[amp_mcq option1=”$${{{s^2} + 1} \\over {s + 3}}$$” option2=”$${1 \\over {s + 3}}$$” option3=”$${{{s^2} + 1} \\over {\\left( {s + 3} \\right)\\left( {s + 2} \\right)}} + {{s + 2} \\over {{s^2} + 1}}$$” option4=”None of the above” correct=”option1″]<\/p>\n
The correct answer is $\\boxed{{s^2} + 1} \\over {\\left( {s + 3} \\right)\\left( {s + 2} \\right)} + {{s + 2} \\over {{s^2} + 1}}$.<\/p>\n
The Laplace transform of the product of two functions $f(t)$ and $g(t)$ is given by the convolution theorem:<\/p>\n
$$L\\left[ {f\\left( t \\right)g\\left( t \\right)} \\right] = {1 \\over {2\\pi i}} \\int_{c-i\\infty}^{c+i\\infty} {F\\left( s \\right)G\\left( {s – z} \\right)dz}$$<\/p>\n
where $F(s)$ and $G(s)$ are the Laplace transforms of $f(t)$ and $g(t)$, respectively, and $c$ is a real number greater than the real parts of all the poles of $F(s)$ and $G(s)$.<\/p>\n
In this case, we have:<\/p>\n
$$L\\left[ {f\\left( t \\right)g\\left( t \\right)} \\right] = {1 \\over {2\\pi i}} \\int_{c-i\\infty}^{c+i\\infty} {{s + 2} \\over {{s^2} + 1}} \\cdot {{s^2} + 1} \\over {\\left( {s + 3} \\right)\\left( {s + 2} \\right)}dz}$$<\/p>\n
We can simplify this integral by using partial fractions:<\/p>\n
$$\\begin{align} Therefore, the Laplace transform of $h(t)$ is $\\boxed{{s^2} + 1} \\over {\\left( {s + 3} \\right)\\left( {s + 2} \\right)}} + {{s + 2} \\over {{s^2} + 1}}$.<\/p>\n","protected":false},"excerpt":{"rendered":" [amp_mcq option1=”$${{{s^2} + 1} \\over {s + 3}}$$” option2=”$${1 \\over {s + 3}}$$” option3=”$${{{s^2} + 1} \\over {\\left( {s + 3} \\right)\\left( {s + 2} \\right)}} + {{s + 2} \\over {{s^2} + 1}}$$” option4=”None of the above” correct=”option1″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[959],"tags":[],"class_list":["post-44474","post","type-post","status-publish","format-standard","hentry","category-signal-processing","no-featured-image-padding"],"yoast_head":"\n
\nL\\left[ {f\\left( t \\right)g\\left( t \\right)} \\right] &= {1 \\over {2\\pi i}} \\int_{c-i\\infty}^{c+i\\infty} {{s + 2} \\over {{s^2} + 1}} \\cdot {{s^2} + 1} \\over {\\left( {s + 3} \\right)\\left( {s + 2} \\right)}dz \\\\
\n&= {1 \\over {2\\pi i}} \\int_{c-i\\infty}^{c+i\\infty} \\left( {{1 \\over {s + 3}} + {1 \\over {s + 2}}} \\right) \\cdot {{s^2} + 1} dz \\\\
\n&= {1 \\over {2\\pi i}} \\left( {1 \\over {2\\pi i}} \\int_{c-i\\infty}^{c+i\\infty} {{1 \\over {s + 3}}dz} + {1 \\over {2\\pi i}} \\int_{c-i\\infty}^{c+i\\infty} {{1 \\over {s + 2}}dz} \\right) \\cdot {{s^2} + 1} \\\\
\n&= {1 \\over {2}} \\left( {1 \\over {s + 3}} \\cdot {{s^2} + 1} + {1 \\over {s + 2}} \\cdot {{s^2} + 1} \\right) \\\\
\n&= {{s^2} + 1} \\over {2\\left( {s + 3} \\right)\\left( {s + 2} \\right)} + {{s + 2} \\over {2\\left( {s^2} + 1} \\right)}} \\\\
\n&= {{s^2} + 1} \\over {\\left( {s + 3} \\right)\\left( {s + 2} \\right)}} + {{s + 2} \\over {{s^2} + 1}}
\n\\end{align<\/em>}$$<\/p>\n