the laplace transform and its inverse

7/29/2019 The Laplace Transform and Its Inverse

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Copyright Ren Barrientos Page 1

THE LAPLACE TRANSFORM AND ITS INVERSE

The Laplace1 Transform is an example of an Integral Transformand an integral transform is a specialkind function whose domain and range are sets of other functions. We call it an integral transformbecause the operation used in the process of mapping one function to another involves an integral. Werefer to functions whose domain and range consists of other functions asfunctionals.

More precisely, let, be a function of the real variables and, and let be a real valuedfunction defined for 0. Then the integral transform offwith kernel, on the interval , isgiven by

, Observe that the product , is a function of both variables s and t. However, only thevariable survives when the integral is performed with respect to t over the interval ,. This newfunction of the variable is denoted by and defined by

, The transform converts a function of

into anther function of

;

:.

The Laplace Trasform is an integral transform whose kernel is the function:

Observe that there is nothing special about the variables

and

. We could have just as easily defined

the Laplace Transform using other variables. For example,

or even usex instead ofs: Customarily, however, the input function is a function of time and the letter is used as theindependent variable of the output function.

Functions of Exponential Order

What functions have a Laplace Transform? Considering its definition, it is clear that

cannot be

unbounded on the interval 0, and furthermore, this product should be well-behaved in order to beintegrable.The two conditions that guarantee the existence of the transform are piecewise continuity andexponential order.

1Pierre-Simon Laplace (1749 - 1827) was a French mathematician who authored Trait de Mcanique Cleste,considered to be his greatest work. His contributions to mathematics, physics, and other branches of knowledge areof great substance and he is considered to be one of the most influential scientific minds of his time.

The Laplace Transform

Letthe functionbe defined on 0,. TheLaplace Transformof, denoted by , is defined by

Provided the integral exists. When it does, the resulting function of the variable s is denoted by and its domainis the set of all for which the improper integral converges. Write .


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Piecewise continuous functions are functions whose set of discontinuities on the given interval is finiteand such that the function has one-sided limit at each of its points of discontinuity, including theintervals endpoints. Hence, piecewise continuous functions have a finite number of jump-discontinuities.

A function is piecewise continuous on

0,if it is piecewise continuous on every finite subinterval of

0, .

A function isofexponential order if there exist nonnegative numbers, , and such that|| for all , that is, the function does not grow any faster that a typical exponentialfunction beyond some point.

Piecewise continuous Not piecewise continuous

Before we establish this theorem, note that both properties are necessary. For if we consider the function (which it is piecewise continuous but not of exponential order), it can easily be establishedthat diverges for all .Pf:

By the piecewise continuity of

and continuity of

for all

and t,

|| exists for all

[from standard theorems in calculus]. Since is of exponential order as , there existnumbers, , and such that || for .If , the piece-wise continuity of ensures that exists and is therefore boundedby some number . Thus,

||

1 ; QED.

Theorem 1

If is piecewise continuous and of exponential order as on 0,, then exists.

Asymptote


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Pf. See optional section.

Linearlty Properties of The Transform

Assume both and have transforms and that is a fixed real number. Then,1 2 Pf:

Similarly, . The proof is left as an exercise.Example 1(the constant unit function) Let 1. Compute.Solution

Applying the definition, 1 1Improper integrals are evaluated through a limiting process:

1 1 l i m

l i m 1 1 lim 1 1 ; for 0

Note that the restrictions >0 is necessary for the convergence of the integral. Thus, we say that thetransform of 1 is 1/ provided 0. This will be the first entry on our table oftransforms:

0

Working with the transform can be very cumbersome. For that reason, we will rely on tables oftransforms extensively. Nevertheless, there will be times when the definition is the only recourse.

Example 2 Use the linearity properties to establish that ; 0 for any constant.Solution 1 1

; 0

Theorem 2

If possesses a Laplace transform , then 0 a s .


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Thus, ; 0Example 3 Let . Compute.Solution

Applying the definition,

This integral may be evaluated using integration by parts: let and Then, , 1

[Remember,s is a parameter, not a variable of integration so it acts as a const ant.]

Thus, 1 1 1 1 1 1

Proceeding to the limit2, l i m

l i m 1

lim 1 0The second entry in our basic table is 0

Example 4 Compute the Laplace Transform of wherea is a constant.Solution

Let and observe that .Once again applying the definition,

But the right hand side of this equation just where is the transform of 1.

Thus, 1| 1 ; 2We have used in order to get rid of the negative sign.


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So we have the third entry of our table:

; Example 5 Let be differentiable and of exponential order on the interval 0,. Determine

.

Solution Let

Then

Thus, 0 This gives us our next very important result:

We denote

by

,following the convention of using capital letters for the transform

of a function. Thus we have This will prove to be one of the central results in the application of the transform to differentialequations. For now we will use it as an aid to find other transforms.

Example 6 Derive the formula for.Solution

Assume suitable differentiability conditions on and let . Then and 0 from previous result

0 by sbustitution 0 0 by sbustitutionwhich gives us our second very important result:

In general, we have the following result:

0 becauseis of exp. order


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We can use the formula for the transform of a derivative to find the transforms of the sine and cosine.

Example 7 Find

sinwherek

is a constant.

SolutionLet sin. Then c os and sin. Therefore, 0 0gives us sin sin 0 By the linearity properties,sin sin Solving forsin: /

We have a new entry for our basic table. In a completely analogous way,

/ The table below is out basic table of transforms and we will use it freely when needed.f(t) F(s)={(t)}

2 1 ; 0

3 tn! ; 0

4 eat ;

5

sin

;

6 cos ; 7 9 10

Using the table and the linearity properties we may find transforms of other, more complex functions.However, we must keep the definition in mind because we will encounter situations where it is needed.

Example 8 Find cos4Solution

Using table entry #6 with 4,cos4 4 16Example 9 Find the Laplace transform of the function 5 4Solution

Applying linearity, 5 4 5 4


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5 1 2 4 3! #3 4Simplifying, 5 2 24

Example 10 Find the Laplace transform of the function

cos4sin3.

SolutionApplying linearity,cos4sin3 cos4 sin3

16 3 9

These may be added to obtain the expression

3 9 4 8 16 9

Example 11 Let cosh3. Find.Solution

Recall the definition of thehyperbolic cosine:

Thus,

cosh3 By the linearity properties,

2 12 12 1 3 1 3 12 1 3 1 3

Thus 9In fact, the hyperbolic functions are used often enough to put them in out list.

[Notice how similar these are to the transforms of the sine and cosine functions].

Example 12 Find the Laplace transform of the function sin.Solution

The definition requires that we evaluate the integral

4 3


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sin This does not look promising. Instead let us see if we can use the formula for the transform of aderivative. Let s i n. Then,

sin cos c o s cos sin2cossinUsing gives us 2cossin sin 0 0By linearity, 2cossin sinSolving forsin:sin 2

from which we obtain

The First Shifting Theorem and the Derivative of a Transform

The Laplace transform has some very interesting and important properties that make it a powerful tool insolving differential equations. First Shifting Theorem is one of them.

Example 13 FindSolution

By definition, we seek

We could combine the exponential functions and obtain

but still this does not look too promising. However, notice that this last integral looks like

with replaced by 2. In other words, If

Then

2

This says that: the transform of is the same as the transform of evaluated at 2. Since

3!We have 2 3! 2


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Therefore, This is the essence of theFirst Shifting Theorem, which we proceed to state:

Alternatively, we may write this result as follows: |To establish this theorem, all we need to do is proceed as in example14. This is done in the optionalsection.

Example 14 Find Solution

|

3! 6 2

Example 15 FindSolution so let . Then 1/. Applying the shifting theorem, |

1

1 1Example 16 Find cos3Solution cos3 cos3|

9 4 4 9

Example 17 Find 1 Solution 1 2 1 2 1 | 2| 1|

2! 2 1 1 2! 3 2 3 1 3

Theorem 3 If , then where is an arbitrary realnumber.


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Example 18 Find cosh2sinh3Solution 10cosh2sinh3 10cosh2| sinh3|

1 0

4 3

9

1 0 1 1 4 3 1 9Example 19 Find 1 sinSolution 1 sin 2 1sin sin 2 sin sin

2

4

2

2

Example 20 Find sin 4SolutionFirst we need touse a reduction formula onsin 4:

sin 4 1cos82 12 1cos8Thus, sin 4 12 1cos8|

12 1 64

Derivatives of Transforms

The other important tool we wish to study comes from considering the derivative of a transform.

Let . Then,

This gives us3

3The essential result for this theorem is Leibnitz rule for differentiating integrals:

, ,

,


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We sometimes write this formula like this:

Example 21 Find c o s Solution c o s 1 cos Sincecos cos 1 1 Example 22 Findin two different ways.Solution

Method I: View this problem as an application of the First Shifting Theorem with .Then,

4 where 1

Method II: UseTheorem4: where 14 1 4

Example 23 Generalize theorem 4 and show that 1 where is a positiveinteger.

SolutionFirst, let us see what we obtain if 2:

1

Based on this computation, we make the following conjecture 1

and proceed to establish it by induction. The statement has already been established for 1.Assume that it is true for , that is, suppose that for any that possesses a transform,

Theorem 4 If , then .


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1 Then

1 1 1 1

QED.

Example 24 Find the Laplace transform of the function 1 0 2

2Solution Although individually 1 and are easy to find (both can be obtained from the table),combined as a piecewise defined function forces us to use the definition. Note that we mustdivide0, into two subintervals: 0,2 and 2, .

1

Notice that neither integral on the right side of the equation is a Laplace Transform because bydefinition, the limits of integration of the transform are0 and infinity.

The first integral is straightforward. The second one may be simplified via the substitution

, . The new limits of integration are again0 and.We have 1 1

The second integral becomes,

2

2

But now using the properties of the exponential function together with linearity we have

2

; 2The first integral is the transform of and the second integral is2 2 1. Thus,

1 2 Finally,

Proper integral Improper integral


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1 1 2

The Unit Step Function

The previous example illustrates the need to develop tools to hadle piecedefined functions. Theunitstepfunction

, defined below, will play a very important role in this regard. It is definned by

0 01 0Graphically,

We will be interested in its transform and its operational properties. Clearly

0 .

However, we will be more interested in theshifted unit step function (called the Heaviside Function4

) defined by 0 1 Example 25 Letabe a fixed positive real number. Find .Solution

Applying the definition of the Transform,

where 0 1 . Therefore,

Let . Then and the integral takes the form

1

But the integral 1 is just1. Thus,

This transform is important and should be added to the basic table of transforms.

The Inverse Laplace Transform

Definition The function is an inverse Laplace transform of the function if . We alsowrite 4Oliver Heaviside (1850-1925) was a self-taught English mathematician, physicist and electrical engineer who madegreat contributions in applied mathematics.

1


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Finding the inverse transform5is like looking at the transform table in reverse from right to left. Thus,if we wan to find / for example, all we need to do is identify the entry on the rightcolumn that resembles/ and then identify the constant(s) involved:

f(t) F(s)= {f(t)}4 eat ;

The function 1 / 2 is of the form1/ with . This function is identified with . Therefore, We call an inverse transform of 1 / 2 .

This is of course a very simple example, but it illustrates the concept. Usually it is much more difficultto find an inverse transform, just as it is harder to integrate than it is to differentiate.

Linearity Properties of the Inverse Transform

In addition to these properties every formula for a transform has a counterpart in the opposite direction:

Alternatively: | And similarly, theorem 4 has its counterpart:

Of these, theorem3is the more useful one.Example 26 Find4/2Solution

The second linearity property allows us to pull the constant4 out:

4

2 4

1

2

Example 27 Find/ 4Solution

The function / 4 may be identified with / in the table with . Theinverse transform is therefore of the formcos:

5Unlike the transform, the inverse transform of a function F(s) is not unique unless certain restrictions are imposed[see example 32].

Let and and let be a real number. Then

Theorem

Theorem


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Example 28 Find1/Solution

The function1/ may be identified with!/ with . Thus, the inverse transform willbe of the form . However, we dont have the ! " in the numerator.The second linearity property allows us to take constants out so we use the same procedure usedwith integration:

1 13! 3!

13! 3! Example 29 Find Solution

applying linearity, 1 2 1 2 1 2 1

But we see that the numerators are all wrong. No problem, as long as all that is missing areconstants, we can use linearity to fix the problem:

1 2 1 13! 3! 4! 4!Each of the inverse transforms on the right corresponds to the form!/; the first one with 3

and the second one with

4. Thus,

! ! Example 30 Find Solution 4 9 9 4 9

The expressions on the right hand side correspond to the transforms ofcos and sinrespectively. However, the second expression on the right is missing a

3 in the numerator and

actually has a 4 that we do not need. Thus, we proceed as follows:Take out the 4 using the second linearity property and simultaneously multiply and divide by3: 9 4 9 3 43 3 9


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Example 31 Find4Solution

This looks very much like a step function situation:

4 4 4 5 Example 32 Find two different functions whose transform is1/2Solution

Obviously 1 2 Thus, is one function whose transform is1/2. Now consider the function

14 1

Certainly . However, it can be shown that

That is, 1 / 2 . In fact, any piecewise function which differs from at finitelymany points will have the same Laplace transform. However, we will assume that we seek inversetransforms that are continuous.

Optional Section (this material may be omitted)

Proof of Theorem 2

If possesses a Laplace transform , then 0 as .Pf.

|| || But if is of exponential0, , there are numbers , 0such that || . Therefore,

||

; Thus, || if and || 0 as .

Proof of Theorem 3

If , then .Pf.

We are given that . Thus,


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CopyrightRen Barrientos Page 17

A Nice Resul t

Let us see what happens when we apply the transform to the complex exponential :

1

1 applying Eulers Formula to and expressing the right hand side in standard form of complexnumbers6: cossin 1

cos sin Identifying the real and imaginary parts:

Complex numbers are a very powerful tool indeed.

6A complex number is in standard form when it is written as . In the present case that may be accomplishedby rationalizing the denominator.

the laplace transform and its inverse

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