# The second shifting theorem and the unit step function

• Rubik
In summary, we use the second shifting theorem to find the Laplace transform of the given function, f(t) = t2, by breaking it into two cases (t < 4 and t ≥ 4) and using the unit step function to account for the change in the function at t = 4. This allows us to express the function in terms of the unit step function and ultimately find its Laplace transform.

## Homework Statement

I am trying to do some revision for an upcoming exam and one question I am trying to figure out is

Use the second shifting theorem to find the Laplace transfrom of the following function:
f(t) = t2, t < 4
t, t ≥ 4

## The Attempt at a Solution

I just don't understand how to get from the question to
f(t) = t2[1 - u(t-4)] + tu(t-4)
I am really struggling with applying the second shifting theorem to express in terms of the unit step function I am failing to see how it works because nothing is explained in basic detail?

Rubik said:

## Homework Statement

I am trying to do some revision for an upcoming exam and one question I am trying to figure out is

Use the second shifting theorem to find the Laplace transfrom of the following function:
f(t) = t2, t < 4
t, t ≥ 4

## The Attempt at a Solution

I just don't understand how to get from the question to
f(t) = t2[1 - u(t-4)] + tu(t-4)
I am really struggling with applying the second shifting theorem to express in terms of the unit step function I am failing to see how it works because nothing is explained in basic detail?

Think of it this way. You start out with f(t) = t2. Then at t = 4 you want to take out the t2 and put in t, so you add the term u(t-4)(-t2+t).

Then put it all together:

f(t) = t2+u(t-4)(-t2+t)

Now, if you wish, you can collect terms on the various powers of t:

f(t) = t2(1-u(t-4)) + tu(t-4)