Finding the Laplace Transform of sin(7t)u(t-3) with Step Response u(t)

[Meadman23]

Homework Statement

Find the laplace transform of:

v(t) = sin(7t)u(t-3)

Homework Equations

u(t) is step response

The Attempt at a Solution

I know that taken alone the laplace of:

sin(7t) is 7/(s^2+49)

u(t-3) is (1/s)*(e^-3s)I don't understand how I would figure out the answer to this problem if I have u(t-3). For some reason, seeing the u(t-3) rather than u(t) is really making things hard to understand since I don't know how u(t-3) would be integrated.I considered setting t-3 =x which would yield:

v(t) =sin(7x+21)u(x)

v(t) = sin(7x)cos(21) + cos(7x)sin(21)

Laplace = [7*cos(21) + s*sin(21)]/[s^2 + 49]

but I don't know if I'm that's the right answer.

 

[mighty2000]

Well, let's break this down step by step. First, let's consider the u(t-3) term. This is just a step response that is delayed by 3 units of time. So, we can rewrite it as u(t) with a shift of 3 units to the right, which means we can use the time-shifting property of the Laplace transform:

L[u(t-a)f(t-a)] = e^(-as)F(s)

In our case, a = 3, so we get:

L[u(t-3)f(t-3)] = e^(-3s)F(s)

Now, we can apply this to our original problem:

L[v(t)] = L[sin(7t)u(t-3)]

= e^(-3s) * L[sin(7t)]

= e^(-3s) * 7/(s^2+49)

= 7e^(-3s)/(s^2+49)

So, the final answer is 7e^(-3s)/(s^2+49). I hope this helps clarify the process for you!