What is the laplace transform of $\delta(t-\pi /6)\sin (t)$
I know that $L\{\delta(t-\pi/6) \}=e^{-s\pi/6}$ I also know that $L\{\sin (t) \}=1/(s^2+1)$
I also know that $L\{(u(t-\pi/6)f(t-\pi/6)\}=e^{-s\pi/6}F(s)$ where $F(s)=L\{f(t)\}$
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if $f(t)$ is continuous then $\delta(t-a) f(t) = f(a) \delta(t-a)$ ... did you ever draw what looks like a $\delta$ distribution ?? – reuns Feb 28 '16 at 23:18
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$\delta (t-t_0) =0 $ when $t\ne t_0$. So I'm assuming it will be $\infty$ at $t_0$ – GRS Feb 28 '16 at 23:23
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The Laplace transform is by definition $$ \int_0^{\infty}\exp(-st)\delta(t-\pi /6)\sin (t)dt=\sin(\frac{\pi}{6})\exp(-s\frac{\pi}{6})=(1/2)\exp(-s\frac{\pi}{6})$$
Silvia
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By definition of Dirac delta integrating over all possible values of t the Dirac delta only selects $t=\frac{\pi}{6}$ as it would be always zero otherwise. – Silvia Feb 28 '16 at 23:15
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It is not an integral over a single point, you are integrating over all possible values of $t$, namely $(0,\infty)$ but then you have the delta $\delta(t-\frac{\pi}{6}$ which is non zero only if $t=\pi/6$ (by definition) that basically selects only this particular value of $t$ ( as it is the only value that would give a non zero contribution to the integral). – Silvia Feb 28 '16 at 23:43
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In THIS ANSWER and THIS ONE, I provided primers on the Dirac Delta.
Using the "sifting property" of the Dirac Delta, for any suitable test function $\phi(t)$, we have
$$\int_{-\infty}^\infty \delta(t-t')\,\phi(t)\,dt=\phi(t') \tag 1$$
Now, let $\phi(t)=e^{-st}\sin(t)u(t)$ and $t'=\pi/6$. Then from $(1)$ we find
$$\begin{align} \int_0^\infty \delta(t-\pi/6)\,e^{-st}\sin(t)\,dt&=\int_{-\infty}^\infty \delta(t-\pi/6)\,e^{-st}\sin(t)u(t)\,dt\\\\ &=e^{-s\pi/6}\sin(\pi/6)u(\pi/6)\\\\ &=e^{-s\pi/6}\sin(\pi/6)\\\\ &=\frac12 e^{-\pi s/6} \end{align}$$
as was to be shown!
Mark Viola
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