hw07_sol

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 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

EECS 16B Homework 7 2023-10-15 11:31:33-07:00 To solve it, we can use the matrix inverse that was provided by the hint to get " c 1 c 2 # = " 1 1 λ 1 λ 2 # − 1 " V s 0 # (39) = 1 λ 2 − λ 1 " λ 2 − 1 − λ 1 1 # " V s 0 # (40) = 1 λ 2 − λ 1 " λ 2 − λ 1 # V s . (41) Thus we have c 1 = λ 2 λ 2 − λ 1 V s , c 2 = − λ 1 λ 2 − λ 1 V s . (42) We have found λ 1 , λ 2 , c 1 , c 2 , so by substituting into eq. ( 10 ) we have solved for x 1 ( t ) = V C ( t ) ! © UCB EECS 16B, Fall 2023. All Rights Reserved. This may not be publicly shared without explicit permission. 5

EECS 16B Homework 7 2023-10-15 11:31:33-07:00 = e λ Δ ⌊ t Δ ⌋ x 0 + b e λ Δ − 1 λ ⌊ t Δ ⌋ − 1 êçæêôæ k = 0 ( e λ Δ ) ( ⌊ t Δ ⌋ − 1 ) − k u c ( k Δ ) . (50) (b) To simplify our (discrete-time) eq. ( 46 ) so we can take Δ → 0, we would like to make some approximations which are valid for small Δ . By using the following two estimates for small Δ : 2 i. t Δ ≈ t Δ ; ii. e λ Δ − 1 λ ≈ Δ ; 3 show that x ( t ) ≈ e λ t x 0 + b e − λ Δ t Δ − 1 êçæêôæ k = 0 e λ ( t − k Δ ) u c ( k Δ ) Δ . (51) Solution: The first estimate justifies getting rid of the “floor” terms. We have a lot of those terms, so it’s good to use it here. Plugging in t Δ ≈ t Δ gives x ( t ) ≈ e λ Δ ⌊ t Δ ⌋ x 0 + b e λ Δ − 1 λ ⌊ t Δ ⌋ − 1 êçæêôæ k = 0 ( e λ Δ ) ( ⌊ t Δ ⌋ − 1 ) − k u c ( k Δ ) (52) ≈ e λ Δ t Δ x 0 + b e λ Δ − 1 λ t Δ − 1 êçæêôæ k = 0 ( e λ Δ ) ( t Δ − 1 ) − k u c ( k Δ ) (53) ≈ e λ t x 0 + b e λ Δ − 1 λ t Δ − 1 êçæêôæ k = 0 e λ t − λ Δ − λ Δ k u c ( k Δ ) (54) ≈ e λ t x 0 + b e λ Δ − 1 λ e − λ Δ t Δ − 1 êçæêôæ k = 0 e λ ( t − k Δ ) u c ( k Δ ) . (55) Then plugging in e λ Δ − 1 λ ≈ Δ gives x ( t ) ≈ e λ t x 0 + b e λ Δ − 1 λ e − λ Δ t Δ − 1 êçæêôæ k = 0 e λ ( t − k Δ ) u c ( k Δ ) (56) ≈ e λ t x 0 + b e − λ Δ t Δ − 1 êçæêôæ k = 0 e λ ( t − k Δ ) u c ( k Δ ) Δ . (57) Note that the dependence of x ( t ) on both x 0 and the input u c is the same; it’s been preserved, and perhaps made more clear, through our approximations. NOTE : This may seem like a long solution, but the main idea is to just use the estimates one by one, and simplify as much as possible. (c) Take the limit of x ( t ) as Δ → 0 , and show that x ( t ) is given by eq. ( 47 ). Recall that the definite integral is defined from Riemann sums as Z t 0 f ( τ ) d τ = lim n → ª§¦ª n − 1 êçæêôæ k = 0 f ( τ ∗ k ) Δ k (58) 2 Both these approximations become equalities in the limit Δ → 0. 3 We can see this approximation using Taylor’s theorem from calculus. © UCB EECS 16B, Fall 2023. All Rights Reserved. This may not be publicly shared without explicit permission. 7

EECS 16B Homework 7 2023-10-15 11:31:33-07:00 where 0 = τ 0 < τ 1 < · · · < τ n = t , τ ∗ k ∈ [ τ k , τ k + 1 ] , and Δ k = τ k + 1 − τ k . The Δ k is the length of the base of the rectangles and the f ( τ ∗ k ) are the heights. As n goes to infinity, the rectangles get skinnier and skinnier, but there are more and more of them. (HINT: Start with eq. ( 51 ) and take limits on both sides. What is n? What is τ k and τ ∗ k ? What is Δ k ? What is f?) (HINT: We chose the form of eq. ( 51 ) carefully; it turns out that Δ k is one particular term involving Δ that goes to 0 as Δ → 0 , and also that it is independent of k.) Solution: We are evaluating lim Δ → 0 x ( t ) = lim Δ → 0   e λ t x 0 + b e − λ Δ t Δ − 1 êçæêôæ k = 0 e λ ( t − k Δ ) u c ( k Δ ) Δ   (59) = e λ t x 0 + b lim Δ → 0   e − λ Δ t Δ − 1 êçæêôæ k = 0 e λ ( t − k Δ ) u c ( k Δ ) Δ   (60) = e λ t x 0 + b lim Δ → 0 e − λ Δ   lim Δ → 0 t Δ − 1 êçæêôæ k = 0 e λ ( t − k Δ ) u c ( k Δ ) Δ   (61) = e λ t x 0 + b lim Δ → 0 t Δ − 1 êçæêôæ k = 0 e λ ( t − k Δ ) u c ( k Δ ) Δ . (62) Here, we want to evaluate the sum on the right side; by pattern matching with the Riemann integration template and the fact that Δ k should shrink to 0 in the limit, we have n = t Δ Δ k = Δ . (63) This implies that τ k = k Δ . (64) To recover τ ∗ k and f from what we already have, one notes that τ ∗ k ∈ [ k Δ , ( k + 1 ) Δ ] and that we must have f ( τ ∗ k ) = e λ ( t − k Δ ) u c ( k Δ ) . (65) From here we see that τ ∗ k = k Δ f ( τ ) = e λ ( t − τ ) u c ( τ ) . (66) We have lim Δ → 0 x ( t ) = e λ t x 0 + b lim Δ → 0 t Δ − 1 êçæêôæ k = 0 e λ ( t − k Δ ) u c ( k Δ ) Δ (67) = e λ t x 0 + b lim n → ª§¦ª n − 1 êçæêôæ k = 0 e λ ( t − τ ∗ k ) u c ( τ ∗ k ) Δ k (68) = e λ t x 0 + b lim n → ª§¦ª n − 1 êçæêôæ k = 0 f ( τ ∗ k ) Δ k (69) = e λ t x 0 + b Z t 0 f ( τ ) d τ (70) = e λ t x 0 + b Z t 0 e λ ( t − τ ) u c ( τ ) d τ (71) © UCB EECS 16B, Fall 2023. All Rights Reserved. This may not be publicly shared without explicit permission. 8