[amp_mcq option1=”ae-at” option2=”$$\\left( {{1 \\over a}} \\right)\\left( {1 – {e^{ – at}}} \\right)$$” option3=”a(1 – e-at)” option4=”1 – e-at” correct=”option1″]<\/p>\n
The correct answer is $\\boxed{\\left( {{1 \\over a}} \\right)\\left( {1 – {e^{ – at}}} \\right)}$.<\/p>\n
The unit impulse response of a linear time invariant system is the unit step function $u(t)$. This means that if the system is excited with a unit impulse, $u(t)$, the output will be a unit step function, $u(t)$.<\/p>\n
The excitation in this question is $e^{-at}u(t)$. This is a decaying exponential function that is multiplied by a unit step function. The output of the system to this excitation can be found using the convolution integral:<\/p>\n
$$y(t) = \\int_{-\\infty}^t h(t-\\tau)e^{-a\\tau}d\\tau$$<\/p>\n
where $h(t)$ is the unit impulse response of the system.<\/p>\n
The unit impulse response of the system is given as $h(t) = u(t)$. Substituting this into the convolution integral gives:<\/p>\n
$$y(t) = \\int_{-\\infty}^t u(t-\\tau)e^{-a\\tau}d\\tau$$<\/p>\n
The convolution of $u(t)$ and $e^{-at}u(t)$ can be found using the following formula:<\/p>\n
$$\\int_{-\\infty}^t u(t-\\tau)e^{-a\\tau}d\\tau = \\left( {{1 \\over a}} \\right)\\left( {1 – {e^{ – at}}} \\right)$$<\/p>\n
Substituting this into the convolution integral gives:<\/p>\n
$$y(t) = \\left( {{1 \\over a}} \\right)\\left( {1 – {e^{ – at}}} \\right)$$<\/p>\n
Therefore, the response of the system to an excitation $e^{-at}u(t)$, $a > 0$ will be $\\left( {{1 \\over a}} \\right)\\left( {1 – {e^{ – at}}} \\right)$.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”ae-at” option2=”$$\\left( {{1 \\over a}} \\right)\\left( {1 – {e^{ – at}}} \\right)$$” option3=”a(1 – e-at)” option4=”1 – e-at” 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-49709","post","type-post","status-publish","format-standard","hentry","category-signal-processing","no-featured-image-padding"],"yoast_head":"\n