# Laplace Transform of Function of t minus a

## Theorem

Let $f$ be a function such that $\laptrans f$ exists.

Let $\laptrans {\map f t} = \map F s$ denote the Laplace transform of $f$.

Let $a \in \C$ or $\R$ be constant.

Let $g$ be the function defined as:

$\map g t = \begin{cases} \map f {t - a} & : t > a \\ 0 & : t \le a \end{cases}$

Then:

$\laptrans {\map g t} = e^{-a s} \map F s$

## Proof 1

 $\ds \laptrans {\map f {t - a} }$ $=$ $\ds \int_0^{\to + \infty} e^{-s t} \map f {t - a} \rd t$ Definition of Laplace Transform $\ds$ $=$ $\ds \int_0^{\to + \infty} e^{-s \paren {t - a} } e^{-a s} \map f {t - a} \rd \paren {t - a}$ $\ds$ $=$ $\ds e^{-a s} \int_0^{\to + \infty} e^{-s \paren {t - a} } \map f {t - a} \rd \paren {t - a}$ $\ds$ $=$ $\ds e^{-a s}\map F s$ Definition of Laplace Transform

$\blacksquare$

## Proof 2

 $\ds \laptrans {\map g t}$ $=$ $\ds \int_0^\infty e^{-s t} \map g t \rd t$ Definition of Laplace Transform $\ds$ $=$ $\ds \int_0^a e^{-s t} \map g t \rd t + \int_a^\infty e^{-s t} \map g t \rd t$ $\ds$ $=$ $\ds \int_0^a 0 \times e^{-s t} \rd t + \int_a^\infty e^{-s t} \map f {t - a} \rd t$ Definition of $\map g t$ $\ds$ $=$ $\ds \int_a^\infty e^{-s t} \map f {t - a} \rd t$ $\ds$ $=$ $\ds \int_0^\infty e^{-s \paren {u + a} } \map f u \rd u$ Integration by Substitution: $t = u + a$ $\ds$ $=$ $\ds e^{-a s} \int_0^\infty e^{-s u} \map f u \rd u$ $\ds$ $=$ $\ds e^{-a s} \map F s$ Definition of Laplace Transform

$\blacksquare$

## Also known as

This property of the Laplace transform operator is sometimes seen referred to as:

the second translation property

or:

the second shifting property.

## Examples

### Example $1$

$\laptrans {\paren {t - 2}^3} = \dfrac {6 e^{-2 s} } {s^4}$

where $t > 2$.

### Example $2$

Let $f: \R \to \R$ be the function defined as:

$\forall t \in \R: \map f t = \begin {cases} \map \cos {t - \dfrac {2 \pi} 3} & : t \ge \dfrac {2 \pi} 3 \\ 0 & : t < \dfrac {2 \pi} 3 \end {cases}$

Then:

$\laptrans {\map f t} = s \exp \dfrac {-2 \pi s} 3 \dfrac 1 {s^2 + 1}$