[amp_mcq option1=”e2tu(t) – 2e-tu(t)” option2=”-e2tu(-t) + 2e-tu(t)” option3=”-e2tu(-t) – 2e-tu(t)” option4=”e2tu(-t) – 2e-tu(t)” correct=”option1″]<\/p>\n
The correct answer is A. e2tu(t) – 2e-tu(t).<\/p>\n
The Laplace transform of a continuous-time signal $x(t)$ is defined as<\/p>\n
$$X(s) = \\int_{0}^{\\infty} x(t) e^{-st} dt$$<\/p>\n
The Fourier transform of a continuous-time signal $x(t)$ is defined as<\/p>\n
$$X(\\omega) = \\int_{-\\infty}^{\\infty} x(t) e^{-j\\omega t} dt$$<\/p>\n
The Laplace transform and Fourier transform are related by the following equation:<\/p>\n
$$X(s) = \\int_{0}^{\\infty} X(\\omega) e^{s\\omega} d\\omega$$<\/p>\n
In this case, we are given that<\/p>\n
$$X(s) = {{5 – s} \\over {{s^2} – s – 2}}$$<\/p>\n
We can use the partial fraction expansion to decompose $X(s)$ as follows:<\/p>\n
$$X(s) = {1 \\over 2} \\left( {1 \\over s – 1} + {2 \\over s + 2} \\right)$$<\/p>\n
The inverse Laplace transform of $1 \/ (s – 1)$ is $e^{t}$, and the inverse Laplace transform of $2 \/ (s + 2)$ is $2e^{-2t}$. Therefore,<\/p>\n
$$x(t) = {1 \\over 2} \\left( e^{t} + 2e^{-2t} \\right)$$<\/p>\n
We can write this as<\/p>\n
$$x(t) = e^{t} – e^{-2t} + e^{-2t} = e^{2t} – 2e^{-t}$$<\/p>\n
Therefore, the correct answer is A.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”e2tu(t) – 2e-tu(t)” option2=”-e2tu(-t) + 2e-tu(t)” option3=”-e2tu(-t) – 2e-tu(t)” option4=”e2tu(-t) – 2e-tu(t)” correct=”option1″]<\/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-50925","post","type-post","status-publish","format-standard","hentry","category-signal-processing","no-featured-image-padding"],"yoast_head":"\n