ECE-0 Exam January 3, 08 Name: (Pleae print clearly.) CIRCLE YOUR DIVISION 0 0 DeCarlo DeCarlo 7:30 MWF :30 TTH INSTRUCTIONS There are multiple choice worth 5 point each and workout problem worth 40 point. Allow 0 minute for workout problem a you can obtain partial credit. Thi i a cloed book, cloed note exam. No crap paper or calculator are permitted. Nothing i to be on the eat beide you no coat, book, phone, etc. Carefully mark your multiple choice anwer on the cantron and alo on the text booklet a you will need to put the cantron inide the tet booklet and turn both in at the end of the exam. No turned in exam mean a zero at leat on workout. When the exam end, all writing i to top. No writing while turning in the exam/cantron or rik an F in the exam. Again, turn in the exam booklet with the cantron inide the exam. There are two exam form and thi i neceary for the proper grading of your cantron. All tudent are expected to abide by the cutomary ethical tandard of the univerity, i.e., your anwer mut reflect only your own knowledge and reaoning ability. A a reminder, at the very minimum, cheating will reult in a zero on the exam and poibly an F in the coure. Communicating with any of your clamate, in any language, by any mean, for any reaon, at any time between the official tart of the exam and the official end of the exam i ground for immediate ejection from the exam ite and lo of all credit for thi exercie. The profeor reerve the right to move tudent around during the exam. Do not open, begin, or peek inide thi exam until you are intructed to do o.
ECE-0 Spring 08 Exam. The repreentation of the ignal, f (t), hown below and whoe value i for t, in term of caled and hifted tep and ramp i: () r(t +) r(t ) + u(t ) () r(t +) r(t ) + 4u(t ) (3) r(t ) r(t +) (4) r(t ) r(t +) + 4u(t +) (5) r(t +) r(t ) + 4u(t) (6) r(t +) r(t ) + u(t ) (7) r(t +) r(t ) (8) r(t +) r(t ) (9) None of above.5.5 f(t) 0.5 0.5 0.5 0 0.5.5.5 time in ec Solution. By inpection, f (t) = r(t +) r(t ) Anwer: (8)
ECE-0 Spring 08 3 Exam. The ignal, f (t), hown below, i zero for t < 0 and jump to at t = 0 ; it value i for t. The Laplace tranform of f (t) i F() = : () + () + (3) + ( e ) (4) + ( e ) (7) + (+ e ) (5) ( e ) (8) + ( e ) (6) (+ e ) (9) None of above.5.5 f(t) 0.5 0 0.5 0 0.5.5.5 time in ec Solution. By inpection, f (t) = u(t) + r(t) r(t ) implie F() = + e. Anwer: (4)
ECE-0 Spring 08 4 Exam 3. In the circuit below, L = H, C = 0.5 F, and G = 0.5 mho. The value of the input impedance, Z in (), at = i: () () (3) 3 (4) 4 (5) 5 (6) 6 (7) 0.5 (8) 0.5 (9) None of above Solution 3. Z in () = L + C + G. Thu at =, Z in () = L + = + = 4 Ω.. Anwer: (4) C + G 4. The ignal below i f (t) = F() = : in(πt) 0 t 0 otherwie. The Laplace tranform of thi ignal i () e π + π () e π + π (4) + e π + π (5) e π π (7) + e π π (8) + e π π (3) + e (6) e π + π π π (9) None of above
ECE-0 Spring 08 5 Exam.5.5 f(t) 0.5 0 0.5 0 0.5.5 time in ec Solution 4. f (t) = in(πt)u(t) + in( π (t ) )u(t ) in which cae F() = + e Anwer: (4) π + π. 5. Suppoe L f (t) { } = ln () + a () (4) + a (5) { } i: + a. Then L te at f (t) ( + a) (3) + a (7) (8) (6) (9) None of above
ECE-0 Spring 08 6 Exam { } = ln { } = Solution 5. L e at f (t) + a. Therefore, L te at f (t) d d ln + a = + a ( + a) =. Anwer (4) + a 6. An integro-differential equation for a linear 0 circuit i given by t 4 v C (τ )dτ 0 + d dt v C (t) + 4v C (t) = u(t) Suppoe v C (0 ) = 0 and!v C (0 ) = V/. Then v C (t) ha a term of the form Kte at u(t) where (K,a) = : () (,) () (,) (3) (4,) (4) ( 4,) (5) (,) (6) (,) (7) (,) (8) (,) (9) None of above Solution 6. The above equation implie that v!! C (t) + 4!v C (t) + 4v C (t) = δ (t). In the -world V C () v C (0 )!v C (0 ) + 4V C () v C (0 ) + 4V C () =. If v C (0 ) = 0, then ( + 4 + 4)V C () = +!v C (0 ) implie V C () = +!v C (0 ) + 4 + 4 v C (t) = ( +!v C (0 ))te t u(t) V. Anwer: (3) = +!v C (0 ) ( + ) which implie that 7. A circuit characterizing the decay of the ordinary attention pan of a tudent in a DeCarlo circuit cla i H()= a term of the form Ae t u(t) where A = : 0. The tep repone of the aociated tranfer function ha ( +)( +3) () 0 () 5 (3) 5 (4) 0 (5) 5 (6) 4 (7) 5 (8) 4 (9) None of above
ECE-0 Spring 08 7 Exam Solution 7. (Tranfer function, tep repone, partial fraction expanion ditinct pole, invere tranform) L[StepRepone]= H() 0 = ( +)( +3) = 5 + + 5 +3. StepRepone = 5e t +5e 3t u(t). ANSWER: (7). 8. In the circuit below, R = 0.5 Ω, g m =, and C = 0.5 F. The Thevenin equivalent open circuit voltage at the output terminal A-B i: V () in () V in V (3) in (4) V in 4V (5) in + + + 4 + 4 + 4 (6) 4V in 8V (7) in (8) 8V in (9) none of above + 4 + 4 + 4 Solution 8. V out = R V out V in I in + C + G g m + CV out g m V out = I in implie (C + G g m )V out = I in + GV in. Thu G C + G g m V in = 0.5 + I in + 0.5 + V in. Concluion: Z th = 4 + 4 and V oc = 8 + 4 V in. ANSWER: (7) 9. Referring again to problem 8, the Thevenin Equivalent impedance een at the A-B terminal i: 4 () () (3) (4) (5) + + + 4 + 4 + 4 4 8 8 (6) (7) (8) (9) none of above + 4 + 4 + 4 Anwer: (6)
ECE-0 Spring 08 8 Exam 0. In the circuit below, R = Ω, C = 0.5 F, and v in (t) = 4δ (t) V and v C (0 ) = 8 V. Then v C (t) ha a term of the form Ke at u(t) where (K,a) = : () (,4) () (6,) (3) (4,4) (4) (4,) (5) (,4) (6) (8, 4) (7) ( 8,4) (8) (8,4) (9) None of above Solution 0. Initial capacitor model, complete repone, impedance, voltage diviion. Uing the current ource model of the capacitor we have that 4 4 V C () = + 4 V in () + Cv C (0 ) + 4 = 4 + 4 V in () + v C (0 ) + 4. Hence, if v in (t) = Kδ (t) V C () = 4K + v C (0 ) ( + 4) ANSWER: (8) = 8 + 4 and v C (t) = 8e 4t u(t) V.. In the circuit below, I in () = 0, L = 0.5 H, Z() = Y() = 0.5 + 9, and i L (0 ) = A. Then i L (t) ha a term of the form K co(ω t)u(t) where (K,ω ) = : () (,3) () (,9) (3) (,9) (4) (,3)
ECE-0 Spring 08 9 Exam (5) ( 6,3) (6) (6,9) (7) (6,3) (8) ( 6,9) (9) None of above Solution. ( Z() + 0.5) = + 9 = + 9, i.e., I L () = Li L (0 ) Z() + 0.5 = 6 + 9 i L (t) = 6co(3t)u(t). Anwer: (5). In the circuit below, I in () =, i L (0 ) = 0, L = 0.5 H and Y() = + 4 + 6. Then v L (t) ha a term of the form Ke at in(ω t)u(t) where (K,a,ω ) = : () (3,,4) () (,,) (3) (3,,) (4) (4,,) (5) (4,,4) (6) (6,,) (7) (8,,) (8) (8,,4) (9) None of above Solution. Here, Y eq () = Y() + = + 4 + 8 = + 4 + 8 Z eq () = ( + ) +. Uing the voltage ource model, V L () = Z eq ()I in () = ( + ) + = 6 ( + ) +. Hence v L (t) = 6e t in(t)u(t) A.
ECE-0 Spring 08 0 Exam Anwer: (6)
ECE-0 Spring 08 Exam Workout Problem: (40 point) Part. (a) (6 pt) Compute the tranfer function H () = V out () V in () of the circuit below in term of the literal G = R, G, C, and. Your final reult hould be of the form: H() = K ( +??)( +??). (b) (4 pt) Compute the value of G = R, G, C, and, o that H () = under the contraint that = F. ( +)( + 4) Part ( pt). (a) (8 pt) Compute the tranfer function H () = V out () V in () in term of the literal G, C, and. Your final reult hould be of the form: H() = K +? +????. of the circuit below (b) (4 pt) Compute the value of G, C, and o that H () = 4 + under the contraint ( + 4) that G = mho.
ECE-0 Spring 08 Exam Part 3. (8 pt) The output of the circuit of Part replace the voltage ource of the circuit of Part and thu drive the circuit of part without any loading effect. Thi i called a cacade connection. (a) What i the new tranfer function, H new (), of the cacaded circuit? (b) Compute the tep repone of the new circuit. (c) Compute the impule repone of the new circuit. Solution Part, (a) and (b): (0 pt) (a) (6 pt) H() = V out () V in () = Y in () Y f () = = G = R + G + G C C R + C = + G C ( R C +) + G + G + G (Thi i the correct form that wa required.)
ECE-0 Spring 08 3 Exam (b) (4 pt) H() = G + G + G C G = + G + G C = ( +)( + 4). Cae : G = R = mho, G = 4 mho, C = F, and = F. Cae : However, you could alo have et + G = + o that G = mho. Then C = 4 and C = 0.5 F. Solution Part, (a) and (b): ( pt) (a) (8 pt) C V in = ( + G)(V out V in ) ((C + ) + G)V in = ( + G)V out. Hence, H() = (C + ) + G + G = (C + ) + G (C + ) + G. (Thi i the correct form that wa required.) (b) (4 pt) H() = (C + ) G + (C + ) ( +) + G = 4 ( + 4). Thu C + = 4 C = 3 in which cae G = G = and G = 4. Set G = mho. Then, C C + 4 C = 0.75 F and = 0.5 F. Solution part 3 (8 pt): (a) (4 pt) H new () = H ()H () = 8. (All or nothing.) ( + 4) 8 (b) (4 pt) Step Repone i the invere tranform of ( + 4) which i Step-Rep = 8te 4t u(t) V. (c) (0 pt) Impule Repone i the invere tranform of H new (). H new () = 8 ( + 4) = A + 4 + B ( + 4) = 8 + 4 + 3. Impule Repone ( + 4) = 8e 4t u(t) + 3te 4t u(t). = 8e 4t u(t) + 3te 4t u(t). You could alo have ued the relationhip: Derivative(tep repone) = impule repone, provided you accounted for the derivative of the tep function properly and explained what you did.
ECE-0 Spring 08 4 Exam (3 pt for B, 4 pt for A, and 3 pt for the correct invere. - for ign error. If the tudent make up a repone with no partial fraction expanion, the grade i zero.)