Finding the Laplace Transform for f(t)=2(1-u(t-1))+t(u(t-1))

[jwang34]

Homework Statement

Given f(t)= 2 if 0<=t<1, and t if t>=1 I need to find the laplace transform.

Homework Equations

Using the second shifting theorem:
f(t)= 2(1-u(t-1))+t(u(t-1)), which is the heaviside function.

The Attempt at a Solution

The first term, I understand. It is 2-2u(t-1) which I got the transform to be:
(2/s)-(2/s)(e^-s). The second term I'm a little fuzzy. I think I need to shift the t, so I guess I would have (t-1)u(t-1). But I'm not sure if this is right. So any help is greatly appreciated.

 

[mighty2000]

Thank you for your question. The Laplace transform of the given function can be found using the definition of the Laplace transform and the properties of the Heaviside function. Let's break it down step by step:

1. First, we can rewrite the function as f(t) = 2(1-u(t-1)) + t(u(t-1)). This can be done by separating the function into two parts: one for t < 1 and one for t >= 1. In the first part, t is always less than 1, so we can replace it with the constant 2. In the second part, t is always greater than or equal to 1, so we can leave it as is.

2. Now, using the definition of the Laplace transform, we can write:
L[f(t)] = ∫f(t)e^-st dt = ∫[2(1-u(t-1)) + t(u(t-1))]e^-st dt.

3. We can use the linearity property of the Laplace transform to split this into two separate integrals:
L[f(t)] = 2∫(1-u(t-1))e^-st dt + ∫tu(t-1)e^-st dt.

4. Now, using the property of the Laplace transform that states L[u(t-a)f(t-a)] = e^-asF(s), where F(s) is the Laplace transform of f(t), we can rewrite the first integral as:
L[f(t)] = 2∫e^-st dt - 2∫u(t-1)e^-st dt + ∫tu(t-1)e^-st dt.

5. The first integral is a standard integral that evaluates to 1/s. The second integral can be rewritten as u(t)e^-st, which evaluates to e^-st for t >= 1 and 0 for t < 1. The third integral can be written as u(t-1)te^-st, which evaluates to te^-st for t >= 1 and 0 for t < 1.

6. Putting it all together, we get:
L[f(t)] = 2/s - 2e^-s/s + 1/s ∫te^-st dt.

7. The final integral can be evaluated using integration by parts, which gives us:
L[f(t)] = 2/s - 2e