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- Thread starter djeitnstine
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- #2

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Your statement is a little confusing. In any case, continuity is not necessary for Laplace transforms, which are integrals from 0 to oo, while Fourier series are obtained by integrals over a finite interval. The function has to meet certain conditions related to integrability, but continuity is not one of them.

- #3

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Yes but why can the integrals simply be added in both cases?

Example:

[tex]f(t)= \left\{^{5, t<1}_{sin(t), t>1}[/tex] So the laplace transform of this would be

[tex]L\left\{ f(t) \right\} = \int_0^1 5e^{-st}dt + \int_1^{\infty} sin(t)e^{-st}dt[/tex]

Why can they just be added?

Example:

[tex]f(t)= \left\{^{5, t<1}_{sin(t), t>1}[/tex] So the laplace transform of this would be

[tex]L\left\{ f(t) \right\} = \int_0^1 5e^{-st}dt + \int_1^{\infty} sin(t)e^{-st}dt[/tex]

Why can they just be added?

Last edited:

- #4

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Example:

[tex]f(t)= \left\{^{5, t<1}_{sin(t), t>1}[/tex] So the laplace transform of this would be

[tex]L\left\{ f(t) \right\} = \int_0^1 5e^{-st}dt + \int_1^{\infty} sin(t)e^{-st}dt[/tex]

Why can they just be added?

Well.

[tex]L\left\{ f(t) \right\} = \int_0^{\infty} e^{-st} f(t) dt [/tex]

[tex]L\left\{ f(t) \right\} = \int_0^1 f(t)e^{-st}dt + \int_1^{\infty} f(t)e^{-st}dt[/tex]

[because the integration operator is LINEAR]

which then becomes

[tex]L\left\{ f(t) \right\} = \int_0^1 5e^{-st}dt + \int_1^{\infty} sin(t)e^{-st}dt[/tex]

- #5

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Hmm. I should be hitting myself in the head.

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