📂Odinary Differential Equations

Inverse Laplace Transform of F(as+b)

공식¹

Assuming that the Laplace transform $\mathcal{L} \left\{ f(t) \right\}= \displaystyle \int _{0} ^\infty e^{-st}f(t)dt =F(s)$ of function $f(t)$ exists as $s>\alpha \ge 0$, the inverse Laplace transform of $F(as+b)$ for constant $a>0 , b$ is as follows.

$$\mathcal{L^{-1}} \left\{ F(as+b) \right\} =\frac{1}{a}e^{-\frac{b}{a}t}f\left(\frac{t}{a}\right)$$

Derivation

1

1. Inverse Laplace transform of $F(ks)$: $$\mathcal{L^{-1}} \left\{ F(ks) \right\} =\dfrac{1}{k}f\left(\frac{t}{k}\right)$$
2. Translation of Laplace transform: $$\mathcal{L^{-1}} \left\{ F(s-c) \right\}=e^{ct}f(t)$$

From 1., the following is obtained.

$$\mathcal{L^{-1}} \left\{ F(as) \right\} =\frac{1}{a}f\left(\frac{t}{a}\right)$$

To derive $F(as+b)$ from $F(as)$, it is sufficient if $s$ $\to$ $s+\frac{b}{a}$. That is, $F$ needs to be translated by $-\frac{b}{a}$ towards the $s$ direction. By 2., the translated $F$ is found as

$$\begin{align*} && \mathcal{L^{-1}} \left\{ F\left( a(s+\frac{b}{a}) \right) \right\} &= \frac{1}{a} e^{-\frac{b}{a}t}f\left(\frac{t}{a}\right) \\ \implies && \mathcal{L^{-1}} \left\{ F( as+b) \right\} &= \frac{1}{a} e^{-\frac{b}{a}t}f\left(\frac{t}{a}\right) \end{align*}$$

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2

By the definition of Laplace transform,

$$\begin{align*} \mathcal{L} \left\{ \frac{1}{a} e^{-\frac{b}{a}t} f\left( \frac{t}{a} \right) \right\} &= \dfrac{1}{a}\int _{0} ^\infty e^{-st}e^{-\frac{b}{a}t}f \left( \frac{t}{a} \right) dt \\ &= \dfrac{1}{a}\int _{0} ^\infty e^{-\left(s+\frac{b}{a}\right)t}f \left( \frac{t}{a} \right) dt \end{align*}$$

Let’s denote it as $\dfrac{t}{a}=\tau$. Then, $\left(s+\frac{b}{a}\right)t=(as+b)\tau$ and, because $dt=ad\tau$,

$$\begin{align*} \mathcal{L} \left\{ \frac{1}{a} e^{-\frac{b}{a}t} f\left( \frac{t}{a} \right) \right\} &= \dfrac{1}{a}\int _{0} ^\infty e^{-\left(s+\frac{b}{a}\right)t}f \left( \frac{t}{a} \right) dt \\ &= \int _{0} ^\infty e^ {-(as+b)\tau}f (\tau) d\tau \\ &= F(as+b) \end{align*}$$

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See also

• Table of Laplace Transforms