applying two Laplace properties on same function

Question

For example, the Laplace transform of

$(t - 3)\cdot u(t-3)$

I'm confused about how to apply the two Laplace properties (multiplication of t and time shift).

Do I apply one property first then the other? But if I do that won't the second property be applied to a function of s?

I must be missing something trivial because no one else seems to have this question, sorry if it's a dumb problem

Answer 1

You can do it in a number of ways.

Method 1: Start with $f_1(t) = u(t)$, then we have $\hat{f_1} (s) = {1 \over s}$.

Now multiply by $t$ to get $f_2(t) = t f_1(t)$, which gives $\hat{f_2}(s) = - f_1'(s) = {1 \over s^2}$.

Finally time shift to get $f_3(t) = f_2(t-3)u(t-3) = f_2(t-2)$, which gives $\hat{f_3}(s) = e^{-3s} \hat{f_2}(s) = { e^{-3 s} \over s^2 }$.

Method 2: Start with $g_1(t) = t$ (implicitly $t u(t)$, assuming the unilateral transform) to get $\hat{g_1}(s) = {1 \over s^2}$.

Then time shift to get $g_2(t) = g_1(t-3)u(t-3)$ which gives $\hat{g_2}(s) = e^{-3s} \hat{g_1}(s) = { e^{-3 s} \over s^2 }$.

Method 3: Start with $h_1(t) = u(t)$, which has $\hat{h_1}(s) = {1 \over s}$.

Then time shift to get $h_2(t) = h_1(t-3)$ which gives $\hat{h_2}(s) = e^{-3s} {1 \over s}$.

Now we need to multiply by $t-3$ which gives $h_3(t) = (t-3) h_2(t) = t h_2(t)-3 h_2(t)$. Then we have $\hat{h_3}(s) = - \hat{h_2}'(s) -3 \hat{h_2}(s) = { e^{-3s} \over s^2}$.

Answer 2

Just apply the definition of a Laplace transform. Note that the function is only nonzero when $t \gt 3$, so you have the LT being

$$\int_3^{\infty} dt \, (t-3) \, e^{-s t} = e^{-3 s} \int_0^{\infty} du \, u \, e^{-s u} = \frac{e^{-3 s}}{s^2}$$