Understanding Laplace Transforms: Solving Problems and Applying Properties

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The discussion revolves around two main questions regarding Laplace transforms. The first question addresses a discrepancy in the results of Laplace transforms, specifically how to simplify the expression 1/s - 1/(s-1) to match the provided answer of 1/s(s-1). The second question explores whether the function f(t) = sinh(t)cosh(t) involves convolution and clarifies the correct Laplace transform of e^-t as 1/(s+1). Participants emphasize the importance of understanding the definitions of hyperbolic functions and suggest referring to resources for further clarification on Laplace transforms. Understanding these concepts is crucial for accurately solving related problems.
LostEngKid
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Hi all, 2 questions here

1) I've been doing some questions on laplace transforms and have been running into some trouble getting my answers into the same form as the answers given with the questions.
For example:

f(t) = 1 - e^(-t)
Using the linearity property i got 1/s - 1/(s-1) which is correct? But the answers give 1/s(s-1). Later in the lecture notes they show how to do the inverse laplace transform of the answer they gave and use partial fractions to get it into the form of the answer I got, i was just wondering what they have done to go from my answer to their answer, what steps am i missing?

2) f(t) = sinhtcosht, does this involve convolution or is that only when doing the inverse laplace transform where L(f)L(g) = f(t)*g(t)?
 
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Its been a while since I've looked at Laplace transforms, but is the LT of e^-t not 1/(s+1) instead of 1/(s-1)?
 
1) Try and simplify your expression using a common denominator of s and (s-1).

2) Do you know the definition of sinh(t) and cosh(t) in terms of exponentials?
 
oops yeah i did mean 1/(s+1) danago

and Pere i wouldn't have a clue about sinh(t) and cosh(t) a exponentials, maybe i should have paid more attention in first year maths
 
LostEngKid said:
oops yeah i did mean 1/(s+1) danago

and Pere i wouldn't have a clue about sinh(t) and cosh(t) a exponentials, maybe i should have paid more attention in first year maths


If you don't know about the definition of sinh(t) I don't see how you could calculate its LT..

<br /> \begin{align*}<br /> \sinh t =&amp; \frac{e^t-e^{-t}}{2}\\<br /> \cosh t =&amp; \frac{e^t+e^{-t}}{2}<br /> \end{align}<br />

For some background reading you could have a look at http://en.wikipedia.org/wiki/Hyperbolic_trigonometric_function" .
 
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There should be a table of Laplace transforms which you can refer to check LT of sinh and cosh. But in any case, Pere Callahan's post is sufficient for you to do it, so long as you know the LT of e^t and e^-t.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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