Find the Laplace transform of the following functions : $$i)\ \sin(t-2π) H(t-2π) \\ ii) \ \sin3t\ \delta(t-π)$$
I could do the first one as follows : $H(t-2π) = u_{2π}(t)$ where $u$ is the unit step funtion.
Let, $f(t) = \sin t$. Then Laplace transform of $f(t)$ is $F(s) = \frac{1}{s^2+1}$.
Then, $$\sin(t-2π) H(t-2π) \\ = f(t-2π) \ u_{2π}(t) \\ = u_{2π}(t)f(t-2π) $$
Now we know Laplace transform of $$u_c(t)f(t-c)$$ is $$e^{-cs}F(s)$$.
Hence the required Laplace transform is $$ e^{-2πs} \frac{1}{s^2+1}$$.
But I haven't found any formula for the second one. Can anyone please help me with that and alsp check if I've done the first one correctly?
Your first formula is correct. Lets use the definition for the second one
$$ \int_0^\infty \sin (3t) \delta(t - \pi) e^{-st} \ dt = \sin(3\pi)e^{-s\pi} = 0, $$ by using the properties of the delta function