Calculate the Laplace transform of $g(t)=e^{-2t} \ f(t-1)$ and $h(t)=e^{-2t} \ f(2t)$

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I have used time delay property:


$$G(s)=e^{-s} \ F(s+2)$$


$$H(s)=\frac{1}{2} \ F(\frac{s+2}{2})$$


Is it correct?

Thanks!

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Your solution is correct. Let $\mathcal{L}[f(t)]=F(s)$, we have

Frequency shifting property

$$\mathcal{L}[e^{at}f(t)]=F(s-a)$$

Time scaling property

$$\mathcal{L}[f(at)]=\frac 1a F\left(\frac sa\right)$$

Delay property $$\mathcal{L}[f(t-a)]=e^{-as} F(s)$$