How do I go about performing the following Laplace transform?

Question

I'm unable to compute the following Laplace transform. How do I deal with cases such as

$$f(t) = \sin(t-3)\theta(t) \quad \text{or} \quad f(t) = \sin(t-3)\theta(t-3),$$

where $\theta(t)$ is the Heaviside function?

Any input at all would be appreciated.

Answer 1

We have to remember the Laplace's delay property:

$$ L[f(t-T)]=e^{-Ts}F(s) $$

So, in the case

$ f(t)=sin(t-3)H(t-3) $ becomes: $$ F(s)=\frac{e^{-3s}}{s^2+1} $$

Using: $L[sin(at)]=\frac{a}{s^2+a^2}$

The first case $f(t)=sin(t-3)H(t)$ is more complex. The function is a sinusoid with a different phase (because it not starts from 3, but 0)
It's equal to: $cos(3)sin(t)-sin(3)cos(t)$

So, the laplace transform is: $$\frac{cos(3)}{s^2+1}-\frac{s*sin(3)}{s^2+1}$$

Using: $L[cos(at)]=\frac{s}{s^2+a^2}$

Answer 2

$$\int_0^\infty f(t) \theta(t-t') dt = \int_0^{t'}f(t) \theta(t-t')dt + \int_{t'}^\infty f(t)\theta(t-t')dt$$ for $t\in (0,t') , \theta(t-t') = 0$ and for $t\ge t',\theta(t-t')=1$, so above integral becomes $$\int_0^\infty f(t) \theta(t-t') dt = \int_{t'}^\infty f(t)dt$$ That would give $$\mathcal L (\sin(t-3) \theta(t)) = \int_0^\infty e^{-st}\sin(t-3) dt$$ and,
$$\mathcal L (\sin(t-3) \theta(t-3)) = \int_3^\infty e^{-st}\sin(t-3) dt$$