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1 Find the Laplace transform of f (t) = 4 sinh 3t18 e 5t(15 POINTS)2. Find the Laplace transform of f(t) = cos2 2t (15 POINTS) NOTE: Used your Laplace transformatio...

Question

1 Find the Laplace transform of f (t) = 4 sinh 3t18 e 5t(15 POINTS)2. Find the Laplace transform of f(t) = cos2 2t (15 POINTS) NOTE: Used your Laplace transformation table; be able to simplify your final answer

1 Find the Laplace transform of f (t) = 4 sinh 3t 18 e 5t (15 POINTS) 2. Find the Laplace transform of f(t) = cos2 2t (15 POINTS) NOTE: Used your Laplace transformation table; be able to simplify your final answer



Answers

Determine the Laplace transform of the given function $f.$ $$f(t)=e^{-2(t-1)} \sin 3(t-1) u_{1}(t)$$.

Hello and welcome the problem. 10 Chapter six Section 3 here were asked to final applause transformation of the given heavy said function. Have you said functions. And um to do this we have to remember the golden rule of each little past transformations and that's that you subsea of T um times F of t minus C is equal to E. To the minus C. S. Times little plus transformation of um F. Of T. All right that is. And of course the left side is a little pot transformation as well. So with this we can just substitute this F. F. T minus C. For one because that's essentially what all of these are being multiplied by. So doing that we're gonna be left with E. To the minus s. Well what's a little past transformation of one? What's one of us? Because the plus transformation T. To the end that we found in previous sections is in factorial Over us to the end plus one. So plugging this in The zero factorial is one over S. To the zero plus one is one. Yes. So I left with one of our S. And well so the whole posture information of you won each of the negative s over us. Likewise we can repeat this process for the others. We get to E. To the minus three S over us And -6 e. to the -4 s. Over us. We can of course combine them to be E. To the minus S. Plus two E. To the minus three S. Find a six e. to the -4 s over S. and that concludes the problem.

Hello and welcome to problem 10 of chapter six. Section three you are asked to find the applause transform of the given function F. T. Where's the linear combination of use of one. Use of three and use it for. So let's get into this. We're going to use the equation we learned at the beginning of this chapter. Which is which is that use of C. T. F. Of T minus C. Is equal to the universal applause transform of E. To the minus C. S. Times F. Of S. Great. And how does that help us? Uh Well what is multiplied by this use of one in this case? Um Actually it's uh in this case it's one in this case it's to this case it's six so we can really sub substitute that in. And um what is the laplace transform of 12 and three? Well it's that constant over S. So it will be one over S. To over s. -6 arrests. So let's let's find out what the final laplace transformation is going to be a capital F. Of S. Well that's gonna be eat the minus S over S. And we'll see argument of linear combination with a plus Uh to E. to the -3 s. and -6 E. To the minus for us for this S right there. And that concludes our problem

Hello and welcome to problem 11 of chapter six. Section three you're asked final pause transform 50 which is the quantity t minus three times you said to minus the quantity t minus two times U. Sub three. And to solve this um we are going to recognize our equation that were given in this chapter which is that the applause transform of you subsea of T times F. Of t minus C. Is equal to capital F. Of S. Times E. To the minus C. S. Great. And from here we can ah just look at what matches up and unfortunately nothing matches up by matching up. I mean a. C. Here and to see their so we're gonna have to force it to match up. And so let's say what would the perfect scenario B. It will look like a t minus two times U. Sub two of t minus a T -3 times use of three two. And what do we have to change to do that? Well in this case we added a U. Sub two. So to keep things equal, we'll subtract that. Use up to. And what do we do in this case? Well we had appear we had plus two times U. Sub two of T. And right here we've plus three times use of three of two. So we're going to subtract a use of three. And from here we can use our simple formula on the first bit and uh we know what the laplace transform of just the um heavy side functions by itself and we can do that for those bits. So the laplace transform Fft. Fft, which is F. F. S. Is equal to um E. To the minus two S. Over s minus E to the -3 s. And we're gonna subtract, okay? So this it should be S squared and we're gonna subtract um S. E. To the minus to us and S. E. To the Um -3. Yes. And that concludes our problem.


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