hw4_sol.pdf - ECE 202 PROBLEM SET 4 SOLUTIONS 1(a See...

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ECE 202: PROBLEM SET 4: SOLUTIONS 1. (a) See Figure 1. Figure 1: 1 (a). (b) For the left node, I S 2 - V C 1 - sV C + Cv C ( 0 - ) - I d = 0 ( s + 1) V C + I d = I S 2 + Cv C ( 0 - ) (1) For the right node, I d + 3 I d - V 2 s - i L (0 - ) s - V 2 1 = 0 ⇒ - 1 s + 1 V 2 + 4 I d = i L (0 - ) s (2) For the constraint equation, I d = V C - V S 1 - V 2 1 V C - V 2 - I d = V S 1 (3) (c) From (1), (2), and (3), we obtain s + 1 0 1 0 - ( 1 s + 1 ) 4 1 - 1 - 1 V C V 2 I d = I S 2 + Cv C (0 - ) i L ( 0 - ) s V S 1 (4) (d) By substituting the inputs into (4), we obtain s + 1 0 1 0 - ( 1 s + 1 ) 4 1 - 1 - 1 V C V 2 I d = 10 0 10 (5)
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By using Cramer’s Rule, I d = det s + 1 0 10 0 - ( s +1) s 0 1 - 1 10 det s + 1 0 1 0 - ( s +1) s 4 1 - 1 - 1 = - 2 s s + 2 5 = - 2 + 4 5 s + 2 5 Hence, i d ( t ) = - 2 δ ( t ) + 4 5 e - 2 5 t u ( t ) A (e) By substituting the inputs into (4), we obtain s + 1 0 1 0 - ( 1 s + 1 ) 4 1 - 1 - 1 V C V 2 I d = 0 0 10 s (6) By using Cramer’s Rule, V 2 = det s + 1 0 1 0 0 4 1 10 s - 1 det s + 1 0 1 0 - ( s +1) s 4 1 - 1 - 1 = - 8 s + 2 5 Hence, v 2 ( t ) = - 8 e - 2 5 t u ( t ) V 2. (a) Since the circuit has stayed there for a long time, the capacitors become open circuit, and the circuit looks like Figure 2. Figure 2: 2 (a). 2
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Hence, v C 1 ( 0 - ) = R 2 + R 3 R 1 + R 2 + R 3 × ( - 30) = - 24 V v C 2 ( 0 - ) = R 3 R 1 + R 2 + R 3 × ( - 30) = - 12 V (b) See Figure 3. Figure 3: 2 (b). (c) For the left node, - 7 . 2 - 3 sV C 1 10 - V C 1 - V out 4 = 0 - 3 s 10 - 1 4 V C 1 + 1 4 V out = 7 . 2 (7) For the right node, V C 1 - V out 4 + ( - 3) - sV out 4 - V out 4 = 0 1 4 V C 1 - s + 2 4 V out = 3 (8) (d) From (7) and (8), we obtain - 3 s 10 - 1 4 1 4 1 4 - s +2 4 V C 1 V out = 7 . 2 3 By MATLAB: syms s % system and input A = [-3*s/10-1/4,1/4;1/4,-(s+2)/4]; b = [7.2;3]; % compute the voltage vector V = A \ b; % Vout is the second element of V 3
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Vout = V(2) % use inverse Laplace transform to find vout vout = ilaplace(Vout) % optional, display vout with decimal numbers vpa(vout,4) V out ( s ) = - 12 (6 s + 17) 6 s 2 + 17 s + 5 v out ( t ) = 24 13 e - 5 2 t u ( t ) - 180 13 e
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