[amp_mcq option1=”v(t) = z(t).i(t)” option2=”$$v\\left( t \\right) = \\int\\limits_0^t {i\\left( \\tau \\right)} z\\left( {t – \\tau } \\right)d\\tau $$” option3=”$$v\\left( t \\right) = \\int\\limits_0^t {i\\left( \\tau \\right)} z\\left( {t + \\tau } \\right)d\\tau $$” option4=”v(t) = z(t) + i(t)” correct=”option4″]<\/p>\n
The correct answer is: $$v\\left( t \\right) = \\int\\limits_0^t {i\\left( \\tau \\right)} z\\left( {t – \\tau } \\right)d\\tau $$<\/p>\n
The Laplace transform of a function $f(t)$ is denoted by $F(s)$ and is defined as:<\/p>\n
$$F(s) = \\int_0^\\infty f(t) e^{-st} dt$$<\/p>\n
The inverse Laplace transform of a function $F(s)$ is denoted by $f(t)$ and is defined as:<\/p>\n
$$f(t) = \\frac{1}{2\\pi i} \\int_{c-i\\infty}^{c+i\\infty} F(s) e^{st} ds$$<\/p>\n
where $c$ is a real number such that all the poles of $F(s)$ lie to the left of $c$.<\/p>\n
In the given question, we are given that $V(s) = Z(s) I(s)$. We can write this as:<\/p>\n
$$V(s) = \\int_0^\\infty z(t) e^{-st} dt \\int_0^\\infty i(t) e^{-st} dt$$<\/p>\n
Using the convolution theorem, we can write this as:<\/p>\n
$$V(s) = \\int_0^\\infty \\left( \\int_0^t z(\\tau) d\\tau \\right) i(t-\\tau) e^{-st} dt$$<\/p>\n
Taking the inverse Laplace transform of both sides, we get:<\/p>\n
$$v(t) = \\int_0^t z(\\tau) i(t-\\tau) d\\tau$$<\/p>\n
This is the convolution of the functions $z(t)$ and $i(t)$.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”v(t) = z(t).i(t)” option2=”$$v\\left( t \\right) = \\int\\limits_0^t {i\\left( \\tau \\right)} z\\left( {t – \\tau } \\right)d\\tau $$” option3=”$$v\\left( t \\right) = \\int\\limits_0^t {i\\left( \\tau \\right)} z\\left( {t + \\tau } \\right)d\\tau $$” option4=”v(t) = z(t) + i(t)” 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-59366","post","type-post","status-publish","format-standard","hentry","category-signal-processing","no-featured-image-padding"],"yoast_head":"\n