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2 Consider the piecewise function0 <t < 1 f(t) = 3 - t 1 <t < 3 3 < t(a) Graph the piecewise function f(t):(b) Write f in terms of Heaviside function...

Question

2 Consider the piecewise function0 <t < 1 f(t) = 3 - t 1 <t < 3 3 < t(a) Graph the piecewise function f(t):(b) Write f in terms of Heaviside functions.(c) Find F(s) (i.e;, find the Laplace transform of f(t)):

2 Consider the piecewise function 0 <t < 1 f(t) = 3 - t 1 <t < 3 3 < t (a) Graph the piecewise function f(t): (b) Write f in terms of Heaviside functions. (c) Find F(s) (i.e;, find the Laplace transform of f(t)):



Answers

Determine a function $f(t)$ that has the given Laplace transform $F(s)$. $$F(s)=\frac{3}{s^{2}}$$

Okay. So for our first question, we've got enough of t zero to t squared between zero and two and everywhere else while its periodic. So this right here implies that periodic with period Big T is equal to two. And this is sort of what the graph might look like. Okay. And so on and so forth. Now, from the room 10.3 point three in the book, the LaPlace transform is going to be given by this. Okay? And so this is equal to Well, big T is equal to two. So this is that This is interval from 0 to 2 of its the minus s t t squared. DT. Now aside, let's figure out what this integral right here is. Well, this is clearly going to use integration by parts. We're gonna want to differentiate this polynomial and integrate this exponential. Okay, so this is going to be given by. So if you like you is equal t squared on D V by D. T is equal to eat the minus esty. So this is U times V, and I'm and and you evaluate that between the bounds and then that's minus V D u d t but the minuses cancel. So you get this. So evaluating this one, this is gonna be minus one over us. E to the minus. Two s times T squared two squared, which is that's four. So remember where evaluating TZ equals two and t Z equals zero No s. Now, finally this. What is this? So this is to over us times the integral from zero to t of tea times eats the minus s t d t. So again we identify you in DVD t like this. So on the other side, we have to do integration by parts again. That's going to be so minus one over S T E to the minus Esty. Serious too minus within the council of the minus one over this. Now let's think to go from 0 to 2. Me to the minus esty T t. So now this is minus one over s times two times eats the minus two s. Plus, I was just gonna be minus over a squared. Need to the minus two s minus one. Okay, so this is equal to this. All right, so what's that going to give us? And so now, combining this all on over. A common denominator of s cubed we get. So we get this when we combine it. And I just factored out into the monies to us and we get this. So therefore, this is what the integral from 0 to 2 of t squared e to the minus s t d t is. So if we go back, the plaice transform of F was given by this. And so if we plug everything in, we get eats the minus two s into minus forest squared minus four s minus two plus two K over s cubes. Times this and this is the answer the LaPlace transform.

The problem of we have but A which is like a place of a 50 and this is equal to to christian 02 infinity. It'll power minus 80 entity DT. So this is written as a minus of ace plus one eat bar minus 80 over esquire and 02 infinity. So it is equal to one upon s square. In part we we have the question is 02 infinity. You'd power minus a sTI into T esquire DT and this is equal to minus S. S quality squired plus two S. T Plus two into 8. Power- SD over SQ. Political image, 02 infinity. So this becomes too upon sq And further we have taxi severe. Lovelace of FFT becomes zero and this is three to infinity. You born minus S T D. T. So this is equal to ID par minus SD over minus three and there's three to infinity. So it is minus one upon is in 20 Plus one upon is into eat power minus trees. So further we have the answer has eat par -3 is over. A is so this is the answer to the problem

So this is problem. Seven were given this periodic function. That's one 14 this range and minus one for T in this range. And we've got that f of t plus two is equal to half of tea. So the period T is equal to. So to find the LaPlace transform, we used their, um 10.3 point three in the book. That's going to be 1/1 minus E to the minus period times s terms, the integral from zero to the period of f of tea, times E to the minus s T d t. Okay, now, this is peace wise defined, so we're gonna have to do the integral like this, split it up into an integral from 0 to 1, and then from 1 to 2 can This is multiplying this entire thing. And so this is just the integral of one times this experiential and then minus this. Okay. And so, evaluating the integral this is going to give us minus one over arrests. Eat the minus. Ste. Evaluated between zero and one. And then this is minus. This integral evaluates between one and two. Okay, so this term right here, what is that going to be? That's going to be negative. One over arrests times E to the minus s and then plus one of rest times one. And then this is minus on. Don't forget, everything's in brackets. Negative. One over us eats the minus two s. Then that's minus. And then therefore, that becomes a plus, one of rest times E to the minus s. Okay, so let me rewrite this, okay? So rewritten it becomes this. And so let's just simplify. So we're going to get with. So let's expand and simplify. So these a role s is, by the way, there's no fives in here, okay? And so get it. All under this common denominator, this becomes so it's one and then minus to the minus s, and then plus heats the minus to us. So that is our answer for the LaPlace transform. What about

For this problem. Well, right, this nominator as something squared, plus some constant. So we're going to be completely square. And if we look at as plus one squared, this is s squared. Plus two s, this one. And if we want to get to this denominator wishing to add one on both sides So as plus one squared plus one, is it? Well, the s squared plus two s plus two. That's our denominator. So we're gonna write this as s minus two, all divided by s plus one squared, just one. And we also know that negative too. There's nothing other than s trust. One minus three since plus one and negative three at too negative too. And we can separate. This is in the two fractions, namely, as plus one over s plus one squared plus one minus three over as plus one squared, just one. And now we're gonna take the inverse plus transform on both sides. So the inverse applies transform our capital F is gonna give us just some smaller half of tea. What would be the inverse of boss Transform s plus one over s plus one squared plus one minus three. China's Anversa plots transform of one over s plus one squared. Just one. Now we can use the believe first shifting fear. Um, since we have an s plus one in both of these spots, you know that we can then write this as e too busy. Negative. Two years since a will be negative. One of the inverse loss transform of s over s squared plus one minus three times. You need to The negative team turns inverse of loss. Transform of one over s squared, plus one. And now we're gonna use the rules for the plush transform of the sign and co sign right this as eat negative t factoring out on eating the negative t China is the universal questions form of this, which is just the co sign of t. So one is one squared. So be it. Just one minus three times. Same deal. One is just one square. Somebody is just one three times sign of tea.


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