ECE-202 FINAL December 13, 2016 CIRCLE YOUR DIVISION

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1 ECE-202 Final, Fall 16 1 ECE-202 FINAL December 13, 2016 Name: (Pleae print clearly.) Student CIRCLE YOUR DIVISION DeCarlo- 8:30-9:30 Talavage-9:30-10: INSTRUCTIONS There are 35 multiple choice worth almot 6 point each. Thi i a cloed book, cloed note exam. No crap paper or calculator are permitted. Promied Table are included with the exam. Carefully mark your multiple choice anwer on the cantron form 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. Nothing i to be on the eat beide you. When the exam end, all writing i to top. No writing while turning in the exam/cantron or rik an F in 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 before the exam tart and during the exam. Do not open, begin, or peek inide thi exam until you are intructed to do o.

2 ECE-202 Final, Fall 16 2 MULTIPLE CHOICE 1. The impule repone of a circuit i h(t) whoe Laplace tranform i L h(t) = 1. Suppoe a ignal f (t) = 2Du(t +1) + Du(t) + Du(t 1) for contant D. Then L f (t)* h(t) = : (1) ( ) D 1+ e (5) 3D + De (8) 2De + D + De (2) D( 1+ e ) (6) 3D + De (9) None of above (3) D De (7) 2De + D + De (4) D De Solution 1. Oberve that h(t) = δ (t) and thu h(t)* f (t) = δ (t)* f (t) = f (t). For t 0, f (t) = 3Du(t) + Du(t 1) whoe Laplace tranform i F() = 3D + D e. Anwer: (6) 2 2. The Laplace Tranform of f (t) i F() =. Let g(t) = tf (t). Then 2 ( + b) L{ e bt g(t) } = : (Follow the problem tatement carefully.) (1) 4 ( + b) 3 2 ( + b) 2 (2) 4( + b) ( + 2b) 3 2 ( + 2b) 2 (3) 2( b) (4) 4( b) (5) (7) 2( b) (6) 4( + b) ( + 2b) ( + 2b) 2 (8) (9) None of above 4( b) ( + b) ( + b) 2

3 ECE-202 Final, Fall 16 3 Solution 2. L{ g(t) } = L{ tf (t)} = d d L e bt 4( b) { g(t) } = G( b) = Anwer: (6) 2 ( + b) 2 = 4 ( + b) 3 2 ( + b) 2. Thu 3. In the circuit below, R = 2 Ω, C = 0.5 F, L = 2 H and β = 0.5. The (Thevenin) input IMPEDANCE at = 1 i Z th (1) = (in ohm): (1) 1 (2) 2 (3) 3 (4) 4 (5) 5 (6) 6 (7) 7 (8) 8 (9) none of above Solution 3. 1 V in = C + R I in + L I in βv C = 1 C + R + L I in β LV C ( ) = 1 C + R + L I in β LV C But, V C = 1 C I in. Hence V in = 1 C + R + β L C + L I in = Z th ()I in. Thu Z th (1) = 3 Ω. Anwer: (4)

4 ECE-202 Final, Fall In the circuit below, uppoe C 1 = 0.4 F, C 2 = 0.1 F, and R = 2 Ω. Then the impule repone ha term of the form Aδ (t) and Be 5t u(t) where ( A, B) = : (1) (5, 20) (2) (0.5,20) (3) (0.5, 20) (4) (5,20) (5) (5, 10) (6) (5,10) (7) (0.5, 10) (8) (0.5,10) (9) None of above Solution 4. H() = V out () ( V in () = C 1 + C 2) + G C 2 + G = 5( +1) + 5 Anwer: (1) = = C 1 + C + ( 2) C 2 G ( ) C 1 + C 2 + G C Conider the circuit below in which R = 5 Ω and L = 1 H. t Suppoe v in (t) = 50 e 5q u(q)dq V and all initial condition are ZERO. Then i L (t) ha 0 a term of the form Ae 5t u(t) where A = : (1) 1 (2) 2 (3) 10 (4) 5

5 ECE-202 Final, Fall 16 5 (5) 5 (6) 6 (7) 2 (8) 18 (9) None of above Solution 5. (Integral Property, Partial Fraction Expanion, Invere Tranform) I L () = V in () L + R = ( + 5) = 50 ( + 5) 2 = 2 10 ( + 5) Anwer: (2) 6. In the circuit below, R = 1 Ω, C = 0.25 F, and v in (t) = 2δ (t) V and v C (0 ) = 4 V. Then v C (t) ha a term of the form Ke 4t u(t) where K = : (1) 12 (2) 2 (3) 6 (4) 4 (5) 4 (6) 6 (7) 12 (8) 8 (9) None of above Solution 6. Initial capacitor model, complete repone, impedance, voltage diviion. Uing the current ource model of the capacitor we have that 4 4 V C () = 1+ 4 V in () + Cv C (0 ) 1+ 4 = V in () + v C (0 ) + 4. Hence V C () = 8 + v C (0 ) ( + 4) = and v C (t) = 12e 4t u(t) V.

6 ECE-202 Final, Fall 16 6 Anwer: (7) 7. In the circuit below, L = 1 H, R = 2 Ω, C = 0.1 F, and I in () = 6. The zero-tate + 2 repone, v out (t) = (in V): (1) 10e t in( 10 t)u(t) (2) 20e t in( 10 t)u(t) (3) 60e t in( 10 t)u(t) (4) 3e t in( 10t)u(t) (5) 20e t in(3t)u(t) (6) 60e t in(3t)u(t) (7) 3e t in(3t)u(t) (8) 10e t in(3t)u(t) (9) None of above 1 2 R + Solution 7: C + L + R V out () = C L + 1 LC + R V out () = I in (). Thu L V out () = 1 C + R L 2 + R L + 1 LC v out (t) = 20e t in(3t)u(t) V. Anwer: (5) + 2 I in () = = 20 3 ( +1) Hence 2 8. Given the pole zero plot below, a bounded-input that will caue the repone to be untable (NOT BIBO-table) i: (1) 5u(t 10) (2) 6r(t 1) (3) 10t co(t + 45 o )u(t) (4) 10co( t +12 o )u(t) (5) 5in(2t)u(t) (6) 5in( 2t)u(t) (7) 10co(2(t 1))u(t) (8) 4in ((t 1) )u(t 1)

7 ECE-202 Final, Fall 16 7 (9) two of above (10) none of above Solution 8. The TWO input are 10co( t +12 o )u(t) and 4in ((t 1) )u(t 1). Anwer: (9) 9. For the circuit below, C 1 = 4 F, C 2 = 1 F, v in (t) = 10u(t) V. For 1 t < 1, the witch i in poition B and v out (0) = 5 V. At t = 1, the witch move to poition A. At t = 2, the witch move to poition B again and remain there forever. Then v out (3) = (in volt): (1) 10 (2) 2 (3) 9 (4) 4 (5) 5 (6) 6 (7) 20 (8) 8 (9) none of above

8 ECE-202 Final, Fall 16 8 Solution 9. For t 2, By conervation of charge, C C 2 5 = ( C 1 + C 2 )v out (3), which implie that v out (3) = C C 2 5 = = 9 V. C 1 + C 2 5 Anwer: (3) 10. (Scaling) In a particular normalized Friend BP filter deign, one et H cir () = K R R 1 R 2 = K Q +1 to determine NORMALIZED value of R 1 and R 2, given that the circuit capacitor are alo et to normalized value of C 1 = C 2 = 1 F. The objective i to realize the BP tranfer function 1000 H BP () = (20) 2 If the only capacitor available for a final deign are 5 mf, then the final caled value of the reitor, R 1, final and R 2, final, are: (1) 1, 100 (2) 0.25, 4 (3) 5, 40 (4) 2.5, 40 (5) 5, 20 (6) 25, 400 (7) 0.125, 2 (8) 10, 200 (9) 400, 25 (10) none of above Solution 10. K m = K m = Anwer (4) 1 Q = 2 R 2 R 2 = 2Q Ω. 1 = 1 R R 1 R 1 = 1 = 1 2 R 2 2Q Ω. K f = ω m. 1 1 =. For thi problem, ω C new K f C new ω m = 20 rad/ and Q = 2 in which cae, m = 10. Therefore R 1, final = K m 2Q = 10 4 = 2.5 Ω and R 2, final = 40 Ω.

9 ECE-202 Final, Fall The tranfer function of a complex circuit i given by H() = V out () V in () = 64 ( 8). The circuit i excited by an input ( + 8)( ) v in (t) = 0.5co(8t 45 o )u(t) V. The magnitude and phae (in degree), repectively, of v out (t) in SSS at i: (1) 16, 0 o (2) 16, 45 o (3) 16, 45 o (4) 8, 45 o (5) 8, 180 o (6) 8, 0 o (7) 8, 45 o (8) 4, 180 o (9) 16, 180 o (10) none of above j8( j8 8) Solution 11. H( j4) = 16 ( j8 + 8)( ) H( j8) = j8( j8 8) H( j8) = ( j8 + 8)( ) = = 0o. Hence v out, (t) = 8co(4t 45 o )u(t) V. Anwer: (4) = 16 and 12. In the circuit below, R = 2 Ω, C = 0.5 F, and K i variable. The complete range of K for which the input ADMITTANCE (not the impedance) i BIBO table i: (1) K > 0 (2) K < 0 (3) K < 1 (4) K > 1 (5) K < 2 (6) K > 2 (7) 0 K 2 (8) none of above Solution 12. I in = 0.5V in + 0.5( V in KI in ) implie (1+ 0.5K)I in = 0.5(1+ )V in (K + 2)I in = ( + 1)V in. Thu Y in () = + 1 K + 2. K Stability implie = 2 K < 0 K > 0.

10 ECE-202 Final, Fall Anwer: (1) 13. In the circuit below, the load i conidered to be at the terminal labeled C-D. Suppoe R = 10 Ω, R 1 = 40 Ω, and R L = 2 Ω. The turn ratio for maximum power tranfer to the load i: (1) 1 (2) 2 2 (3) 8 (4) 4 (5) 0.5 (6) 0.25 (7) (8) 2 1 (9) 2 2 (10) none of above Solution 13. The problem reduce to finding to match R 1 / / R = 8 Ω. 8 = a 2 R L a 2 = 4 a = 2. ANSWER: (8) 14. Referring again to problem 13, uppoe v (t) = V rm co(ωt) V where V rm = 60 V. At maximum power tranfer the rm magnitude of v L (t) = V L,rm co(ωt) V i V L,rm = (in V): (1) 10 (2) 20 (3) 25 (4) 40 (5) 50 (6) 12 (7) (8) 2 2 (9) 50 2 (10) none of above

11 ECE-202 Final, Fall Solution 14. Replacing the front end by it Thevenin equivalent, the voltage acro the primary i = 24 V. At maximum power tranfer, thi voltage i reflected to 50 the econdary a 1 24 = 12 V. a Anwer: (6) 15. In the circuit below, L 1 = 5 H, and L 2 = M = 4 H. R = 3 Ω, R L = 4 Ω. The dot i in poition B. The impedance een by the voltage ource at = 1 i: (1) 1 (2) 2 (3) 3 (4) 4 (5) 5 (6) 6 (7) 7 (8) 8 (9) 9 (10) none of above Solution 15. Uing the T-equivalent and the fact that L 2 M = 0, the mutually coupled inductor reduce to a left erie branch of value L 1 M = 1 and a hunt branch, M = 4 H, in parallel with R L = 4. Thu Z in = R + (L 1 M ) ANSWER: (6) = ;Z in (1) = Conider the circuit below in which the dot i in poition B. Suppoe L 1 = 10 H, L 2 = 10 H, the coupling coefficient k = 0.5, and R = 20 Ω. The repone, v out (t), of the circuit with tranfer function, H() = V out () V in () i: to the input v in (t) = 4u(t) V (1) 2e 2t u(t) (2) 2e 2t u(t) (3) 4e 2t u(t) (4) 4e 2t u(t) (5) 8e t u(t) (6) 8e t u(t) (7) 8e 2t u(t) (8) 8e 2t u(t)

12 ECE-202 Final, Fall (9) 2e t u(t) (10) None of above Solution 16. I 2 () = 0 V out () = MI 1 (). V in () = RI 1 () + L 1 I 1 () = ( L 1 + R) I 1 () I 1 () = V in () L 1 + R. Hence H() = V out () V in () = M L 1 + R = v out (t) = 2e 2t u(t) V. ANSWER: (1) = It follow that V 2 out () = + 2 and 17. In the circuit below, R = 1.5 Ω, L 1 = 4 H, L 2 = 4 H, and M = 2 H. The zero-tate repone in the -domain, I 2 (), to the input v in (t) = 24δ (t) V, at = 0.5, i: (1) 1 (2) 2 (3) 2 (4) 4 (5) 5 (6) 4 (7) 5 (8) 8 (9) 8 (10) none of above Solution 17. Since V 2 = 0 = MI 1 L 2 I 2. Thu, MI 1 = L 2 I 2 I 1 = L 2 M I 2 = 2I 2.

13 ECE-202 Final, Fall Further, V 1 = V in RI 1 V in = R I 1 +V 1 = 2R I 2 + 2L 1 I 2 MI 2 = ( 2R + (2L 1 M )) I 2 Therefore I 2 () = ANSWER: (4) V in = I 2 (0.5) = In the circuit below, R = 2 Ω and R L = 1 Ω. The ource voltage i v in (t) = 10co(5t)u(t) V. The value of C for maximum power tranfer to the load at the ource frequency i (in F): (1) 0.1 (2) 0.2 (3) 0.02 (4) 0.25 (5) 0.05 (6) 0.5 (7) (8) 400 (9) none of above 1 L R Solution 18. Y in ( jω ) = jcω + = jω C jlω + R L L 2 ω 2 + L 2 + R L L 2 ω 2 + R. Thu 2 L R G eq = L L 2 ω 2 + R = = L L = L2 L = 0.2 H. Thu C = 0.2 = 0.1 F. 1+1 ANSWER (1) 19. A capacitor i put in parallel with the inductor in the circuit below and at the ame time, inductor of appropriate value are put in erie with the capacitor. The reulting filtering action i of what type: (1) Low Pa (2) Band Reject (3) Band Pa (4) High Pa (5) cannot be determined (6) none of above

14 ECE-202 Final, Fall Anwer: (2) 20. The circuit below conit of the SERIES interconnection of three 2-port, the middle 2-port being an ideal tranformer with turn ratio 1:a a indicated. The top 2-port ha h- parameter given by a matrix H 1, the tranformer ha h-parameter matrix, H 2, and finally, the bottom 2-port, H 3. The h-parameter of the interconnected 2-port are: (1) H 1 + H 2 + H 3 (2) H 1 + H 2 + ah 3 (3) H 1 + H 2 H 3 (4) H 1 + H 2 ah 3 (5) ah 1 + H 2 + H 3 (6) ah 1 + H 2 + H 3 (7) H 1 + H 2 H 3 (8) none of above Solution 20. The problem, like reality TV, i total nonene. Anwer: (8)

15 ECE-202 Final, Fall Conider the circuit below a a cacade of three 2-port. Suppoe Z() = 1 Ω, the dot of the firt ideal tranformer i in poition A, a = 2, and b = 2. The value of the t- parameter t 12 i (in tandard unit): (1) 1 (2) 2 (3) 0 (4) 0.25 (5) 0.5 (6) 0.5 (7) 1 (8) none of above 1 1 Solution 21. For the erie reitor, T 2 =. For the firt and econd ideal tranformer, T 1 = b 0 a = and T 3 = = a 0. Hence b Anwer: (4) T = T 1 T 2 T 3 = = = For the interconnected 2-port below, Y 1 = = 1 and Z() = Z = 1, the value of the overall y-parameter ( y 21 (), y 22 ()) = (in mho): =1 = [Y()] 1. At (1) (2, 3) (2) (2, 2) (3) ( 1,3) (4) ( 2,4) (5) (0,4) (6) (2,4) (7) ( 1,1) (8) (1, 2) (9) none of above

16 ECE-202 Final, Fall Solution 22. (Parallel connection of two port. y-parameter calculation) Y 1 =. The aociated 2-port y-parameter are Y new = Anwer: (4) + =. Thu ; Y new (1) = The next 4 quetion deal with the 2-Port below for which the tage have y-parameter matrice: Y 1 = Z 1 1 = mho and Y 2 = mho Suppoe that R = 0.5 Ω, R L1 = 1 40 Ω, and R L 2 = 0.1 Ω. The formula for the generic gain of a 2-port modeled by y-parameter i: y 21 y 22 + Y L.

17 ECE-202 Final, Fall The input admittance, Y in2 = (in mho): (1) 10 (2) 20 (3) 30 (4) 40 (5) 50 (6) 5 40 (7) 0.1 (8) 8 (9) none of above Solution 23. Y in2 = = = 10 mho Y L The input admittance, Y in = (in mho): (1) 10 (2) 20 (3) 30 (4) 40 (5) 50 (6) 5 40 (7) 0.1 (8) 8 (9) none of above Solution 24. Y L1 / /Yin2 = = 50 mho. Y in1 = = 8 Anwer: (8) 25. The gain G V1 = V 1 V = : (1) 1 (2) 0.2 (3) 0.3 (4) 0.4 (5) 0.5 (6) 0.6 (7) 0.7 (8) 0.8 (9) none of above Solution 25. G V1 = V 1 V = = 0.2.

18 ECE-202 Final, Fall The gain, G V 2 = V out V 1 of the 2-port i: (1) 1.6 (2) 2 (3) 0.4 (4) 4 (5) 0.8 (6) 6 (7) 1.2 (8) 8 (9) none of above Solution 26. Gtage1= = 2, Gtage1= = 4. G V 2 = V out = 8. V For the circuit below L 1 = 2 H, L 2 = 4 H, M = 2 H, and the dot i in poition B. The h-parameter, h 21 and h 22, in tandard unit at = 1 are: (1) (2, 4) (2) ( 2, 4) (3) (0.25, 0.5) (4) ( 0.25, 0.5) (5) ( 0.5, 0.25) (6) (0.5, 0.25) (7) (0.25, 0.5) (8) (2, 2) (9) (2, 4) (10) none of above Solution 27. h-parameter matrix i: V 1 I 2 = h 11 h 12 h 21 h 22 I 1 V 2 The econd mutually coupled inductor equation i: V 2 = MI 1 + L 2 I 2 = 2I 1 + 4I 2. Rearranging we have: 4I 2 = V 2 + 2I 1 I 2 = 1 4 V I 1. At = 1, I 2 = 0.5I V 2 = h 21 I 1 + h 22 I 2. Anwer: (6)

19 ECE-202 Final, Fall Refer again to the circuit of problem 27. The dot i again in poition B. The parameter y 11 and y 21 at = 0.5 are repectively: (1) (1, 1) (2) (2, 2) (3) (2, 1) (4) (2, 1) (5) (4, 2) (6) (4, 2) (7) (8, 4) (8) (8, 4) (9) (2, 4) (10) none of above Solution 28. V 1 V 2 = and y 21 = 1. Anwer: (3) I 1 I 2 I 1 I 2 = V 1 V 2. At = 0.5, y 11 = Conider the circuit below in which a = 0.5, C 2 = 8 F, C 1 = 2 F, and R = 1 Ω. Suppoe v in (t) = 10u(t) V. Then v 1 (t) ha a term of the form Ke at u(t) where (K,a) = : (1) (10, 2.5) (2) ( 10, 2.5) (3) ( 10, 0.25) (4) (10, 0.25) (5) (10,1) (6) ( 10,1) (7) (10,0.1) (9) ( 10,0.1) (9) none of above

20 ECE-202 Final, Fall Solution C 2,reflected = 4 8 = 1 2. Two 2-farad cap in parallel have a C eq = 4 F. 1 V 1 () = 4 1 V 1 in () = = = 2.5 ( ). = Anwer: (3) 30. A one port whoe terminal voltage and current label (in tandard unit and with atifaction of the paive ign convention) are V 1 () and I 1 () in the famou -world. Penny Amp and Aher Volt CORRECTLY wrote the loop equation for thi one port a V = I 1 I 2 I 3 The Thevenin input impedance een at the terminal at = 2 i (in Ω): (1) 11 (2) 12 (3) 13 (4) 14 (5) 15 (6) 16 (7) 17 (8) 18 (9) 19 (10) none of above Solution 30. Therefore, Z th (2) = 13. V 1 = Z I th () = = = 8 1 Anwer: (5) 31. In the circuit below, k = 6 72 L 2 = 9 H, L 1 = 8 H. The circuit i reonant at ω r = 5 rad/. The correponding value of C in F i:

21 ECE-202 Final, Fall (1) 0.1 (2) 0.2 (3) (4) 0.25 (5) (6) 0.01 (7) 0.02 (8) (9) 0.05 (10) none of above Solution 31. M = k L 1 L 2 = = 6 H. L eq1 = M / /(L 2 M ) = 6 / /3 = 2; L eq = (L 1 M ) + L eq1 = 4 H. ω r 2 = 1 LC C = 1 Lω r 2 = = = 0.01 Anwer: (6) 32. A 2-port with tandard labeling i repreented by the h-parameter matrix [h ij ] = h 11 h 12 h 21 h 22 With port 1 open-circuited, a negative tep voltage i applied at port 2, i.e., v 2 (t) = u(t) V. The voltage meaured at port 1 i v 1 (t) = r(t) V. The h-parameter h 12 = : (1) 1 (2) 1 (3) 1 2 (4) 1 2 (5) (6) (7) 2 (8) 2 (9) none of above

22 ECE-202 Final, Fall Solution 32: h 12 = V 1 V 2 I 1 =0 = R() U () = 1. Anwer: (1) 33. The circuit hown below conit of a 100 Ω reitor in parallel with a REAL capacitor, C = 50 mf, having a Q cap (ω ) = 10 at ω = 10 rad/ec. (Replace the capacitor by it non-ideal parallel equivalent.) The REAL part of the input ADMITTANCE at = jω of thi combination (given in mho or Siemen) i: (1) 0.01 (2) 0.02 (3) 0.03 (4) 0.04 (5) 0.05 (6) 0.06 (7) 0.07 (8) 0.08 (9) none of above Solution 33: 10 = ω RC = R R = = 20 Ω. G eq = = 0.06 mho. Anwer: (6) 34. In the circuit below, let v in (t) = V rm co(2π f 0 t)u(t) V where V rm = 100 V. Suppoe further that R = 25 Ω, R L = 10 Ω, and R tran = 60 Ω. Suppoe a = b = 2. The value of V 1,rm = (in Volt)= (1) 10 (2) 20 (3) 30 (4) 40 (5) 50 (6) 60 (7) 100 (8) 80 (9) none of above

23 ECE-202 Final, Fall Solution 34. Z in = 1 ( a 2 R tran + b2 R L ) = V 1,rm = = 50 V. Anwer: (5) 2 ( ) = 25 Ω. Therefore 35. The plot below i of a BP tranfer function of the form H BP () = value of (K,b) = : K 2 + a + b. The (1) (10, ) (2) (10, ) (3) (1000, ) (4) (100, ) (5) (100, ) (6) (1000, ) (7) (7.07, ) (8) (7.07, ) (9) none of above

24 ECE-202 Final, Fall Solution 35. B ω = 100 rad/ and ω p = ω m = 800 rad/. Obviouly, b = Anwer (3) K B ω = 10 K = Table 13.1 LAPLACE TRANSFORM PAIRS Item Number f(t) L[f(t)] = F() 1 Kδ(t) K 2 Ku(t) or K K 3 r(t) t n u(t) n! n+1 5 e at u(t) 1 + a 6 te at u(t) 1 ( + a) 2 7 t n e at u(t) n! ( + a) n+1 8 in(ωt)u(t) ω 2 + ω 2 9 co(ωt)u(t) 2 + ω 2 10 e at in(ωt)u(t) ω ( + a) 2 + ω 2 11 e at co(ωt)u(t) ( + a) ( + a) 2 + ω 2 12 tin(ωt)u(t) 2ω ( 2 + ω 2 ) 2

25 ECE-202 Final, Fall tco(ωt)u(t) 2 ω 2 ( 2 + ω 2 ) 2 14 in(ωt + φ)u(t) in(φ) + ω co(φ) 2 + ω 2 15 co(ωt + φ)u(t) co(φ) ω in(φ) 2 + ω 2 16 e at [in(ωt) ωtco(ωt)]u(t) 2ω 3 [( + a) 2 + ω 2 ] 2 17 te at in(ωt)u(t) + a 2ω [( + a) 2 + ω 2 ] e at C 1 co(ωt) + C 2 C 1 a ω in(ωt) u(t) C 1 + C 2 ( + a) 2 + ω 2 2 A 2 + B 2 e at co ωt tan 1 B A A + jb + a + jω + A jb + a jω 20 2 A 2 + B 2 t e at co ωt tan 1 B A A + jb ( + a + jω) 2 + A jb + a jω ( ) 2 Table 13.2 LAPLACE TRANSFORM PROPERTIES Property Linearity Tranform Pair L[a 1 f 1 (t) + a 2 f 2 (t)] = a 1 F 1 () + a 2 F 2 () Time Shift L[f(t T)u(t T)] = e T F(), T > 0 Multiplication by t Multiplication by t n L[tf(t)u(t)] = d d F() L[t n f (t)] = ( 1) n d n F() d n Frequency Shift L[e at f(t)] = F( + a) Time Differentiation L d dt f (t) = F() f(0 )

26 ECE-202 Final, Fall Second-Order Differentiation nth-order Differentiation Time Integration Time/Frequency Scaling L d2 f (t) dt 2 = 2 F() f (0 ) f (1) (0 ) L d n f (t) dt n = n F() n 1 f (0 ) n 2 f (1) (0 ) f (n 1) (0 ) (i) L t (ii) L f (q)dq = t 0 F() f (q)dq = 0 + L[f(at)] = 1 a F a F() f (q)dq