Purdue University School of Electrical and Computer Engineering ECE 20200 : Linear Circuit Analysis II Summer 2014 Instructor: Aung Kyi San Instructions: Midterm Examination I July 2, 2014 1. Wait for the BEGIN signal before opening this booklet. 2. Enter your name, student ID number, e-mail address and your full signature in the space provided on this page. 3. You have 90 minutes to complete all 5 questions contained in this exam. When the end of the exam is announced, you must stop writing immediately. Anyone caught writing after the exam is over will get a grade of zero. 4. Read the questions carefully. Unless otherwise stated, you must fully justify your answers. You may use any method you want unless you are asked to use a specific method. 5. This booklet contains 15 pages including the Laplace Transform Tables. 6. Notes, books, cell phones, pagers and any other electronic communication device are strictly forbidden. However, you are allowed to use a one-line calculator. Name: Student ID: Email: Signature: -1-
(Total 20 pts) 1. (a) (8 pts) If f 1 (t) = r(2 t)u(1 t), f 2 (t) = δ(t 2) cos then find the Laplace transform of f 3 (t). ( π 2 t ) u(t), and f 3 (t) = f 1 (t)+f 2 (t 1), -2-
(b) Consider the circuit given below. (i) (1 pt) Express V out (s) in terms of V in (s). (ii) (5 pts) If V in (s) = 4s2 + ( 3 + 3 ) s + 2, then find the partial fraction expansion of V s 3 + s 2 out (s). + s -3-
(iii) (2 pts) Find v out (t), the inverse Laplace transform of V out (s). (c) (4 pts) Find the inverse Laplace transform of the function G(s) = s + 2 s 2 + 7s + 12 + s + 3 (s + 1) 2 + 7(s + 1) + 12. -4-
(Total 20 pts) 2. (a) Consider the RLC circuit given below with R = 0.25Ω, C = 2F and L = 0.5H. Suppose v in (t), v C (t) and i L (t) are all zero for t < 0. (i) (2 pts) Show that the integro-differential equation of the circuit is given by 2v c (t) + dv c(t) dt + t v c (q)dq = t v in (q)dq (ii) (6 pts) By taking Laplace transform of the equation obtained in part (i), find V c (s) if v in (t) = te t u(t) V. -5-
(iii) (2 pts) Find v c (t). -6-
(b) Compute the input impedance of each of the circuits given below. (i) (6 pts) (ii) (4 pts) For this circuit, assume that i L (0 ) = 0 A and v C (0 ) = 1 V. -7-
(Total 20 pts) 3. Consider the circuit below. (a) (7 pts) Compute H 1 (s) = V (s) V in (s) and H 2(s) = V out(s) V (s). Then find H(s) = V out(s) V in (s). Give your answers in terms of R 1, R 2, R 3, L, C 1, and C 2. -8-
( 1 (b) (3 pts) If H(s) = s + 1 C 2. ) ( ) 4, find a set of values of R s 2 1, R 2, R 3, L, C 1, and + 4s + 4 (c) (5 pts) Given H(s) in part (b), find the impulse response of the circuit. (d) (5 pts) Given H(s) in part (b), find v out (t) if v in (t) = δ(t) + 2u(t) V. -9-
(Total 20 pts) 4. (a) (6 pts) In the circuit below, i in (t) = 4u(t) A and i L (0 ) = 2 A. Find i L (t) for t 0. 5(2s + 7) (b) (6 pts) If F (s) = s(s 2 + 2s + 5). Find f(0+ ) and f( ) using the Initial Value Theorem and Final Value Theorem respectively. -10-
(c) (8 pts) For the circuit shown below, C 1 = 1 F, C 2 = 4 F, v out (0 ) = 0, and v in (t) = 10u(t) V. The switch moves from position A to position B at t = 1 s, back to position A at t = 2 s, and then back to position B at t = 3 s, where it remains forever. Find v out (3.6). -11-
(Total 20 pts) 5. Consider the circuit given below. The switch moves from position A to position B at t = 0. Assume i L (0 ) = 0. (a) (1 pt) Find v C (0 ). (b) (3 pts) Draw the equivalent s-domain circuit for t 0. (c) (9 pts) Write three nodal equations for the circuit of part (b) only in terms of V C (s), V out (s), I L (s), the answer from part (a) for the intial condition v C (0 ) and the input current, I in (s). Simplify your equations. -12-
(d) (2 pts) Put equations in matrix form. (e) (5 pts) Compute the zero-input response v out (t) for t 0. -13-
Table 12.1 LAPLACE TRANSFORM PAIRS Item Number f(t) L[f(t)] = F(s) 1 Kδ(t) K 2 Ku(t) or K K s 3 r(t) 1 s 2 4 t n u(t) n! s n+1 5 e at u(t) 1 (s + a) 6 te at u(t) 1 (s + a) 2 7 t n e at u(t) n! (s + a) n+1 8 sin(ωt)u(t) ω s 2 + ω 2 9 cos(ωt)u(t) s s 2 + ω 2 10 e at sin(ωt)u(t) ω (s + a) 2 + ω 2 11 e at cos(ωt)u(t) (s + a) (s + a) 2 + ω 2 12 tsin(ωt)u(t) 2ωs (s 2 + ω 2 ) 2 13 tcos(ωt)u(t) s 2 ω 2 (s 2 + ω 2 ) 2 14 sin(ωt + φ)u(t) ssin(φ) + ω cos(φ) s 2 + ω 2 15 cos(ωt + φ)u(t) s cos(φ) ω sin(φ) s 2 + ω 2 16 e at [sin(ωt) ωtcos(ωt)]u(t) 2ω 3 [(s + a) 2 + ω 2 ] 2-14-
17 te at sin(ωt)u(t) s + a 2ω [(s + a) 2 + ω 2 ] 2 18 e at C 1 cos(ωt) + C 2 C 1 a ω sin(ωt) u(t) C 1 s + C 2 s + a ( ) 2 + ω 2 Table 12.2 LAPLACE TRANSFORM PROPERTIES Property Linearity Transform Pair L[a 1 f 1 (t) + a 2 f 2 (t)] = a 1 F 1 (s) + a 2 F 2 (s) Time Shift L[f(t T)u(t T)] = e st F(s), T > 0 Multiplication by t Multiplication by t n L[tf(t)u(t)] = d ds F(s) L[t n f (t)] = ( 1) n d n F(s) ds n Frequency Shift L[e at f(t)] = F(s + a) Time Differentiation Second-Order Differentiation nth-order Differentiation Time Integration Time/Frequency Scaling L d dt f (t) = sf(s) f(0 ) L d2 f (t) dt 2 = s2 F(s) sf (0 ) f (1) (0 ) L dn f (t) dt n = sn F(s) s n 1 f (0 ) s n 2 f (1) (0 ) K f (n 1) (0 ) (i) L t (ii) L f (q)dq = F(s) + s t 0 0 f (q)dq = F(s) s L[f(at)] = 1 a F s a f (q)dq s -15-