Introduction to the Laplace Transform

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CHAPTER 1

INTRODUCTION TO THE LAPLACE TRANSFORM

1.1. The Laplace Transform: Preliminary definitions and notation

Pierre-Simon Laplace (1749-1827)

Laplace was a French mathematician, astronomer, and physicist who applied theNewtonian theory of gravitation to the solar system (an important problem of his day). He played aleading role in the development of the metric system. The main idea behind the LaplaceTransformation is that we can solve an equation (or system of equations) containing differentialand integral terms by transforming the equation in "t-space" to one in "s-space". This makes the

problem much easier to solve.In mathematics, the Laplace transform is a widely used integral transform. Denoted

, it is a linear operatorof a functionf(t) with a real argument t(t 0), that transforms itto a function F(s) with a complex argument s. This transformation is essentiallybijective for themajority of practical uses; the respective pairs off(t) andF(s) are matched in tables. The Laplacetransform has the useful property that many relationships and operations over the originals f(t)correspond to simpler relationships and operations over the images F(s).1 The Laplace transformhas many important applications throughout the sciences.

The Laplace transform can be used to solve differential equations. Besides being a different

and efficient alternative to variation of parameters and undetermined coefficients, the Laplacemethod is particularly advantageous for input terms that are piecewise-defined, periodic orimpulsive.

The Laplace transform is related to the Fourier transform, but whereas the Fouriertransform resolves a function or signal into its modes ofvibration, the Laplace transform resolves afunction into its moments. Like the Fourier transform, the Laplace transform is used for solvingdifferential and integral equations. In physics and engineering, it is used for analysis oflinear time-invariantsystems such as electrical circuits, harmonic oscillators, optical devices, and mechanicalsystems. In this analysis, the Laplace transform is often interpreted as a transformation from thetime-domain, in which inputs and outputs are functions of time, to the frequency-domain, wherethe same inputs and outputs are functions ofcomplexangular frequency, in radians per unit time.Given a simple mathematical or functional description of an input or output to a system, theLaplace transform provides an alternative functional description that often simplifies the process ofanalyzing the behavior of the system, or in synthesizing a new system based on a set ofspecifications.

1KORN, G.A.; KORN, T.M. (1967), Mathematical Handbook for Scientists and Engineers (2nd ed.), McGraw-HillCompanies, 8.1

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History

The Laplace transform is named in honor ofmathematician and astronomerPierre-Simon Laplace,who used the transform in his work onprobability theory. From 1744, Leonhard Eulerinvestigatedintegrals of the form

as solutions of differential equations but did not pursue the matter very far. Joseph Louis Lagrangewas an admirer of Euler and, in his work on integratingprobability density functions, investigatedexpressions of the form

which some modern historians have interpreted within modern Laplace transform theory.

These types of integrals seem first to have attracted Laplace's attention in 1782 where he wasfollowing in the spirit of Euler in using the integrals themselves as solutions of equations.However, in 1785, Laplace took the critical step forward when, rather than just looking for asolution in the form of an integral, he started to apply the transforms in the sense that was later tobecome popular. He used an integral of the form:

akin to a Mellin transform, to transform the whole of a difference equation, in order to look forsolutions of the transformed equation. He then went on to apply the Laplace transform in the same

way and started to derive some of its properties, beginning to appreciate its potential power.Laplace also recognised that Joseph Fourier's method ofFourier series for solving the diffusionequation could only apply to a limited region of space as the solutions were periodic. In 1809,Laplace applied his transform to find solutions that diffused indefinitely in space.

Formal definition

The Laplace transform of a functionf(t), defined for all real numbers t 0,and taking complexvalues is the functionF(s), defined by:2

The parameters is a complex number:

with real numbers and .

2 ABELL, Martha; BRASELTON,J.P, Differential Equations with MAPLE V, AP Professional,1994, p.302

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The meaning of the integral depends on types of functions of interest. A necessarycondition for existence of the integral is that must be locally integrable on [0,). For locallyintegrable functions that decay at infinity or are ofexponential type, the integral can be understoodas a (proper) Lebesgue integral. However, for many applications it is necessary to regard it as aconditionally convergent improper integral at . Still more generally, the integral can be

understood in a weak sense, and this is dealt with below.The definition of the Laplace transform is difficult to apply in most cases. Hence, we can

discuss about the linearity property that enables us to use the found transforms to find theLaplace transform of other functions.

THEOREM Linearity Property

Let a and b be complex constants, and suppose that the Laplace

transform of the functionsf andg exists. Then:

L{af(t)+ bg(t)}=aL{f(t)}+bL{g(t)}

Example : Calculate (a)L{6} and (b)L{5-2e-t

}Solution: Using the fact that L{f}(s) = 0 e-stdt =1S, we have:

for (a)

L{6}=6L{1}=g(1S,)= 6S,

And for (b)

L{5-2e-t} = 5L{1}-2 L{ e-t}= 5(1S,)- 2(1S-(-1),))= 5S, - 2S+1,

One can define the Laplace transform of a finite Borel measure by the Lebesgue integra.3

An important special case is where is aprobability measure or, even more specifically, the Diracdelta function. In operational calculus, the Laplace transform of a measure is often treated asthough the measure came from a distribution function. In that case, to avoid potential confusion,one often writes

3FELLER, William (1971),An introduction to probability theory and its applications. Vol. II., Second edition, NewYork: John Wiley & Sons, XIII.1

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where the lower limit of 0 is short notation to mean

This limit emphasizes that any point mass located at 0 is entirely captured by the Laplace

transform. Although with the Lebesgue integral, it is not necessary to take such a limit, it doesappear more naturally in connection with the LaplaceStieltjes transform.

Probability theory

In pure and applied probability, the Laplace transform is defined by means of anexpectation value. IfXis a random variable withprobability density function , then the Laplacetransform of is given by the expectation

By abuse of language, this is referred to as the Laplace transform of the random variable Xitself.

Replacing s by t gives the moment generating function ofX. The Laplace transform hasapplications throughout probability theory, including first passage times of stochastic processessuch as Markov chains, and renewal theory.

Exponential Order, Jump Discontinuities and Piecewise Continuous Functions

In calculus, we saw that in some cases, improper integrals diverge, meaning that the limitof the definite integral does not exist. Hence, we may believe that Laplace transform may not existfor some function. Therefore, we present the following definitions and theorems so that we canbetter understand the types of functions for which the Laplace transform exists.4

DEFINITIONExponential Order

A functionf is of exponential orderb if there are numbers b, M> 0, and T >0, such that

f(t) Mebt, fort>T

Note : This definition is the same as showing that limtxf(t)ebt is bounded.

DEFINITIONJump Discontinuity

A function fhas a jump discontinuity at t=c, on the closed interval [a,b] if the one-sidelimits limtc-ft and limtc+ft are finite, but unequal values. f has a jump discontinuity at t=aif limta+ft is a finite value.fhas a jump discontinuity at t=b if limtb-ft is a finite value

DEFINITIONPiecewise Continuous

A functionf is piecewise continuous on the finite interval [a,b] iffis continuous at everypoint in [a,b]except at finitely many points at whichf has a jump discontinuity.

4 ABELL, Martha; BRASELTON,J.P, Differential Equations with MAPLE V, AP Professional,1994, p.305-306

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A functionf is piecewise continuous on [0,) iff is piecewise continuous on [0,N], for allN>0.

THEOREM Sufficient condition for existence of L{f(t)}

Suppose that f is a piecewise continuous function on the interval [0,) and that it is of

exponential orderb fort>T. ThenL{f(t)} exists fors>b.Example: Find the Laplace transform of

f(t) = -1, 0t41, t>4

Solution: Becausef is a piecewise continuous function on [0,) and of exponential order, L{f(t)}exists. We simply use the definition and evaluate the integral using the sum of two integrals.

L{f(t)} =0xfte-stdt=04(-1)e-stdt4e-stdt= e-stst=4t=0+ limM-e-stst=Mt=4=

= 1s(e-4s-1)- 1slimM(e-Ms-e-4s)= 1s(2e-4s-1)

1.2. Properties and theorems

The Laplace transform has a number of properties that make it useful for analyzing lineardynamical systems. Most of the properties follow directly from our knowledge of integral. One ofthe property is the shifting, or translation property.

THEOREM: Shifting property

IfL{f(t)}=F(s) exists fors>b, then L{eatf(t)} =F(s-a), for s>a+b.

Another property is that the differentiation and integration become multiplication and

division, respectively, bys (similarly to logarithms changing multiplication of numbers to additionof their logarithms). Because of this property, the Laplace variable s is also known as operatorvariable in the L domain: either derivative operator or (fors1) integration operator. Thetransform turns integral equations and differential equations to polynomial equations, which aremuch easier to solve. Once solved, use of the inverse Laplace transform reverts back to the timedomain.

Given the functionsf(t) andg(t), and their respective Laplace transformsF(s) and G(s):

The following table is a list of properties of unilateral Laplace transform:5

5 KORN, G.A.; KORN, T.M. (1967), Mathematical Handbook for Scientists and Engineers (2nd ed.), McGraw-HillCompanies, p. 226227

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Properties of the unilateral Laplace transform

Time domain `s` domain Comment

LinearityCan be proved using basic rulesof integration.

Frequency

differentiation is the first derivative of .

Frequency

differentiation

More general form, nth

derivative of F(s).

Differentiation

is assumed to be adifferentiable function, and itsderivative is assumed to be ofexponential type. This can thenbe obtained by integration byparts

Second

Differentiation

is assumed twicedifferentiable and the secondderivative to be of exponentialtype. Follows by applying theDifferentiation property to

.

GeneralDifferentiation

is assumed to be n-timesdifferentiable, with nth

derivative of exponential type.Follow by mathematicalinduction.

Frequency

integration

6


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Integration

u(t) is the Heaviside stepfunction. Note (u *f)(t) is theconvolution ofu(t) andf(t).

Scaling where a is positive.

Frequency

shifting

Time shiftingu(t) is the Heaviside stepfunction

Multiplicationthe integration is done along thevertical lineRe() = c that liesentirely within the region ofconvergence ofF.

Convolution

(t) andg(t) are extended byzero fort< 0 in the definitionof the convolution.

PeriodicFunction

f(t) is a periodic function ofperiod Tso that

. This is the result of the timeshifting property and thegeometric series.

Initial value theorem:

Final value theorem:

, if all poles ofsF(s) are in the left half-plane.

The final value theorem is useful because it gives the long-term behaviour without havingto performpartial fraction decompositions or other difficult algebra. If a function's polesare in the right-hand plane (e.g. et or sin(t)) the behaviour of this formula is undefined.

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In order to use the Laplace transform to solve differential equations, we will need to be able tocompute the Laplace transform of the derivatives of an arbitrary function, provided the Laplacetransform of such a function exists.

THEOREM:Laplace Transform of the First Derivative

Suppose that f is a piecewise continuous function on the interval [0,) and that it is ofexponential orderb fort>T. Then fors>b,

L{f (t)}=sL{f(t)}-f(0)

Using induction, we can easily verify the following corollary.

COROLLARY:Laplace Transform of Higher Derivatives

More generally, iff(i)(t)is a continuous function on [0,) for i =0,1,.,n-1 and f (n)(t) ispiecewise continuous on [0,) and of exponential orderb , then fors>b,

L{ f(n)(t) }=snL{f(t)}- sn-1f(0)--sf(n-2)(0)-f(n-1(0))

THEOREM: Derivatives of the Laplace Transform

Suppose that F(s)=L{f(t)}, where f is piecewise continuous function on [0,) and ofexponential orderb . Then fors>b,

L{tnf(t)}= (-1)ndn Fdsn(s)

Proof of the Laplace transform of a function's derivative

It is often convenient to use the differentiation property of the Laplace transform to find thetransform of a function's derivative. This can be derived from the basic expression for a Laplacetransform as follows:

yielding

and in the bilateral case,

8


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The general result

wherefn is the n-th derivative off, can then be established with an inductive argument.

Evaluating improper integrals

Let , then (see the table above)

or

Let we get the identity

For example,

1.1. Inverse Laplace transform

The inverse Laplace transform is given by the following complex integral, which is knownby various names (the Bromwich integral, the Fourier-Mellin integral, and Mellin's inverseformula):

where is a real number so that the contour path of integration is in the region of convergence ofF(s). An alternative formula for the inverse Laplace transform is given by Post's inversion formula.

Region of convergence

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If is a locally integrable function (or more generally a Borel measure locally of boundedvariation), then the Laplace transformF(s) of converges provided that the limit

exists. The Laplace transform converges absolutely if the integral

exists (as proper Lebesgue integral). The Laplace transform is usually understood as conditionallyconvergent, meaning that it converges in the former instead of the latter sense.

The set of values for which F(s) converges absolutely is either of the form Re{s} > a orelse Re{s} a, where a is an extended real constant, a . (This follows from thedominated convergence theorem.) The constant a is known as the abscissa of absolute

convergence, and depends on the growth behavior of(t).6

Analogously, the two-sided transformconverges absolutely in a strip of the form a < Re{s} < b, and possibly including the linesRe{s} = a or Re{s} = b.7 The subset of values ofs for which the Laplace transform convergesabsolutely is called the region of absolute convergence or the domain of absolute convergence. Inthe two-sided case, it is sometimes called the strip of absolute convergence. The Laplace transformis analytic in the region of absolute convergence.

Similarly, the set of values for whichF(s) converges (conditionally or absolutely) is known as theregion of conditional convergence, or simply the region of convergence (ROC). If the Laplacetransform converges (conditionally) at s =s0, then it automatically converges for all s withRe{s} > Re{s0}. Therefore the region of convergence is a half-plane of the form Re{s} > a,

possibly including some points of the boundary line Re{s} = a. In the region of convergence Re{s}> Re{s0}, the Laplace transform of can be expressed by integrating by parts as the integral

That is, in the region of convergence F(s) can effectively be expressed as the absolutelyconvergent Laplace transform of some other function. In particular, it is analytic.

A variety of theorems, in the form of PaleyWiener theorems, exist concerning therelationship between the decay properties of and the properties of the Laplace transform withinthe region of convergence.

In engineering applications, a function corresponding to a linear time-invariant (LTI)system is stable if every bounded input produces a bounded output. This is equivalent to the

6WIDDER, David Vernon (1941), The Laplace Transform, Princeton Mathematical Series, v. 6, PrincetonUniversity Press, Chapter II, 1

7Ibidem, Chapter VI, 2

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absolute convergence of the Laplace transform of the impulse response function in the regionRe{s} 0. As a result, LTI systems are stable provided the poles of the Laplace transform of theimpulse response function have negative real part.

Relationship to other transforms

LaplaceStieltjes transformThe (unilateral) LaplaceStieltjes transform of a function g: R Ris defined by the LebesgueStieltjes integral

The functiongis assumed to be ofbounded variation. Ifgis the antiderivative of:

then the LaplaceStieltjes transform ofgand the Laplace transform of coincide. In general, theLaplaceStieltjes transform is the Laplace transform of the Stieltjes measure associated tog. So inpractice, the only distinction between the two transforms is that the Laplace transform is thought ofas operating on the density function of the measure, whereas the LaplaceStieltjes transform isthought of as operating on its cumulative distribution function.8

Fourier transform

The continuous Fourier transform is equivalent to evaluating the bilateral Laplace transform withcomplex arguments = i ors = 2fi :

This expression excludes the scaling factor , which is often included in definitions of the

Fourier transform. This relationship between the Laplace and Fourier transforms is often used todetermine the frequency spectrum of a signal ordynamical system.

8 FELLER, William (1971),An introduction to probability theory and its applications. Vol. II., Second edition, NewYork: John Wiley & Sons,p. 432

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The above relation is valid as stated if and only if the region of convergence (ROC) ofF(s)contains the imaginary axis, = 0. For example, the function f(t) = cos(0t)u(t) has a Laplacetransform F(s) =s/(s2 + 02) whose ROC is Re(s) > 0. Therefore, substituting s = i in F(s) doesnot yield the Fourier transform off(t) = cos(0t).

However, a relation of the form

holds under much weaker conditions. For instance, this holds for the above example provided thatthe limit is understood as a weak limit of measures (see vague topology). General conditionsrelating the limit of the Laplace transform of a function on the boundary to the Fourier transformtake the form ofPaley-Wiener theorems.

Mellin transform

The Mellin transform and its inverse are related to the two-sided Laplace transform by a simple

change of variables. If in the Mellin transform

we set = e-t we get a two-sided Laplace transform.

Z-transform

The unilateral or one-sided Z-transform is simply the Laplace transform of an ideally sampledsignal with the substitution of

where is the sampling period (in units of time e.g., seconds) and is thesampling rate (in samples per second orhertz)

Let

be a sampling impulse train (also called a Dirac comb) and

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be the continuous-time representation of the sampled

are the discrete samples of .

The Laplace transform of the sampled signal is

This is precisely the definition of the unilateral Z-transform of the discrete function

with the substitution of .

Comparing the last two equations, we find the relationship between the unilateral Z-transform andthe Laplace transform of the sampled signal:

The similarity between the Z and Laplace transforms is expanded upon in the theory of time scalecalculus.

Borel transform

The integral form of the Borel transform

is a special case of the Laplace transform for an entire function ofexponential type, meaning that

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for some constantsA andB. The generalized Borel transform allows a different weighting functionto be used, rather than the exponential function, to transform functions not of exponential type. Nachbin's theorem gives necessary and sufficient conditions for the Borel transform to be welldefined.

Laplace Transform of an Integral

The Laplace transform of the derivatives of a given function can be found from the Laplacetransform of the function. Similarly, the Laplace transform of the integral of a given function canalso be obtained from the Laplace transform of the function, as stated in the following theorem.

THEOREM:Laplace Transform of an Integral

Suppose that F(s)= L{f(t)}, where f is a piecewise continuous function on [0,) and ofexponential orderb. then fors>b,

L0tf()d=L{f(t)}s

In other words,

L-1L{f(t)}s= 0tf()d

Example: ComputeL-11s(s+2)

Solution: in this case, 1s(s+2) = 1s+2s, soL{f(t)}=1s+2. Therefore,f(t)=L-11s+2=e-2t. with theprevious theorem, we then have:

L-11}s(s+2) = 0te-2d=1-e-2t2

1.2. Properties and theorems applications

The Laplace transform is used frequently in engineering and physics; the output of a lineartime invariant system can be calculated by convolving its unit impulse response with the inputsignal. Performing this calculation in Laplace space turns the convolution into a multiplication; thelatter being easier to solve because of its algebraic form. For more information, see control theory.

The Laplace transform can also be used to solve differential equations and is used extensively inelectrical engineering. The Laplace transform reduces a lineardifferential equation to an algebraicequation, which can then be solved by the formal rules of algebra. The original differentialequation can then be solved by applying the inverse Laplace transform. The English electrical

engineer Oliver Heaviside first proposed a similar scheme, although without using the Laplacetransform; and the resulting operational calculus is credited as the Heaviside calculus.

Example #1: Solving a differential equation

In nuclear physics, the following fundamental relationship governs radioactive decay: the numberof radioactive atomsNin a sample of a radioactive isotope decays at a rate proportional toN. Thisleads to the first order linear differential equation

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where is the decay constant. The Laplace transform can be used to solve this equation.

Rearranging the equation to one side, we have

Next, we take the Laplace transform of both sides of the equation:

where

and

Solving, we find

Finally, we take the inverse Laplace transform to find the general solution

which is indeed the correct form for radioactive decay.

Example #2: Deriving the complex impedance for a capacitor

In the theory ofelectrical circuits, the current flow in a capacitoris proportional to the capacitanceand rate of change in the electrical potential (in SI units). Symbolically, this is expressed by thedifferential equation

where Cis the capacitance (in farads) of the capacitor, i = i(t) is the electric current (in amperes)through the capacitor as a function of time, and v = v(t) is the voltage (in volts) across the terminalsof the capacitor, also as a function of time.

Taking the Laplace transform of this equation, we obtain

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where

and

Solving forV(s) we have

The definition of the compleximpedanceZ(in ohms) is the ratio of the complex voltage Vdividedby the complex currentIwhile holding the initial state Vo at zero:

Using this definition and the previous equation, we find:

which is the correct expression for the complex impedance of a capacitor.

Example #3: Method of partial fraction expansion

Consider a linear time-invariant system with transfer function

The impulse response is simply the inverse Laplace transform of this transfer function:

To evaluate this inverse transform, we begin by expanding H(s) using the method ofpartialfraction expansion:

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The unknown constantsPandR are the residues located at the correspondingpoles of the transferfunction. Each residue represents the relative contribution of that singularity to the transferfunction's overall shape. By the residue theorem, the inverse Laplace transform depends only uponthe poles and their residues. To find the residue P, we multiply both sides of the equation by s + to get

Then by lettings = , the contribution fromR vanishes and all that is left is

Similarly, the residueR is given by

Note that

and so the substitution ofR andPinto the expanded expression forH(s) gives

Finally, using the linearity property and the known transform for exponential decay (see Item #3 inthe Table of Laplace Transforms, above), we can take the inverse Laplace transform ofH(s) toobtain:

which is the impulse response of the system.

Example #4: Mixing sines, cosines, and exponentials

Time function Laplace transform

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Starting with the Laplace transform

we find the inverse transform by first adding and subtracting the same constant to the numerator:

By the shift-in-frequency property, we have

Finally, using the Laplace transforms for sine and cosine (see the table, above), we have

Example #5: Phase delay

Time function Laplace transform

Starting with the Laplace transform,

we find the inverse by first rearranging terms in the fraction:

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We are now able to take the inverse Laplace transform of our terms:

This is just the sine of the sum of the arguments, yielding:x(t) = sin(t+ ).

We can apply similar logic to find that

CHAPTER 2

APPLICATIONS OF LAPLACE TRANSFORMS

Circuit Equations

There are two (related) approaches:1. Derive the circuit (differential) equations in the time domain, then transform these ODEs to

thes-domain;

2. Transform the circuit to thes-domain, then derive the circuit equations in thes-domain(using the concept of "impedance").

We will use the first approach. We will derive the system equations(s) in the t-plane, thentransform the equations to thes-plane. We will usually then transform back to the t-plane.

EXAMPLE 1

Consider the circuit when the switch is closed at t= 0 with VC(0) = 1.0 V. Solve for the current i(t)in the circuit.

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Multiplying throughout by :

Taking Laplace transform:

Now in this example, we are told .

So

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That is: Therefore:

NOTE:

CollectingIterms and subtracting from both sides:

Finding the inverse Laplace transform gives us

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EXAMPLE 2

Solve fori(t) for the circuit, given that V(t) = 10 sin5tV,R = 4 W andL = 2 H.

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So

25 =A(s2 + 25) + (Bs + C)(s + 2)

We need to solve forA, B and C.

First, lets = 2 and this gives25 = 29A

ThusA = 25/29

Next, we equate coefficients ofs2:

gives

Equating coefficients of :

gives

So

So we have

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A

EXAMPLE 3

In the circuit shown below, the capacitor is uncharged at time t= 0. If the switch is then closed,find the currents i1 and i2, and the charge on Cat time tgreater than zero.

NOTE: We could either:

Set up the equations, take Laplace of each, then solve simultaneously Set up the equations, solve simultaneously, then take Laplace.

It is easier in this example to do the second method. In many examples, it is easier to do the firstmethod.

For the first loop, we have:

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For the second loop, we have:

Substituting (2) into (1) gives:

Next we take the Laplace Transform of both sides.Note:

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In this example, . So

Now taking Inverse Laplace:

And using result (2) from above, we have:

For charge on the capacitor, we first need voltage across the capacitor:

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So, since , we have:

Graph ofq(t):

EXAMPLE 4

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In the circuit shown, the capacitor has an initial charge of 1 mC and the switch is in position 1 longenough to establish the steady state. The switch is moved from position 1 to 2 at t= 0. Obtain thetransient current i(t) fort> 0.

Position 1, after a `long time': A

Position 2: ( )

We apply emf, and consider the sum of the potential difference across elements.

In position 2, there is no emf.

Finding Laplace Transform:

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Multiplying by :

Solving forIand completing the square on the denominator gives us:

So the transient current is:

We could transform the trigonometric part of this to asingle expression:

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2 cos 222.2t 0.45 sin 222.2t=R cos(222.2t+ )

So

EXAMPLE 5

The system is quiescent. Find the loop current i2(t).

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Quiescent implies i1, i2 and their derivatives are zero fort= 0, ie

i1(0) = i2(0) = i1'(0) = i2'(0) = 0.

For loop 1:

For loop 2:

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Substituting our result from (1) gives:

Taking Laplace transform:

Let

So

SoTaking Inverse Laplace:

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So

EXAMPLE 6

Consider a series RLC circuit where R = 20 W, L = 0.05 H and C = 10-4 F and is

driven by an alternating emf given by E = 100 cos 200t. Given that both the

circuit current i and the capacitor charge q are zero at time t = 0, find an

expression for i(t) in the region t> 0.

We use the following:

and obtain:

After multiplying throughout by 20, we have:

Taking Laplace transform and using the fact that i(0) = 0:

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Using Scientific Notebook to find the partial fractions:

So

So

+ cos200t 2 sin 200t

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NOTE: Scientific Notebook can do all this for us very easily. In one step, we have:

+ cos200t 2 sin 200t

Transient part:

Steady state part:

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EXAMPLE 7

A rectangular pulse vR(t) is applied to the RC circuit shown. Find the response, v(t).

Graph ofvR(t):

Note: v(t) = 0 V for all t< 0 s implies v(0-) = 0 V.

Now

To solve this, we need to work in voltages, not current.

We start with .

The voltage across a capacitor is given by .

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It follows that .

So for this example we have:

Substituting known values:

Then

Taking Laplace:

Since , we have:

So, taking inverse Laplace

NOTE: For the part: , we use:

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So we have:

Solution Using Scientific Notebook

1. To find the Inverse Laplace:

2. To solve the original DE:

Exact solution forv(t):

To see what this means, we could write it as follows:

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To see what our expression forv(t) means, we graph it as follows:

BIBLIOGRAPHY

ABELL, Martha; BRASELTON, J.P, Differential Equations with MAPLE V, APProfessional, 1994

BRACEWELL, R. N. (2000), The Fourier Transform and Its Applications (3rd ed.),Boston: McGraw-Hill, ISBN0071160434 .

FELLER, William (1971),An introduction to probability theory and its applications. Vol.II., Second edition, New York: John Wiley & Sons, MR0270403 .

KORN, G.A.; KORN, T.M. (1967), Mathematical Handbook for Scientists and Engineers(2nd ed.), McGraw-Hill Companies, ISBN0-0703-5370-0 .

WIDDER, David Vernon (1941), The Laplace Transform, Princeton Mathematical Series,v. 6, Princeton University Press, MR0005923

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Internet

Introduction to the Laplace Transform

Documents

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