$${1 over {{s^2}}},{s over {{s^2} + 1}}$$
$${1 over s},{1 over {{s^2} + 1}}$$
$${1 over {{s^2}}},{1 over {{s^2} + 1}}$$
$$s,{s over {{s^2} + 1}}$$
Answer is Right!
Answer is Wrong!
The correct answer is A.
The Laplace transform of a function $f(t)$ is defined as
$$Lf(t): s = \int_0^\infty f(t) e^{-st} dt$$
The Laplace transform of $tu(t)$ is
$$Ltu(t): s = \int_0^\infty t e^{-st} dt = \frac{1}{s^2}$$
The Laplace transform of $u(t)sin(t)$ is
$$Lu(t)sin(t): s = \int_0^\infty u(t)sin(t) e^{-st} dt = \frac{s}{s^2 + 1}$$
Here is a brief explanation of each option:
- Option A is correct because it is the Laplace transform of $tu(t)$.
- Option B is incorrect because it is the Laplace transform of $u(t)$.
- Option C is incorrect because it is the Laplace transform of $u(t)cos(t)$.
- Option D is incorrect because it is the Laplace transform of $s$.