[amp_mcq option1=”-1″ option2=”0″ option3=”1″ option4=”Unbounded” correct=”option3″]<\/p>\n
The correct answer is $\\boxed{\\text{C}}$.<\/p>\n
The final value theorem states that the final value of a system is equal to the limit of its Laplace transform as $s$ approaches infinity, provided that the limit exists. In this case, we have<\/p>\n
$$\\lim_{s \\to \\infty} Y\\left( s \\right) = \\lim_{s \\to \\infty} {1 \\over {s\\left( {s – 1} \\right)}} = \\lim_{s \\to \\infty} {1 \\over s} \\cdot \\lim_{s \\to \\infty} {1 \\over {s – 1}} = 0 \\cdot 1 = 0.$$<\/p>\n
Therefore, the final value of the signal is $0$.<\/p>\n
Option A is incorrect because the final value of the signal is not $-1$.<\/p>\n
Option B is incorrect because the final value of the signal is not $0$.<\/p>\n
Option D is incorrect because the final value of the signal is bounded.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”-1″ option2=”0″ option3=”1″ option4=”Unbounded” correct=”option3″]<\/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-52401","post","type-post","status-publish","format-standard","hentry","category-signal-processing","no-featured-image-padding"],"yoast_head":"\n