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 $F(s)=L\{f(t)\}$

  • $\begingroup$ 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 ?? $\endgroup$ – reuns Feb 28 '16 at 23:18
  • $\begingroup$ $\delta (t-t_0) =0 $ when $t\ne t_0$. So I'm assuming it will be $\infty$ at $t_0$ $\endgroup$ – GRS Feb 28 '16 at 23:23

The Laplace transform is by definition $$ \int_0^{\infty}\exp(-st)\delta(t-\pi /6)\sin (t)dt=\sin(\frac{\pi}{6})\exp(-s\frac{\pi}{6})=(1/2)\exp(-s\frac{\pi}{6})$$

  • $\begingroup$ Thanks, but could you explain how did you integrate that? $\endgroup$ – GRS Feb 28 '16 at 23:05
  • $\begingroup$ 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. $\endgroup$ – Silvia Feb 28 '16 at 23:15
  • $\begingroup$ Wouldn't an integral over a single point be $0$? $\endgroup$ – GRS Feb 28 '16 at 23:33
  • $\begingroup$ 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). $\endgroup$ – Silvia Feb 28 '16 at 23:43

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!


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