# Laplace transform of $f\left(\frac{t - b}{a}\right)$

Consider the function $f\left(\frac{t - b}{a}\right)$. We want want to calculate its Laplace transform. There are two approaches:

• Firstly,

1. let $g(t) = f\left(\frac ta\right)$.
2. Then $\mathcal{L}\left\{f\left(\frac{t-b}{a}\right)\right\} = \mathcal{L}\left\{g(t - b)\right\} = e^{-bs}G(s)$ and $G(s) = \mathcal{L}\{g(t)\} = \mathcal{L}\left\{f\left(\frac ta\right)\right\} = |a|F(as)$.
3. Therefore $\mathcal{L}\left\{f\left(\frac{t - b}{a}\right)\right\} = |a|e^{-bs}F(as).$
• Secondly,

1. let $h(t) = f\left(\frac{t - b}{a}\right)$.
2. Then $\mathcal{L}\left\{f\left(\frac{t-b}{a}\right)\right\} = \mathcal{L}\left\{h\left(\frac ta\right)\right\} = |a|H(as)$ and $H(s) = \mathcal{L}\{h(t)\} = \mathcal{L}\left\{f\left(\frac{t - b}{a}\right)\right\} = e^{-\left(\frac ba\right) s}F(s)$.
3. Therefore $\mathcal{L}\left\{f\left(\frac{t - b}{a}\right)\right\} = |a|e^{-bs}F(as).$

So far so good. However, let $f(t) = e^t$. Then $$F(s) = \frac{1}{s - 1}$$ and $$\mathcal{L}\left\{f\left(\frac{t - b}{a}\right)\right\} = \frac{ae^{-\left(\frac ba\right)s}}{as - 1}$$ from Wolfram Alpha.

• The problem comes from the fact that we introduce new data by shifting, i.e., the function is not zero for $t < 0$. What can we do then? Can we not use the time delay property?

• What about the Fourier transform? The arguments above are valid for the Fourier transform for $s = j\omega$ and you do not have the requirement of $f(t < 0) = 0$ when applying the shift property. But when you set $f(t) = \cos(t)$ you get: $$\mathcal{F}\left\{f\left(\frac{t - b}{a}\right)\right\}\bigg\vert_{\omega \ge 0,\ a > 0} = \frac{1}{2}e^{\left(\frac ba\right)j\omega}\delta(\omega - 1/a)$$ So it looks like it should be $e^{-\left(\frac ba\right)s}$ instead of $e^{-bs}$. But where have I made a mistake?

• If you are working with the bilateral Laplace transform, there should be a step function somewhere, making the function zero not only for $t<0$ but for $t<b$, given that $b>0$. – Tendero Apr 5 '17 at 15:56

## 1 Answer

Your two alternate derivations of the Laplace transform (specifically the bilateral I'm referring to) of the signal $f((t - b)/a)$ seems right, both resulting in, the same , as a check.

$$\mathcal{L}\{f((t - b)/a)\} = |a|e^{-bs}F(as)$$

However, when you assume the signal $f(t) = e^t u(t)$ whose Laplace transform is $$F(s) = \frac{1}{s - 1}$$

your conclusion is wrong $$\mathcal{L}\{f((t - b)/a)\} = |a|e^{-bs}F(as) \neq \frac{ae^{-(b/a)s}}{as - 1}$$

which should be corrected as (according to your own derivation) $$\mathcal{L}\{f((t - b)/a)\} = |a|e^{-bs}\frac{1}{as - 1}$$

So is this the problem you are referring to ?