[amp_mcq option1=”Cannot be determined” option2=”Is zero” option3=”Is unity” option4=”Is infinite” correct=”option4″]<\/p>\n
The correct answer is $\\boxed{\\text{A. Cannot be determined}}$.<\/p>\n
The Laplace transform of a function $f(t)$ is defined as<\/p>\n
$$L[f(t)] = \\int_0^\\infty f(t) e^{-st} dt$$<\/p>\n
The Laplace transform of a constant function is $1$. Therefore, if $sL[f(t)] = \\frac{\\omega}{\\left(s^2 + \\omega^2\\right)}$, then<\/p>\n
$$\\int_0^\\infty f(t) e^{-st} dt = \\frac{\\omega}{\\left(s^2 + \\omega^2\\right)}$$<\/p>\n
This equation can be solved for $f(t)$ in terms of the inverse Laplace transform, which is defined as<\/p>\n
$$f(t) = \\frac{1}{2\\pi i} \\int_{c-i\\infty}^{c+i\\infty} L[f(t)] e^{st} ds$$<\/p>\n
where $c$ is a real number greater than any of the poles of $L[f(t)]$.<\/p>\n
In this case, the poles of $L[f(t)]$ are at $s = \\pm i\\omega$. Therefore, the inverse Laplace transform is<\/p>\n
$$f(t) = \\frac{1}{2\\pi i} \\int_{-i\\omega}^{i\\omega} \\frac{\\omega}{\\left(s^2 + \\omega^2\\right)} e^{st} ds$$<\/p>\n
This integral cannot be evaluated in closed form. Therefore, the value of $f(t)$ cannot be determined.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”Cannot be determined” option2=”Is zero” option3=”Is unity” option4=”Is infinite” correct=”option4″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[959],"tags":[],"class_list":["post-47303","post","type-post","status-publish","format-standard","hentry","category-signal-processing","no-featured-image-padding"],"yoast_head":"\n