- #1

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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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- Thread starter iScience
- Start date

- #1

- 466

- 5

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..

- #2

lurflurf

Homework Helper

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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

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

or go ahead and use convolution

- #3

Ray Vickson

Science Advisor

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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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