How do you find the Laplace transform of the Dirac Delta function multiplied by a cos function with a phase shift?

Question

More specifically,

I am having trouble understanding how to find the Laplace Transform of the following problem:

$$5cos(6\pi t + \pi/6) \delta(t-0.4) $$

I know that the Laplace transform of

$$ \delta(t-0.4) $$ would be $${e}^{-0.4s}$$

but I don't know understand how to proceed in regards to this specific problem...

Also I already submitted my homework so NO, this is NOT me asking you to do my homework assignment for me for any wise guys out there who will feel the need to chastise me - I am asking for a walkthrough because I just don't get it even after reading through my notes. Thank you in advance!

Answer 1

From the definition of Laplace Transform and the Dirac Delta distribution: $$ \int^{\infty}_{-\infty}f(x)\delta(x-a)dx=f(a) $$ we have $$ \int^\infty_05\cos(6\pi t + \pi/6) \delta(t-0.4)e^{-st}dt\\\Longrightarrow5\cos(6\pi (0.4) + \pi/6)e^{-0.4s} $$

Answer 2

We have

$$\int_0^{\infty} f(t)\delta (t-t_0) dt = f(t_0)$$

So,

$$\begin{align}\mathcal {L} \left\{5\cos\left(6\pi t +\frac\pi6\right)\delta(t-0.4)\right\} &= \int_0^{\infty}5\cos\left(6\pi t +\frac\pi6\right)e^{-st}\delta(t-0.4)dt \\&= 5\cos\left(2.4\pi +\frac\pi6\right)e^{-0.4s} \\&= 5\cos\left(\frac{77\pi}{30}\right)e^{-0.4s}\end{align}$$