Laplace Transform: Difference between revisions

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:<math>F(s) = \mathcal{L} \left\{cos(\omega t)\right\}=\int_0^{\infty} e^{-st} cos(\omega t) \,dt = </math> <math> \frac {s}{s^2+\omega^2}</math>
:<math>F(s) = \mathcal{L} \left\{cos(\omega t)\right\}=\int_0^{\infty} e^{-st} cos(\omega t) \,dt = </math> <math> \frac {s}{s^2+\omega^2}</math>


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


:<math>F(s) = \mathcal{L} \left\{t sin(\omega t)\right\}=\int_0^{\infty} e^{-st} t sin(\omega t) \,dt = </math> <math> \frac {2 \omega s} {(s^2+\omega^2)^2} </math>
:<math>F(s) = \mathcal{L} \left\{t sin(\omega t)\right\}=\int_0^{\infty} e^{-st} t sin(\omega t) \,dt = </math> <math> \frac {2 \omega s} {(s^2+\omega^2)^2} </math>
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:<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 = </math> <math> 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}} </math> for n=1,2,...
:<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 18:23, 11 January 2010

Standard Form:

Sample Functions: