DifferentialEq: Laplace

Loading...

Do curso por Georgia Institute of Technology

Fundamentos da Engenharia - Revisão

183 classificações

Georgia Institute of Technology

Fundamentos da Engenharia - Revisão

183 classificações

The purpose of this course is to review the material covered in the Fundamentals of Engineering (FE) exam to enable the student to pass it. It will be presented in modules corresponding to the FE topics, particularly those in Civil and Mechanical Engineering. Each module will review main concepts, illustrate them with examples, and provide extensive practice problems.

Na lição

Mathematics

This module reviews the basic principles of mathematics covered in the FE Exam. We first review the equations and characteristics of straight lines, then classify polynomial equations, define quadric surfaces and conics, and trigonometric identities and areas. In algebra we define complex numbers and logarithms, and show how to manipulate matrices and determinants. Basic properties of vectors with their manipulations and identities are presented. The discussion of series includes arithmetic and geometric progressions and Taylor and Maclaurin series. Calculus begins with definitions of derivatives and gives some standard forms and computation of critical points of curves, then presents grad, del and curl operators on scalar and vector functions. Differential equations are calcified and to methods to solve linear, homogenous equations are presented. Fourier series and transforms are defined along with standard forms, and finally Laplace transforms and their inverse are discussed. In all cases, basic ideas and equations are presented along with sample problems that illustrate the major ideas and provide practice on expected exam questions.Time: Approximately 4.5 hours | Difficulty Level: Medium

Analytic Geometry and Trigonometry: Straight Lines 8:13

Analytic Geometry and Trigonometry: Polynomials and Conics9:23

Analytic and Geometry and Trigonometry: Trigonometry7:41

Algebra and Linear Algebra: Complex numbers and logarithms5:39

Algebra and Linear Algebra: Matrices and determinants7:29

Vectors: Basic Definitions and operations13:05

Vectors: Examples8:18

Series: Arithmetic and geometric progressions 10:13

Calculus: Derivatives and curvature10:32

Calculus: Integration5:38

Calculus: Gradient, divergence and curl7:37

DifferentialEq: Classification6:11

DifferentialEq: Solutions7:58

DifferentialEq: Fourier series7:09

DifferentialEq: Laplace7:51

Conheça os instrutores

Dr. Philip Roberts
Professor
School of Civil and Environmental Engineering

The last topic I want to discuss in differential equations and

transforms is Laplace transforms.

Laplace transforms transforms a function of time

into a function of complex frequency.

And it's very important and useful.

Especially for solving nonhomogeneous differential equations.

Because sometimes you can convert them into relatively simple

algebraic equations which can be solved.

The definition of the LAPLACE TRANSFORM is given in the extract here

from the Reference handbook and here are the definitions again.

So f of s, the Laplace transform,

is the integral from 0 to infinity of f of t e to the minus st, dt.

And the inverse transformation f of t 1 over 2 pi g.

Integral from sigma minus J W, to sigma plus J W of F of S, E to the S T.

Where again, J is the square root of minus one in that transformation.

But again normally most of these can be looked up in

tables such as those given in the reference handbook.

For example the free transform of a delta function, or

an impulse, is at the origin of T=0

is = to one the free transform of a step function

which looks like this at T=0 is equal to one over S.

The fourier transform of a ramp function,

starting at time zero is equal to one over S squared, etc.

And a couple of other things from this first, the step function is one over S.

And if the function is transformed or shifted in time by an amount,

tau, then the fourier transform of that function,

is given by e to the minus tau s multiplied by the LaBlas function.

Which is the last equation right here.

So to see how we use that let's compute the Laplace Transform of

the rem function of f of t is equal to 2t which of these is it?

Well, this is one of the standard Laplace Transforms.

In the previous table, and we can just look it up right there.

But let's compute it from first principles.

So the function here is just a linear function, a ramp function like this.

A function of t versus t.

And we utilize our basic definition of the Laplace transform.

Which is F of S is equal to integral from zero to infinity

times the given function multiplied by e to the minus st, dt.

So in this case the function of t here, f of t is 2t so

the integral becomes 2t e to the minus st dt.

And now the question is just to evaluate that integral.

So the factor 2 is constant, I can take that out.

And the integral here I might recognize, is a standard integral which

I obtained from from the previous tables we've discussed in the reference handbook.

And it appears in the handbook like this.

The integral of xe to the axdx is equal to e to the ax of a squared times ax minus 1.

So in this case we recognize that we replace x by t,

and a, the exponent here, with minus s.

So therefore the integral becomes e to the minus st of s squared

multiplied by S T plus 1, with a factor 2 in front.

So, therefore the integral is this evaluated at the limits of 0 and

infinity, and substituting in, first of all, T equals infinity.

E to the minus infinity is 0 so that term goes out.

And the second term evaluated at T equals 0,

that term goes out, leaves me with minus 2 over s squared.

So the solution is, of the integral, is equal to two over s squared and the answer

is a, which we could have found by just looking it up in the standard tables.

Next we'll do the Laplace transform,

the step function of height one at t equals zero.

And a step function of height 2 at t equals 1 second.

Which of these is it?

So here's a situation.

We have here function of time versus time.

So we have a step function of height 1 first of all, at the origin.

And then a second step function of height 2, height equal to 2,

displaced a horizontal distance of t = 1 second.

What is the Laplace transform of that combined function?

Well, here we're going to use two pieces of information from the table.

First of all, the transform of a function shifted in time by a time tau.

Where tau in this case is one second, is this expression, is equal to e to

the minus tau s multiplied by the Laplace transform of the original function.

And furthermore, the transform of a unit step

at time t = 0 is 1/s, which we obtain from the table.

So therefore, the transform of a step of height 2 is 2 times that,

and at time tau = 1, Is e to the minus s,

times 1 over s, multiplies by 2.

So, this term, 1 over s is the Laplace transform of this function,

and this term Is the Laplace transform of this step function.

And finally we can simply add these together by superposition

to find the Laplace transform of the combined function.

Therefore the Laplace transform combined is equal to that, which I can read,

it's a little more tidily in that form therefore

we see that the answer is A.

Now finally to conclude this topic of mathematics

I'll mention that there are a number of topics that I didn't cover.

Because I think they're of lesser importance in the effie exam.

Although of course they should be covered in a more comprehensive class.

And in particular, we didn't talk about putting linear equations into matrix form

and solving by Cramer's rule.

We didn't look at a number of topics involving numerical analysis particularly

difference equations, numerical methods, numerical integration.

And solving ordinary differential equations by numerical solutions.

And finally, we didn't cover Centroids and Moments of Inertia.

Although I put this in parenthesis, because those will be covered later on in

two topics, topic six, which is statics, and topics eight.

Mechanics of materials we'll talk about those topics in more detail.

So this concludes the module on mathematics.

Explore nosso catálogo

Registre-se gratuitamente e obtenha recomendações, atualizações e ofertas personalizadas.

O Coursera proporciona acesso universal à melhor educação do mundo fazendo parcerias com as melhores universidades e organizações para oferecer cursos on-line.

© 2018 Coursera Inc. Todos os direitos reservados.

Baixar na App Store

Baixar no Google Play