202Ex3Sp08

202Ex3Sp08 - EE-202 Exam III April 10, 2008 Name:

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Unformatted text preview: EE-202 Exam III April 10, 2008 Name: __________________________________ (Please print clearly) Student ID: _________________ CIRCLE YOUR DIVISION Morning 8:30 MWF Afternoon 12:30 MWF INSTRUCTIONS There are 13 multiple choice worth 5 points each and there is 1 workout problem worth 35 points. 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 EE-202, Ex 3 Sp 08 page 2 MULTIPLE CHOICE. 1. In terms of the voltages and currents defined in the figure below, the correct expression for V2 (s) is: (1) L1sI1 + MsI 2 (5) !MsI1 ! L2 sI 2 (2) L1sI1 ! MsI 2 (6) MsI 2 ! L2 sI 2 (3) MsI1 + L2 sI 2 (7) MsI1 ! L2 sI 2 (4) !MsI1 + L2 sI 2 (8) None of above 2. Suppose R = 4 Ω. The real coil shown below is characterized by an inductance of 2 H and a Q = 25 at 200 rad/s. The input impedance, Zin (s) : (1) 16 + 2s (5) 12 + 2s (2) 20 + 2s (6) 4 + 2s (3) 404 + 2s (7) 16 – 2s (4) 8 + 2s (8) None of these EE-202, Ex 3 Sp 08 page 3 3. If R = 0.6 Ω, L = 0.1 H, and C = 0.1 F, then the resonant frequency ! r " 0 in rad/sec of the circuit sketched below is: (1) 10 (2) 100 (3) 64 (4) 8 (5) 36 (6) 6 (7) 136 (8) none of above EE-202, Ex 3 Sp 08 page 4 THE CIRCUIT BELOW IS FOR PROBLEMS 4 – 6. The band pass circuit shown below has transfer function R2 s Vout (s) L H (s) = = R + R2 Vin (s) 1 s2 + 1 s+ L LC It is known that ! m = 10 rad/s, R1 = 0.2 , Bw = 5 , and H max = 0.8 . 4. The exact half power frequencies are !1 and ! 2 (in rad/s) (1) 5, -5 (5) 7.5, 12.5 (2) 5, 15 (6) 95, 105 (3) 0, 20 (7) 97.5, 102.5 (4) 8.5, 12.5 (8) None of these 5. The value of R2 is (in ohms): (1) 1 (5) 16 (6) 0.8 (2) 2 (7) 8 (3) 0.2 (8) None of these (4) 4 6. The values of L and C respectively are (in H and F): (1) 0.05, 0.2 (5) 0.25, 0.04 (2) 0.2, 0.05 (6) 0.01, 1 (3) 1, 0.01 (7) None of these (4) 0.1, 0.1 EE-202, Ex 3 Sp 08 page 5 7. In the circuit below, L = 0.2 H, C = 0.05 F, Rs = 0.25 Ω, R p = 16 , then the approximate half power frequency !1 (rad/s) and the bandwidth Bw (in rad/s) are: (1) 7.5, 5 (5) 8.75, 2.5 (2) 7.5, 2.5 (6) 8.75, 1.25 (3) 7.5, 1.25 (7) None of these (4) 8.75, 5 8. Referring to the circuit of problem 7, the approximate value of H max is: (1) 1 (5) 5 (2) 2 (6) 0.25 (3) 2.5 (7) None of these (4) 4 EE-202, Ex 3 Sp 08 page 6 THE FOLLOWING IS FOR PROBLEMS 9, AND 10. A 3rd order Butterworth HP filter has 3 dB down point at ! c = 10 rad/s. The 3rd order Butterworth NLP prototype circuit is given in the figure below and has transfer function 1 V (s) 1 LC1C2 H cir ( s ) = out = = 3 2 Vin ( s ) s 3 + 1 s 2 + C1 + C2 s + 1 s + 2s + 2s + 1 C1 LC1C2 LC1C2 9. The values of L in H and C1 and C2 in F to realize the 3 dB NLP 3rd order prototype are: (1) 1, 2, 0.5 (2) 1.334, 0.5, 2 (3) 1.334, 0.5, 1.5 (4) 0.5, 2, 1 (5) 1, 0.5, 2 (6) 0.75, 1.334, 1 (7) none of above 10. Given the appropriate transformation of the NLP circuit to HP form, the value of C HP in F to realize the 3rd order HP filter in which ! c,HP = 10 rad/s and the source resistance is 3 Ω is: 1 40 4 (6) 90 (1) (2) 0.75 (3) 0.25 (4) 1 120 (5) 9 40 (7) none of above EE-202, Ex 3 Sp 08 page 7 CIRCUIT FOR PROBLEMS 11, 12, AND 13. A Sallen and Key circuit with parameter choices as indicated is shown below The circuit transfer function is H cir (s) = 11. The value of the circuit Q is: (1) 0.5 (2) 2 (7) none of above (3) 0.04 12.5 2 . If this circuit is to realize H LP (s) = 2 . 1 2 s + 2s + 25 s + s +1 Q (4) 0.25 (5) 5 (6) 2.5 12. The final values of C1 and C2 to realize H LP (s) are C1 f and C2 f in F which are: (1) 0.2, 0.4 (2) 1, 0.5 (3) 0.2, 0.1 (4) 0.2, 0.08 (5) 0.2, 0.8 (6) 0.2, 0.5 (7) 0.2, 0.1 (8) none of above 13. If input attenuation is used to adjust the DC gain using a combo of RA and RB as shown below, then the values of RA and RB (in ohms) are respectively: (1) 2, 2 (5) 1.25, 4 (2) 4, 1.333 (6) 4, 1.25 (3) 1.333, 4 (7) 3, 1.5 (4) 4, 3 (8) none of above EE-202, Ex 3 Sp 08 page 8 WORKOUT PROBLEM (35 PTS) V 2s + 0.1 Use the observable canonical form biquad realization technique to H nuts (s) = out = 2 as Vin s + 2s + 4 follows: (i) (5 pts) Construct the differential equation (time domain) in vout (t) and vin (t) associated with H nuts (s) . (s 2 ! ! + 2s + 4 Vout = ( 2s + 0.1)Vin !!!!!!!out + 2 vout + 4vout = 0.1vin + 2 vin v ) (ii) (5 pts) Use the D k and D !k notation as per the class examples to put the differential equation of part (i) into the proper form for constructing the observable canonical biquad realization. D 2 vout + 2Dvout + 4vout = 0.1vin + 2Dvin implies vout = D !2 ( 0.1vin ! 4vout ) + D !1 ( 2vin ! 2vout ) (iii) (6 pts) Given your (correct) answer to (ii), define the variable x1 as per the class room derivation, and construct and draw an op amp circuit for vout (t) in terms of vin (t) and x1 (t) . x1 = vout = D !2 ( 0.1vin ! 4vout ) + D !1 ( 2vin ! 2vout ) No need for any op amp circuit. (iv) (9 pts) Given your (correct) definition of x1 , properly define the x2 , and construct and draw an op amp circuit whose output is x1 (t) . Dx1 = D !1 ( 0.1vin ! 4vout ) + ( 2vin ! 2vout ) = x2 + 2vin ! 2x1 . Thus ! ! " x1 = !x1 = ! " x2 ! 2 " vin ! 2 " !x1 (v) (10 pts) Construct the op amp circuit whose output is ±x2 (t) depending on the sign needed for the input to the circuit for x1 (t) . EE-202, Ex 3 Sp 08 page 9 x2 = D !1 ( 0.1vin ! 4vout )!!"!!Dx2 = 0.1vin ! 4x1 EE-202, Ex 3 Sp 08 page 10 EE-202, Ex 3 Sp 08 page 11 EE-202, Ex 3 Sp 08 page 12 Original Circuit Exact Equivalent Circuit at ω0 Approximate Equivalent circuit, for high Q, (QL > 6 and QC > 6) and ω within (1 ± 0.05) ! 0 ...
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