ee16b_hw3_sol.pdf - HW 3 18:59:02-07:00 EECS 16B Fall 2019 Designing Information Devices and Systems II UC Berkeley HW 3 This homework is due on 09 23

# ee16b_hw3_sol.pdf - HW 3 18:59:02-07:00 EECS 16B Fall 2019...

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HW 3 @ 2019-09-25 18:59:02-07:00 EECS 16B Designing Information Devices and Systems II Fall 2019 UC Berkeley HW 3 This homework is due on 09, 23, 2019, at 11:59PM. Self-grades are due on 09, 25, 2019, at 11:59PM. 1. RLC Responses: Initial Part Consider the following circuit like you saw in lecture: - + V s t = 0 t = 0 C + - V C I L R + - V R L + - V L Assume the circuit above has reached steady state for t < 0. At time t = 0, the switch changes state and disconnects the voltage source, replacing it with a short. In this problem, the current through the inductor and the voltage across the capacitor are the natural physical state variables since these are what correlate to how energy is actually stored in the system. (A magnetic field through the inductor and an electric field within the capacitor.) (a) Write the system of differential equations in terms of state variables x 1 ( t ) = I L ( t ) and x 2 ( t ) = V C ( t ) that describes this circuit for t 0 . Leave the system symbolic in terms of V s , L , R , and C . Solution: For this part, we need to find two differential equations, each including a derivative of one of the state variables. First, let’s consider the capacitor equation I C ( t ) = C d dt V C ( t ) . In this circuit, I C ( t ) = I L ( t ) , so we can write I C ( t ) = C d dt V C ( t ) = I L ( t ) (1) d dt V C ( t ) = 1 C I L ( t ) . (2) If we use the state variable names, we can write this as d dt x 2 ( t ) = 1 C x 1 ( t ) , (3) so now we have one differential equation. © UCB EECS 16B, Fall 2019. All Rights Reserved. This may not be publicly shared without explicit permission. 1

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HW 3 @ 2019-09-25 18:59:02-07:00 For the other differential equation, we can apply KVL around the single loop in this circuit. (Alterna- tively, we could just solve it directly and substitute in for the desired voltage on the capacitor, which is a state variable.) Going clockwise, we have V C ( t )+ V R ( t )+ V L ( t ) = 0 . (4) Using Ohm’s Law and the inductor equation V L = L d dt I L ( t ) , we can write this as V C ( t )+ RI L ( t )+ L d dt I L ( t ) = 0 , (5) which we can rewrite as d dt I L ( t ) = - R L I L ( t ) - 1 L V C ( t ) . (6) If we use the state variable names, this becomes d dt x 1 ( t ) = - R L x 1 ( t ) - 1 L x 2 ( t ) , (7) and we have a second differential equation. To summarize the final system is d dt x 1 ( t ) = - R L x 1 ( t ) - 1 L x 2 ( t ) (8) d dt x 2 ( t ) = 1 C x 1 ( t ) . (9) (b) Write the system of equations in vector/matrix form with the vector state variable ~ x ( t ) = " x 1 ( t ) x 2 ( t ) # . This should be in the form d dt ~ x ( t ) = A ~ x ( t ) with a 2 × 2 matrix A . Solution: By inspection from the previous part, we have " d dt x 1 ( t ) d dt x 2 ( t ) # = " - R L - 1 L 1 C 0 #" x 1 ( t ) x 2 ( t ) # , (10) which is in the form d dt ~ x ( t ) = A ~ x ( t ) , with A = " - R L - 1 L 1 C 0 # . (11) (c) Find the eigenvalues of the A matrix symbolically. (Hint: the quadratic formula will be involved.) Solution: To find the eigenvalues, we’ll solve det ( A - λ I ) = 0. In other words, we want to find λ such that det ( A - λ I ) = det " - R L - λ - 1 L 1 C - λ # (12) = - λ - R L - λ + 1 LC (13) = λ 2 + R L λ + 1 LC = 0 . (14) © UCB EECS 16B, Fall 2019. All Rights Reserved. This may not be publicly shared without explicit permission. 2
HW 3 @ 2019-09-25 18:59:02-07:00 The Quadratic Formula gives λ = - 1 2 R L ± 1 2 r ( R L ) 2 - 4 LC . (15) (d) Under what condition on the circuit parameters R , L , C are there going to be a pair of distinct

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