My question is about something incomplete in my lesson. I learnt about the modulation and the dilatation of a function in the Laplace transform but what if I have both in the same time?
For the modulation, I have the following relation:
$$
\mathcal{L}[e^{-\gamma t} f(t)] = \mathcal{L}[f](z + \gamma)
$$
For the dilatation, I have the following relation:
$$
\mathcal{L}[f(\lambda t)] = \frac{1}{\lambda} \mathcal{L}[f](\frac{z}{\lambda})
$$
What if I have the following Laplace transform : $\mathcal{L}[e^{\gamma t} f(\frac{t}{\lambda})]$
Is the answer $\lambda \mathcal{L}[f](\lambda z – \gamma)$ or $\lambda \mathcal{L}[f](\lambda(z – \gamma))$?
Best Answer
Define modulation and dilation operators: $M_{\gamma}f(t) = e^{-\gamma t} f(t)$ and $D_{\lambda}f(t) = f(\lambda t)$. Then $t \mapsto e^{\gamma t} f(t/\lambda)$ can be expressed as either $M_{-\gamma} (D_{1/\lambda} f)$ (dilation then modulation) or $D_{1/\lambda}(M_{-\gamma \lambda} f)$ (modulation then dilation). Using either form, we obtain the same result. $$\mathcal L[M_{-\gamma} (D_{1/\lambda} f)](z) = \mathcal L[D_{1/\lambda} f](z \ \gamma) = \lambda \mathcal L[f] (\lambda(z - \gamma))$$ $$\mathcal L[D_{1/\lambda}(M_{-\gamma \lambda} f)](z) = \lambda \mathcal L[M_{-\gamma \lambda} f](\lambda z) = \lambda \mathcal L[f] (\lambda (z - \gamma))$$