Hello wonderful mathematics people, this is Anna Cox from Kellogg Community College. The convolution of two functions. The convolution F star G of the piecewise continuous functions F&G is defined for when T is greater than or equal to 0. As follows, F star G of T equal the integral zero to T of F Tau times G of T minus Tau D Tau. The idea is that this is a new type of product of F&G, so that it's transform is the product of transforms of F and Gie. If we can find a transformation of F and of G, we might be able to come up with a product that's easier to deal with. So the Laplace of F star G is just the Laplace of F times the Laplace of G. The convolution is also commutative. The commutative property holds here. So an example cosine T star sine T. We can think of that as the integral of 0 to T of cosine Tau sine of T minus Tau D Tau. Now we have an identity from our trick days of cosine A sine B equal 1/2, sine A+B minus sine a -, b that all quantity of putting in the proof. Just so that you can remember it, if we thought about our sine alpha plus beta or sine A+B and our sine A -, b formulas, we'd have sine A, cosine B plus cosine A, sine B and sine a cosine B minus cosine A, sine B. If I subtracted these two formulas, the sine A cosine BS would cancel, leaving us two cosine A sine B equaling sine A+B minus sine A -, B. Divide each side by two and there's the formula. So if we go back to the original, we get the integral of 0 to T of 1/2 sine of the two angles added together. So the Taos cancel, so sine T minus sine of the two angles subtracted. So we get T minus or we get 2 Tau minus TDT. So continuing on, we can see that that would evaluate into if we're integrating in terms of D Tau, we're going to have 1/2. This first term didn't have a Tau in it, so Tau sine T this next one, this two Tau in the inside, so derivative the inside is 2, so we're going to divide by 1/2 and we're doing the integral. So negative sine turns it into positive cosine, the two Tau minus T from zero to T. Now when we stick T in for the Taos, we get 1/2 T sine T minus 0 sine T I did the first term 1st and then plus 1/2 cosine two t -, t or just t -, 1/2 cosine 0 - t So cosine of negative T is just really the cosine of T because it's an even function. So the last two terms cancel, so we just get 1/2 T sine T. So cosine T star sine T is just really 1/2 T sine T. There's a theorem that says the convolution property. Suppose that F of T&G of T are piecewise continuous for T greater than or equal to 0, and that the absolute value of both functions are bounded by some function ME to the CT as T goes to Infinity. Then the Laplace transforms of the convolution FT star G of T exist for S greater than C, and the Laplace of those products, those star functions, is just the Laplace of each piece individually, so that the inverse Laplace of F of s * g of S is just equal to FT star G of T. Or we could think of that as the integral of 0 to T of F Tau times G of T minus Tau D Tau. So if we thought about F of T equaling sine 2T and G of T equaling E to the T, we'd get sine 2T star E to the T, so zero to T sine of two Tau E to the T minus Tau DT D Tau. So then I can take out an E to the T. So we get E to the negative Tau sine of two Tau DT. Now we're going to do this by integration by parts twice to figure that out. So if we do integration by parts twice, we can see that we really get in the end are zero to T either the negative Tau sine of two Tau D Tau equaling if we divide by 5 negative 1/5 E to the negative Tau sine 2 Tau -2 fifths E to the negative Tau cosine 2 Tau from zero to T. So when I put in the zero and the T the bounds, I'm going to end up with that just coming out to be putting in my zero to T right here I get that all evaluating to 2/5 E to the t -, 2/5 cosine two t -, 1/5 sine 2T. So this was the original. This is what it was equivalent to. And then doing the integration by parts on the last page, we got this. So I'm going to take out this portion. I'm going to be replacing what it equals and evaluating for the bounds differentiation of transforms. If F of T is piecewise continuous for T greater than or equal to 0, and the absolute value of that function is less than or equal to some MECT to the CT as T goes to Infinity, then the Laplace of negative T times that function equals F prime of the S for S greater than some C. Thus the function equals the Laplace inverse of F of S. Or we could think of that as -1 / t Laplace inverse of F prime of S. Hence the Laplace of T to the NF of T is really just equal to -1 to the N power of F to the derivative of N of S for N equal 123, etcetera. So if we have the Laplace of t ^2 sine of KT using that formula from a moment ago, we're just going to have -1 to the two -1 ^2, the 2nd derivative because of the two here of our sine KT in terms of our s s. So that's just going to be K / s ^2 + K ^2. So the first derivative in terms of S of that is just -2 SK over s ^2 + K ^2 ^2. And then when we do the 2nd derivative of that, the derivative of the top times the bottom minus the derivative of the bottom times the top all over the bottom squared is going to simplify into SKS squared minus two K ^3 / s ^2 + K ^2. All quantity cubed. Back here I cancelled one of the s ^2 + K ^2 in each term. So the Laplace inverse of F of S equaling -1 / t Laplace inverse of F prime of S. So if we thought about the Laplace inverse of tangent inverse of 1 / s, we use that formula above and we get -1 / t the Laplace inverse of the derivative in terms of S of tangent inverse of 1 / s. Now the derivative of tangent inverse of 1 / s is going to be -1 / s ^2 / 1 plus the quantity 1 / s ^2. If we just multiply the top and bottom by s ^2, we're going to get -1 / s ^2 + 1. And the fact that we want the inverse Laplace tells us this is really our sine T formula with the negative so negative sine T. So when we finish that up, we just get sine t / t. Now if we don't remember the derivative of this inverse trig, I showed it here. So if we want the derivative of the inverse trig, let X equal the inside piece and then we're going to find DXDS. So if I take the tangent of each side, take the derivative of each side secant squared XDX equal -1 / s ^2, DS secant squared by our trig identities is just one plus tangent squared X. But tangent X was 1 / s, so that's 1 + 1 / s ^2. And that all should have been multiplied by the DX. So DXDS -1 / s ^2 / 1 + 1 / s ^2. So the derivative in terms of S if we get rid of the axes of tangent inverse 1 / s is just this formula multiplied top and bottom by s ^2. So the integration of transformation. Suppose that F of T is piecewise continuous for T greater than or equal to 0, and that F of T satisfies the limit as T goes to 0 of F of T over T exists and is finite, and that the function, the absolute value of the function is less than or equal to some ME to the CT. As T goes to Infinity. Then the Laplace of F of t / t is just the integral S to Infinity of F gamma, no F Sigma D Sigma for S greater than C So that also tells us that F of T equals the Laplace inverse of F of S which would just equal T Laplace inverse of the integral S to Infinity of F Sigma D Sigma. So we can actually back here the limit as T goes to 0. It has to exist, but we really care about it existing from the right hand side because we're going to look at the absolute value of that. But it really does need to exist. So if we look at an example, the Laplace of sine HT, the Laplace of the hyperbolic sine of t / t. So the first thing is to prove that the limit exists. So the limit as T goes to zero. Our definition of hyperbolic sine is E to the t -, e to the negative t / 2. So if we do Lopital's rule there we get E to the t + e to the negative t / 2. When we stick T going to 0 in, we get 1 + 1 / 2 equaling 1. So that exists. Then the Laplace of the hyperbolic sine of t / t is going to equal the definition of the integral as to Infinity of the Laplace of the hyperbolic sine of TD Sigma. So S to Infinity of one. The definition of the Laplace hyperbolic sine T is just one over in this case Sigma squared -1 D Sigma. Evaluating that using our partial fraction concept, we get our one over Sigma squared -1 and we saw for A&B. So we get the integral of S to Infinity of 1/2 one over Sigma -1 -, 1 over Sigma plus 1D Sigma. Evaluating that those are just our natural log rules, so we would get 1/2 natural log of the absolute value of Sigma -1 over Sigma plus one, putting in our bounds of Infinity to 1. So 1/2 of the natural log of one is just zero minus 1/2 natural log of s -, 1 S plus one divided take the negative up into the exponent. So that just takes the reciprocal. So a final answer of 1/2 natural log s + 1 / s - 1. Thank you and have a wonderful day.