Can someone show me how to show that:

f(t) = (t, 0 < t < 2

(4-t, 2 < t < 4

(0, t > 4

is equal to

f(t) = t - 2(t-2)U(t-2) + (t-4)U(t-4)

and determine its Laplace transform.

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- Nov 23rd 2010, 01:13 AMhunterage2000step functions and Laplace Transforms
Can someone show me how to show that:

f(t) = (t, 0 < t < 2

(4-t, 2 < t < 4

(0, t > 4

is equal to

f(t) = t - 2(t-2)U(t-2) + (t-4)U(t-4)

and determine its Laplace transform. - Nov 23rd 2010, 04:39 AMFernandoRevilla
Have you tried using the definition?:

$\displaystyle \mathcal{L}\left\{{f(t)}\right\}=\displaystyle\int _0^{+\infty}e^{-st}f(t)\;dt=\displaystyle\int_0^{2}e^{-st}t\;dt+\ldots$

Regards.

Fernando Revilla - Nov 23rd 2010, 04:59 AMAckbeet
What ideas have you had for the first part so far?

- Nov 23rd 2010, 05:18 AMhunterage2000
I would have put

f(t) = tU(t-2) - 4-t-U(t-4)

and not sure how it would be cleaned up.

The laplace transform is a mystery. - Nov 23rd 2010, 06:16 AMAckbeet
The way to think of this problem is that the Heaviside step functions (the U(t-a)'s) allow you to "switch on" certain portions of the function at different times. So, let's take your candidate. When t<2, both step functions are zero, because the arguments to both step functions are negative. You're left with f(t) = -4-t for t<2. That does not match up with the definition in the OP. So this can't be correct.

Technically, the function in the OP is undefined for t<0, so you don't really need a step function U(t) multiplying the initial t.

Keep in mind that the step functions are cumulative. As t marches past the various cutoffs, more and more step functions will be active. So, for example, when 2<t<4, you've got the initial t, but also the -2(t-2) multiplying the U(t-2). That coefficient of the step function is carefully chosen not only to cancel out the first piece (because it's still active), but also to get the desired 4-t.

Does this help?

As for the Laplace Transform, I would just use its linearity and the various properties and theorems.