Hello, my name is Daniel Friedrich, and this is part five of week one of the Laplace transform topic of Engineering Mathematics 2A. In this part, I'm going to introduce the frequency shift property. After this part. And with a bit of practice, you will be able to use the frequency shift property to calculate the Laplace transform of functions with exponential factors. To be able to do this, you need to know how to calculate the Laplace transform of simple functions and how to use Laplace transform tables. In addition, you should know the properties of integration and differentiation. In the previous part, we introduced the first two properties, the linearity property and the derivative of transform property. The linearity property basically says that the Laplace transform is a linear operator. And if we have a linear sum of functions in the time domain, we get a linear sum of their Laplace transforms in the Laplace domain. The derivative of transform property links multiplication with t in the time domain, to differentiation with respect to s in the Laplace or frequency domain. The next property which we're handling in this part is the frequency shift property. Let us consider a function in the time domain f(t), which has a Laplace transform, capital F(s) in the Laplace or frequency domain. This Laplace transform exists when the real part of S is larger than sigma c, which is the abscissa of convergence. The frequency shift property and enables us to find the Laplace transform when we multiply this time domain function with an exponential. So the Laplace transform of the exponential of alpha(t) times f(t), where alpha is a constant, is equal to the Laplace transform capital F of s minus alpha. If we consider the  Laplace domains, other name is frequency domain, then the frequency s has shifted by minus alpha. The next thing to consider is the abscissa of convergence. So the original function had the abscissa of convergence sigmaC  So in this case, we're multiplying the function by an exponential. For cases where alpha is positive, the exponential is fast growing. When alpha is negative, the exponential is very fast decreasing. So we need to add the real part of alpha to the original abscissa of convergence. At the bottom of the slide, I give alternative notation for this. So we have the Laplace transform of the exponential of alphat times f(t), And in square brackets, the Laplace transform of f(t), where S tilde is shifted to s minus alpha. And then we get in square brackets, capital F(S tilde). when S tilde is shifted to s minus alpha. Let us use the frequency shift property. We want to determine the Laplace transform of small g(t), which is equal to e^(-3t)cos(3t). Let us define cos(3t) = f(t) Then we get the expression at the end of the first slide here. Anytime you see an expression like this, where you have a function in the time domain which is multiplied by an exponential, then you should immediately think of the frequency shift property. Because this is directly applicable here. So we can look for the Laplace transform of the cosine in the Laplace transform tables. So we know the Laplace transform of cos(3t), which we call capital F(s), is equal to s/(s^2 + 9) with an abscissa  of convergence of 0, because cosine function doesn't grow, just oscillates between minus 1 and 1. Now we can apply the frequency shift property. So we want to find the Laplace transform of  e^(-3t)*cos(3t) which is equal to capital F(s + 3) Because this is a minus 3 here. Or we can write it as capital [F(s)] from s to (s + 3) In this case This exists when the real part of s is (larger than three minus). When the real part of s is larger than 0 minus 3. Why minus 3?  We multiply the cosine with the fast decreasing function. So our region of convergence actually shifts left into the negative complex plane. When we have this, we can now replace the s in our capital F(s) with (s + 3). So on top we just replace s with (s + 3),  at the bottom we need to be careful  that we put parentheses around this because we need to have  (s + 3)^2, not just s^2 or 3^2. We need to have the whole  (s + 3) term needs to be squared. We can simplify this and then we get our Laplace transform of the function g (t), which is e^(-3t)*cos(3t). Let us briefly look at the proof of the frequency shift property. And again, like most or, I think even all of the Laplace  transform properties we use the definition directly. So let us insert e^(alpha*t) times f(t) into Laplace transform definition, Which gives us this second term here. Now we can group the two exponential terms together. So we get f(t) is equal to e^-(s - alpha*t) And this is just the definition of capital F, where we have replaced s with (s - alpha). The last thing to check is the convergence. We know that our function f(t) has abscissa of convergence of sigmaC. So capital F(s - alpha), has the convergent sigmaC.  When we consider the real part of (s - alpha), Complex numbers are linear so we can split this real part of s minus real part of alpha and move the real part of alpha to the right-hand side. And then we get that the real part of s needs to be larger than sigma C plus the real part of alpha. Before we finish for this part, a word of warning. The Laplace transform is not multiplicative. Let us consider an exam question from a previous year. Students were asked to solve a differential equation where the forcing or right hand side term, which was f (t) is equal to 3t*e^(2t). A small but not vanishing number of students split this function into two parts. They look for the Laplace transform of t and for the Laplace transform of  e^(2t) separately, you are not allowed to do this. The Laplace transform is not multiplicative. Let us consider this. If we split this, we would get 3 times the Laplace transform of t, times the Laplace transform of the exponential. Both terms we can find in the table will be  3*(1/s^2)(1/s-2) If we use a frequency shift or derivative of transform property, we get the Laplace transform is  3 / (s-2)^2 And these two terms are clearly not the same. This one is the correct one. You're not allowed to do. This. Laplace transform is not multiplicative. So over to you. Try out the frequency shift property on the example question, f (t)= e^2t*sinh(4t) The four possible solutions are given at the bottom of the slide. And as a multiple choice question on the discussion board  head there to answer this question and to discuss your solution, and your approach to solution with the class. If you have any questions about this or any other part, also head over to the discussion board. In the next part, I'm going to introduce a time shift property of the Laplace transform. Thank you.