- #1
faust9
- 692
- 2
Ok, using the definition of Laplace transforms to find [itex]\L\{f(t)\}[/itex]
Given:
[tex]f(t)=\{^{\sin{t}, 0\le{t}<{\pi}}_{0, t\ge{\pi}}[/tex]
So, this is what I did:
[tex]\L\{\sin t\}=\int^{\pi}_{0} e^{-st}\sin t dt+\int^{\infty}_{\pi} e^{-st}(0)dt[/tex]
[tex]=\int^{\pi}_{0} e^{-st}\sin t dt[/tex]
[tex]=\frac{-e^{-st}\sin{t}}{s}]^{\pi}_{0}+\frac{1}{s}\int^{\pi}_{0} e^{-st}\cos t dt[/tex]
[tex]=\frac{-e^{-st}\sin{t}}{s}]^{\pi}_{0}+\frac{1}{s}(\frac{-e^{-st}\cos{t}}{s}]^{\pi}_{0}-\frac{1}{s}\int^{\pi}_{0} e^{-st}\sin t dt)[/tex]
[tex]=\frac{-se^{-s\pi}}{s^2}+\frac{1}{s}(\frac{-1}{s}-\frac{1}{s}\L\{\sin t\})[/tex]
[tex]\L\{\sin t\}(\frac {s^2+1}{s^2})=\frac{-se^{-s\pi}}{s^2}-\frac{1}{s^2}[/tex]
[tex]\L\{\sin t\}=\frac {-se^{-s\pi}-1}{s^2+1}[/tex]
Which I know is wrong because the Laplace for sin t should be:
[tex]\L\{\sin t\}=\frac {1}{s^2+1}[/tex]
I know my limits of integration will affect the problem but the restricted limits should only add a term to the numerator [itex]e^{-s\pi}[/itex] I believe.
Where did I mess up?
Thanks a lot.
Given:
[tex]f(t)=\{^{\sin{t}, 0\le{t}<{\pi}}_{0, t\ge{\pi}}[/tex]
So, this is what I did:
[tex]\L\{\sin t\}=\int^{\pi}_{0} e^{-st}\sin t dt+\int^{\infty}_{\pi} e^{-st}(0)dt[/tex]
[tex]=\int^{\pi}_{0} e^{-st}\sin t dt[/tex]
[tex]=\frac{-e^{-st}\sin{t}}{s}]^{\pi}_{0}+\frac{1}{s}\int^{\pi}_{0} e^{-st}\cos t dt[/tex]
[tex]=\frac{-e^{-st}\sin{t}}{s}]^{\pi}_{0}+\frac{1}{s}(\frac{-e^{-st}\cos{t}}{s}]^{\pi}_{0}-\frac{1}{s}\int^{\pi}_{0} e^{-st}\sin t dt)[/tex]
[tex]=\frac{-se^{-s\pi}}{s^2}+\frac{1}{s}(\frac{-1}{s}-\frac{1}{s}\L\{\sin t\})[/tex]
[tex]\L\{\sin t\}(\frac {s^2+1}{s^2})=\frac{-se^{-s\pi}}{s^2}-\frac{1}{s^2}[/tex]
[tex]\L\{\sin t\}=\frac {-se^{-s\pi}-1}{s^2+1}[/tex]
Which I know is wrong because the Laplace for sin t should be:
[tex]\L\{\sin t\}=\frac {1}{s^2+1}[/tex]
I know my limits of integration will affect the problem but the restricted limits should only add a term to the numerator [itex]e^{-s\pi}[/itex] I believe.
Where did I mess up?
Thanks a lot.