EE-202 Exam III April 6, 207 Name: (Please print clearly.) Student ID: CIRCLE YOUR DIVISION DeCarlo--202 DeCarlo--2022 7:30 MWF :30 T-TH INSTRUCTIONS There are 3 multiple choice worth 5 points each and one workout problems 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 peak inside this exam until you are instructed to do so.
EE-202 Ex 3 Sp 7 page 2 MULTIPLE CHOICE CIRCUIT FOR PROBLEMS -3. The transfer function of the Sallen and Key circuit below is 2 Hcir() s = under the conditions that C 2 = C2 = F, R = Ω, and R2 =. Your room s + s+ mate in basket weaving II has designed a circuit to filter out the neighbors base speakers based on a 27 normalized filter transfer function (incorrectly designed) given by H NLP (s) = s 2. In the + 0.5s + 36 final design, ω p = 600 rad/s, i.e., after appropriate frequency scaling of the normalized design.. The value of needed for the S&K circuit design is: () (2) 2 (3) 0 (4) 4 (5) 2 (6) 6 (7) 9 (8) 8 (9) None of these Solution. ω m = 6 rad/s. B ω = 0.5 = ω m = 6 = 2. => (5) 2. In the final realization (after all scalings), the value of C,new is to be 6 R,new = (in ohms): () 0 (2) 20 (3) 30 (4) 40 (5) 50 (6) 60 (7) 20 (8) 80 (9) None of these mf. Thus the final value of
EE-202 Ex 3 Sp 7 page 3 Solution 2. C C 6 C K m 600 = K m 600 K m = C,old C,new K f = R,new = K m = 20 Ω. => (7) 6 0 3 600 = 0 Hence, 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 value for R A (in ohms) BEFORE MAGNITUDE SCALING BY K m, is R A = : () 2 (2) 32 (3) 6 (4) 3 8 (5) 8 3 (6) 6 (7) 24 (8) 9 (9) None of these Solution 3. Circuit dc gain is 2. Desired filter design gain is 0.75. 2K = 3 4 K = 3 8. R B R A + R B = 3 8 and R = 2 = 3 8 R A R A = 32 Ω. => (2) 4. Consider h(t) and f (t) shown below for which K = 2, and T =. Suppose y(t) = h(t)* f (t) then y(0) = : () (2) 2 (3) 3 (4) 4 (5) 0 (6) 6 (7) 6 (8) 8 (9) None of these
EE-202 Ex 3 Sp 7 page 4 Solution 4. Leaving h(t) alone and flipping f (t), then h(τ ) f (0 τ ) is simply twice the first two seconds of h(τ ) scaled by 2. The resultant area is then 3. 5. Suppose f (t) = r(t ) and h(t) = cos(t)u(t). Then y(t) = h(t)* f (t) is: () ( cos(t) )u(t) (2) ( sin(t) )u(t) (3) sin(t )u(t ) (4) cos(t )u(t ) (5) ( cos(t ) )u(t ) (6) ( sin(t ) )u(t ) (7) cos(t )u(t ) (8) ( cos(t ) )u(t ) (9) none of above
EE-202 Ex 3 Sp 7 page 5 Solution 5.!! f (t) = δ (t ). t 0 t sin(τ )dτ = cos(τ ) 0 t 0 t h(τ )dτ = sin(τ ) 0 = sin(t)u(t). = ( cos(t) )u(t). Hence, y(t) = ( cos(t ) )u(t ) 6. The 3 db NLP Butterworth transfer function is H 3dBNLP (s) = s 2 is to be realized by the + 2s + circuit below whose transfer function is H cir (s) = Cs Ls + +. The circuit below is to realize a LP Cs Butterworth filter having 3 db down point at ω c = 500 2 rad/s. If the source resistance is to be scaled to 0 Ω, the value of L in mh is: () (2) 2 (3) 0 (4) 0. (5) 2.5 (6) 20 (7) 0.2 (8) none of above Solution 6. H cir (s) = mh. LC s 2 + L s + LC in which case L = 2 H and C = 2 F. L new = 0 000 or 0 7. The third order 3dBNLP Butterworth transfer function is H3dBNLP () s =. This is to 3 2 s + 2s + 2s+ be realized by the circuit below whose transfer function, when C = C 2 = C, is H cir s ( ) = V out ( s) V in ( s) = s 3 + 2 C s2 + The values of L (in H) and C (in F) are respectively: LC 2 2 LC + C 2 s + 2 LC 2
EE-202 Ex 3 Sp 7 page 6 (), (2) 2, 2 (3), 2 (4) 2, (5), 2 (6) 2, (7) 2, 2 (8) none of above Solution 7. s 3 + 2s 2 + 2s += s 3 + 2 C s2 + 2C + L LC 2 s + 2 LC 2. Thus C = F and L = 2 H. ANSWER: (4) 8. Reconsider the circuit design of problem 7. The circuit is to be modified so that it realizes a 3 rd order Butterworth HP filter with ω c,min = 00 rad/s. The final capacitor value in the HP design is to be 0.5 mf. Then the final values of the two inductors are (in H): () (2) 2 (3) 0. (4) 0.2 (5) 0.5 (6) 0.002 (7) 0.004 (8) 0.00 (9) none of above Solution 8. s s means capacitors go to inductors of value C and inductors go to capacitors of value L. Hence C HPfinal = C HPinitial K m K f K m = C HPinitial C HPfinal K f = L HP, final = 0 00 = 0.. 0.5 = 0. Hence, 0.5 0 3 00 The next 3 questions relate to the formulas in the table below. Choose wisely. Original Circuit Exact Equivalent Circuit at ω 0 Approximate Equivalent circuit, for high, ( L > 6 and C > 6) and ω within ( ± 0.05) ω 0
EE-202 Ex 3 Sp 7 page 7 ω0l ω = R L ( 0 ) s C ( ω ) = ω R C 0 0 p L Rp ( ω0) = ω L 0 C ( ω0 ) = ω RC 0 s 9. Consider the circuit below in which R s = 0.2 Ω, L = 5 H, C = 5 00 F, and R p = 80 Ω. Recall that coil (ω 0 ) = ω 0L. The approximate value of the parallel resistance in an approximately equivalent R s parallel RLC circuit is (in Ohms): () 0 (2) 20 (3) 25 (4) 4
EE-202 Ex 3 Sp 7 page 8 (5) 5 (6) 6 (7) 2 (8) 8 (9) none of above Solution 9, 0,. ω 0 = LC =0 rad/s. coil = ω 0 L R s = 0. Thus the equivalent resistance of the approximate parallel RLC is R eq = 80 / /20 = 6 Ω. B w = R eq C = 00 6 5 = 5 =.25. Hence 4 = ω p = 0 B ω.25 = 4 0 = 8, ω = 0.25 = 9.375 rad/s. 5 2 0. Reconsider problem 9. The of the approximating parallel RLC is: () (2) 2 (3) 3 (4) 4 (5) 5 (6) 6 (7) 0 (8) 8 (9) none of above. Reconsider problem 9 again. The approximate value of the LOWER half power frequency is (in rad/s): () 8.75 (2) 0.625 (3) 9.375 (4).25 (5) 2 (6) 9.75 (7) 0.25 (8) 8 (9) none of above 2. In the circuit below, R s = 2 Ω and R L = Ω. The source voltage is v in (t) = 0cos(0t)u(t) V. The value of C for maximum power transfer to the load at the source frequency is (in F): () (2) 0.02 (3) 0.2 (4) 0.0 (5) 0.05 (6) 0.5 (7) 0.005 (8).25 (9) none of above
EE-202 Ex 3 Sp 7 page 9 L R Solution 2. Y in ( jω ) = jcω + = jω C jlω + R L L 2 ω 2 + L 2 + R L L 2 ω 2 + R. Thus 2 L R G eq = L L 2 ω 2 + R = 2 = 0.5 2 L L 2 00 + 00 = L2 L = 0. H. Thus C = 0. = 0.05 F. + ANSWER (5) 3. A circuit has impulse response h(t) = u(t + ) u(t 2). If it is excited by the input v in (t) = 2u(t + 2) V, then y(t) = h(t) * v in (t) is: () 2r(t + ) 2r(t 2) (2) 2u(t + 3) 2u(t) (3) 2r(t 2) 2r(t + 3) (4) 2u(t + ) 2u(t 2) (5) 2r(t + 3) + 2r(t) (6) 2r(t + 3) 2r(t) (7) 2r(t + 4) 2r(t) (8) none of above SOLUTION 3. y(t) = 2u(t + ) * u(t + 2) 2u(t 2) * u(t + 2) = 2r(t + 3) 2r(t). Answer: (6) WORKOUT (35 points) Realize the 2 nd order band reject amplifier transfer function H(s) = V out (s) V in (s) = 2(s2 + 4) s 2 using the OBSERVABLE canonical form with a minimum number of op + 2s + 5 amps. In the observable canonical form, ± vout () 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. Further, any inverting of the input to obtain v in (t) is an immediate 5 points plus any other deductions for incorrect procedures. 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) (8 pts) Using the D k and D k notation as per the class examples, develop from the transfer function (s-domain) a differential equation (in the time domain) in vout () t and vin() t associated with H() s that has the form
EE-202 Ex 3 Sp 7 page 0 2 v () t = (??) v () t + D (????) + D (????) Specify all terms. (THIS IS a 0 point or 7 points answer.) out in (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 () t in terms of v t and x () t. Clearly label all inputs, outputs, resistors, and capacitors. Again, ALL op amps are to in () 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 () t must not be present in your feedback loops for the observable form. IF v () 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, out out out 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 7 page SOLUTION WORKOUT: H(s) = V out (s) V in (s) = 2(s2 + 4) s 2 + 2s + 5 (a) (8 pts) ( s 2 + 2s + 5)V out (s) = 2s 2 V in (s) + 8V in (s) D 2 v out + 2Dv out + 5v out = 2D 2 v in + 8v in Hence, v out = 2v in 2D v out + D 2 8v in 5v out ( ) = 2v in + x (b) (7 pts) v out = 2v in x (3 pts). Realize this with the usual summing amplifier (4 pts). (c) (0 pts) x = 2D v out + D 2 ( 8v in 5v out ) = 2D (2v in + x ) + D 2 ( 2v in 5x ). Equivalently Dx = 4v in 2x + x 2 (4 pts) Thus Realize this as an op amp circuit. (4 pts) x =!x = 4 v in 2 x ( x 2 ) (2 pts)
EE-202 Ex 3 Sp 7 page 2 (d) (0 pts) x 2 = D ( 2v in 5x )!x 2 = 2v in 5x. (3 pts) Hence x 2 =!x 2 pts) = 2 v in 5 x (2 pts) Since x2 is needed in (c), we must realize with another op amp. (5
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