How Do You Use the Unit Step Function to Find the Laplace Transform of g(t)?

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The discussion focuses on using the unit step function to express g(t) = 1 for 0 < t < 1 and g(t) = 0 for t > 1. The correct representation of g(t) in terms of the unit step function is u(t) - u(t-1). To find the Laplace transform, the relevant formula is L[f(t-a)u(t-a)] = e^{-as}F(s), where F(s) is the Laplace transform of f(t). The Laplace transform of g(t) is derived as L(g(t)) = 1/s^2 - e^{-s}(1/s^2 + 1/2). Understanding the manipulation of unit step functions is crucial for solving this problem effectively.
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Homework Statement



This has to do with the unit step function. The question is;

Sketch the function g(t) = 1 for 0<t<1 and is 0 for t>1.

Express g(t) in terms of the unit step function and hence or otherwise show that

L(g(t)) = 1/s^2 - e^-s (1/s^2 + 1/2)

Homework Equations





The Attempt at a Solution



I sketched the graph (see attachment below). I am guessing the unit step function is t(u(t) - u(t-1)).

I tried one way of getting that answer like this t( u(t-1) u(t+1) - u(t-1) ). Obviously it didnt work out. I think I need to get u(t) in the form u(t-a) so I can use the table of Laplace transforms and just read out of it. How do I manipulate this kind of functions?

Thanks.
 

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  • Unit Step function.JPG
    Unit Step function.JPG
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This is from my notes:

http://img508.imageshack.us/img508/1289/unitstepdk5.jpg

In this case, the function should be u(t) - u(t-1).

The Laplace transform you should be using is L[f(t-a)u(t-a)] = e^{-as}F(s) \ \mbox{where} \ F(s) = L[f(t)]

Note that u(t-a) = f(t-a)u(t-a) where f(t) = 1. f is a constant function.

So you only need to find the laplace transform of 1 and add in the e^(-as) factor in front of it.
 
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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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