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.
