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1 EE 202 Exam III April Name: (Please print clearly) Student ID: CIRCLE YOUR DIVISION Morning 7:30 MWF Furgason INSTRUCTIONS Afternoon 3:30 MWF DeCarlo There are 10 multiple choice worth 5 points each and there are two workout problems worth 55 points. There are 105 points on the exam but the maximum you can receive is minimum[your Score, 100] This is a closed book, closed notes exam. No scrap paper or calculators are permitted. A transform table will be handed out separately. Carefully mark your multiple choice answers on the scantron form. Work on multiple choice problems and marked answers in the test booklet will not be graded. Nothing is to be on the seat beside you. When the exam ends, all writing is to stop. This is not negotiable. No writing while turning in the exam/scantron or risk an F in the exam. 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. Do not open, begin, or peek inside this exam until you are instructed to do so

2 EE-202, Ex 3 Sp 11 page 2 MULTIPLE CHOICE 1. Coupled inductors are connected to a circuit represented by the box in the figure below where L 1 = 9 H, L 2 = 1 H, and M = 1.5 H. Primary is open circuited. If i 2 (t) = 0.4sin(10t) A, then v 1 (t) is (in volts): (1) 6sin(10t) (2) 6cos(10t) (3) 6sin(10t) (4) 6cos(10t) (5) 4sin(10t) (6) 4 cos(10t) (7) 4sin(10t) (8) 4 cos(10t) (9) None of these Solution 1. v 1 (t) = M di 2 = d sin(10t) = 6 cos(10t) V. dt dt Answer (2) No Comma : Answer (6) 2. In the circuit below, R = 2 Ω, L 1 = 6 H, L 2 = 1 H, and M = 2 H. The step response of the circuit below, v 1 (t), (in volts) is: (1) 2δ(t) 4e 2t u(t) (2) 2δ(t) + 4e 2t u(t) (3) 2e 2t u(t) (4) 2e 2t u(t) (5) 2(1 e 2t )u(t) (6) 2(1 e 2t )u(t) (7) 4e 2t u(t) (8) 4e 2t u(t) (9) None of these SOLUTION 2. V 1 (s) = L 1 si 1 (s) MsI 2 (s) = MsI 2 (s) V 2 (s) = V in (s) RI 2 (s) and V 2 (s) = MsI 1 (s) + L 2 si 2 (s) = L 2 si 2 (s). Thus V in (s) = (L 2 s + R)I 2 (s) I 2 (s) = V in (s) L 2 s + R. Thus V 1 (s) = MsI 2 (s) = Ms L 2 s + R V 2s in (s) = s + 2 V in (s) = M L 2 s + R = 2 s + 2. v 1(t) = 2e 2t u(t) V. Answer: (3) No Comma: Answer (6)

3 EE-202, Ex 3 Sp 11 page 3 3. If G = 2 S and C = 0.2 F, then the value of L in H which makes the circuit resonant at ω r = 5 rad/s is: (1) 0.01 (2) 0.02 (3) 0.2 (4) 0.04 (5) 0.05 (6) 0.4 (7) 0.08 (8) none of above 1 Solution 3. Z in ( jω) = jωl + G + jωc = jω L C G 2 + ω 2 C 2 + G G 2 + ω 2 C 2. Hence C L = G 2 + ω 2 C 2 = = 1 = 0.04 H. 25 ANSWER: (4) No Comma: Answer (4) 4. Reconsider the circuit of problem 4. At resonance, the impedance reduces to a pure resistance, R, which is equal to (in Ω): (1) 0.1 (2) 0.2 (3) 2.5 (4) 0.5 (5) 5 (6) 0.25 (7) 0.4 (8) none of above G Solution 4. R = G 2 + ω 2 C 2 = 2 = 0.4 Ω. 5 ANSWER: (7) No Comma: Answer (5)

4 EE-202, Ex 3 Sp 11 page 4 Questions 5 and 6 are in regards to the band pass magnitude response shown below The associated transfer function has the classic usual form as taught in class 5. In the band pass magnitude response shown above, the half power frequencies (i.e., the frequencies where the magnitude is 3dB down from its maximum value) are approximately (in rad/s): (1) 725, 900 (2) 725, 880 (3) 775, 825 (4) 750, 850 (5) 630, 870 (6) 700, 925 (7) 650, 950 (8) none of above Answer: (4) No Comma: Answer (3) 6. The Q of the associated band pass circuit is (approximately): (1) 1 (2) 2 (3) 3 (4) 4 (5) 5 (6) 6 (7) 7 (8) 8 (9) none of above Answer: Q = 800 = 8. Answer: (8) No Comma: Answer (8)

5 EE-202, Ex 3 Sp 11 page 5 CIRCUIT FOR PROBLEMS 7, 8, AND 9. The transfer function of the Sallen and Key circuit below 2 is H cir (s) = s under the conditions that C 1 = C 2 = 1 F, R 1 = Q Ω, and R 2 = 1. This circuit Q s + 1 Q is to realize H LP (s) = 18 s 2 + 3s The value of Q needed for the first phase of a design realization is: (1) 1 3 (2) 2 (3) 3 (4) 1 12 (5) 12 (6) 6 (7) 1 36 (8) none of above Solution 7. 3 = 6 Q = 2. Answer (2) No Comma: Answer (6) Q 8. The FINAL value of C 1 needed to realize H LP (s) is C 1 f = (in F): (1) 1 (2) 1 3 (5) 0.5 (6) 1 6 (9) none of above (3) 1 12 (7) 1 24 (4) 1 36 (8) 1 72 Solution 8. C 1 final = C old 6 = 1 6 F. Answer: (6) No Comma: Answer (4)

6 EE-202, Ex 3 Sp 11 page 6 9. If input attenuation is used to adjust the DC gain using a combo of R A and R B as shown below, then the values of R A and R B (in ohms) are respectively: (1) ( 4, 4) (2) (5) 5, 10 3 (6) 8 3,8 (3) 16, 16 7 (4) 8, ,16 (7) 10, 2.5 ( ) (8) none of above Solution 9. R B R A + R B = 0.25 R 1 = 2 = R A R B R A + R B = 0.25R A R A = 8 Ω. Thus R B = R B R B = 8 3 Ω. ANSWER (4) No Comma: Answer (4) 10. The circuit below is a NLP prototype ( R s = R L = 1 Ω, C = 1 F and L = 0.5 H) that is to be converted to a HP filter with the A max point at ω p = 10 4 rad/s. The capacitor in the final HP circuit is to be C HP, final = 1 µf, then the value of L HP, final for the HP circuit is: (1) 1 mh (2) 20 mh (3) 0.1 mh (4) 40 mh (5) 5 mh (6) 0.05 mh (7) 2 mh (8) 10 mh (9) none of above Solution 10. ANSWER: (2) No Comma: Answer (2) >> Kf = 1e4; C = 1; L = 0.5; >> CNHP = 1/L CNHP = 2 >> LNHP = 1/C LNHP = 1 >> Km = CNHP/(Kf*1e-6) Km = 200 >> LHP = Km * LNHP/Kf LHP =

7 EE-202, Ex 3 Sp 11 page 7 Original Circuit Exact Equivalent Circuit at ω 0 Approximate Equivalent circuit, for high Q, (Q L > 6 and Q C > 6) and ω within (1 ± 0.05)ω 0

8 EE-202, Ex 3 Sp 11 page 8 Workout Problem 1 (25 pts) The following circuit is more or less a band pass circuit whose form (because it contains a practical inductor and capacitor) is too complex for easy analysis. L = 0.2 H, C = 0.05 F, R 1 = R 2 = 0.1 Ω, and R 3 = 80 Ω. (a) (4 pts) Determine the appropriate and approximate ω 0 at which the BP analysis is to be done. (b) (6 pts) Compute the inductor-q Q L (ω 0 ), and capacitor-q Q C (ω 0 ) using the formulas on the included table preceding this problem. Be sure to check that you are using the correct formula before you begin. (c) (5 pts) Next, showing your work, construct an approximately equivalent parallel RLC circuit as shown below. Specify R new, L new, and C new. (d) (10 pts) Compute, showing your work, by for example obtaining the transfer function of the circuit in part (c): approximate values for ω m, B ω, Q cir, the half power frequencies ω 1, and ω 2, and H m. SOLUTION WORKOUT 1: L = 0.2; C = 0.05; RsL = 0.1; RsC = 0.1; R3 = 80; w0 = 1/sqrt(L*C) QC = 1/(w0*RsC*C) QL = w0*l/rsl RparaL = RsL*QL^2 RparaC = RsC*QC^2 Rpar = 1/(1/R3 + 1/RparaL + 1/RparaC) Bw = 1/(Rpar*C) Qcir = w0/bw w1 = w0 - Bw/2 w2 = w0 + Bw/2 wm = w0 Hm = 1/C/(1/(Rpar*C))

9 EE-202, Ex 3 Sp 11 page 9 w0 = 10 QC = 20 QL = 20 RparaL = 40 RparaC = 40 Rpar = 16 Bw = Qcir = 8 w1 = w2 = wm = 10 Hm = 16

10 EE-202, Ex 3 Sp 11 page 10 WORKOUT PROBLEM 2 (30 PTS) This problem is to be done using the OBSERVABLE canonical form using at most 5 op amps. In the observable canonical form, ±v out (t) appears as the output of the final summing op amp and is NOT fed back to the other op amps. If you choose to do the controllable canonical form, then you will receive a maximum of 60% credit because it is a much less complex development. Construct the observable canonical form biquad realization of H LP (s) = V out (s) V in (s) = 0.3s2 + 5 s 2 + 2s + 5. (a) (6 pts) Using the D k and D k notation as per the class examples, construct a differential equation (time domain) in v out (t) and v in (t) associated with H LP (s) that has the form v out (t) = Kv in (t) + D 1 (????) + D 2 (????) Specify all terms. (b) (6 pts) Given your (correct) answer to (a), define the variable x 1 (t) as per the derivation of the observable canonical form. Then construct and draw an op amp circuit for v out (t) in terms of v in (t) and x 1 (t). Clearly label all inputs, outputs, resistors, and capacitors. ALL op amps are to have the +terminal grounded. (c) (8 pts) Given your (correct) definition of x 1 in part (ii), form Dx 1, properly define the new variable x 2 (t), and construct and draw an op amp circuit whose output is x 1 (t). Clearly label all inputs, outputs, resistors, and capacitors. ALL op amps are to have the +terminal grounded. NOTE: v out (t) must not be present in your feedback loops for the observable form. (d) (10 pts) Construct an op amp circuit whose output is ±x 2 (t) depending on the sign needed for the input to the circuit for x 1 (t). Clearly label all inputs, outputs, resistors, and capacitors. ALL op amps are to have the +terminal grounded. Do you need two op amps (total) to complete your circuit? If so, be sure to draw it. SOLUTION WORKOUT 2: (a) ( s 2 + 2s + 5)V out (s) = ( 0.3s 2 + 5)V in (s) implies D 2 v out (t) + 2Dv out (t) + 5v out (t) = 0.3D 2 v in (t) + 5v in (t) implies v out (t) = 0.3v in (t) 2D 1 v out (t) + D 2 ( 5v in (t) 5v out (t)) (b) v out (t) = 0.3v in (t) 2D 1 v out (t) + D 2 ( 5v in (t) 5v out (t))= 0.3v in (t) + x 1 (t). Thus v out (t) = 0.3v in (t) x 1 (t)

11 EE-202, Ex 3 Sp 11 page 11 (c) x 1 (t) = 2D 1 v out (t) + D 2 ( 5v in (t) 5v out (t)) implies Dx 1 (t) = 2v out (t) + D 1 ( 5v in (t) 5v out (t))= 2v out (t) + x 2 (t) implies Dx 1 (t) = 0.6v in (t) 2x 1 (t) + x 2 (t). Thus x 1 (t) = 0.6 v in (t) dt 2 x 1 (t) dt ( x 2 (t))dt (d) x 2 (t) = D 1 ( 5v in (t) 5v out (t)) implies Dx 2 (t) = 5v in (t) 5v out (t) = 3.5v in (t) 5x 1 (t). Thus x 2 (t) = 3.5 v in (t) dt 5 ( x 1 (t)) dt requiring an op amp to change the sign on x 1 (t).

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