Section 5 Laplace Transforms
SECTION 5: LAPLACETRANSFORMSMAE 3401 – Modeling and Simulation
2Introduction – TransformsThis section of notes contains an introductionto Laplace transforms. This should mostly be areview of material covered in your differentialequations course.K. WebbMAE 3401
Transforms3 What is a transform?A mapping of a mathematical function from one domain toanother A change in perspective not a change of the function Why use transforms? Some mathematical problems are difficult to solve in theirnatural domainTransform to and solve in a new domain, where the problem issimplified Transform back to the original domain K. WebbTrade off the extra effort of transforming/inverse‐transforming for simplification of the solution procedureMAE 3401
Transform Example – Slide Rules4 Slide rules make use of a logarithmic transform Multiplication/division of large numbers is difficultTransform the numbers to the logarithmic domain Add/subtract (easy) in the log domain to multiply/divide(difficult) in the linear domain Apply the inverse transform to get back to the originaldomain Extra effort is required, but the problem is simplifiedK. WebbMAE 3401
5K. WebbLaplace TransformsMAE 3401
Laplace Transforms6 An integral transform mapping functions from the timedomain to the Laplace domain or s‐domain Time‐domain functions are functions of time, Laplace‐domain functions are functions of K. Webbis a complex variableMAE 3401
Laplace Transforms – Motivation7 We’ll use Laplace transforms to solve differentialequations Differential equations in the time domain Apply the Laplace transform Transform to the s‐domainDifferential equations become algebraic equations difficult to solveeasy to solveTransform the s‐domain solution back to the time domainTransforming back and forth requires extra effort, butthe solution is greatly simplifiedK. WebbMAE 3401
Laplace Transform8 Laplace Transform:(1) Unilateral or one‐sided transform Lowerlimit of integration is Assumed that the time domain function is zero for allnegative time, i.e.K. WebbMAE 3401
9Laplace Transform PropertiesIn the following section of notes, we’ll derive afew important properties of the Laplacetransform.K. WebbMAE 3401
Laplace Transform – Linearity10 Say we have two time‐domain functions:and Applying the transform definition, (1) (2) The Laplace transform is a linear operationK. WebbMAE 3401
Laplace Transform of a Derivative11 Of particular interest, given that we want to use Laplacetransform to solve differential equations Use integration by parts to evaluate LetandthenandK. WebbMAE 3401
Laplace Transform of a Derivative120 00The Laplace transform of the derivative of afunction is the Laplace transform of that functionmultiplied by minus the initial value of thatfunction(3)K. WebbMAE 3401
Higher‐Order Derivatives13 The Laplace transform of a second derivative is0 0In general, the Laplace transform of theof a function is given by0K. Webb0 (4)derivative0(5)MAE 3401
Laplace Transform of an Integral14 The Laplace Transform of a definite integral of afunction is given by(6) Differentiation in the time domain corresponds tomultiplication by in the Laplace domainIntegration in the time domain corresponds todivision by in the Laplace domainK. WebbMAE 3401
15Laplace Transforms of CommonFunctionsNext, we’ll derive the Laplace transform ofsome common mathematical functionsK. WebbMAE 3401
Unit Step Function16 A useful and common way of characterizing a linearsystem is with its step response The system’s response (output) to a unit step inputThe unit step function or Heaviside step function:K. WebbMAE 3401
Unit Step Function – Laplace Transform17 Using the definition of the Laplace transform111 011The Laplace transform of the unit step(7) Note that the unilateral Laplace transform assumes thatthe signal being transformed is zero for K. WebbEquivalent to multiplying any signal by a unit stepMAE 3401
Unit Ramp Function18 The unit ramp function is a useful input signal forevaluating how well a system tracks a constantly‐increasing inputThe unit ramp function:K. WebbMAE 3401
Unit Ramp Function – Laplace Transform19 Could easily evaluate the transform integral Requires Alternatively, recognize the relationship betweenthe unit ramp and the unit step Unit integration by partsramp is the integral of the unit stepApply the integration property, (6)11 1 (8)K. WebbMAE 3401
Exponential – Laplace Transform20 Exponentials are common components of theresponses of dynamic systems01(9)K. WebbMAE 3401
Sinusoidal functions21 Another class of commonly occurring signals, whendealing with dynamic systems, is sinusoidal signals –bothand Recall Euler’s formula From which it follows thatK. WebbMAE 3401
Sinusoidal functions2212sin121212121210211021122(10)K. WebbMAE 3401
Sinusoidal functions23 It can similarly be shown that(11) Note that for neithernoris thefunction equal to zero foras the Laplacetransform assumesReally, what we’ve derived is1K. Webb sinand1 cosMAE 3401
24K. WebbMore Properties and TheoremsMAE 3401
Multiplication by an Exponential,25 We’ve seen that What if another function is multiplied by thedecaying exponential term? This is just the Laplace transform ofreplaced bywith(12)K. WebbMAE 3401
Decaying Sinusoids26 The Laplace transform of a sinusoid issin And, multiplication by an decaying exponential,, results in a substitution offor , soandK. WebbMAE 3401
Time Shifting27 Consider a time‐domainfunction, To Laplace transformwe’ve assumed0 for0, or equivalentlymultiplied by 1 To shiftby an amount,, in time, we must alsomultiply by a shifted stepfunction, 1K. WebbMAE 3401
Time Shifting – Laplace Transform28 The transform of the shifted function is given by 1 Performing a change of variables, letand The transform becomes 1 A shift by in the time domain corresponds to multiplication byLaplace domain 1K. Webbin the(13)MAE 3401
Multiplication by time,29 The Laplace transform of a function multiplied by time:(14) Consider a unit ramp function: 1 11Or a parabola: In general!K. WebbMAE 3401
Initial and Final Value Theorems30 Initial Value Theorem Can determine the initial value of a time‐domain signal orfunction from its Laplace transform0 lim (15)Final Value Theorem Can determine the steady‐state value of a time‐domainsignal or function from its Laplace transform K. Webblim (16)MAE 3401
Convolution31 Convolution of two functions or signals is given by Result is a function of time is flipped in time and shifted byMultiply the flipped/shifted signal and the other signalIntegrate the result from0 May seem like an odd, arbitrary function now, but we’ll latersee why it is very important Convolution in the time domain corresponds to multiplicationin the Laplace domain K. Webb(17)MAE 3401
Impulse Function32 Another common way to describe a dynamic systemis with its impulse response System output in response to an impulse function inputImpulse function defined by0,01 Aninfinitely tall, infinitelynarrow pulseK. WebbMAE 3401
Impulse Function – Laplace Transform33 To derive, consider the following function10, Can think of,00 oras the sum of two step functions:1 11The transform of the first term is1 111Using the time‐shifting property, the second term transforms to1K. Webb1MAE 3401
Impulse Function – Laplace Transform34 In the limit, as 0, , solim lim1 Apply l’Hôpital’s rulelim1lim The Laplace transform of an impulse function is one1K. Webb(18)MAE 3401
Common Laplace Transforms35sin111cos10!01011 1sincosK. Webb MAE 3401
36Inverse Laplace TransformWe’ve just seen how time‐domain functions can betransformed to the Laplace domain. Next, we’ll look athow we can solve differential equations in the Laplacedomain and transform back to the time domain.K. WebbMAE 3401
Laplace Transforms – Differential Equations37 Consider the simplespring/mass/damper system fromthe previous section of notesState equations are:(1)(2) Taking the displacement of the mass as the output(3) Using (2) and (3) in (1) we get a single second‐order differentialequation(4)K. WebbMAE 3401
Laplace Transforms – Differential Equations38 We’ll now use Laplace transforms todetermine the step response of the system1N step force input0 ,1 ,1 1 (5)For the step response, we assume zero initial conditions0 000 and00(6)Using the derivative property of the Laplace transform, (4)becomes0001(7)K. WebbMAE 3401
Laplace Transforms – Differential Equations39 The input is a step, so (7) becomes1 (8)Solving (8) for11/ (9)Equation (9) is the solution to the differential equation of (4),given the step input and I.C.’s K. WebbThe system step response in the Laplace domainNext, we need to transform back to the time domainMAE 3401
Laplace Transforms – Differential Equations40/ The form of (9) is typical of Laplacetransforms when dealing with linearsystems (9)A rational polynomial inHere, the numerator is 0th‐orderRoots of the numerator polynomial,the functionRoots of the denominator polynomial,of the functionK. Webb, are called the zeros of, are called the polesMAE 3401
Inverse Laplace Transforms41/ To get (9) back into the time domain, weneed to perform an inverse Laplacetransform (9)An integral inverse transform exists, but we don’t use itInstead, we use partial fraction expansionPartial fraction expansion Idea is to express the Laplace transform solution, (9), as a sum ofLaplace transform terms that appear in the tableProcedure depends on the type of roots of the denominator polynomial K. WebbReal and distinctRepeatedComplexMAE 3401
Inverse Laplace Transforms – Example 142 Consider the following system parameters11610 Laplace transform of the step response becomes(10) Factoring the denominator(11) In this case, the denominator polynomial has three real, distinct roots0,K. Webb2,8MAE 3401
Inverse Laplace Transforms – Example 143 Partial fraction expansion of (11) has the form(12) ,, and, areCan already see the form of the time‐domain function The numerator coefficients,called residuesSum of a constant and two decaying exponentialsTo determine the residues, multiply both sides of (12) by the denominator ofthe left‐hand side121108816282Collecting terms, we have1K. Webb108216(13)MAE 3401
Inverse Laplace Transforms – Example 144 Equating coefficients of powers of on bothsides of (13) gives a system of three equationsin three unknowns:0:108:16120Solving for the residues gives0.06250.08330.0208 The Laplace transform of the step response is. .(14)Equation (14) can now be transformed back to the time domain using theLaplace transform tableK. WebbMAE 3401
Inverse Laplace Transforms – Example 145 The time‐domain step response of the system is the sum of a constantterm and two decaying exponentials:0.06250.0833 Step response plotted in MATLAB Characteristic of a signal havingonly real poles 0.0208(15)No overshoot/ringingSteady‐state displacement agreeswith intuition K. Webb1 force applied to a 16 /springMAE 3401
Inverse Laplace Transforms – Example 146 Go back to (10) and apply the initial valuetheorem0limlim 110160 Which is, in fact our assumed initialcondition Next, apply the final value theorem to the Laplace transform stepresponse, (10) lim110lim 0.0625166.25This final value agrees with both intuition and our numerical analysisK. WebbMAE 3401
Inverse Laplace Transforms – Example 247 Reduce the damping and re‐calculate the stepresponse1168 Laplace transform of the step response becomes(16) Factoring the denominator(17) In this case, the denominator polynomial has three real roots, two of whichare identical0,K. Webb4,4MAE 3401
Inverse Laplace Transforms – Example 248 Partial fraction expansion of (17) has theform(18) Again, find residues by multiplying both sides of (18) by the left‐hand side denominator14184164Collecting terms, we have1K. Webb8416(19)MAE 3401
Inverse Laplace Transforms – Example 249 Equating coefficients of powers of on bothsides of (19) gives a system of three equationsin three unknowns:0:8:16401Solving for the residues gives0.06250.06250.2500 The Laplace transform of the step response is. .(20)Equation (20) can now be transformed back to the time domain using theLaplace transform tableK. WebbMAE 3401
Inverse Laplace Transforms – Example 250 The time‐domain step response of the system is the sum of a constant, adecaying exponential, and a decaying exponential scaled by time:0.06250.0625 Step response plotted in MATLAB Again, characteristic of a signalhaving only real poles Similar to the last case A bit faster – slow pole atwas eliminatedK. Webb0. 25(21)2MAE 3401
Inverse Laplace Transforms – Example 351 Reduce the damping even further and gothrough the process once again1164 Laplace transform of the step response becomes(22) The second‐order term in the denominator now has complex roots, so wewon’t factor any further The denominator polynomial still has a root at zero and now has two rootswhich are a complex‐conjugate pair0,K. Webb23.464,23.464MAE 3401
Inverse Laplace Transforms – Example 352 Want to cast the partial fraction terms intoforms that appear in the Laplace transformtableSecond‐order terms should be of the form(23) This will transform into the sum of damped sine and cosine termscos sinTo get the second‐order term in the denominator of (22) into the form of(23), complete the square, to give the following partial fraction expansion.K. Webb(24)MAE 3401
Inverse Laplace Transforms – Example 353 Note that the and terms in (23) and(24) are the real and imaginary parts ofthe complex‐conjugate denominator roots2, 3.464Multiplying both sides of (24) by the left‐hand‐side denominator,equate coefficients and solve for residues as before:0.06250.06250.0361 Laplace transform of the step response is.K. Webb.(25)MAE 3401
Inverse Laplace Transforms – Example 354 The time‐domain step response of the system is the sum of a constant andtwo decaying sinusoids:0.06250.0625 Step response andindividual componentsplotted in MATLAB Characteristic of a signalhaving complex poles K. Webbcos 3.4640.0361sin 3.464(26)Sinusoidal terms result inovershoot and (possibly)ringingMAE 3401
Laplace‐Domain Signals with Complex Poles55 The Laplace transform of the step response in the last example hadcomplex poles A complex‐conjugate pair:Results in sine and cosine terms inthe time domaincos Imaginary part of the roots, Frequency of oscillation of sinusoidalcomponents of the signalReal part of the roots, , sinRate of decay of the sinusoidalcomponentsMuch more on this laterK. WebbMAE 3401
K. Webb MAE 3401 7 Laplace Transforms –Motivation We’ll use Laplace transforms to solve differential equations Differential equations in the time domain difficult to solve Apply the Laplace transform Transform to the s‐domain Differential equations becomealgebraic equations easy to solve Transfo
and Laplace transforms F(s) Z 0 f(t)e st dt. Laplace transforms are useful in solving initial value problems in differen-tial equations and can be used to relate the input to the output of a linear system. Both transforms provide an introduction to a more general theory of transforms, which are u
Exact Differential Equations . – The definition of the Laplace transform. We will also compute a couple Laplace transforms using the definition. Laplace Transforms – As the previous section will demonstrate, computing Laplace transforms directl
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
the Laplace transform Laplace transform of the solution Solution L L 1 Algebraic solution, partial fractions Bernd Schroder Louisiana Tech University, College of Engineering and Science Laplace Transforms for Systems of Differential EquationsFile Size: 306KB
No matter what functions arise, the idea for solving differential equations with Laplace transforms stays the same. Time Domain (t) Transform domain (s) Original DE & IVP Algebraic equation for the Laplace transform Laplace transform of the solu
equations with Laplace transforms stays the same. Time Domain (t) Transform domain (s) Original DE & IVP Algebraic equation for the Laplace transform Laplace transform of the solution L L 1 Algebraic solution, partial fractions Bernd Schroder Louisiana Tech University, College of Engineering and Science
the Laplace transform, as shown in Table 5.3, we can deal with many ap-plications of the Laplace transform. We will first prove a few of the given Laplace transforms and show how they can be used to obtain new trans-form pairs. In
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