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Questlon 5Use the Laplace transformation palrs and propertles to (ind the bilateral Laplace transform of the signal I(t) wheres() =ite "u(t n2/"u(-t-21. I...

Question

Questlon 5Use the Laplace transformation palrs and propertles to (ind the bilateral Laplace transform of the signal I(t) wheres() =ite "u(t n2/"u(-t-21. If the transformation does not exist, state the reason. If it exists, sketch its ROC; You must mention the pairts) and property number(s) used In the solutlon;

Questlon 5 Use the Laplace transformation palrs and propertles to (ind the bilateral Laplace transform of the signal I(t) where s() =ite "u(t n2/"u(-t-21. If the transformation does not exist, state the reason. If it exists, sketch its ROC; You must mention the pairts) and property number(s) used In the solutlon;



Answers

Use properties of the Laplace transform and the table of Laplace transforms to determine $L[f]$. $$f(t)=e^{3 t} \cos 5 t-e^{-t} \sin 2 t$$

For this problem. We're going to start by taking the Flashdance perform of F. And when we do so forget does applause transformer the right side here. And we have a Constance. We can pull that outside. This is two times of loss. Transform of the heavy side of five of tea times T minus five. And now we didn't recognize that we can use the second I believe shifting the're, um which says they believe the boss transformer, The heavy sided. A times a function of T minus A. This will just be each of the negative A s time still lost transform of our function f And we can notice here that we have a heavy side of five and a function effort e legal the tea such that if we have if don t minus A which is five music will the t minus five, which is what we have up here. Tens will just be two times e to the negative five s, since a is five times little plus transform of f which is just tea. Another boss transform of just t is the class transform of tea to the first power which would be one factorial over s to the one plus one, which is to to this is engines to ease the negative five s all divided by a squared.

When we take the flash transform both sides here you have to the class transforming in England You're the boss Transform of you to the negative five t divided by the square GTI And here we can use the first shifting through which says that it would take the class transform of eating to the 18 time some function FFT This would be the plotters transform of the function fft just evaluated to add s minus a some is that this will be the boss transform of one over the square root of tea as we know from this rule will be a pie over s calls with 1/2 power. But we have to subtract a which in this case here would be negative five So ask minus negative five yes, plus five all square root.

For this problem. We're going to use the convolution. Param shown down here and we're gonna write are integral here as the convolution Two functions. Where were one F tilda, which is equal doing t squared and another gene is equal to to the t then By this definition, F. Tilda the convolution about Tilda and G will therefore be the integral from zero to t of F Tilda, which is T square and 17 minus town squared times, GM, tilt of town, which is just be to the tower, you tell. And this is what we have for our initial integral. So therefore, this integral. There's just the convolution of Tilda and G. So therefore, if we take the loss, transform of our original equation f also just be the loss transform of are of Tilda Consulate E g. And by this rule here, we know that this will just be the class transform Uh, F Tilda times in the last transform of G. So we'll have the passion swarms Tell them which was t squared times the transformer gene, which was e to the team. And we know that the most transformative racism manager is given by this sort of preposterous. For much he squared will be two factorial overestimate two plus one, which is three times of the questions from each of the tea. Well, here is just one so given by this rule and serve, you won Over s minus one. And now we can just simplify this a little bit. Two factorial is too. And most boy, these two fractions together we'll get to over s to the fourth minus sq.

The first step to this problem is going to be rewriting are starting equation. We shall be two plus two times you to the negative. T you're rewriting one as e over tea. Was that every site at one of tea? And when we do this, we can make a Colin factor inside of the parentheses here and removed the factor of E whenever you need to get the business too. Plus two over e thank you to the negative. T minus one minus you of the heavy side at one of tea. By doing this, we can Now, when we go to take the flash transform, we're gonna get the LaPlace transform with two. She was a constant. Just pull it outside. So this will be able Fox Transformer of one. But one is nothing other than Team 20 with power plus two over it many times to look lost. Transform of the heavy side at one of t times, usually negative. T minus one minus The boss transform of two over e times Negative. You will be too negative, too. Out front, seven times the heavy side function that one. And now we can look at each one of these individually, so we can use our rules. So you know that the class transform of TV's to some manager and it's just this so therefore this across France will be two times. And his 00 factorial zero factory was just one over s to the zero plus one is just one power plus to overeating, times of questions, form of this. But here we can use the second shifting zero up here are a people toe. One in this function over here is mean to the negativity. So therefore, are the loss transform will be each than a s where there's one since meeting to the s negative. Negative, yes. Times old class transformer F, which is just given by this rule up here where Eddie is negative one. So this will be one over s pose one minus two times of class transforming heavy side function. Just this where is one to minus two times each of the negative s times, one of us. And now we can just reduce this. We're simplify this the other this to over s plus two times Keating to the negative one times near the negative s or two times you need to the negative as plus one. All over s plus one minus two You to the negative s progress.


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