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A CT system is described by the following differential equation. d²y (t) dy (t) +5.4- dt? +2y(t)=x(t) dt Find the output yt) for the input x (t) =2.4e-tu(t) Note that the system is initially at rest. y(t) =2e-0.4t_1.5e-5t+4, t20 y(t) =-e-2t (t-1) +2e-t, t20 y(t) =-e-2* (t+1) +e-t, t20 y(t) =-e-2t (2t+1) +1, t20 y(t)=0.2e-0.4t +1.6e-5t_2.4e-t, t20 y(t) =0.87e-0.4t+0.13e-St-e-t, t20 y(t) =-2.17e0.4t +0.17e-5t+2, t20 y(t) =-e-2t (t+2)+3, t20

A CT system is described by the following differential equation. d²y (t) dy (t) +5.4- dt? +2y(t)=x(t) dt Find the output yt) for the input x (t) =2.4e-tu(t) Note that the system is initially at rest. y(t) =2e-0.4t_1.5e-5t+4, t20 y(t) =-e-2t (t-1) +2e-t, t20 y(t) =-e-2* (t+1) +e-t, t20 y(t) =-e-2t (2t+1) +1, t20 y(t)=0.2e-0.4t +1.6e-5t_2.4e-t, t20 y(t) =0.87e-0.4t+0.13e-St-e-t, t20 y(t) =-2.17e0.4t +0.17e-5t+2, t20 y(t) =-e-2t (t+2)+3, t20

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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A CT system is described by the following differential equation. d²y (t) dy (t) dt? +5.4- +2y(t)=x(t) dt Find the output y(t) for the input x (t)=2.4e-tu (t) Note that the system is initially at rest. y (t) =2e-0.4t-1.5e-5t+4, t20 y(t) =-e-2t (t-1) +2e-t, t20 y(t)=-e-2t (t+1) +e-t, t20 y (t) =-e-2t (2t+1)+1, t20 y (t) =0.2e-0.4t+1.6e-5t_2.4e-t, t20 y (t) =0.87e-0.4t +0.13e¬5t_e-t, t20 y (t) =-2.17e-0.4t+0.17e-5t+2, t20 y (t) =-e-2t ( t+2) +3, t20
Transcribed Image Text:A CT system is described by the following differential equation. d²y (t) dy (t) dt? +5.4- +2y(t)=x(t) dt Find the output y(t) for the input x (t)=2.4e-tu (t) Note that the system is initially at rest. y (t) =2e-0.4t-1.5e-5t+4, t20 y(t) =-e-2t (t-1) +2e-t, t20 y(t)=-e-2t (t+1) +e-t, t20 y (t) =-e-2t (2t+1)+1, t20 y (t) =0.2e-0.4t+1.6e-5t_2.4e-t, t20 y (t) =0.87e-0.4t +0.13e¬5t_e-t, t20 y (t) =-2.17e-0.4t+0.17e-5t+2, t20 y (t) =-e-2t ( t+2) +3, t20
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