Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Piecewise, continuous and periodic functions. We're going to use a unit step function here. The unit step function is usually used for a concept of time, so something doesn't happen until some given time a. So our definition for a unit step function, it's, it's going to be 0 if T is less than a or one if T is greater than or equal to a. We have two different notations U sub a of T or U of the quantity t -, a that t -, a is just telling us we're shifting the starting time to whatever a was. So if we look at the Laplace of those functions, we know that that would really just be by definition the integral zero to Infinity of E to the negative St. times EU of T minus ADT. And this has to be when T is greater than or equal to A. If T was less than or equal to a, it would have a different concept, it would take on a different definition. So in this case we have 0, the integral 0 to a of E to the negative St. And when we look at that, we know that our U of AT function is just 0 for that location times DT plus then if we take that A to Infinity, E to the negative St. our U of t -, a function, there is one DT. If we think about the limit as B goes to Infinity of that integral evaluated, we end up seeing that that's really just 1 / s E to the negative AS. So that's going to be an important piece we're going to use. If we have Laplace of the function U of T, we can see that that equals 0 to Infinity of E to the negative Stu of T is just one DT because there was no shift, so it was defined. So when we evaluate that we get 1 / s So these two relationships, this t -, a versus just T tells us there is some sort of a relationship between those two functions. So we can see that when we have the t -, a, the two Laplace's just really have an E to the negative AS that differs between them. So the theorem of translation on the T axis, if the Laplace of F of T exists for some S greater than the constant, then the Laplace of the unit function times the F of t -, a or the shift equals E to the negative AS of F of S. And the inverse Laplace is E to the negative ASF of S equaling U to the T minus AF to the t -, a for some S greater than or equal to C + a because of that shift. So if we go back and we use our unit function concept, we get 0 for T less than A1 for T greater than or equal to a. So if we thought about our U of t -, a times that function with that translation or that shift of a, we could see that it's really going to be 0 when it's T less than a or F of t -, a when T is greater than or equal to a. So let's do the proof here. We have E to the negative ASF of S equaling E to the negative AS the integral zero to Infinity of E to the negative S Tau F Tau D Tau. So here we could think of that as zero to Infinity of E to the negative SA plus Tau F Tau D Tau. This is just a translation and I all I did was I multiplied that E to the negative AS in. So next we're going to substitute T for a plus Tau. And if we have T equaling A+ Tau, we know that t -, a would be Tau and also DT equal D Tau. So if we change our lower bounds, we would get T equaling a + 0 and the upper bound would be T equaling A+ Infinity, which is obviously Infinity. So now we get a new function we're going to pull out that eat in the negative AS because that's just a constant of the F of S and then that's the integral A. Now our new lower bound to Infinity of E to the negative STF of T minus ADT. So next, by the unit step function, we get that E to the negative ASF of S equaling the zero to AE to the negative St. Now remember here, our U of T -, a is just 0. So the second piece, it's more interesting because that U of t -, a is actually 1. So if I put those two pieces together, that's really just our zero to Infinity of E to the negative St. our U of t -, a once again, zero from zero to a or one from A to Infinity. So I can now merge those two pieces of F of T minus ADT. So our E to the negative ASFS equals the Laplace of UT minus a times FT minus a translation there. So let's do an example. If we're given F of T equaling 1/2 T squared. Now we have to know that the Laplace of F to the T to the N power is N factorial over S to the n + 1. Using that concept, we could think of the 1/2 being in front, just being a constant. So the 1/2 of Laplace t ^2 would be 1/2 * 2 factorial over s ^3, which would simplify just to 1 / s ^3. So with a translation of U * t minus AU of t -, a times that 1/2, Now the quantity would be t -, a ^2. So it's going to be 0 if we have T less than A and it's going to be 1 half t -. A squared squares on the outside. Square was on the outside for T greater than or equal to a. So the Laplace inverse of E to the negative St. over s ^3. This E to the negative St. is coming from that translation is just UT minus a * 1 half t -, a ^2. So looking at another example, the Laplace of G of T if G of t = 0 when T is less than 3 and t ^2 when T is greater than or equal to three. So our UTGT would be zero t ^2 t less than 3T greater than or equal to three. But because this is a three, we actually are going to need to get U of t - 3 and F of t - 3. So we need the new function to be shifted 3 to the right. So if we thought about some new function, let's say F of t -, 3, we could then think of that as t -, 3 + 3 to get back that G of T equation to get some equivalencies. Now what we're really going to do is if we think about t -, 3, we could call it a new variable, say a. If we let a be t -, 3, then over here would be a + 3 ^2. Now for simplicity sake and because we're using TS in our functions and our definitions, we're actually going to use T Again, this is just a shift. If you thought about if I add 3 here, I have to add 3 to this T. So these are equivalent equations. So now we get our F of S equaling the Laplace of. If I distribute this out, I get t ^2 + 6 T +9. Once again, using the formula we talked about just a moment ago, we get 2 factorial over s ^3 + 6 times the one factorial over s ^2 + 9 * 0 factorial over S or that simplifies to 2 / s ^3 + 6 / s ^2 + 9 / s. So the Laplace of G of T would equal the Laplace of this transformation shifting of three. So U of t - 3 F of t - 3, which would equal E to the -3 S because that -3 there of the F of S. And we just found the F of S being this equation here. Putting it all together. Another example, the Laplace of F of T given F of T equaling cosine 2T if zero is less than or equal to T less than 2π and 0 if T is greater than 2π. Now this is a little different because of the unit function usually has the less than 2π portion of 0 and the greater than 2π is 1. So we're going to have the shift here U of t - 2π because the 2π here and we get zero and one. But we need to figure out how we can define this F of T in terms of multiplying by a shift. So we're going to think of 1 -, U of t -, 2π because then if I have 1 -, 0, we'd get one and we get 1 - 1, we would get 0. So here that's going to be what I want. If I just multiply this equation 1 -, U of t -, 2π times my F of T, so one times the cosine 2T and 0 * 0. So now our F of T is going to be 1 minus the UT -2π cosine 2T when T is less than 2π. Now we're going to say time is got to be greater than 0. So that's going to be understood here. And then zero of T is greater than 2π. So now we're going to just distribute out. We get cosine 2 TU of t -, 2π, cosine 2T. So the Laplace of F of T would be the Laplace of cosine 2T minus the Laplace of UT -2π cosine 2T. And our Laplace cosine 2T is just s / s ^2 plus whatever the coefficient on the T was squared minus this U2 - 2π t - 2π is a shift, so it's E -2π S and we could actually think of this cosine as 2 * t - 2π. And because it's periodic, we know that that's really the equivalent of cosine 2T. So when we do that, we get our S over the S square plus 4. But this piece was necessary to have it be the transformation. So in it then equals 1 -, e to the -2 Pi s * s / s ^2 + 4. Another example, this one is actually going to be a spring and we have M equal 3 hat 32K equal 4. And we're going to have an external force from zero to 2π of cosine 2T. And then at 2π it's going to turn off. So there won't be an external force. So we have our F of T equation being given as X double prime plus 4X and our initial conditions X0X prime 0 equaling zero. So the previous example, our F of T was cosine 2T and we found our F of S equation also. So now taking that F of T equation up here, by our definition this X double prime is going to be s ^2 X of S minus SX, not no minus s * X of 0 -, X prime 0 plus this 4 * X of S equaling that whole F of S equation, because that's the Laplace of one side equaling the Laplace of the other. If I factor out an X of S and divide by what was left, which is an s ^2 + 4, all that really happens is like an s ^2 + 4 quantity squared in the bottom, and I'm going to separate the two terms in the numerator. So if we keep going on this, we're going to recall that the inverse Laplace of E to the negative AS of F of S is just a transformation, it's a shift. And then the Laplace inverse of S over the quantity s ^2 + K ^2 ^2 is a formula, and that equals 1 / 2 KT sine KT. So when we're looking at this X of S equaling these equations, this s / s ^2 + 4 ^2, we're going to use this formula. So 1 / 2 * 2 T sine 2T minus. In this case, we have our shift because of this E to the -2π S so UT -2 Pi 1 / 2 * 2 t - 2π because it's got to be the same parenthesis here. And then sine of the two t - 2π, here's two t - 4 Pi. The period of this is going to be 2π / 2 or π. So this -4 Pi is just a repeat so that we can think about going away because it's just repeating. So then we're going to factor out what they have in common, which is going to be a 1/4 sine 2T. When I factor that out, I see that there's left AT minus UT -2π * t -, 2π. So our X of T definition, if we just think about taking out these shifts, is going to be T less than 2π. So if AT is less than 2π, we know that this U of t -, 2π is going to be 0. So we get T times the 1/4 sine 2T. Now if T is greater than or equal to 2π this U turns into one and we get t -, t So those cancel and a -, a negative 2π. So 2π / 4 is π halves sine 2T. Now we have another theorem. Periodic functions F of T plus P equal F of T. This is a theorem on transforms of periodic functions. So let F of T be periodic with period P and piecewise continuous for T greater than or equal to 0. Then the transform F of S equals the Laplace F of T exists for S greater than 0 and is given by F of S equaling 1 / 1 -, e to the negative PS from zero to P on the integral of E to the negative STFTDT. If we do our proof, we have F of S equaling the integral zero to Infinity of E to the negative STFTDT. Now what we're going to think about is we're going to think about taking that and making them into tiny, tiny, tiny slices, because we're going to take each of those slices and add them up eventually. So if I'm taking slices now, I'm going to look at just one specific slice from north to north plus one. We think about our norm of our partition here. So we're just going to take one of those increments. When we do this, we have that E to the negative STFTDT. Now we're going to substitute T equaling Tau plus NP. So T minus NP is Tau. And we could know that our DT would equal our D Tau. And if we change our lower bounds and our upper bounds, we'd have NP minus NP because the T here equaling Tau and then the n + 1 * P minus NP equaling Tau for the upper bound. So our new bounds are going to be 0 to P E to the negative S quantity T Tau plus NPF of Tau plus NPD Tau. So I can pull out that E to the negative NPS because that's just a constant. We're going zero to P on the integral E to the negative S Tau F Tau because it was periodic. This NP is really the same thing as just the Tau D Tau. So when we think about that Sigma notation, we can realize that if I stick in 0, E to the zero is 1, hence the one here. And then when I stick in north equaling one, I get E to the negative PS and N equal 2 I get E to the -2 PS. So this actually turns into a geometric series. This piece here. And remember, geometric series is 1 + X + X ^2 plus... equaling 1 / 1 -, X So this equation is really going to turn into this first piece of that parenthesis is 1 / 1 -, e to the negative PS because that's what we multiplied to get from one term to the next. And then the integral zero to PE to the negative S Tau F Tau D Tau. So let's look at an example. If we look at this example, we have F of T equaling -1. We're going to use the step function here. And the notation for step function is a bracket with a line on the inside. So this is just going to be our fraction of t / a where that represents the greatest integer not exceeding t / a. So over here is a picture. We can see it's periodic because it's going to repeat every two A. So our F of S is going to equal 1 / 1 -, e to the -2 A S0 to 2A of E to the negative STFTDT. But this FT we're going to split up and to 0 to a it's going to be 1 and A to 2A it's going to be -1. Then if we evaluate each of those pieces, we get the -1 / s E to the negative St. and we stick in our upper bound minus our lower bound for both those pieces. And now we're going to do a bunch of algebra. So I'm going to factor out the 1 / s because that's in common. And I grouped my like terms together. And then we're actually going to factor this parenthesis. And we're also going to factor the denominator as the difference of two squares. So when we look at this this denominator here, I'm going to pull the 1 / s out in front the denominator 2 difference of two squares. E to the negative A s * E to the negative AS is, E to the -2 AS and then this term here factors 1 -, e to the negative AS 1 -, e to the negative AS. So if I think about now, I can have a term in the numerator and denominator cancel. Simplifying to this now this looks kind of similar to a formula we might know, but it's not quite there. So what we're going to do is we're going to multiply the top and the bottom by E to the A s / 2 and E to the A s / 2. When we do that, we can now see that we get 1 / s times this hyperbolic tangent formula of A s / 2. Thank you and have a wonderful day.