Find the Laplace transform of u(t-2)

The Laplace transform of u(t-2) is equal to e^-2s/s and it is denoted by L{u(t-2)} = e^-2s/s. Before we find the Laplace of u(t-2), the shifted unit step function by 2, let us first recall the definition of u(t-2):

$u(t-2)= \begin{cases} 0 & \text{ if } t<2 \\ 1 & \text{ if } t \geq 2. \end{cases}$

u(t-2) Laplace Formula

For any a, as the Laplace of u(t-a) is L{u(t-a)} = e^-as/s, so the formula of the Laplace of u(t-2) is given by

L{u(t-2)} = e-2s/s.

We now learn how to find the Laplace transform of u(t-2).

What is the Laplace of u(t-2)?

Answer: The Laplace of u(t-2) is e-2s/s.

By the definition of Laplace transforms, we will find the Laplace of u(t-2). The definition states that the Laplace of a function f(t) is equal to the integral below:

L{f(t)} = $\int_0^\infty$ e^-st f(t) dt.

Thus, the Laplace of u(t-2) is

L{u(t-2)} = $\int_0^\infty$ e^-st u(t-2) dt.

As u(t-2) = 0 if t<2 and u(t-2) = 1 if t≥2, it follows that

L{u(t-2)}= $\int_2^\infty$ e^-st dt

= $\Big[ \dfrac{e^{-st}}{-s}\Big]_2^\infty$

= lim_t→∞ $\dfrac{e^{-st}}{-s}$ + e^-2s/s

= 0 + e^-2s/s as we know that lim_t→∞ e^-st = 0.

= e^-2s/s.

So the Laplace transform of u(t-2) is e^-2s/s. That is, L{u(t-2)} = e^-2s/s, and it is obtained by the definition of Laplace transforms.

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