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):

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

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## 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.

**Have You Read These Laplace Transforms:**

Laplace transform of unit step function

Find the Laplace transform of u(t-1)

## FAQs

**Q1: What is the Laplace transform of u(t-2)?**

Answer: The Laplace transform of u(t-2) is e^{-2s}/s. Mathematically, L{u(t-2)} = e^{-2s}/s.