Question #277189

find laplace transform : sin wt (0 < t< pi/w)


1
Expert's answer
2021-12-13T17:02:26-0500

sinωtωs2+ω2sin\omega t \to \frac{\omega}{s^2+\omega^2}


Laplace transform:


L(f(t))=0π/ωestf(t)dt=0π/ωestsinωtdt=L(f(t))=\int^{\pi/\omega}_0 e^{-st}f(t)dt=\int^{\pi/\omega}_0e^{-st}sin\omega tdt=


=est(ssinωt+ωcosωt)s2+ω20π/ω=ωeπs/ωωs2+ω2=ωeπs/ω+ωs2+ω2=-\frac{e^{-st}(ssin\omega t+\omega cos\omega t)}{s^2+\omega^2}|^{\pi/\omega}_0=-\frac{-\omega e^{-\pi s/\omega}-\omega}{s^2+\omega^2}=\frac{\omega e^{-\pi s/\omega}+\omega}{s^2+\omega^2}


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