[amp_mcq option1=”$${{a – b} \\over s}$$” option2=”$${{{e^z}\\left( {a – b} \\right)} \\over s}$$” option3=”$${{{e^{ – as}} – {e^{ – bs}}} \\over s}$$” option4=”$${{{e^{ – \\left( {a – b} \\right)}}} \\over s}$$” correct=”option4″]<\/p>\n
The correct answer is $\\boxed{{e^{ – \\left( {a – b} \\right)}}} \\over s}$.<\/p>\n
The bilateral 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
In this case, $f(t)$ is a step function that is $1$ for $a \\le t \\le b$ and $0$ otherwise. The Laplace transform of a step function is<\/p>\n
$$L\\delta(t-a): s = {1 \\over s} e^{-as}$$<\/p>\n
where $\\delta(t)$ is the Dirac delta function. The Laplace transform of a product of two functions is the convolution of their Laplace transforms. In this case, the Laplace transform of $f(t)$ is<\/p>\n
$$Lf(t): s = \\int_a^b {1 \\over s} e^{-st} dt = {e^{-as} – e^{-bs} \\over s}$$<\/p>\n
Therefore, the bilateral Laplace transform of $f(t)$ is<\/p>\n
$$Lf(t): s = {e^{-\\left( {a – b} \\right)}} \\over s}$$<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”$${{a – b} \\over s}$$” option2=”$${{{e^z}\\left( {a – b} \\right)} \\over s}$$” option3=”$${{{e^{ – as}} – {e^{ – bs}}} \\over s}$$” option4=”$${{{e^{ – \\left( {a – b} \\right)}}} \\over s}$$” 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-56755","post","type-post","status-publish","format-standard","hentry","category-signal-processing","no-featured-image-padding"],"yoast_head":"\n