# Laplace transform of squared fading

Say we have a formula for Laplace

$$\prod \mathbb E_h[\exp(-sP_jh_{xj}l(xj))]$$

If we want to find the $\mathscr L_{h_{xj}}$ then

$$\prod[\mathscr L_{xj}\exp(-sP_jl(xj))]$$

the channel $h_{xj}$ is exponential distribution with mean = 1, (i.e $h$ ~ $(1)$)

my understanding that if we want to find the Laplace transform then we calculate as follows

$$\mathscr L_{h_{xj}} =\int_0^\sim e^{-st}e^{-sP_jl(xj)t}dt$$ $$\mathscr L_{h_{xj}} =\int_0^\sim e^{-(s+P_jl(xj))t} dt$$ $$\mathscr L_{h_{xj}} = -\frac{1}{(s+P_jl(xj))}e^{-(s+P_jl(xj))}|_0^\sim$$ $$\mathscr L_{h_{xj}} = \frac{1}{(s+P_jl(xj))}$$

However , the correct answer is $$\mathscr L_{h_{xj}} =\frac{1}{(1+sP_jl(xj))}$$ can someone please point me the mistake and explain the correct step?

• Maybe $e^{-st}e^{sP_jl(xj)t}=e^{-(s-P_jl(xj))t}$ and not $e^{-(s+P_jl(xj))t}$ is where it goes wrong? I don't exactly follow your notation, so that is just a guess. Also, the correct answer looks the same as yours, did you forget to change it (add a minus sign)? – mickep Mar 27 '15 at 9:34
• hi, thx for the input, the answer is completely different where the coefficient of the right answer (denominator) is $1+sP_jl(xj)$, while my answer is $s+P_jl(xj)$, i'm still confused here – m3wIsw3m Mar 27 '15 at 11:18