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 19:26, 11 January 2010

Standard Form:

F(s)=ℒ{f(t)}=∫0∞e−stf(t)dt

Sample Functions:

F(s)=ℒ{1}=∫0∞e−stdt= 1s
F(s)=ℒ{tn}=∫0∞e−sttndt= n!sn+1
F(s)=ℒ{eat}=∫0∞e−steatdt= 1s−a
F(s)=ℒ{sin(ωt)}=∫0∞e−stsin(ωt)dt= ωs2+ω2
F(s)=ℒ{cos(ωt)}=∫0∞e−stcos(ωt)dt= ss2+ω2
F(s)=ℒ{tng(t)}=∫0∞e−sttng(t)dt= (−1)ndnG(s)dsn forn= 1,2,...
F(s)=ℒ{tsin(ωt)}=∫0∞e−sttsin(ωt)dt= 2ωs(s2+ω2)2
F(s)=ℒ{tcos(ωt)}=∫0∞e−sttcos(ωt)dt= s2−ω2(s2+ω2)2
F(s)=ℒ{g(t)}=∫0∞e−stg(t)dt= 1aG(sa)
F(s)=ℒ{eatg(t)}=∫0∞e−steatg(t)dt=G(s−a)
F(s)=ℒ{eattn}=∫0∞e−steattndt= n!(s−a)n+1 forn= 1,2,...