Introduction to Laplace Transforms for Engineers

1 Introduction to Laplace Transforms for Dodson, School of Mathematics, Manchester University1 What are Laplace Transforms , and Why?This is much easier to state than to motivate! We state the definition in two ways,first in words to explain it intuitively, then in symbols so that we can 1 Givenf,a function of time, with valuef(t) at timet,the Laplace transform offisdenoted fand it gives an average value offtaken over all positive values oftsuchthat the value f(s) represents an average offtaken over all possible time intervalsof 2L[f(t)] = f(s) = 0e stf(t)dt,fors >0.( )A short table of commonly encountered Laplace Transforms is given in Section that this definition involves integration of a product so it will involve frequentuse of integration by parts see Appendix Section for a reminder of the formulaand of the definition of an infinite integral like ( ).This immediately raises the question of why to use such a procedure.

2 In fact thereason is strongly motivated by real engineering problems. There, typically we en-counter models for the dynamics of phenomena which depend on rates of change offunctions, eg velocities and accelerations of particles or points on rigid bodies, whichprompts the use of ordinary differential equations (ODEs). We can use ordinary cal-culus to solve ODEs, provided that the functions are nicely behaved which meanscontinuous and with continuous derivatives. Unfortunately, there is much interest inengineering dynamical problems involving functions that input step change or spikeimpulses to systems playing pool is one example. Now, there is an easy way tosmooth out discontinuities in functions of time: simply take an average value overall time. But an ordinary average will replace the function by a constant, so we usea kind of moving average which takes continuous averages over all possible very neatly deals with the discontinuities by encoding them as a smoothfunction of interval amazing thing about using Laplace Transforms is that we can convert a wholeODE initial value problem into a Laplace transformed version as functions ofs,12 Introduction to Laplace Transformssimplify the algebra, find the transformed solution f(s),then undo the transform toget back to the required solutionfas a function , it turns out that the transform of a derivative of a function is a simplecombination of the transform of the function and its initial value.

3 So a calculus prob-lem is converted into an algebraic problem involving polynomial functions, which is one further point of great importance: calculus operations of differentiationand integration are linear. So the Laplace Transform of a sum of functions is thesum of their Laplace Transforms and multiplication of a function by a constant canbe done before or after taking its this course we find some Laplace Transforms from first principles, ie from thedefinition ( ), describe some theorems that help finding more Transforms , then useLaplace Transforms to solve problems involving Finding Laplace TransformsWe have three methods to find f(s) for a givenf(t);From the definition:Here we use ( ) directly: egForf(t) = 1,L[1] = 0e stdt=[ 1se st] 0= (t) =t,L[t] = 0e stt dt=[ 1se stt] 0+ 01se st= (t) =dydt,L[dydt]= 0e stdydtdt=[e sty] 0+ 0se sty dt= y(0)+s y(s).Forf(t) =eat, aconstant,L[eat] = 0e steatdt= 0e (s a)tdt=[ 1s ae (s a)t] 0=1s a, s > a property:There are a number of powerful theorems about the propertiesof Transforms : egL[af+bg] =aL[f] +bL[g]L[3t+ 4] = 31s2+ [cosat+isinat] =L[eiat] by [eiat] =1s ia=ss2+a2+ias2+ , equating real and imaginary parts and using linearityL[cosat] =ss2+ Dodson3L[sinat] =as2+ can apply the convolution property from the table to findL 1[f(s)s].

4 L 1[f(s)] =f(t),andL 1[1s] = 1 =g(t),soL 1[f(s)s]= t0f( )d .From a list:Computer algebra packages like Mathematica, Matlab and Mapleknow Laplace Transforms of all the functions you are likely to encounter,so you have access to these online, and the packages have also an inversionroutine to find a functionffrom a given f .There are books with long listsof Transforms of known functions and compositions of functions; we give somein Section , which you should read through, eg some that are harder tocalculate:L[tn] =n!sn+1, n= 0,1,2, .. ,L[t1/2] =12( s3)1/2,L[t 1/2] =( s)1 Exercises1. Use the definition to prove the Shift Theorem:L[eatf(t)] = f(s a).2. Deduce thatL[e2tcos 4t] =s 2s2 4s+ Check thatL[e 3tt3] =6(s+ 8) Finding inverse Transforms using partial frac-tionsGiven a functionf,oft,we denote its Laplace Transform byL[f] = f; the inverseprocess is written:L 1[ f] = common situation is when f(s) is a polynomial ins,or more generally, a ratioof polynomials; then we use partial fractions to simplify the expressions.

5 Givenan expression for a Laplace transform of the formN/Dwhere numeratorNanddenominatorDare both polynomials ofs, possibly in the form of factors, andNmay be constant; use partial fractions:4 Introduction to Laplace Transforms (i)ifNhas degree equal to or higher thanD, divideNbyDuntil the remainderis of lower degree thanD(ii)For every linear factor like (as+b) inD, write a partial fraction of the formA/(as+b)(iii)For every repeated factor like (as+b)2inDwrite two partial fractions ofthe formA/(as+b) andB/(as+b)2. Similarly for every repeated factor like(as+b)3inDwrite three partial fractions of the formA/(as+b),B/(as+b)2andC/(as+b)3; and so on.(iv)For quadratic factor (as2+bs+c) write a partial fraction (As+B)/(as2+bs+c).For repeated quadratic factors write a series of partial fractions as in (iii), but withnumerators of the form (As+B) and successive powers of the quadratic factor asthe a little more algebra you should in this way be able to write the originalexpression as a sum of simpler Transforms , which are found in your table.

6 You thenadd their inverse Transforms together, to get the inverse of the original Exercises1. Show thatL 1[12s+ 3]=12e 3 Given f(s) =1 +s(s+ 3)(s 2)show that f(s) =25s+ 3+35s thatf(t) =25e 3t+ Given f(s) =1 +ss2+s+ 1complete the square in the denominator to obtain1 +ss2+s+ 1=1 +s(s+12)2+34=(s+12) +12(s+12)2+ ( 32) the Shift Theorem and the table of Transforms to deduceL 1[(s+12) +12(s+12)2+ ( 32)2]=e t/2cos 32t+1 3e t/2sin 1:Solution of the ODE problem2y y= sint, y(0) = 1in that this last expression can be simplified to2 32e t/2cos( 32t 6)by using the phase relationship between the sine and cosine Solving ODEs and ODE SystemsThe application of Laplace Transform methods is particularly effective forlinearODEs with constant coefficients, and forsystemsof such ODEs. To transform anODE, we need the appropriate initial values of the function involved and initialvalues of its derivatives.

7 We illustrate the methods with the following Exercises1. For the ODE problem2dydt y= sint, y(0) = 1.( )(a) obtain the transformed version as2(s y 1) y=1s2+ 1.(b) Rearrange to get y(s) =2s2+ 3(2s 1)(s2+ 1)=A2s 1+Bs+Cs2+ 16 Introduction to Laplace Transforms (c) Show thatA=145, B= 25, C= 15,and take the inverse transform toobtain the final solution to ( ) asy(t) =75et/2 25cost For the system of ODEsdydt dxdt+y+ 2x=et( )dydt+dxdt x=e2t( )Initial data :x(0), y(0) = 1,( )(a) transform to obtain(s y y0) (s x x0) + y+ 2 x=1s 1( )(s y y0) + (s x x0) x=1s 2.( )(b) Rearranging,(s+ 1) y (s 2) x=1s 1+ 1 1 =1s 1( )s y+ (s 1) x=1s 2+ 1 + 1 =2s 3s 2.( )(c) To eliminate y,multiply ( ) bysand ( ) by (s+ 1) then subtract,and deduce as follows((s 1)(s+ 1) +s(s 2)) x=(2s 3)(s+ 1)s 2 ss 1,( ) x(s) =2s3 4s2+ 3(s 1)(s 2)(2s2 2s 1).( )Then, by partial fractions, x(s) =1s 1+1s 2 s 12(s 12)2 ( 32)2 1 3 32(s 12)2 ( 32)2.

8 ( )(d) From the table of Transforms , we can findx(t) asx(t) =et+e2t et/2cosh( 32t) 1 3et/2sinh( 32t).(e) You can findy(t) by differentiating and substitutingdxdtin either of thesystem equations. Quicker here is to subtract the second equation fromthe first to obtain 2dxdt+y+x+ 2x=et e2tsoy(t) = 2dxdt 3x+et trajectory curve of the ODE system inx, y-space, with timeincreasing along the curve from left to right, is shown in Figure 2:Solution of the ODE system to ; here we plot the trajectory of thesystem inx, y-space, with parametertrepresenting time increasing along the curvefrom left to Impulse problemsLaplace transform methods are particularly valuable in handling differential equa-tions involving impulse and step functions. The problem in the Exercise belowrepresents the dynamics of a point, initially at rest, moving away from the originalong they-axis under a constant acceleration of value 10 for 0 t <1 and an extraimpulse acceleration of size 10 is applied att= is like a simple rocket boost,but can you solve it any other way?

9 We use the Dirac impulse function (t a)which is nonzero att=a,but zero elsewhere while having unit total area under it: (t a) = 0 if (t6=a) and (t a)dt= 1.( ) ExercisesConsider the ODE initial value problem given byy = 10 + 10 (t 1), y(0) =y (0) = 0.( )1. Begin by sketching the graph of the acceleration,y ,to show the step Transforming according to the table, to gets2 y sy(0) y (0) =10s+ 10e sso,rearranging y(s) =10s3+10e ss2= 52s3+ 10e s1s23. From the table use the Delay property to deduce thaty(t) = 5t2+ 10(t 1)H(t 1)8 Introduction to Laplace 3:Solution of x+ 3 x+ 2x=H(t), x(0) = x(0) = 0in ( );x(t)isasymptotic tox=12,why is that obvious?4. By interpreting the step functionH(t 1) up to and aftert= 1,show that theimpulse att= 1 produces what you would expect: a discontinuity invelocityatt= the full solution:y(t) = 5t2fort 1 (so herey (t) = 10t)y(t) = 5t2+ 10(t 1) fort >1 (so herey (t) = 10t+ 10).

10 6 Step Input problemsHere we consider the following initial value problem which involves a step inputfunction typical of many control-type problems: x+ 3 x+ 2x=H(t), x(0) = x(0) = 0.( )Here we have a second order ODE representing a system that is at rest until timet= 0,when a unit step inputH(t) is applied; we seek the outputx(t). Exercises1. Transform ( ) to get(s2+ 3s+ 2) x(s) =1s, x(s) =1s(1(s+ 2)(s+ 1))=1s( 1(s+ 2)+1(s+ 1)). Dodson92. Apply the Integration property to obtainx(t) = t0( e 2 +e )d ( )=[12e 2 e ]t0( )=12e 2t e t 12+ 1( )=12e 2t e t+12.( )3. Show that the solutionx(t) is asymptotic tox=12,cf Figure 3; why is thisobvious as a steady state solution?7 Integration by Parts and Infinite IntegralsRecall the product rule for differentiation, wheny=uv:d(uv)dt=vdudt+udvdtSuppose we integrate both sides with respect tot: d(uv)dtdt= vdudtdt+ udvdtdtOn the left, we are integrating the derivative, so we get back to the original function,which isuv.