[amp_mcq option1=”sF(s) – f(0)” option2=”$${1 \\over s}F\\left( s \\right)$$” option3=”$$\\int\\limits_0^s {F\\left( \\tau \\right)} d\\tau $$” option4=”$${1 \\over s}\\left[ {F\\left( s \\right) – f\\left( 0 \\right)} \\right]$$” correct=”option4″]<\/p>\n
The correct answer is $\\boxed{{1 \\over s}\\left[ {F\\left( s \\right) – f\\left( 0 \\right)} \\right]}$.<\/p>\n
The Laplace transform of a function $f(t)$ is defined as<\/p>\n
$$Lf(t): s = \\int_0^\\infty f(t) e^{-st} dt$$<\/p>\n
The Laplace transform of the integral $\\int_0^t f(\\tau) d\\tau$ is then given by<\/p>\n
$$\\begin{align} L\\left \\int_0^t f(\\tau) d\\tau \\right: s &= \\int_0^\\infty \\int_0^t f(\\tau) e^{-st} d\\tau dt \\\\ &= \\int_0^\\infty \\frac{t}{t+s} f(t) dt \\\\ &= \\frac{1}{s} \\int_0^\\infty f(t) \\left( 1 – e^{-st} \\right) dt \\\\ &= \\frac{1}{s} \\left[ F(s) – f(0) \\right] \\end{align}$$<\/p>\n
where $F(s)$ is the Laplace transform of $f(t)$.<\/p>\n
Option A is incorrect because it does not take into account the initial condition $f(0)$.<\/p>\n
Option B is incorrect because it is the Laplace transform of $f(t)$, not the integral $\\int_0^t f(\\tau) d\\tau$.<\/p>\n
Option C is incorrect because it is the Laplace transform of the function $t f(t)$, not the integral $\\int_0^t f(\\tau) d\\tau$.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”sF(s) – f(0)” option2=”$${1 \\over s}F\\left( s \\right)$$” option3=”$$\\int\\limits_0^s {F\\left( \\tau \\right)} d\\tau $$” option4=”$${1 \\over s}\\left[ {F\\left( s \\right) – f\\left( 0 \\right)} \\right]$$” 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-42881","post","type-post","status-publish","format-standard","hentry","category-signal-processing","no-featured-image-padding"],"yoast_head":"\n