f ' (t ) 0e st f ' (t ) dt df (t ) st ste st dt ef(t) f(t) sedt 00 dt 0leading to:L f ' (t ) 0 00e st f ' (t ) dt e st f (t ) f (t ) se st dt f (0) s e st f (t )dt f (0) sL f (t ) 0or in a simplified form:L[f’(t)] s L[f(t)] – f(0)(5.12) Likewise, we may find the Laplace transform of second order derivative of function f(t) to be:(5.13)L[f’’(t)] s2 L[f(t)] – sf(0) – f’(0) A recurrence relation for Laplace transform of higher order (n) derivatives of function f(t)may be expressed as:L[fn(t)] snL[f(t)] - sn-1 f(0) – sn-2f’(0) – sn-3f’’(0) - .fn-1(0)(5.14)16

Example 5.12:Find the Laplace transform of the second order derivative of function: f(t) t SintThe second order of derivative of f(t) meaning n 2 in Equation (5.14), or as inEquation (5.13):L[f’’(t)] s2 L[f(t)] – sf(0) – f’(0)We thus have: d 2 f t 2' 0 L sLft sf0 f 2dt f ' t Since(5.13)df t d t Sin t t Cos t Sin tdtdtWe thus have:L f " (t ) s 2 L f (t ) s f (0) f ' (0) s 2 L t Sin t s(t Sin t ) t 0 (t Cos t Sin t ) t 0 s 2 L t Sin t 17

Solution of DEs Using Laplace Transform (P.155) One popular application of Laplace transform is solving differential equations However, such application MUST satisfy the following two conditions: The variable(s) in the function for the solution, e.g., x, y, z, t must coverthe range of (0, ).That means the solution function, e.g., u(x) or u(t) MUST also be VALIDfor the range of (0, ) ALL appropriate conditions for the differential equation MUST be available The solution procedure is as follows:(1) Apply Laplace transform on EVERY term in the DE(2) The Laplace transform of derivatives results in given conditions, such as f(0),f’(0), f”(0), etc. as shown in Equation (5.14)(3) After apply the given values of the given conditions as required in Step (2),we will get an ALGEBRAIC equation for F(s) as defined in Equation (5.1):L f t 0e st f t dt F s (5.1)(4) We thus can obtain an expression for F(s) from Step (3)(5) The solution of the DE is the inverse of Laplace transformed F(s), i.e.,:f(t) L-1[F(s)]18

Example 5.13:Solve the following DE with given conditions:d 2 y (t )dy (t ) t 2 5y(t) eSin t2dtdtwherey(0) 0and0 t (a)(b)y’(0) 1Solution:(1) Apply Laplace transform to EVERY term in the DE: d 2 y (t ) dy (t ) t 25()L L Lyt LeSin t 2 dt dt whereL y (t ) 0(c) y (t ) e st dt Y ( s )Use Equation (5.12) and (5.13) in Equation (c) will result in: s2 Y ( s ) sy (0) y ' (0) 2 sY ( s ) y (0) 5Y ( s ) 11 ( s 1) 2 1 s 2 2 s 2(d)(2) Apply the given conditions in Equation (b) in Equation (d) s 02 1 0Y ( s ) sy (0) y ' (0) 2 sY ( s ) y (0) 5 Y ( s ) 11 ( s 1) 2 1 s 2 2 s 219

(3) We can obtain the expression:s 2 2s 3Y ( s) 2s 2s 2 s 2 2s 5 (e)(4) The solution of the DE in Equation (a) is the inverse Laplace transform ofY(s) in Equation (e), i.e. y(t) L-1[Y(s)], or: s 2 2s 3y t L Y s L 2 2s2s2s2s5 1 1 (5) The inverse Laplace transform of Y(s) in Equation (e) is obtained by using either“Partial fraction method” or “convolution theorem.” The expression of Y(s) can beshown in the following form by “Partial fractions:”121121Y (s) 2 3 2 3 2 23 s 2s 2 3 s 2s 5s 2s 2 s 2s 5The inversion of Y9s) in the above form is:y (t ) L 1 [Y ( s )] 1 1 12 1 112 1 1 1 1 L 2LLL 222 s 2s 5 33 s 2s 2 33 ( s 1) 4 ( s 1) 1 Leading to the solution of the DE in Equation (a) to be:y (t ) L 1 Y ( s ) 1 t2 1 t1e Sin t e Sin 2t e t Sin t Sin 2t 33 2320

Chapter 6Review of Fourier Series and Its Applicationsin Mechanical Engineering AnalysisCONDENSED VERSIONChapter Learning Objectives:1) Fourier series is a mathematical expression used to describe PERIODICALPHENOMENA in real world2) Learn how to derive Fourier series with the function that represents the oneperiod of a given periodical phenomenon3) How Fourier series CONVERGE (in other words, how many terms in the“infinite series” are required in the Fourier series to “converge” to the valuesof the function that represents the value of the given Periodic Phenomenon.4) How Fourier series converge the values of a given non-continuous, orpiece-wise continuous functions in given periods21

Periodic Physical Phenomena:Motions of poniesForces on the needle22

Machines with Periodic Physical PhenomenaA stampingmachine involvingcyclic punchingof sheet metalsSheet metalMass, Mx(t)ElasticfoundationIn a 4-stroke internalcombustion engine:Cyclic gas pressureson cylinders,and forces on connectingrod and crank shaft23

Mathematical expressions for periodical signals from an oscilloscopeby Fourier series:24

FOURIER SERIES – The mathematicalrepresentation of periodic physical phenomena Mathematical expression for periodic functions: If f(x) is a periodic function with variable x in ONE period 2L Then f(x) f(x 2L) f(x 4L) f(x 6L) f(x 8L) . f(x 2nL)where n any integer numberf(x)Period: ( -π, π) or (0, 2π)-3π-2π-π0π2π3πx(a) Periodic function with period (-π, π)Period 2L:-3Lt-4L -2L-L t-2Lf(t)0Lt2L(b) Periodic function with period (-L, L)3Lt25

Mathematical Expressions of Fourier Series (P.163) Required conditions for Fourier series: The mathematical expression of the periodic function f(x) in one periodmust be available The function in one period is defined in an interval (c x c 2L)in which c 0 or any arbitrarily chosen value of x, and L half period The function f(x) and its first order derivative f’(x) are either continuousor piece-wise continuous in c x c 2L The mathematical expression of Fourier series for periodic function f(x) is:a0 n xn x a n Cos bn Sinf ( x) f x 2 L f x 4 L .L2 n 1 L (6.1)where ao, an and bn are Fourier coefficients, to be determined by the following integrals:1an L1bn L c 2 Lc c 2 Lcn xdxLn xf ( x) SindxLf ( x) Cosn 0,1, 2, 3,.n 1, 2, 3,.Examples 6.1-6.4 for continuous functions.Examples 6.5 for piece-wise continuous function(6.2a)(6.2b)26

Special Example (Problem 6.4 and Problem (3) of Final exam S09)Derive a function describing the position of the sliding block M in one period in a slidemechanism as illustrated below. If the crank rotates at a constant velocity of 5 rpm.(a) Illustrate the periodic function in three periods, and(b) Derive the appropriate Fourier series describing the position ofthe sliding block x(t) in which t is the time in minutes27

Solution:(a) Illustrate the periodic function in three periods:Determine the angular displacement of the crank: t 5t Based on one revolution (θ 2π) corresponds12 to 1/5 min. We thus have θ 10πtFor N 5 rpm, we have:5xRθABXt)ROne half revolution(one stroke)Dead-end B:Dead-end A:x 0x 2Rt 0 mint 1/10 minPosition of the sliding block along the x-direction can be determined by:x R – RCosθx(t) R – RCos(10πt) R[1 – Cos(10πt)]0 t 1/10 minorSTRONG Recommendation:Make sure you know how to derive this function x(t)corresponding to the angular position of the crank!!28

We have now derived the periodic function describing the instantaneous position of thesliding block as:x(t) R[1 – Cos(10πt)]RAx2,4,6,.strokesxθ(a)0 t 1/5 min1,3,5,. strokesBHalf revolutionDead-end A:x 2Rt 1/10 minDead-end B:x 0t 0 minGraphical representation of function in Equation (a) can be produced as:x(t)x(t) R[1 – Cos(10πt)]2RR0Θ Time t 0π/2Time, t (min)π 3π/2 2π1/201/10 minOne revolution2nd period(one period)3rd period29

(b) Formulation of Fourier Series:We have the periodic function: x(t) R[1 – Cos(10πt)] with a period: 0 t 1/10 minIf we choose c 0 and period 2L 1/10 (L 1/20), we will have the Fourier series expressedin the following forms by using Equations (6.1) and (6.2a,b):with aon tn t x t a n Cos b n Sin2 n 1 L 1 / 20L 1 / 20 ao a n Cos 20n t b n Sin 20n t 2 n 111 10an x t Cos 20n tdt1 020(b)(c)We may obtain coefficient ao from Equation (c) to be ao 0:The other coefficient bn can be obtained by:1100b n 20 x t Sin 20n t dt 20 1 / 100R 1 Cos10 t Sin 20n t dt(d)Make sure that you know how to obtain the integrals in Equations (c) and (d)30

Convergence of Fourier SeriesWe have learned the mathematical representation of periodic functions by Fourier seriesIn Equation (6.1):a0 n xn x f ( x) a n Cos bn Sin f x 2 L f x 4 L .2 n 1 L L(6.1)This form requires the summation of “INFINITE” number of terms, which is UNREALISTIC.The question is “HOW MANY” terms one needs to include in the summation in order toreach an accurate representation of the required periodic function [i.e.

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

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