{"id":43661,"date":"2024-04-15T21:32:38","date_gmt":"2024-04-15T21:32:38","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=43661"},"modified":"2024-04-15T21:32:38","modified_gmt":"2024-04-15T21:32:38","slug":"the-response-of-an-initially-relaxed-linear-constant-parameter-network-to-a-unit-impulse-applied-at-t-0-is-4e-2tut-the-response-of-this-network-to-a-unit-step-function-will-be","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/the-response-of-an-initially-relaxed-linear-constant-parameter-network-to-a-unit-impulse-applied-at-t-0-is-4e-2tut-the-response-of-this-network-to-a-unit-step-function-will-be\/","title":{"rendered":"The response of an initially relaxed linear constant parameter network to a unit impulse applied at t = 0 is 4e-2tu(t). The response of this network to a unit step function will be"},"content":{"rendered":"

[amp_mcq option1=”2[1 – e-2t]u(t)” option2=”4[e-t – e-2t]u(t)” option3=”sin2t” option4=”(1 – 4e-4t)u(t)” correct=”option1″]<\/p>\n

The correct answer is A. 2[1 – e-2t]u(t).<\/p>\n

The response of an initially relaxed linear constant parameter network to a unit impulse applied at t = 0 is 4e-2tu(t). This means that the network’s output is 4e-2t for all values of t after t = 0.<\/p>\n

The response of this network to a unit step function is the integral of the response to a unit impulse. The integral of 4e-2t is 2(1 – e-2t). Therefore, the response of this network to a unit step function is 2(1 – e-2t)u(t).<\/p>\n

Option B is incorrect because it does not include the factor of 2. Option C is incorrect because it is not a step function. Option D is incorrect because it is not a decaying exponential.<\/p>\n","protected":false},"excerpt":{"rendered":"

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