Inverse laplace transform of this simple function?

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  • #1
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what is the inverse laplace transform of (2s)(1/(s-2))?

could i use the identity ∫f(T)g(t-T)dT=F(s)G(s)?

i was hesitant so i figured i'd just ask before i continue..
 

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  • #2
lurflurf
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write

$$\frac{s}{s-2}=\frac{s-2+2}{s-2}=1+\frac{2}{s-2}$$

or go ahead and use convolution
 
  • #3
Ray Vickson
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what is the inverse laplace transform of (2s)(1/(s-2))?

could i use the identity ∫f(T)g(t-T)dT=F(s)G(s)?

i was hesitant so i figured i'd just ask before i continue..

You could use the property
[tex] \cal{L}(f^\prime)(s) = s\, \cal{L}(f)(s) - f(0+)[/tex]
Be careful, though: that is for a LT defined as
[tex] \cal{L}(f)(s) = \int_{0+}^{\infty} e^{-st} f(t) \, dt. [/tex]
If, instead, you prefer to define your LT as
[tex] \cal{L}(f)(s) = \int_{0-}^{\infty} e^{-st} f(t) \, dt, [/tex]
then you will need
[tex]\cal{L}(f^\prime)(s) = s\, \cal{L}(f)(s) - f(0-)[/tex]
 

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