Question

3. Employ nodal analysis and Laplace transform techniques to calculate v(0 in the circuit shown in Figure 3. There is no energy circuit. initially stored in the 4H 5u(t) 162 1+4u(t) Figure 3
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Answer 1

Given ckt is the above ckt converters s - parallel ckt we get for Kirchhoff law V_x(s) - s/s/2/s + v_x(s)/1 + v_x(s) - 5/s/4s = 0 s/2 v_x(s) + v_x(s) + v_x(s)/4s - 5/2 - 5/4s^2 = 0 v_x(s)[s/2 + 1 + 1/4s] = 5/2 + 5/4s^2 v_x(s) = [2s^2 + 4s + 1/4s] = 10s^2 + 54s^2 therefore v_x(s) = 1s^2 + 5/s(2s^2 + 4s + 1) therefore by partial fractions 10s^2 + 5/s(2s^2 + 4s + 1) = A/s + B/(2s^2 + 4s + 1) put s = 1 rightdoublearrow 15/7 = +A + B/7 rightdoublearrow 7A + B = 15 put s = 1 rightdoublearrow 15/1 = A - B rightdoublearrow A + B = -15 \r\n 6A = 30 rightdoublearrow A = 5 and B = -20 V_x(s) = 5/3 - 20/2s^2 + 4s + 1 2s^2 + 4s + 1 = 0 rightdoublearrow s = -4 plusminus squareroot 16 - 8/2 times 2 = -4 plusminus squareroot 8/4 s rightdoublearrow -1 plusminus 1/squareroot 2