The Laplace transform of u(t-1) is equal to e-s/s, that is, L{u(t-1)} = e-s/s. Note that u(t-1) is the shifted unit step function by 1 and it is defined as follows.
u(t-1) = 0 if t<1
u(t-1) = 1 if t≥1.
Let us now learn how to find the Laplace transform of u(t-1).
Table of Contents
Laplace of u(t-1)
We will find the Laplace of u(t-1) by definition. The definition of Laplace transforms says that the Laplace of a function f(t) is given by the following integral formula
L{f(t)} = $\int_0^\infty$ e-st f(t) dt.
So the Laplace of u(t-1) is
L{u(t-1)} = $\int_0^\infty$ e-st u(t-1) dt.
By the definition of u(t-1) given above, this integral will be equal to
L{u(t-1)}= $\int_1^\infty$ e-st dt
= $\Big[ \dfrac{e^{-st}}{-s}\Big]_1^\infty$
= limt→∞ $\dfrac{e^{-st}}{-s}$ + e-s/s
= 0 + e-s/s as we know that limt→∞ e-st = 0.
= e-s/s.
So the Laplace transform of u(t-1) is equal to e-s/s, that is, L{u(t-1)} = e-s/s, and this is obtained by the definition of Laplace transforms.
More Laplace Transforms:
Laplace of unit step function, L{u(t)}
FAQs
Answer: The Laplace transform of u(t-1) is e-s/s, that is, L{u(t-1)} = e-s/s.
