The second shifting property of Laplace transforms is used to find the Laplace of a function multiplied by a unit step function. Let us discuss the second shifting property, its proof and some solved examples.

Before we proceed, let us recall the definition of a unit step function. It is defined as follows:

u(t-a) = $\begin{cases} 1 & \text{ if } t>a \\ 0 & \text{ if } t < a \end{cases}$

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## Second Shifting Property Statement

Let L{f(t)} = F(s). For a>0 we have

L{f(t-a) u(t-a)} = e^{-as} F(s) |

**Proof:**

As L{f(t)} = F(s), by definition of Laplace transforms we have that

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

Now, L{f(t-a) u(t-a)}

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

= $\int_a^\infty$ e^{-st} f(t-a) dt, by the definition of unit step function.

We now make a change of variable. Let us put

z=t-a.

So dz=dt.

t | z |

a | 0 |

∞ | ∞ |

Therefore, from above we obtain that

L{f(t-a) u(t-a)} = $\int_0^\infty$ e^{-s(z+a)} f(z) dz

= e^{-as} $\int_0^\infty$ e^{-sz} f(z) dz

= e^{-as} F(s).

This proves the second shifting property of Laplace transforms whose formula states that L{f(t-a) u(t-a)} = e^{-as} F(s) for a>0.

**Read Also: **First shifting property of Laplace transforms

Laplace Transform: Definition, Table, Formulas, Properties

## Solved Examples

**Question:** Find the Laplace of (t-2) u(t-2), that is, find L{(t-2) u(t-2)}.

**Solution:**

Put a=2 in the second shifting property. Thus,

L{(t-2) u(t-2)} = e^{-2s} F(s)

where F(s)= L{t} = 1/s^{2}.

So, L{(t-2) u(t-2)} = $\dfrac{e^{-2s}}{s^2}$

So the Laplace of (t-2) u(t-2) is equal to e^{-2s}/s^{2}.

Laplace Transform of Derivatives | Laplace Transform of Integrals

## FAQs

**Q1: What is the second shifting property of Laplace transforms?**

Answer: The second shifting property of Laplace transforms states that if L{f(t)} = F(s) then L{f(t-a) u(t-a)} = e^{-as} F(s) for a>0. Here u(t-a) is the unit step function.