EE-202 Exam III April 15, 2010

Transcription

1 EE-0 Exam III April 5, 00 Name: SOLUTION (No period) (Please print clearly) Student ID: CIRCLE YOUR DIVISION Morning 8:30 MWF Afternoon 3:30 MWF INSTRUCTIONS There are 9 multiple choice worth 5 points each and there are two workout problems worth 55 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

2 EE-0, Ex 3 Sp 09 page MULTIPLE CHOICE. The resonant frequency of the circuit ( R = 4 Ω, C = 8 F, L = 0.5 H, L = H, M = 3 8 H) below is (in rad/s): () 3 () 56 (3) 5 (4) 5 (5) 6 (6) 64 (7) 8 (8) none of above Solution. L = L + L M = = 0.5 H. eq ω r = = 56 = 6 r/s. Ans. (5) LeqC. Assume k = 0.5, L = H, L = H, and M = k L L. The zero-state response i out (t) to the input v in (t) = 6u(t) V at t = is (in A): () 0 () (3) 0 (4) 6 (5) 3 (6) 6 (7) 8 (8) none of above Solution. M = k L L = 0.5 = 0.5 H. 0 = V = 0.5sIout + si I = 0.5Iout. Vin ( s) = siout ( s) 0.5 si( s) = [ s 0.5 s] Iout ( s) = 0.75sIout. Thus V ( ) 8 ( ) in s Iout s = i ( ) 8 ( ) 0.75 out t tu t s = s =. At t = s, iout() = 6. Ans. (4)

3 EE-0, Ex 3 Sp 09 page 3 3. Suppose L = 4 H, L = 3 H, L 3 = H, and M = H. The value of L eq = (in H): () () (3) 3 (4) 4 (5) 5 (6) 6 (7) 8 (8) none of above M Solution 3. V = L si + MsI and V = L3sI = MsI + LsI I = I. Thus ( L + L ) V M Leq s = = L s = ( 4 ) s = 3s. Ans. (3) I L + L If R = Ω and L = 0. H, then the value of C in F which makes the circuit resonant at ω r = 0 rad/s is: () 0.0 () 0.0 (3) 0. (4) 0. (5) 5 (6) 6 (7) 8 (8) none of above Solution 4. in ( ) L R Y jω = jcω + = jω C + jlω + R L ω + R L ω + R L 0. C = = = = 0.0 F. Ans. () L ω R Thus

4 EE-0, Ex 3 Sp 09 page 4 5. Reconsider the circuit of problem 4. At resonance, the impedance reduces to a pure conductance, G eq, which is equal to (in S): () 0. () 0. (3) 0.3 (4) 4 (5) 5 (6) (7) 0.4 (8) none of above Solution 5. in ( ) L R Y jω = jcω + = jω C + jlω + R L ω + R L ω + R R Geq = = = 0.4 S. Ans. (7) L ω + R 5. Thus CIRCUIT FOR PROBLEMS 6, 7, AND 8. The transfer function of the Sallen and Key circuit below 0 is H cir (s) = s +. This circuit is to realize H LP (s) = Q s + s + s The value of Q needed for the first stage of realization is: () () (3) 0.04 (4) 0. (5) 5 (6) 5 (7) 8 (8) none of above Ans. (5) 7. The final value of C needed to realize H LP (s) is C f = (in F): () 0. () 0. (3) 0.04 (4) 0. (5) 0.5 (6) 0.5 (7) 0.5 (8) none of above Ans. () or (4) SOLUTION 6-7. ω p = 5 r/s and Q = 5 implies that C f = = 0. F. 5

5 EE-0, Ex 3 Sp 09 page 5 8. 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: () (, ) () (.5,5) (3) ( 5,.5 ) (4) 4, 4 3 (5).5, 5 3 (6) 5 3,.5 (7) 4000 (8) none of above 4 5 R α = = 0.4. B = α = 0.4. = RAα = 0.4RA RA =.5Ω. RA + RB α 4 5 = RA =.5 = Ω. Ans. (5) α 6 3 SOLUTION 8. R B 9. Suppose the circuit below is a NLP prototype ( C = F and L = H) that is to be a HP filter with 3 db down point at ω c = 000 rad/s. The capacitor in the final HP circuit is to be µf, then the value of L,HP for the HP circuit in H is: () () 4 (3) (4) 40 (5) 6 (6) 8 (7) 000 (8) none of above SOLUTION 9. L C Cold 6 = = = = 0. Thus K m = 500. K K 000 K 000K f m m m Km 500 = Lold = = = H. Ans. (6) K f

6 EE-0, Ex 3 Sp 09 page 6 Workout Problem (5 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 = H, C = 0.0 F, R = R = 0. Ω, R 3 = 5 Ω. (a) (6 pts) Determine the appropriate ω 0, inductor-q Q L (ω 0 ), and capacitor-q Q C (ω 0 ). (b) (6 pts) Using the table provided on approximate equivalents for the practical inductors and capacitors, construct an approximately equivalent parallel RLC. Specify R, L, and C. (c) (3 pts) Compute the transfer function associated with your answer to (b) and determine: (i) the circuit Q. (d) approximate values for H m, B ω, ω m, ω, and ω. 0 : (a) ω 0 = = 0 r/s, L 0 LC ( ) ω L Q R 00 ω = = ; Q C ( ω0) = = 00 ω0cr (b) ( pt) L L = 0. H. ( pt) C C = 0. F. (4 pts) 3 3 R = R / /000 / /000 = R / /500 = 00 Ω. (c) ω m s C H ( s) = = s + s + s R C L C 00s 00 + s + ω p B = ω p = 0 r/s, B ω = r/s, = Q = 0, Hm = R = 00, and ω, = ω ω m = 9.5,0.5 r/s Q

7 EE-0, Ex 3 Sp 09 page 7 WORKOUT PROBLEM (30 PTS) This problem is to be done using the OBSERVABLE canonical form using at most 5 op amps. Either the class derivation or the derivation in the notes is acceptable. NOTE : The class derivation was in error, so it is preferred that you use derivation in the notes in which one substitutes for v out (t) to obtain the final form; in the class derivation, I did not substitute for ±v out (t), but fed this quantity back. NOTE : 5 bonus points for the correct derivation as per the notes. NOTE 3: If you use the controllable canonical form instead, your maximum points for a completely correct answer is 0. Construct observable canonical form biquad realization of H BR (s) = V out V in = s + 8 s + 4s + 8. (i) (8 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 BR (s) that has the form v out (t) = Kv in (t) + [???????????? ] where K is a constant you must specify. (ii) (6 pts) Given your (correct) answer to (ii), define the variable x (t) as per the class room derivation. 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. (iii) (8 pts) Given your (correct) definition of x in part (ii), form Dx, properly define the x, and construct and draw an op amp circuit whose output is ±x (t) depending on the sign needed for your answer to (ii). Clearly label all inputs, outputs, resistors, and capacitors. ALL op amps are to have the +terminal grounded. (iv) (8 pts) Construct the op amp circuit whose output is ±x (t) depending on the sign needed for the input to the circuit for x (t). Clearly label all inputs, outputs, resistors, and capacitors. ALL op amps are to have the +terminal grounded.

8 EE-0, Ex 3 Sp 09 page 8 Step : D v out = D v in 4Dv out + 8v in 8v out implies v out = v in 4D v out + 8D Thus v out = v in 4D v out + 8D ( v in v out )= v in + x This last equation is realized as v out = v in x with the circuit ( ) v in v out Step : x = 4D v out + 8D ( ) implies v in v out ɺx = Dx = 4v out + 8D ( v in v out )= 4v out + x = 8v in 4x + x It follows that x = ɺx = 4 v out + x = 4 v out ( x ) ( ) = 8 v in 4 x x This can be realized by the circuit:

9 EE-0, Ex 3 Sp 09 page 9 Step 3. 8D ( v in v out )= x ɺx = Dx = 8v in 8v out = 8v in 8( v in + x )= 8v in 8x Thus x = ɺx = 8 v in 8 x. This can be realized by the circuit:

10 EE-0, Ex 3 Sp 09 page 0 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