Laplace Transform: Difference between revisions

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:<math>F(s) = \mathcal{L} \left\{g(t)\right\}=\int_0^{\infty} e^{-st} g(t) \,dt = </math> <math> \frac {1} {a} G \left(\frac {s} {a}\right)</math>
:<math>F(s) = \mathcal{L} \left\{g(t)\right\}=\int_0^{\infty} e^{-st} g(t) \,dt = </math> <math> \frac {1} {a} G \left(\frac {s} {a}\right)</math>


:<math>F(s) = \mathcal{L} \left\{e^{at} g(t)\right\}=\int_0^{\infty} e^{-st} e^{at} g(t) \,dt = </math> <math> G(s-a) </math>
:<math>F(s) = \mathcal{L} \left\{e^{at} g(t)\right\}=\int_0^{\infty} e^{-st} e^{at} g(t) \,dt = G(s-a) </math>


:<math>F(s) = \mathcal{L} \left\{e^{at} t^n\right\}=\int_0^{\infty} e^{-st} e^{at} t^n \,dt = </math> <math> \frac {n!} {(s-a)^{n+1}} \mbox{ for}~n\ \mbox{= 1,2,...}</math>
:<math>F(s) = \mathcal{L} \left\{e^{at} t^n\right\}=\int_0^{\infty} e^{-st} e^{at} t^n \,dt = </math> <math> \frac {n!} {(s-a)^{n+1}} \mbox{ for}~n\ \mbox{= 1,2,...}</math>

Revision as of 18:26, 11 January 2010

Standard Form:

Sample Functions: