EE-202 Exam III April 3, 205 Name: (Please print clearly.) Student ID: CIRCLE YOUR DIVISION DeCarlo-7:30-8:30 Furgason 3:30-4:30 DeCarlo-:30-2:30 202 2022 2023 INSTRUCTIONS There are 2 multiple choice worth 5 points each and one workout problem worth 40 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 5 page 2 MULTIPLE CHOICE CIRCUIT FOR PROBLEMS -3. The transfer function for the Sallen and Key circuit shown below is H cir (s) = s 2 +. An ill designed NLP transfer function by a student in the Leadership Tech Q s + Program at an engineering education program is H NLP (s) = 20 s 2 + 2.0s + 25.. The value of Q needed for this filter is: () (2) 2 (3) 2.5 (4) 4 (5) 5 (6) 0.5 (7) 0.25 (8) None of these Solution. ω m = 5 rad/s. B ω = 2 = ω m Q = 5 Q Q = 2.5. 2. In order to realize H NLP (s), your lab partner has found a 5 mf capacitor for the final value of C and every imaginable resistor on Planet Hollywood. For such a circuit design, the final value of R,new = R 2,new is (in ohms): () 00 (2) 200 (3) 40 (4) 400 (5) 250 (6) 80 (7) 20 (8) 800 (9) None of these Solution 2. K m = C,old C,new K f = 5 5 0 3 5 = 200 Hence, R,new = 200 Ω.
EE-202 Ex 3, Sp 5 page 3 3. If input attenuation is used to adjust the DC gain of the filter above using a voltage divider consisting of R A and R B as shown below, the appropriate values for R A and R B (in ohms) are respectively: (), (2) 3,.5 (3) 5,.25 (4) 4,.334 (5).25, 5 (6) 2, 2 (7).334, 4 (8).5, 3 (9) None of these Solution 3. R B = 20 R A + R B 25 = 4 5 and = R A R B 4 R R A + R A B 5 = R A = 5 =.25 Ω and 4 R B = 0.8R A + 0.8R B implies R B = 4R A = 5 Ω. 4. The second order 3dBNLP Butterworth transfer function is H 3dBNLP (s) = s 2 +. This is to be 2s + realized by the circuit below. If L = 2 2 H and C = 0.5 2 F, then the value of R = (in Ω): () 0.5 (2) 2 (3) 2 (4) (6) 0.2 (7) 2 (9) none of above 2 0.2 (7) 2 2 (8) 4 2 (5) 2 2
EE-202 Ex 3, Sp 5 page 4 Solution 4. H cir (s) = R L = R 2 2 = 2 R = 4 Ω. Cs R + = LC Cs + Ls s 2 + R L s + LC 5. The transfer function of the circuit below is = s 2 + 2 L s +. If L = 2 H then H cir s ( ) = V out ( s) V in ( s) = s 3 + C + C 2 C C 2 LC C 2 s 2 + C + C 2 + L 2 s + LC C 2 LC C 2 The objective is to realize a LP circuit for which H 3dBNLP (s) = s 3 + 2s 2 + 2s + and ω c = 4000 rad/s when K m = 00. Set C = C 2 = C. The value of C final is in µf : () (2) 2 (3) 0.5 (4) 4 (5) 5 (6) 0.2 (7) 2.5 (8) none of above Solution 5. By pattern recognition C + C 2 = 2C C C 2 C = 2 C = F. 2 C final = 00 4000 = 0.25 0 5 = 2.5 µf. 6. The circuit below is a 3 db NLP EQUIVALENT of a desired HP filter whose 3 db down point is to be 00 rad/s. If the LARGEST capacitor in the final high pass design is to be 5 µf, then the value of the needed magnitude scale factor for the HP design is K m = : () 0 (2) 20 (3) 000 (4) 40
EE-202 Ex 3, Sp 5 page 5 (5) 50 (6) 00 (7) 0,000 (8) none of above Solution 6. Ls = L s C s C = F and C 2 = 3 2 F. Thus C HP, final =.5 K m 00 = 30 0 6 implies.5 K m = = 000 4 5 0 7. If R = 4 Ω and L = 0.2 H, then the value of C in F which makes the circuit resonant at ω r = 0 rad/s is: () 0. (2) 0.02 (3) 0.0 (4) 0.2 (5) 5 (6) 6 (7) 8 (8) none of above Solution 7. Y in ( jω ) = jcω + jlω + R = jω C L L 2 ω 2 + R 2 + R L 2 ω 2 + R. Thus 2 L C = L 2 ω 2 + R = 0.2 2 0.04 00 +6 = 0.2 = 0.0 F. 20 8. Reconsider the circuit of problem 7. At resonance, the impedance reduces to a pure conductance, G eq, which is equal to (in S): () 0. (2) 0.02 (3) 0.3 (4) 0.04 (5) 0.5 (6) 0.2 (7) 0.4 (8) 0.05 (9) none of above
EE-202 Ex 3, Sp 5 page 6 Solution 8. Y in ( jω ) = jcω + jlω + R = jω C L L 2 ω 2 + R 2 + R L 2 ω 2 + R. Thus 2 R G eq = L 2 ω 2 + R = 4 = 0.2 S. 2 0.04 00 +6 9. Consider the RLC circuit below in which R = 0. Ω, L = 0.2 H, C = 0.05 F, and R Cp = 40 Ω. Recall that the Q of a lossy capacitor near resonance is Q cap (ω 0 ) = R Cp Cω 0. The approximate value of Q cir of the overall RLC circuit is: () (2) 20 (3) 5 (4) 0.0 (5) 0 (6) 00 (7) 6 (8) none of above Solution 9. ω 0 = 0 rad/s. Q cap (ω 0 ) = 40 0.05 0 = 20. R Cs = 40 400 = 0. Ω. R eq = 0.2 Ω. B w = R eq L =. Q cir = ω m B ω = 0. 0. Consider the circuit below in which R s = Ω, L = 0.5 H, C = 200 F, and R p = 00 Ω. Recall that 3 Q coil (ω 0 ) = ω 0 L. The approximate value of the parallel resistance in an approximately equivalent R s parallel RLC circuit is (in Ohms): () (2) 25 (3) 30 (4) 20 (5) 5 (6) 5 (7) 0 (8) 00 (9) none of above
EE-202 Ex 3, Sp 5 page 7 Solution 0. Q coil (ω 0 ) = ω 0 L 20 0.5 = = 0. R R s pl = 00 = 00 Ω. R eq = 3 00 + 00 = 25 Ω.. Reconsider the circuit of problem 0. The approximate value of the higher half power frequency is (in rad/s): () 0 (2) 20 (3) 24 (4) 22 (5) 2 (6) 6 (7) 4 (8) 8 (9) none of above Solution. B w = R eq C = 0.25 = 8. ω,2 = ω m B w = 20 4 = 6,24. 2 2. The circuit shown below consists of a 00 Ω resistor in parallel with a REAL 600 µf capacitor having a Q cap (ω ) = 0 at ω = 000 rad/sec. (Replace the capacitor by its non-ideal parallel equivalent.) The REAL part of the input ADMITTANCE at s = jω of this combination (given in mhos or Siemens) is: () 0.0 (2) 0.02 (3) 0.3 (4) 0.04 (5) 0.05 (6) 200 (7) 33.33 (8) 0.03 (9) None of these
EE-202 Ex 3, Sp 5 page 8 Solution 2: 0 = ω RC = 0 3 6 0 4 R R = 00 6 Ω. G eq = 00 + 6 = 0.07 mhos. 00 ANSWER: (9)
EE-202 Ex 3, Sp 5 page 9 Original Circuit Exact Equivalent Circuit at ω 0 Approximate Equivalent circuit, for high Q, (Q L > 6 and Q C > 6) and ω within ( ± 0.05)ω 0
EE-202 Ex 3, Sp 5 page 0 WORKOUT PROBLEM (40 POINTS) (2 ND ORDER TRANSFER FUNCTION REALIZATION USING CANONICAL OP AMP CIRCUITS) 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 any other op amps. If you do this or if you choose to do the controllable canonical form, then you will receive a maximum of 60% credit (of what you do correctly) because it is a much less complex development. Construct the OBSERVABLE canonical form biquad realization of the not so good LP filter transfer function H NLP (s) = V out (s) V in (s) = 0.25s2 + 2s + 4 s 2 + 4s +2. For this NLP filter, ω p = 3 rad/s. ALL op amps are to have the +terminal grounded. CLEARLY LABEL EACH PART LISTED BELOW. IF YOUR WORK IS UNREADABLE OR FAMOUSLY UNCLEAR, YOUR GRADE IS REDUCIBLE. DO NOT LEAVE OUT STEPS. (a) (7 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 NLP (s) that has the form v out (t) = (??)v in (t) + D (????) + D 2 (????) Specify all terms. (THIS IS a 0 point or 7 points answer.) (b) (7 pts) Given your (correct) answer to (a), DEFINE the variable x (t) as per the class 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 (t). Clearly label all inputs, outputs, resistors, and capacitors. ALL op amps are to have the +terminal grounded. (c) (0 pts) Given your (correct) definition of x in part (b), form the equation for Dx, properly define the new variable x 2 (t), write down the appropriate integral equation, and construct and draw an op amp circuit whose output is ±x (t) as needed. 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. IF v out (t) is present, the maximum achievable point value is half of what you do correctly. (d) (0 pts) Construct an op amp circuit whose output is ±x 2 (t). Clearly label all inputs, outputs, resistors, and capacitors. Be sure to draw all needed op amps to complete your design for this part you can only use variables from previously designed circuits and v in.
EE-202 Ex 3, Sp 5 page (e) (6 pts) The smallest resistor is to be 3 kω. The proper LP ω p is to be 3000 rad/sec. Determine the magnitude scale factor K m (3 pts) and the frequency scale factor K f (3 pts). Workout (a) ( s 2 + 4s +2)V out = ( 0.25s 2 + 2s + 4)V in D 2 v out + 4Dv out +2v out = 0.25D 2 v in + 2Dv in + 4v in v out = 0.25v in + D 2v in 4v out ( ) + D 2 ( 4v in 2v out ) (b) v out = 0.25v in + x v out = 0.25v in x (needs x and v in ) ( ) + D 2 ( 4v in 2v out ) x = D 2v in 4v out (c) Dx = 2v in 4v out + D (4v in 2v out ) = 2v in v in 4x + x 2 = v in 4x + x 2 where x 2 = D 4v in 2v out ( ). Thus, integrating both sides and multiplying by ( ) 4D ( x ) D ( x 2 ) (needs x and v in and x 2 ) x = D v in Thus one has to build an inverter to attach to the op amp in part (b). Failure to have the inverter is -3.
EE-202 Ex 3, Sp 5 page 2 (d) Dx 2 = 4v in 2v out = v in 2x. Thus, integrating both sides and multiplying by we have. ( ) x 2 = D ( v in +2( x )) (needs x and v in ) x 2 = D v in 2x Thus one has to build an inverter to attach to the answer to part (c). Failure to build the inverter is 3. CONTROLLABLE FORM IS BELOW FOR AUXILIARY EQUATION:
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