Laplace Transform: Difference between revisions

From Class Wiki
Jump to navigation Jump to search
No edit summary
No edit summary
Line 41: Line 41:
:<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\{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 = sG(s) - g(0) </math>
:<math>F(s) = \mathcal{L} \left\{g'(t)\right\}=\int_0^{\infty} e^{-st} g'(t) \,dt = sG(s) - g(0) </math>
 
:<math>F(s) = \mathcal{L} \left\{g''(t)\right\}=\int_0^{\infty} e^{-st} g''(t) \,dt = s^2 \cdot G(s) - s \cdot g(0) - g'(0) </math>
 
:<math>F(s) = \mathcal{L} \left\{g^{(n)}(t)\right\}=\int_0^{\infty} e^{-st} g^{(n)}(t) \,dt = s^n \cdot G(s) - s^{n-1} \cdot g(0) - s^{n-2} \cdot g'(0) - ... - g^{(n-1)}(0) </math>

Revision as of 19:37, 11 January 2010

Standard Form:

F(s)={f(t)}=0estf(t)dt

Sample Functions:

F(s)={1}=0estdt= 1s
F(s)={tn}=0esttndt= n!sn+1
F(s)={eat}=0esteatdt= 1sa
F(s)={sin(ωt)}=0estsin(ωt)dt= ωs2+ω2
F(s)={cos(ωt)}=0estcos(ωt)dt= ss2+ω2
F(s)={tng(t)}=0esttng(t)dt= (1)ndnG(s)dsn forn= 1,2,...
F(s)={tsin(ωt)}=0esttsin(ωt)dt= 2ωs(s2+ω2)2
F(s)={tcos(ωt)}=0esttcos(ωt)dt= s2ω2(s2+ω2)2
F(s)={g(t)}=0estg(t)dt= 1aG(sa)
F(s)={eatg(t)}=0esteatg(t)dt=G(sa)
F(s)={eattn}=0esteattndt= n!(sa)n+1 forn= 1,2,...
    Normal   0               false   false   false      EN-US   X-NONE   X-NONE                                                     MicrosoftInternetExplorer4
F(s)={tet}=0esttetdt= 1(s+1)2
F(s)={1et/T}=0est(1et/T)dt= 1s(1+Ts)
F(s)={eatsin(ωt)}=0esteatsin(ωt)dt= ω(sa)2+ω2
F(s)={eatcos(ωt)}=0esteatcos(ωt)dt= sa(sa)2+ω2
F(s)={u(t)}=0estu(t)dt= 1s
F(s)={u(ta)}=0estu(ta)dt= eass
F(s)={u(ta)g(ta)}=0estu(ta)g(ta)dt=easG(s)
F(s)={g(t)}=0estg(t)dt=sG(s)g(0)
F(s)={g(t)}=0estg(t)dt=s2G(s)sg(0)g(0)
F(s)={g(n)(t)}=0estg(n)(t)dt=snG(s)sn1g(0)sn2g(0)...g(n1)(0)