Laplace Transform: Difference between revisions

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:<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)=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,...
    Normal   0               false   false   false      EN-US   X-NONE   X-NONE                                                     MicrosoftInternetExplorer4
F(s)=ℒ{te−t}=∫0∞e−stte−tdt= 1(s+1)2
F(s)=ℒ{1−e−t/T}=∫0∞e−st(1−e−t/T)dt= 1s(1+Ts)
F(s)=ℒ{eatsin(ωt)}=∫0∞e−steatsin(ωt)dt= ω(s−a)2+ω2
F(s)=ℒ{eatcos(ωt)}=∫0∞e−steatcos(ωt)dt= s−a(s−a)2+ω2
F(s)=ℒ{u(t)}=∫0∞e−stu(t)dt= 1s
F(s)=ℒ{u(t−a)}=∫0∞e−stu(t−a)dt= e−ass
F(s)=ℒ{u(t−a)g(t−a)}=∫0∞e−stu(t−a)g(t−a)dt=e−asG(s)
F(s)=ℒ{g′(t)}=∫0∞e−stg′(t)dt=sG(s)−g(0)
F(s)=ℒ{g″(t)}=∫0∞e−stg″(t)dt=s2⋅G(s)−s⋅g(0)−g′(0)
F(s)=ℒ{g(n)(t)}=∫0∞e−stg(n)(t)dt=sn⋅G(s)−sn−1⋅g(0)−sn−2⋅g′(0)−...−g(n−1)(0)