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EDUCATION HOLE PRESENTS Engineering Mathematics – II Unit-III

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EDUCATION HOLE PRESENTS Engineering Mathematics – II

Unit-III

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Laplace Transform ................................................................................................................... 2

Laplace transforms of derivatives and integrals ................................................................................ 3

Existence theorem ........................................................................................................................... 6

Initial Value Theorem ....................................................................................................................... 8

Final Value Theorem ........................................................................................................................ 9 Unit step function ............................................................................................................................................... 10

Convolution Theorem .................................................................................................................... 11

Dirac- delta function .............................................................................................................. 12

Laplace transform of periodic function .................................................................................. 13 Inverse Laplace Transforms ................................................................................................................................ 15 Application to solve simple linear and simultaneous differential equations ...................................................... 16

Laplace Transform The Laplace transform is a widely used integral transform in mathematics with many applications in physics and engineering. It is a linear operator of a function f(t) with a real argument t (t ≥ 0) that transforms f(t) to a function F(s) with complex argument s, given by the integral

This transformation is bijective for the majority of practical uses; the most-common pairs of f(t) and F(s) are often given in tables for easy reference. The Laplace transform has the useful property that many relationships and operations over the original f(t) correspond to simpler relationships and operations over its image F(s). It is named after Pierre-Simon Laplace who introduced the transform in his work on probability theory.

The Laplace transform is related to the Fourier transform, but whereas the Fourier transform expresses a function or signal as a series of modes of vibration (frequencies), the Laplace transform resolves a function into its moments. Like the Fourier transform, the Laplace transform is used for solving differential and integral equations. In physics and engineering it is

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used for analysis of linear time-invariant systems such as electrical circuits, harmonic oscillators, optical devices, and mechanical systems. In such analyses, the Laplace transform is often interpreted as a transformation from the time-domain, in which inputs and outputs are functions of time, to the frequency-domain, where the same inputs and outputs are functions of complex angular frequency, in radians per unit time. Given a simple mathematical or functional description of an input or output to a system, the Laplace transform provides an alternative functional description that often simplifies the process of analyzing the behavior of the system, or in synthesizing a new system based on a set of specifications.

Laplace transforms of derivatives and integrals

The Laplace transform is an integral transform perhaps second only to the Fourier transform in its utility in solving physical problems. The Laplace transform is particularly useful in solving linear ordinary differential equations such as those arising in the analysis of electronic circuits.

The (unilateral) Laplace transform (not to be confused with the Lie derivative, also commonly denoted ) is defined by

(1)

where is defined for (Abramowitz and Stegun 1972). The unilateral Laplace transform is almost always what is meant by "the" Laplace transform, although a bilateral Laplace transform is sometimes also defined as

(2)

(Oppenheim et al. 1997). The unilateral Laplace transform is implemented in Mathematica as LaplaceTransform[f[t], t, s].

The inverse Laplace transform is known as the Bromwich integral, sometimes known as the Fourier-Mellin integral (see also the related Duhamel's convolution principle).

A table of several important one-sided Laplace transforms is given below.

conditions

1

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In the above table, is the zeroth-order Bessel function of the first kind, is the delta function, and is the Heaviside step function.

The Laplace transform has many important properties. The Laplace transform existence theorem states that, if is piecewise continuous on every finite interval in satisfying

(3)

for all , then exists for all . The Laplace transform is also unique, in the sense that, given two functions and with the same transform so that

(4)

then Lerch's theorem guarantees that the integral

(5)

vanishes for all for a null function defined by

(6)

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The Laplace transform is linear since

(7)

(8)

(9)

The Laplace transform of a convolution is given by

(10)

Now consider differentiation. Let be continuously differentiable times in . If , then

(11)

This can be proved by integration by parts,

(12)

(13)

(14)

(15)

Continuing for higher-order derivatives then gives

(16)

This property can be used to transform differential equations into algebraic equations, a procedure known as the Heaviside calculus, which can then be inverse transformed to obtain the solution. For example, applying the Laplace transform to the equation

(17)

gives

(18)

(19)

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which can be rearranged to

(20)

If this equation can be inverse Laplace transformed, then the original differential equation is solved.

The Laplace transform satisfied a number of useful properties. Consider exponentiation. If for (i.e., is the Laplace transform of ), then for

. This follows from

(21)

(22)

(23)

The Laplace transform also has nice properties when applied to integrals of functions. If is piecewise continuous and , then

Existence theorem

Let f(x,y) be a real valued function which is continuous on the rectangle

. Assume f has a partial derivative

with respect to y and that is also continuous on the rectangle R. Then there exists an

interval (with ) such that the initial value problem

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has a unique solution y(x) defined on the interval I.

Note that the number h may be smaller than a. In order to understand the main ideas behind this theorem, assume the conclusion is true. Then if y(x) is a solution to the initial value problem, we must have

It is not hard to see in fact that if a function y(x) satisfies the equation (called functional equation)

on an interval I, then it is solution to the initial value problem

Picard was among the first to look at the associated functional equation. The method he developed to find y is known as the method of successive approximations or Picard's iteration method. This is how it goes:

Step 1. Consider the constant function

Step 2. Once the function is known, define the function

Step 3. By induction, we generate a sequence of functions which, under the assumptions made on f(x,y), converges to the solution y(x) of the initial value problem

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Initial Value Theorem

The initial value theorem states

To show this, we first start with the Derivative Rule:

We then invoke the definition of the Laplace Transform, and split the integral into two parts:

We take the limit as s goes to infinity:

Several simplifications are in order. In the left hand expression, we can take the second term out of the limit, since it doesn't depend on 's.' In the right hand expression, we can take the first term out of the limit for the same reason, and if we substitute infinity for 's' in the second term, the exponential term goes to zero:

The two f(0-) terms cancel each other, and we are left with the Initial Value Theorem

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Final Value Theorem

The final value theorem states that if a final value of a function exists that

However, we can only use the final value if the value exists (function like sine, cosine and the ramp function don't have final values). To prove the final value theorem, we start as we did for the initial value theorem, with the Laplace Transform of the derivative,

We let s→0,

As s→0 the exponential term disappears from the integral. Also, we can take f(0-) out of the limit (since it doesn't depend on s)

We can evaluate the integral

Neither term on the left depends on s, so we can remove the limit and simplify, resulting in the final value theorem

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Unit step function

The unit step function is defined as

Some notes about this function:

• Most references use u(t) instead of γ(t). However, u(t) has some other common uses, so we will use γ(t) to avoid confusion (and because it's Laplace Transform Γ(s) looks a little like a step input).

• Some references change the definition of the step so that the bottom inequality becomes a strict inequality (t>0) and either leave γ(0) undefined or say that γ(0)=�.

• Although there is a discontinuity at t=0, we draw a vertical line to help guide the eye and to indicate that the blue line is a single function.

To find the Laplace Transform, we apply the definition.

so

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Aside: Convergence of the Laplace Transform

Careful inspection of the evaluation of the integral performed above:

reveals a problem. The evaluation of the upper limit of the integral only goes to zero if the real part of the complex variable "s" is positive (so e-st→0 as s→∞). In this case we say that the "region of convergence" of the Laplace Transform is the right half of the s-plane (since s is a complex number, the right half of the plane corresponds to the real part of s being positive). As long as the functions we are working with have at least part of their region of convergence in common (which will be true in the types of problems we consider), the region of convergence holds no particular interest for us. Since the region of convergence will not play a part in any of the problems we will solve, it is not considered further.

Convolution Theorem

The convolution theorem states (if you have not studied convolution, you can skip this theorem)

note: we assume both f(t) and g(t) are causal.

We start our proof with the definition of the Laplace Transform

From there we continue:

We can change the order of integration.

Now, we pull f(λ) out because it is constant with respect to the variable of integration, t

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Now we make a change of variables

Since g(u) is zero for u<0, we can change the lower limit on the inner integral to 0-.

We can pull e-sλ out (it is constant with respect to integration).

We can separate the integrals since the inner integral doesn't depend on λ.

We can change the lower limit on the first integral since f(λ) is causal.

Finally we recognize that the two integrals are simply Laplace Transforms.

The Theorem is proven

Dirac- delta function When we first introduced Heaviside functions we noted that we could think of them as switches changing the forcing function, g(t), at specified times. However, Heaviside functions are really not suited to forcing functions that exert a “large” force over a “small” time frame. Examples of this kind of forcing function would be a hammer striking an object or a short in an electrical system. In both of these cases a large force (or voltage) would be exerted on the system over a very short time frame. The Dirac Delta function is used to deal with these kinds of forcing function. There are many ways to actually define the Dirac Delta function. To see some of these definitions visit Wolframs MathWorld. There are three main properties of the Dirac Delta function that we need to be aware of. These are,

1.

2.

3.

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At the Dirac Delta function is sometimes thought of has having an “infinite” value. So, the Dirac Delta function is a function that is zero everywhere except one point and at that point it can be thought of as either undefined or as having an “infinite” value.

Note that the integrals in the second and third property are actually true for any interval containing , provided it’s not one of the endpoints. The limits given here are needed to prove the properties and so they are also given in the properties. We will however use the fact that they are true provided we are integrating over an interval containing

. This is a very strange function. It is zero everywhere except one point and yet the integral of any interval containing that one point has a value of 1. The Dirac Delta function is not a real function as we think of them. It is instead an example of something called a generalized function or distribution. Despite the strangeness of this “function” it does a very nice job of modeling sudden shocks or large forces to a system.

Before solving an IVP we will need the transform of the Dirac Delta function. We can use the third property above to get this.

Note that often the second and third properties are given with limits of infinity and negative infinity, but they are valid for any interval in which is in the interior of the interval.

Laplace transform of periodic function If function f(t) is:

Periodic with period p > 0, so that f(t + p) = f(t), and

f1(t) is one period (i.e. one cycle) of the function, written using Unit Step functions,

then

L{f(t)}=L{f1(t)}1−e−sp

NOTE: In English, the formula says:

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The Laplace Transform of the periodic function f(t) with period p, equals the Laplace Transform of one cycle of the function, divided by (1−e−sp).

Find the Laplace transforms of the periodic functions shown below:

(a)

(b) Saw-tooth waveform:

(c) Full-wave rectification of sin t:

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Inverse Laplace Transforms

Finding the Laplace transform of a function is not terribly difficult if we’ve got a table of transforms in front of us to use as we saw in the last section. What we would like to do now is go the other way. We are going to be given a transform, F(s), and ask what function (or functions) did we have originally. As you will see this can be a more complicated and lengthy process than taking transforms. In these cases we say that we are finding the Inverse Laplace Transform of F(s) and use the following notation.

As with Laplace transforms, we’ve got the following fact to help us take the inverse transform.

Fact

Given the two Laplace transforms F(s) and G(s) then

for any constants a and b.

So, we take the inverse transform of the individual transforms, put any constants back in and then add or subtract the results back up.

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Application to solve simple linear and simultaneous differential equations