ECE-202 FINAL April 30, 2018 CIRCLE YOUR DIVISION

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1 ECE 202 Final, Spring 8 ECE-202 FINAL April 30, 208 Name: (Please print clearly.) Student CIRCLE YOUR DIVISION DeCarlo- 7:30-8:30 DeCarlo-:30-2: INSTRUCTIONS There are 34 multiple choice worth 6 points each. Exam point total is 204; ur max possible points is 200. This is a closed book, closed notes exam. No scrap paper or calculators are permitted. Promised Tables are included with the exam. Carefully mark your multiple choice answers on the scantron form and also on the text booklet as you will need to put the scantron inside the test booklet and turn both in at the end of the exam. Nothing is to be on the seat beside you. When the exam ends, all writing is to stop. No writing while turning in the exam/scantron or risk an F in the exam. There are two exam forms and this is necessary for the proper grading of your scantron. All students are expected to abide by the customary ethical standards of the university, i.e., your answers must reflect only your own knowledge and reasoning ability. As a reminder, at the very minimum, cheating will result in a zero on the exam and possibly an F in the course. Communicating with any of your classmates, in any language, by any means, for any reason, at any time between the official start of the exam and the official end of the exam is grounds for immediate ejection from the exam site and loss of all credit for this exercise. The professor reserves the right to move students around before the exam starts and during the exam. Do not open, begin, or peek inside this exam until you are instructed to do so.

2 ECE 202 Final, Spring 8 2. For the 2-port interconnection below, C = 2 F and the inner 2-port labeled N has z-parameter matrix z ij (s) = + s s Ω. For the overall 2-port N* the y-parameter y s s 22 (s) at s = is (in mho): () (2) 2 (3) 3 (4) 4 (5) 5 (6) 6 (7) 7 (8) 0 (9) none of above Solution. Valid parallel connection. y ij (s) = + s s s s = s s s + s. For the two port with just the series capacitor, I = CsV CsV 2 = I 2. Therefore y ij (s) = cap Cs Cs y ij (s) = Cs Cs N * Cs Cs ANSWER: (4) Cs Cs. Thus + s s s + s = (C )s + ( C +)s (C +)s (C +)s + 2. Consider the two-port interconnection below that is assumed to have the standard labeling. The s s y-parameter matrix of the top two-port is Y = ; suppose Z s + s (s) = s, Z 3 (s) =. The z-parameter, z (s), at s = of the interconnected two port is: () (2) 2 (3) 3 (4) 4 (5) 0 (6) (7) 2 (8) 4 (9) none of above

3 ECE 202 Final, Spring 8 3 Solution 2. s Z = s Z (s) + Z 3 (s) Z bot = Z 3 (s) s + s Z 3 (s) Z 3 (s) = + s s s s = s +.. Thus, Z overall = Z + Z bot = 2(s +) s + s 2 s. 2(s +) s + s 2 s = s= ANSWER: (5) V 3. Recall the h-parameter matrix, I 2 = h h 2 I. In the circuit below, R = 8 Ω h 2 h 22 V 2 and b = The h-parameters, h and h 2, are respectively (in standard units): () (8, 4) (2) ( 2, 4) (3) (2, 4) (4) (2, 0.25) (5) (2, 8) (6) (2, 8) (7) (2, 6) (8) (8, 4) (9) none of above Solution 3. V 2 = bvˆ = b( V RI ) = bv + bri implies bv = V 2 bri implies V = b V 2 + RI = h 2 V 2 + h I. Hence V = 8I 4V 2. Answer: (8)

4 ECE 202 Final, Spring For the two port configuration below, having y-parameters as indicated, the voltage gain = V s = + Y in = + G V V Y s s + 2. Then y = (in S): () s + 2 (2) 2 s + 2 (3) 8.5 s + 2 (4) 7.5 s + 2 (5) 5 s + 2 (6) 0 s + 2 (7) 8.5 s + 2 (8) None of above Solution 4. Thus Y in = 2 s + 2 = y y 2 y 2 = y y 22 + Y 8 L s + 2. Thus y = 0. Answer: (6) s Referring to problem 4, the response, v (t), to the input V s (s) = 6 s Ke at u(t) where (K,a) = () ( 2, 3) (2) (2, 2) (3) ( 2, 2) (4) (4, 3) (5) ( 4, 3) (6) (4, 2) (7) (2, 3) (8) (3,2) (9) none of above, has a term of the form Solution 5. Y in = 2 s + 2. G V = Y s = Y s + Y in Answer: () s + 2 = s + 2 s + 3. Hence V s + 2 (s) = 6 s(s + 3) = 4 s + 2 s Consider the two port below.

5 ECE 202 Final, Spring 8 5 The 2-port N a has y-parameters [y a ], and the 2-port N b has z-parameters [z b ] as given below (with standard port labeling and units): Then Z ina = : y a = ; z b = () 0 (2) 20 (3) (open circuit) (4) 2 5 (5) 5 (6) 0 (short circuit) (7) 0.2 (8) none of above Solution 6. Z inb = = 20 Ω. 20 / /20 = 0 Ω. Y ina = = 0.2 mho. Thus Z ina = 0.2 = 5 Ω. Answer: (5) 7. Referring again to the circuit of problem 6, the voltage gain V b V a () (2) 2 (3) (4) 2 is: (5) 0.5 (6) 0.5 (7) 4 3 (8) None of above Solution 7. G V 2b = y 2 y 22 + Y L = = Answer: (3)

6 ECE 202 Final, Spring In the circuit below, v in (t) = 00cos(0t)u(t) V, C = F, L = 2 H, L 2 = M = H. The magnitude and phase (in degrees) of i out,ss (t) = K cos(0t +ϕ) are (K,ϕ) = : () (,90 o ) (2) (0.,90 o ) (3) (, 90 o ) (4) (0., 90 o ) (5) (0, 90 o ) (6) (00, 90 o ) (7) (0,90 o ) (8) (00,90 o ) (9) none of above Solution 8. Using the T-equivalent circuit for the coupled inductors yields L eq = H since L 2 M = 0 and M is shorted out on the T-equivalent. Thus I out (s) = H(s)V in (s) implies s H(s) = L eq s + = s 2 + Cs Answer: (5) implies H( j0) = j0 00 = j0 99 j0.. Thus (K,ϕ) = (0, 90o ). 9. In the circuit below, L = L 2 = 2 H, M = H, and R = 8 Ω. The impulse response is: () e 8t u(t) (2) 2e 8t u(t) (3) 8e 8t u(t) (4) 2e 4t u(t) (5) 4e 4t u(t) (6) 6e 8t u(t) (7) 8e 4t u(t) (8) none of above Solution 9. L eq = L + L 2 2M = 2 H. Thus H(s) = 4e 4t u(t). ANSWER: (5) R L eq s + R = 4. Hence impulse response is s + 4

7 ECE 202 Final, Spring Find C so that the voltage gain is zero at ω 0 = 0 rad/s. Assume L = 2.5 H, L 2 = 2 H, M = H, R s = R = R 2 = 2 Ω. () 40 (2) 2 (3) 0.25 (4) 400 (5) 0. (6) 2 (7) 8 (8) 0.0 (9) none of above Solution 0. (Resonance, π-equivalent circuit, mutually coupled inductors.) Using the π-equivalent circuit for the inductors, the top most branch of the π is L top = L L 2 M 2 = 4 H. The voltage gain M is zero at ω 2 0 = L top C C = 2 L top ω = 4 00 = 400 F 0 ANSWER: (4). The transfer function that best meets the phase response plot below is H(s) = : () s +000 s +00 (5) s +00 s (2) s 000 s +00 s 25 (6) s +000 (3) s 00 s (7) s 50 s +000 (4) s 800 s +0 s +50 (8) s +000

8 ECE 202 Final, Spring 8 8 ANSWER: (2) 2. The input impedance Z in (s) for the circuit below is: () s (5) 6s 3s + 2.5s (2) 3s + 2 (3) 3s s (6) (7) 3s + 2 3s (4) 4s 2s + 2 (8) none of above

9 ECE 202 Final, Spring 8 9 Solution 2. M = 2 H. Using the T-equivalent circuit, one obtains a parallel connection of two 2 H inductors in series with a 2 H inductor making L eq = 3 H. Thus Z in (s) = ANSWER: (5) 6s 3s The zero-state response i out (t) to the input v in (t) = 6e 2t u(t) V has a term of the form Ke 2t u(t) where K = : () (2) 2 (3) (4) 4 (5) 2 (6) 4 (7) 6 (8) 6 (9) none of above Solution 3. V 2 (s) = 0 = si (s) + 2sI 2 (s). I 2 (s) = 0.5I (s). V (s) = 2sI (s) si 2 (s) = ( 2 0.5)sI (s) =.5sI (s). I out (s) = I (s) =.5s V in (s) = 4 s(s + 2) = 2 s + 2. Answer: (2) s Consider the circuit below which is approximately a BP circuit. Suppose L = 2 H, C = 2 F, and R = Ω. After approximating the non-ideal inductor with a parallel RL circuit, the Q of the 2 new approximate circuit is: () (2) 2 (3) 2 (4) 4 (5) 2 (6) 6 (7) 0.25 (8) none of above

10 ECE 202 Final, Spring 8 0 Solution 4. Q L (ω ) = ω L R = 2. R parallel 44 2 = 2. H(s) = C s s 2 + R parallel C s + LC Answer. (3) = 2s s 2 + s +44. Hence B w = rad/s, Q = 2 = Referring again to the circuit of problem 4, the maximum absolute value of the transfer function of the (new approximate) circuit is: () (2) 2 (3) (4) 4 2 (5) 2 (6) 6 (7) 0.25 (8) none of above Solution 5. Finally, H m = 2. Answer (5) 6. Given that H(s) = V out V in is BIBO stable is: in the circuit below, the COMPLETE range of g m for which the circuit () R > g m (2) R < g m (3) R < g m (4) R > g m (5) g m < RC (6) g m > RC (7) (9) none of the above RC > g m (8) RC < g m

11 ECE 202 Final, Spring 8 Solution 6. CsV out g m V out + R V out V in H(s) = R Cs g m + R ( ) = 0 implies Cs g m + R. Stability requires that R > g m. Answer: (4) V out = R V in implies 7. Consider the circuit below in which a = 0.5, C = 0.5 F, C 2 = 0.5 F, R s = Ω, and the dot is in position A. Suppose V in (s) = 0 s + 5. Then v 2 (t) the form ( K e 0t + K 2 e 5t )u(t) where (K, K 2 ) = : () (6, 6) (2) ( 6, 6) (3) ( 4, 4) (4) (4, 4) (5) (4,4) (6) (8,8) (7) ( 8,8) (8) (8, 8) (9) none of above

12 ECE 202 Final, Spring 8 2 Solution 7. 8 V (s) = s + 0 s C 2 s reflected to the primary side becomes a 2 C 2 s = 8 s V in (s) =. By voltage division, 8 s +0 V in. Thus V 2 (s) = s +0 0 s + 5 = 40 (s +0)(s + 5) = 8 s +0 8 s + 5 Answer: (8) 8. In the circuit below, the load is considered to be at the terminals labeled A-B. Suppose R s = 8 Ω, R = 40 Ω, and R L = 0.4 Ω. The turns ratio for maximum power transfer to the load is: () (2) 2 (3) 0 (4) 4 (5) 5 (6) 0. (7) 0.2 (8) 0.25 (9) none of above Solution 8. The problem reduces to 0 = a 2 R L a 2 = = 00 4 a = 5. Answer: (5) 9. In the circuit below, let v in (t) = V rms cos(2π f 0 t)u(t) V where V rms = 00 V. Suppose further that R s = 5 Ω, R L = 4 Ω, and R trans = 00 Ω. Suppose a = b = 0. The value of V,rms = (in Volts) = () 0 (2) 20 (3) 30 (4) 40 (5) 50 (6) 60 (7) 00 (8) (9) none of above

13 ECE 202 Final, Spring 8 3 Solution 9. Z in = a 2 Answer: (5) ( R trans + b2 R L ) = ( ) = 5 Ω. Therefore V 00,rms = 00 = 50 V Referring again to problem 9, the power lost in the transmission line due to R trans is (in watts): () 0 (2) 20 (3) 25 (4) 50 (5) 00 (6) 200 (7) 250 (8) 2500 (9) none of above Solution 20. The current into the first transformer is I = 00 = 0 Arms. The current through the 0 transmission line is Arms. The power lost in the transmission line is 00 = 00 watts. Answer: (5) 2. In the circuit below, M = 4 H, L = 8 H, L 2 = 4 H, and R = 8 Ω. Suppose V in (s) = s. If one computes the Thevenin equivalent circuit, then the open circuit voltage in the s-world has value at s = 0.25 as V OC (s = 0.25) = : () 0.2 (2) 0.2 (3) 0.4 (4) 0.4 (5) 0.5 (6) 0.5 (7) 0.8 (8) 0.8 (9) none of above

14 ECE 202 Final, Spring 8 4 Solution 2. Replacing the coupled inductors by the T-equivalent, we have by voltage division V OC (s) = Answer: (4) Ms Ms + (L M )s + R = Ms L s + R s = M L s + R = 4 8s + 8. Thus V OC (s = 0.25) = The resonant frequency of the circuit below ( R = 4 Ω, C = 00 F, L = 0.25 H, L 2 = H, M = 0.5 H) is (in rad/s): () 0 (2) 5 (3) 20 (4) 400 (5) 0 (6) (9) none of above (7) (8) 75 Solution 22. L eq = L + L 2 2M = = 0.25 and ω r = 400 = 20 rad/s. Answer: (3)

15 ECE 202 Final, Spring The transfer function of the circuit below is: H(s) = 2s s In the realization achieved by a student named Henrietta Farad, C = C 2 = F. Remembering her 7:30 am 202 lecture, she reasoned that she had to magnitude scale to obtain proper impedance levels. The parts store However, only had, mf capacitors, named after her favorite youngest daughter, Millie, but all manner of resistor values. The magnitude scaled resistor value is in kω: () (2) 2 (3) 2.5 (4) 4 (5) 5 (6) 0.5 (7) 0 (8) 8 (9) none of above Solution 23. C sv in = (C 2 s + G)(V out V in ) ((C + C 2 )s + G)V in = (C 2 s + G)V out. Hence, H(s) = (C + C 2 )s + G C 2 s + G Hence R new = 4 kω. Answer (4). = 2s s Thus R = 4 Ω. C new = 0 3 = = K m 000. K m = 000.

16 ECE 202 Final, Spring The convolution, y(t) = h(t)* f (t), where h(t) = 4u(t + 2) and f (t) = r(t + 2) 2r(t), results in y( 3) = ( y(t) evaluated at t = 0 ): () (2) 2 (3) 3 (4) 4 (5) 5 (6) 6 (7) 7 (8) 8 (9) none of above Solution 24. From class notes and HW, Hence, y( 3) = 2 0 = 2. Answer: (2) y(t) = h(t)* f (t) = 4u(t + 2)*r(t + 2) 4u(t + 2)*r(t) (t + 4)2 (t + 2)2 = 4 u(t + 4) 8 u(t + 2) 2 2 = 2(t + 4) 2 u(t + 4) 4(t + 2) 2 u(t + 2) 25. For the circuit shown below v (0 ) = 2 V. The initial voltages on all other capacitors at 0 are ZERO. At time t = both switches are flipped to positions A. Then v (2) = (in V): () (2) 2 (3) 3 (4) 4 (5) 5 (6) 6 (7) 20 (8) 50 Solution 25. The two 5F caps on the left are charged to 2 V at 0 +. The two 5F caps on the right split the voltage across the source and hence each is charged to 0 V. In the s-domain there are 3 current sources in parallel all pointing up having values from left to right of 0e s, 20e s, and 50e s. C eq = 20 F. V (s) = 20s 40e s = 2 s e s implies v (2) = 2 V. Answer: (2)

17 ECE 202 Final, Spring 8 7 6s 26. A circuit has H(s) = (s + 6) If it is excited by the input v in (t) = 36r(t) V, then the value of the output for very very large t is: () 2 (2) 2 (3) 3 (4) 4 (5) 5 (6) 6 (7) (8) 8 (9) none of above SOLUTION. lim s 0 s 6s (s + 6) s 2 = 3. Answer (3) 27. The signal below is f (t) = cos π 2 t 0 t 0 otherwise p(s) The Laplace transform of this signal is F(s) = s 2 + π 2 2 where p(s) = : () s (2) 0.5π (3) s 0.5π (4) 0.5πe s ( ) (6) ( s 0.5πe s ) (7) 0.5π ( s e s ) ( ) (9) None of above (5) s + 0.5πe s (8) s 0.5πe 2s

18 ECE 202 Final, Spring Solution 27. f (t) = cos(0.5πt)u(t) + sin( 0.5π (t ) )u(t ) in which case s F(s) = s 2 + π πe s 2 s 2 + π 2 = s + 0.5πe s. Answer: (5) 2 s 2 + π Suppose L f (t) { } = ln { } is: s a. Then L te at f (t) () s 2a (2) (4) (s 2a) 2 (5) s 2a s 2a (7) s 2 (8) s 2 (3) s (6) s (9) None of above

19 ECE 202 Final, Spring 8 9 { } = ln s Solution 28. L e at f (t) Answer (6). Therefore, L teat f (t) { } = d ln( s ) ds ( ) = s 2 s = s. 29. An integro-differential equation for a linear 202 circuit is given by t 9 v C (τ )dτ 0 + d dt v C (t) + 6v C (t) = 3u(t) Suppose v C (0 ) = 0 and!v C (0 ) = 9 V/s. Then v C (t) has a term of the form Kte at u(t) where (K,a) = : () (3,2) (2) ( 3,2) (3) (3,3) (4) ( 3,3) (5) ( 3,6) (6) (6,9) (7) ( 6,3) (8) (6,3) (9) None of above Solution 29. The above equation implies that v!! C (t) + 6!v C (t) + 9v C (t) = 3δ (t) In the s-world s 2 V C (s) sv C (0 )!v C (0 ) + 6sV C (s) v C (0 ) + 9V C (s) = 3. If v C (0 ) = 0, then ( s 2 + 6s + 9)V C (s) = 3+!v C (0 ) implies V C (s) = 3+!v C (0 ) s 2 + 6s + 9 v C (t) = ( 3+!v C (0 ))te 3t u(t) V. Answer: (7) ( ) = 3+!v C (0 ) (s + 3) 2 which implies that 30. In the circuit below, I in (s) = 0, L = 0.25 H, Z(s) = Y(s) = 0.25s + 8 s, and i L (0 ) = 6 A. Then i L (t) has a term of the form K cos(ω t)u(t) where (K,ω ) = : () (,4) (2) (2,4) (3) (8,4) (4) (6,4) ( ) (6) ( 2, 8) (7) ( 6, 8) (8) ( 8, 8) (5), 8 (9) None of above

20 ECE 202 Final, Spring 8 20 Solution 30. Source is set to zero. Use voltage source model of the inductor whose value is Li L (0 ). Thus, the current through the inductor is this voltage divided by the total impedance. ( Z(s) s) = 0.5s + 8 s = 0.5s2 + 8 = s2 +6, i.e., I s 2s L (s) = i L (t) = 8cos(4t)u(t) A. Answer: (3) Li L (0 ) Z(s) s = 4 = 8 s + 0.5s 8s s For h(t) and f (t) sketched below, the convolution y(t) = h(t)* f (t) produces the value y(3) = : () (2) 2 (3) 3 (4) 4 (5) (6) 2 (7) 3 (8) 4 (9) none of above Solution 32. Using graphical convolution, with a flip and shift of f (t), means that the leading nonzero edge of f (3 τ ) is at τ = 2 yielding the area of f (3 τ )h(τ ) to be 2. Answer (2). 32. Suppose f (t) = cos(t)u(t) and h(t) is sketched below for which K = T =. Let y(t) = h(t)* f (t). Then y( ) = :

21 ECE 202 Final, Spring 8 2 () (2) 2 (3) 0 (4) 2sin(2) 2sin( 4) (5) 2sin(2) (6) 4sin(2) (7) 0.5sin(2) (8) 4sin(2) 4sin( 4) (9) None of these Solution 32. h(t) = r(t +) r(t). Hence h''(t) = δ (t +) δ (t). The integral of f (t) = cos(t)u(t) is g (t) = sin(t)u(t) and the integral of g (t) is g 2 (t) = ( cos(t) )u(t). Therefore y(t) = h(t)* f (t) = g 2 (t +) g 2 (t) = ( cos(t +) )u(t +) ( cos(t) )u(t). Hence, y( ) = 0. Answer: (3) 33. Recall, the third order 3dBNLP Butterworth transfer function is H 3dBNLP (s) = s 3 + 2s 2 + 2s +. You are to realize a third order LP Butterworth with 3dB down point ω c = 0 rad/s (except for dc gain) by the circuit below whose transfer function, when C = C 2 = C, is H cir ( ) ( ) = ( s) = V out s V in s s C s2 + LC 2 2 LC + C 2 s + 2 LC 2 Unfortunately, the only available inductor is 0.4 H. The resulting value of C (in F) is: () (2) 0.2 (3) 0.05 (4) 0.4 (5) 0.5 (6) (7) 0. (8) 0.25 (9) none of above

22 ECE 202 Final, Spring 8 22 Solution 33. s 3 + 2s 2 + 2s += s C s2 + 2C + L LC 2 s + 2. Thus C = F and L = 2 H. To 2 LC L obtain K m we have that L final = K old m 0 K m = 0L final = = 2. Thus L old 2 C final = C old = K m K f 0 2 = = 0.05 F. 20 ANSWER: (3) 34. The switch in the circuit of figure below has been open for a very long time. At t = 0, the switch closes, and stays closed. Assume R = Ω, R 2 = Ω, L = H, and v in (t) = 0u( t) +0u(t) V. Then for t 0, v out (t) has a term of the form Ke at u(t) where (K,a) = : () (5,) (2) ( 5,2) (3) (0,) (4) ( 0,2) (5) (5,) (6) ( 5,) (7) ( 5,) (8) ( 0,) (9) none of above

23 ECE 202 Final, Spring 8 23 Solution 34. i L (0 ) = 2 I L (s) = s + i L (0 ) + s + 0 s = i L (0 ) s + 0 s s = 5 s s Answer: (6)