hw07_sol

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EECS 16B Designing Information Systems and Devices II UC Berkeley Fall 2023 Homework 7 This homework is due on Saturday, October 14, 2023, at 11:59PM. Self- grades and HW Resubmissions are due on Saturday, October 21, 2023, at 11:59PM. 1. Alternative “second order” perspective on solving the RLC circuit In Homework 6, we solved an RLC circuit by setting state variables x 1 ( t ) = V C ( t ) and x 2 ( t ) = I L ( t ) , and using these to build a linear first-order vector differential equation. In this problem, we will see how to solve the same system by picking different state variables x 1 ( t ) = V C ( t ) and x 2 ( t ) = d d t V C ( t ) , getting a linear second order scalar differential equation, and solving that differential equation. Consider the following circuit like you saw in lecture, discussion, and previous homeworks: − + V s t = 0 t = 0 C + − V C I L R + − V R L + − V L As before, assume that the system 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. Suppose now we insisted on expressing everything in terms of one waveform V C ( t ) instead of two of them (voltage across the capacitor and current through the inductor). This is called the “second- order” point of view, because we will end up using second derivatives. For this problem, use R for the resistor, L for the inductor, and C for the capacitor in all the expressions. (a) Write the current I L ( t ) through the inductor in terms of the voltage V C ( t ) across the capaci- tor. Solution: The current I L ( t ) through the inductor L must be the same as the current I C ( t ) through C , which is C d d t V C ( t ) . Hence, we can write I L ( t ) = C d d t V C ( t ) . (1) (b) Now, notice that the voltage drop across the inductor involves d d t I L ( t ) . Write the voltage drop across the inductor, V L ( t ) , in terms of the second derivative of V C ( t ) . Solution: The voltage drop is V L ( t ) = L d d t I L ( t ) = LC d d t d d t V C ( t ) = LC d 2 d t 2 V C ( t ) . (2) 1

EECS 16B Homework 7 2023-10-15 11:31:33-07:00 (c) Show that a differential equation governing V C ( t ) is d 2 d t 2 V C ( t ) + R L d d t V C ( t ) + 1 LC V C ( t ) = 0. (3) Solution: Note that the current passing through the resistor is I R ( t ) = − V C ( t ) + V L ( t ) R = C d d t V C ( t ) . (4) or equivalently, V L ( t ) + RC d d t V C ( t ) + V C ( t ) = 0. (5) Plugging in V L ( t ) , we have LC d 2 d t 2 V C ( t ) + RC d d t V C ( t ) + V C ( t ) = 0. (6) Finally, dividing by LC , d 2 d t 2 V C ( t ) + R L d d t V C ( t ) + 1 LC V C ( t ) = 0. (7) (d) Recall that previously in class, we solved a second-order differential equation of the form d 2 y ( t ) d t 2 + a d y ( t ) d t + by ( t ) = 0 (8) by converting it into a matrix differential equation d d t " x 1 ( t ) x 2 ( t ) # = " 0 1 − b − a # | {z } A " x 1 ( t ) x 2 ( t ) # (9) where x 1 ( t ) : = y ( t ) and x 2 ( t ) : = d y ( t ) d t . It turned out that, if A has two distinct eigenvalues, the solution to this homogeneous differential equation have the form " x 1 ( t ) x 2 ( t ) # = " c 1 e λ 1 t + c 2 e λ 2 t c 3 e λ 1 t + c 4 e λ 2 t # . (10) where λ 1 , λ 2 are the eigenvalues of A , and c 1 , c 2 , c 3 , c 4 are constants determined by the initial conditions and the coefficients a , b in the differential equation. We would like to use this to construct a solution for V C ( t ) . Show that, if we identify y ( t ) : = V C ( t ) , then x 1 ( t ) = V C ( t ) x 2 ( t ) = d d t V C ( t ) , (11) and that the matrix A for our RLC circuit is A = " 0 1 − 1 LC − R L # . (12) Then, show that the two eigenvalues of A are λ 1 = − R 2 L + 1 2 r R 2 L 2 − 4 LC , λ 2 = − R 2 L − 1 2 r R 2 L 2 − 4 LC . (13) © UCB EECS 16B, Fall 2023. All Rights Reserved. This may not be publicly shared without explicit permission. 2

EECS 16B Homework 7 2023-10-15 11:31:33-07:00 NOTE : From this part onwards, we will assume that the circuit parameters R , L , C are chosen so that the eigenvalues of A are distinct. Solution: We have x 1 ( t ) = y ( t ) = V C ( t ) (14) x 2 ( t ) = d d t y ( t ) = d d t V C ( t ) . (15) Examining the coefficients in eq. (3), we see a = R L (16) b = 1 LC . (17) Thus A = " 0 1 − 1 LC − R L # . (18) The characteristic polynomial of A is 1 p A ( λ ) : = det ( A − λ I ) (19) = det " − λ 1 − 1 LC − R L − λ #! (20) = ( − λ ) − R L − λ − 1 · − 1 LC (21) = λ 2 + R L λ + 1 LC . (22) The solutions to p A ( λ ) = 0 are obtained by the quadratic formula to be λ 1 = − R 2 L + 1 2 r R 2 L 2 − 4 LC λ 2 = − R 2 L − 1 2 r R 2 L 2 − 4 LC . (23) (e) Now, we solve for x 1 ( t ) = V c ( t ) by determining c 1 and c 2 and plugging those, along with λ 1 and λ 2 , into eq. ( 10 ). Note that determining c 3 and c 4 isn’t necessary to find x 1 ( t ) , but we need them to set up a system of equations to solve for c 1 and c 2 . Show that c 3 = λ 1 c 1 , c 4 = λ 2 c 2 . (24) Then use the initial conditions of the RLC circuit to show that c 1 = λ 2 λ 2 − λ 1 V s , c 2 = − λ 1 λ 2 − λ 1 V s . (25) (HINT: This part is complicated, so here’s some scaffolding to start you off. First, differentiate the expres- sion we have x 1 ( t ) = c 1 e λ 1 t + c 2 e λ 2 t to get a form for x 2 ( t ) , and match coefficients of e λ 1 t and e λ 2 t to get the desired expressions for c 3 and c 4 . Next, use the initial conditions for RLC to see what V C ( 0 ) are and 1 Notice that it looks very similar to the original differential equation. This is not an accident, and holds more generally, but that is outside the scope of this problem. © UCB EECS 16B, Fall 2023. All Rights Reserved. This may not be publicly shared without explicit permission. 3

EECS 16B Homework 7 2023-10-15 11:31:33-07:00 d d t V C ( t ) t = 0 are. This corresponds to x 1 ( 0 ) and x 2 ( 0 ) . Plug t = 0 into the "sum of exponentials" form for x 1 and x 2 . This will get you two equations, one for each x i , for c 1 and c 2 , which you can then solve.) (HINT: The following matrix inverse formula may be useful: " 1 1 λ 1 λ 2 # − 1 = 1 λ 2 − λ 1 " λ 2 − 1 − λ 1 1 # (26) ) Solution: By definition, x 2 ( t ) = d d t V C ( t ) = d d t x 1 ( t ) (27) so if x 1 ( t ) = c 1 e λ 1 t + c 2 e λ 2 t (28) then x 2 ( t ) = d d t x 1 ( t ) (29) = d d t c 1 e λ 1 t + c 2 e λ 2 t (30) = λ 1 c 1 e λ 1 t + λ 2 c 2 e λ 2 t . (31) But we know that x 2 ( t ) = c 3 e λ 1 t + c 4 e λ 2 t . (32) Thus by pattern matching the coefficients of e λ 1 t and e λ 2 t , we get c 3 = λ 1 c 1 c 4 = λ 2 c 2 . (33) Now to solve for c 1 and c 2 . Recall that in steady state, a capacitor looks like an open circuit, so V C ( 0 ) = V s . By definition, V C ( t ) = x 1 ( t ) , so x 1 ( 0 ) = V s . Plugging in, we have V s = x 1 ( 0 ) = c 1 e λ 1 · 0 + c 2 e λ 2 · 0 = c 1 + c 2 . (34) Now we have one equation in the variables c 1 and c 2 . To solve the system we need two equations. This motivates looking at x 2 ( 0 ) = λ 1 c 1 e λ 1 · 0 + λ 2 c 2 e λ 2 · 0 = λ 1 c 1 + λ 2 c 2 . (35) To find the physical value of x 2 ( 0 ) = d d t V C ( t ) t = 0 , note that in steady state there is no change in any state variable by definition, so d d t V C ( t ) t = 0 = 0. (An alternate physically motivated ar- gument is to note that inductor current in steady state is I L = 0, and it cannot change infinitely fast, so at time 0 we have I L ( 0 ) = 0. Since I L ( t ) º·¶º d V C ( t ) d t , we also have d V C ( t ) d t t = 0 = 0.) Hence x 2 ( 0 ) = 0. This sets up the system of equations c 1 + c 2 = V s (36) λ 1 c 1 + λ 2 c 2 = 0. (37) There are several ways we can solve this system, and one way is to note that this is a matrix- vector equation of the form " 1 1 λ 1 λ 2 # " c 1 c 2 # = " V s 0 # . (38) © UCB EECS 16B, Fall 2023. All Rights Reserved. This may not be publicly shared without explicit permission. 4

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