Laplace Transform: Difference between revisions
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:<math>F(s) = \mathcal{L} \left\{u(t-a)\right\}=\int_0^{\infty} e^{-st} u(t-a) \,dt = </math> <math> \frac {e^{-as}} {s} </math> | :<math>F(s) = \mathcal{L} \left\{u(t-a)\right\}=\int_0^{\infty} e^{-st} u(t-a) \,dt = </math> <math> \frac {e^{-as}} {s} </math> | ||
:<math>F(s) = \mathcal{L} \left\{u(t-a) g(t-a)\right\}=\int_0^{\infty} e^{-st} u(t-a) g(t-a) \,dt = | :<math>F(s) = \mathcal{L} \left\{u(t-a) g(t-a)\right\}=\int_0^{\infty} e^{-st} u(t-a) g(t-a) \,dt = e^{-as} G(s) </math> | ||
:<math>F(s) = \mathcal{L} \left\{g'(t)\right\}=\int_0^{\infty} g'(t) \,dt = | :<math>F(s) = \mathcal{L} \left\{g'(t)\right\}=\int_0^{\infty} g'(t) \,dt = sG(s) - g(0) </math> |
Revision as of 19:28, 11 January 2010
Standard Form:
Sample Functions:
Normal 0 false false false EN-US X-NONE X-NONE MicrosoftInternetExplorer4