EECE 290 Exam 3 with Solution, Spring 2016

– + vSRC 1 F 1  – + vC 1  t = 0 6 µF 2 µF + – V0 2 2 -2 A 1 -1 t t EECE 290 Quiz 3, Apr 16, 2016 4% (Primal Question) A. Determine vC(t), t  0+, assuming vSRC(t) = (t) V and the capacitor to be initially uncharged. Solution: If the impulse is K(t) V, it will cause a current impulse of K(t) A through the capacitor, depositing a charge of K C on the capacitor and resulting in vC(0+) = K V;  = 0.5 s, so that V, t is in s. Version 1: K = 1, = V Version 2: K = 2, = 2 V Version 3: K = 3, = 3 V Version 4: K = 4, = 4 V Version 5: K = 5, = 5 V. 4% (Primal Question) B. The switch is closed at t = 0, with the 2 F capacitor initially charged to V0 = 2 V, and the 6 F capacitor uncharged. Determine the common voltage across the capacitors at t = 0+, and the strength of the impulse that flows. Solution: The initial capacitor charge is 2V0 C, and the final voltage is 2V0/8 = V0/4 V. The charge on the 6 F capacitor is 1.5V0 C, which is the strength of the current impulse. Version 1: V0 = 2 V, v(0+) = V0/4 = 0.5 V, q = 1.5V0 = 3 C Version 2: V0 = 4 V, v(0+) = V0/4 = 1 V, q = 1.5V0 = 6 C Version 3: V0 = 6 V, v(0+) = V0/4 = 1.5 V, q = 1.5V0 = 9 C Version 4: V0 = 8 V, v(0+) = V0/4 = 2 V, q = 1.5V0 = 12 C Version 5: V0 = 10 V, v(0+) = V0/4 = 2.5 V, q = 1.5V0 = 15 C. 4% (Primal Question) C. Determine the maximum value of the convolution integral when the two pulses are convolved together, assuming A = 1. Solution: The maximum area under the product is 2A×2 = 4A. Version 1: A = 1, ym = 4A = 4 Version 2: A = 1.5, ym = 4A = 6 1 – + K(t) V 1 k iL 2u(t) A 1 H Ku(t) A 1 F – + 5 V 1 H 2  iR 2 A Version 3: A = 2, ym = 4A = 8 Version 4: A = 2.5, ym = 4A = 10 Version 5: A = 3, ym = 4A = 12. 4% (Primal Question) D. Derive the ILT of . Solution: The PFE of is . The ILT is A(1 – cost)u(t). Version 1: A = 1, ILT = A(1 – cost) = (1 – cost)u(t) Version 2: A = 2, ILT = A(1 – cost) = 2(1 – cost)u(t) Version 3: A = 3, ILT = A(1 – cost) = 3(1 – cost)u(t) Version 4: A = 4, ILT = A(1 – cost) = 4(1 – cost)u(t) Version 5: A = 5, ILT = A(1 – cost) = 5(1 – cost)u(t). 4% 1. Determine iL(0+), assuming IL0 = 2 A and K = 1 Vs. Solution: The current source in parallel with the ideal voltage source is redundant. The impulse appears across the inductor establishing a flux linkage of K Wb-T and causing a current of K/1 = K A at t = 0+. This adds to IL0, so that iL(0+) = K + 2 A. For t  0+, the voltage source acts as a short circuit, thereby diverting the current source. Version 1: K = 1 Vs, iL(0+) = K + 2 = 3 A Version 2: K = 2 Vs, iL(0+) = K + 2 = 4 A Version 3: K = 3 Vs, iL(0+) = K + 2 = 5 A Version 4: K = 4 Vs, iL(0+) = K + 2 = 6 A Version 5: K = 5 Vs, iL(0+) = K + 2 = 7 A. 4% 2. Determine iR(t), t  0+, assuming at t = 0- an initial voltage of 5 V across the capacitor and an initial circulating current of 2 A, with K = 1 A. Solution: When the current step is applied, the current through the inductor is not forced to change. iR(0+) = (K + 2) A. As t  , the inductor acts as a short- circuit, so that the source current flows through the short circuit and iR  0.  = 0.5 s, so that 2 – + K(t) V 0.5 H 1 F 1  iC – + V0 1 F 1 F 2  2 F t = 0 – + V0 iR A, t is in s. Version 1: K = 1 C, = A, t is in s Version 2: K = 2 C, = A, t is in s Version 3: K = 3 C, = A, t is in s Version 4: K = 4 C, = A, t is in s Version 5: K = 5 C, = A, t is in s. 4% 3. Determine iC(0+), assuming no initial current in the inductor, an initial V0 = 1 V, and K = 1 Vs. Solution: Initially, the inductor acts as an open circuit, the capacitor voltage is not forced to change at t = 0+, and an initial current of 1 A flows downwards through the resistor. The voltage impulse appears across the inductor, establishing a flux linkage of K Wb-T and causing a current of 2K A to flow through the inductor. It follows that iC(0+) = (2K – 1) A. Version 1: K = 1 Vs, iC(0+) = (2K – 1) = 1 A Version 2: K = 2 Vs, iC(0+) = (2K – 1) = 3 A Version 3: K = 3 Vs, iC(0+) = (2K – 1) = 5 A Version 4: K = 4 Vs, iC(0+) = (2K – 1) = 7 A Version 5: K = 5 Vs, iC(0+) = (2K – 1) = 9 A. 4% 4. Determine iR(t), t  0+, assuming an initial voltage V0 = 4 V, and no initial charge in the other two capacitors. Solution: When the switch is closed, the voltages across the two 1 F capacitors are forced to become equal, that across the 2 F is not forced to change because of iR, which is finite. Charge is conserved between the two 1 F capacitors at t = 0+, so that the voltage across them at t = 0+ is V0/2 V. iR(0+) = (V0/4) A. The equivalent capacitance in series with the resistor is 1 F, so that  = 2 s. Hence, A, t in s. Version 1: V0 = 4 V, = A, t in s Version 2: V0 = 8 V, = A, t in s Version 3: V0 = 12 V, = A, t in s Version 4: V0 = 16 V, = A, t in s 3 4 2 -2 f(t) t A 1 -1 f(t) t A/2 1 -1 f(1)(t) t -A(t) Version 5: V0 = 20 V, = A, t in s. 4% 5. f(t) is convolved with the impulses ((t + 1) + (t – 1)). Determine the value of the convolution integral at t = 0. Solution: When f(t) is convolved with (t + 1), it is shifted by 1 unit to the left. Its value at t = 0 is 2. When convolved with (t – 1), it is shifted by 1 unit to the right. Its value at t = 0 is also 2. Hence y(0) = 4. If the impulses have a strength of K, y(0) = 4K. Version 1: K = 1, y(0) = 4K = 4 Version 2: K = 1.25, y(0) = 4K = 5 Version 3: K = 1.5, y(0) = 4K = 6 Version 4: K = 1.75, y(0) = 4K = 7 Version 5: K = 2, y(0) = 4K = 8. 4% 6. Determine the LT of the derivative of f(t), assuming A = 1. Solution: The derivative of f(t) is a shown. The LT, neglecting values for t < 0 is: . As a check, the LT of f(t) is: = = . Multiplying by s and subtracting A/2, gives , the same as above. Version 1: A = 1, Version 2: A = 2, Version 3: A = 4, Version 4: A = 6, Version 5: A = 8, . 4 1 1 -1 -1  t t+1 1 1 -1 -1  1-t t t - 1 1 1 -1 -1  t t - 1 2-t 2 +v2 – – +v3 – + v1 t = 0 C1 C2 C3 1 1 1 -1 -1 f(t) g(t) t, s t, s 1 20% 7. Given C1 = 8 F, initially charged to V10 = 6 V, C2 = 6 F, initially charged to V20 = 4 V, and C3 = 3 F, initially charged to V30 = 2 V. The switch is closed at t = 0. Determine the final values of v2 and v3. Solution: Because of the short circuit by the switch, C2 and C3 discharge independently of C1. The initial charges are q2 = 24 C and q3 = 6 C. The equivalent series capacitor is 2 F and has an initial voltage of 6 V, so that its charge is 12 C. When this capacitor discharges, 12 C flow CCW. When these flow through C2 and C3, q3f = -6 C, so that v3f = -2 V, and q2f = 12 C, so that v2f = 2 V. 20% 8. Derive y(t) = f(t)g(t) for all t. Solution: When g(t) is folded and shifted, it is seen that: 1. For t  -1, y(t) = 0 2. For -1  t  0, y(t) = . In terns of the area of the triangle, y(t) = . At the limits of this interval, y(-1) = 0, and y(0) = 1/2. 3. For 0  t  1, y(t) = = . In terms of areas, the area of the trapezoid is . The area of the triangle is . Adding these two gives . At the limits of this interval, y(0) = 1/2, and y(1) = -1/2. 4. For 1  t  2, y(t) = . In terms of areas, the area of the trapezoid is 5 – + vSRC 1 F 1  iSRC – + VC0 . At the limits of this interval, y(1) = -1/2, and y(2) = 0. 5. For t  2, y(t) = 0. 20% 9. Given vSRC(t) = V, with an initial voltage VC0 = 1 V at t = 0-. (a) Express KCL at the upper node in terms of vSRC(t) and iSRC(t) and take the LT of both sides of the equation (4 grades); (b) derive Isrc(s) as a rational function of s (8 grades); (c) obtain iSRC(t) by ILT (8 grades). Solution: (a) iSRC = vSRC/1 + 1×dvSRC/dt; taking the LT of both sides: Isrc(s) = Vsrc(s)(s + 1) – 1. (b) ; substituting for Vsrc(s), Isrc(s) = . (c) Taking the ILT, iSRC(t) = A. 6