Laplace transform

Daniel S

Hey,
Having a bit of a problem with Q1 a (i):

When I draw the graph, I get f(t)=U2(t-2) which is correct right? How do I go about integrating that between 2 and infinity? I've gone through the Khan Academy video's on this topic, but he seems to have left this out.

EDIT: Do I just use Ua(t) = (e^-as)/s ? Would that mean the answer is (e^-2s)/s?

Michael Collins

The unit step u(t-a) is zero for t < a and one for t > a, but this is not what you are given in Q1(a)(i).

You are told the f(t) is the function that is zero for t < 2, and for t > 2 it is equal to t-2, this is the equation of a (half) line in the t-f(t) plane. It is a half line because it only extends in one direction (i.e. towards the north-east). Can you draw this function?

You are on the right track with the step function though. To get the Laplace transform you do need to make use of the unit step: instead of writing f(t) as a piecewise defined function (i.e. separate conditions using "if" statements), you can write it on a single line using the fact that the unit step u(t-a) will force f(t) to be zero for t < a when written like

f(t) = (t-2)u(t-2).

This is exactly the same function as given on the exam paper.

So now can you find this in the Laplace transform table?

See this threat for more information: http://www.boards.ie/vbulletin/showthread.php?p=81419876

Note: you've confused the two different ways of writing the unit step function, you can use a subscript like

[latex] \displaystyle u_{a}(t), [/latex]

or you can use the functional form like

[latex] \displaystyle u(t-a), [/latex]

both of which mean exactly the same thing, i.e.

[latex] u_{a}(t) = u(t-a)[/latex]

which both define a function that is zero for t < a and one for t > a.