Solution Manual of Fundamental of Electric Circuits, 2nd Ed. by Alexander & Sadiku
Oct 29, 2014
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Sol u t i onSol u t i onChapter 1, Solution 1 (a)q = 6.482x1017 x [-1.602x10-19 C] = -0.10384 C (b) q = 1. 24x1018 x [-1.602x10-19 C] = -0.19865 C (c) q = 2.46x1019 x [-1.602x10-19 C] = -3.941 C (d) q = 1.628x1020 x [-1.602x10-19 C] = -26.08 C Chapter 1, Solution 2 (a)i = dq/dt = 3 mA (b)i = dq/dt = (16t + 4) A (c)i = dq/dt = (-3e-t + 10e-2t) nA (d)i=dq/dt =1200 120 cos tpA (e)i =dq/dt = +e tt 480 50 1000 50 ( cos sin )A u t Chapter 1, Solution 3 (a)C 1) (3t + = + =q(0) i(t)dt q(t)(b)mC 5t) (t2+ = + + =q(v) dt s) (2t q(t)(c)( ) q(t) 20 cos10t / 6 q(0) (2sin(10 / 6) 1) C t u = + + = + + (d) C 40t) sin 0.12 t (0.16cos40 e30t -+ =+= + = t) cos 40 - t 40 sin 30 (1600 900e 10q(0) t 40 sin 10e q(t)-30t30t - Chapter 1, Solution 4 ( ) mC 4.698 = == = = 06 . 0 cos 165 t 6 cos65dt t 6 5sin idt q100 Chapter 1, Solution 5 C mC ) e 1 (21e21- mC dt e idt q4202t - 2t -490 = == = = Chapter 1, Solution 6 (a) At t = 1ms,mA 40 = = =280dtdqi (b) At t = 6ms,mA 0 = =dtdqi (c) At t = 10ms,mA 20 - = = =480dtdqi Chapter 1, Solution 7
< 0,,0 -t) ( u = 0 ) ( v = 30 10 20 Rth= + = ,3 ) 1 . 0 )( 30 ( C Rth= = = | | + =t -e ) ( v ) 0 ( v ) ( v ) t ( v= ) t ( v V e 103 t - 3 t -e 1031 -) 1 . 0 (dtdvC ) t ( i |.|
\|= == ) t ( i A e31 -3 t - Chapter 7, Solution 49. For 0 < t < 1,,0 ) 0 ( v = 8 ) 4 )( 2 ( ) ( v = = 10 6 4 Req= + = ,5 ) 5 . 0 )( 10 ( C Req= = = | | + =t -e ) ( v ) 0 ( v ) ( v ) t ( v( )V e 1 8 ) t ( v5 t - = For t > 1, ( )45 . 1 e 1 8 ) 1 ( v-0.2= = ,0 ) ( v = | | + =) 1 t ( -e ) ( v ) 1 ( v ) ( v ) t ( vV e 45 . 1 ) t ( v5 ) 1 t ( - = Thus, = ) t ( v( )> < < 1 t , V e 45 . 11 t 0 , V e 1 85 ) 1 t ( -5 t - Chapter 7, Solution 50. For the capacitor voltage, | | + =t -e ) ( v ) 0 ( v ) ( v ) t ( v0 ) 0 ( v = For t < 0, we transform the current source to a voltage source as shown in Fig. (a). 1 k 1 k+ v + 30 V2 k(a)V 15 ) 30 (1 1 22) ( v =+ += = + = k 1 2 || ) 1 1 ( Rth 41104110 C R3 - 3th= = = ( )0 t , e 1 15 ) t ( v-4t> = We now obtainifrom v(t).Consider Fig. (b). xix1/4 mFv 1 k1 kiT30 mA2 k (b)T xi mA 30 i =But dtdvCRvi3T+ =( )A e -15)(-4) ( 1041mA e 1 5 . 7 ) t ( i4t - 3 - 4t -T + =( )mA e 1 5 . 7 ) t ( i-4tT+ = Thus, mA e 5 . 7 5 . 7 30 ) t ( i-4tx == ) t ( ix( ) 0 t , mA e 3 5 . 7-4t> Chapter 7, Solution 51. Consider the circuit below. fter the switch is closed, applying KVL gives R - di=LR i t = 0 + + v VS AdtL Ri VS+ =di|.|
\| =RVi R -dtdiLSor dtL R V iS tegrating both sides,IntLR -Ri lnI0 = |.
\ V) t ( i S | |=|.|
\| t -R V IR V ilnS 0S =t -S 0SeR V IR V ior |.|
\| + =t -S0SeRVIRV) t ( iwhich is the same as Eq. (7.60). hapter 7, Solution 52. C A 21020) 0 ( i = = ,A 2 ) ( i = | | + =t -e ) ( i ) 0 ( i ) ( i ) t ( i = ) t A 2 ( i hapter 7, Solution 53. (a)Before t = 0, C =+=2 325i A 5After t = 0, =t -e ) 0 ( i ) t ( iL224R= = = ,5 ) 0 ( i == ) t ( i A e 52 t - (b)Before t = 0, the inductor acts as a short circuit so that the 2 and 4 resistors are short-circuited. = ) t ( i A 6After t = 0, we have an RL circuit. = e ) 0 ( i ) t ( i ,23R = = t -L= ) t ( i A e 63 t 2 - Chapter 7, Solution 54. (a)Before t = 0, i is obtained by current division or =+= ) 2 (4 44) t ( i A 1After t = 0, | | + =t -e ) ( i ) 0 ( i ) ( i ) t ( ieqRL= , = + = 7 12 || 4 4 Req 2175 . 3= = 1 ) 0 ( i = , 76) 2 (3 43) 2 (12 || 4 412 || 4) ( i =+=+= t 2 -e76176) t ( i |.|
\| + == ) t ( i ( ) A e 6712t -(b)Before t = 0,=+=3 210) t ( i A 2After t = 0,5 . 4 2 || 6 3 Req= + =945 . 42RLeq= = = 2 ) 0 ( i =To find, consider the circuit below, at t= when the inductor becomes a short circuit, ) ( i v 2 24 V6 + + i 2 H10 V 3 9 v3v6v 242v 10= =+ A 33v) ( i = = 4 t 9 -e ) 3 2 ( 3 ) t ( i + == ) t ( i A e 34 t 9 -Chapter 7, Solution 55. For t < 0, consider the circuit shown in Fig. (a). +v 4io io + io 0.5 H + 0.5 H + v + i 3 8 2 2 24 V20 V(a)(b) 24 i 0 i 4 24 i 3o o o= = += =oi 4 ) t ( v V 96 A 482vi = = For t > 0, consider the circuit in Fig. (b). [ ] + =t -e ) ( i ) 0 ( i ) ( i ) t ( i48 ) 0 ( i = ,A 22 820) ( i =+= = + = 10 8 2 Rth, 201105 . 0RLth= = = -20t -20te 46 2 e ) 2 48 ( 2 ) t ( i + = + == = ) t ( i 2 ) t ( v V e 92 4-20t+ Chapter 7, Solution 56. = + = 10 5 || 20 6 Req,05 . 0RL= = [ ] + =t -e ) ( i ) 0 ( i ) ( i ) t ( i i(0) is found by applying nodal analysis to the following circuit. 0.5 H + 12 5 20 6 + v vx i 20 V 2 A 12 v6v20v12v5v 202xx x x x= + + =+A 26v) 0 ( ix= =4 5 || 20 =6 . 1 ) 4 (6 44) ( i =+= -20t 0.05 t -e 4 . 0 6 . 1 e ) 6 . 1 2 ( 6 . 1 ) t ( i + = + = Since, 20t -e -20) ( ) 4 . 0 (21dtdiL ) t ( v = == ) t ( v V e 4 --20t Chapter 7, Solution 57. At, the circuit has reached steady state so that the inductors act like short circuits. = 0 t + 6 i 5 i1i220 30 V 3103020 || 5 630i = =+= ,4 . 2 ) 3 (2520i1= = ,6 . 0 i2 =A 4 . 2 ) 0 ( i1= , A 6 . 0 ) 0 ( i2= For t > 0, the switch is closed so that the energies inLand flow through the closed switch and become dissipated in the 5 and 20 resistors. 1 2L1t -1 1e ) 0 ( i ) t ( i= , 2155 . 2RL111= = = = ) t ( i1A e 4 . 2-2t 2t -2 2e ) 0 ( i ) t ( i= , 51204RL222= = = = ) t ( i2A e 6 . 0-5t Chapter 7, Solution 58. For t < 0,0 ) t ( vo=For t > 0,,10 ) 0 ( i = 53 120) ( i =+= = + = 4 3 1 Rth, 16144 1RLth= = = [ ] + =t -e ) ( i ) 0 ( i ) ( i ) t ( i= ) t ( i( )A e 1 5-16t+ ( )16t - 16t -oe -16)(5) (41e 1 15dtdiL i 3 ) t ( v + + = + == ) t ( voV e 5 15-16t Chapter 7, Solution 59. Let I be the current through the inductor. For t < 0,, 0 vs = 0 ) 0 ( i =For t > 0,6 3 || 6 4 Req= + = ,25 . 065 . 1RLeq= = = 1 ) 3 (4 22) ( i =+= [ ] + =t -e ) ( i ) 0 ( i ) ( i ) t ( i-4te 1 ) t ( i = ) -4)(-e )( 5 . 1 (dtdiL ) t ( v4t -o= == ) t ( voV e 6-4t Chapter 7, Solution 60. Let I be the inductor current. For t < 0,0 ) 0 ( i 0 ) t ( u = =For t > 0, = = 4 20 || 5 Req,248RLeq= = = 4 ) ( i = | | + =t -e ) ( i ) 0 ( i ) ( i ) t ( i( )2 t -e 1 4 ) t ( i = 2 t -e21 -) 4 - )( 8 (dtdiL ) t ( v |.|
\|= == ) t ( v V e 16-0.5t Chapter 7, Solution 61. The current source is transformed as shown below. 4 20u(-t) + 40u(t) + 0.5 H 8142 1RL= = = ,5 ) 0 ( i = ,10 ) ( i = | | + =t -e ) ( i ) 0 ( i ) ( i ) t ( i= ) t ( i A e 5 10-8t 8t -e ) 8 - )( 5 - (21dtdiL ) t ( v |.|
\|= == ) t ( v V e 20-8t Chapter 7, Solution 62. 16 || 32RLeq= = = For 0 < t < 1, so that0 ) 1 t ( u = 0 ) 0 ( i = , 61) ( i = ( )t -e 161) t ( i = For t > 1, ( )1054 . 0 e 161) 1 ( i1 -= =216131) ( i = + = 1) - -(te ) 5 . 0 1054 . 0 ( 5 . 0 ) t ( i + =1) - -(te 3946 . 0 5 . 0 ) t ( i =Thus, = ) t ( i( )> < < 1 t A e 3946 . 0 5 . 01 t 0 A e 161-1) (t -t - Chapter 7, Solution 63. For t < 0,,1 ) t - ( u = 2510) 0 ( i = = For t > 0,,0 -t) ( u = 0 ) ( i = = = 4 20 || 5 Rth, 8145 . 0RLth= = = | | + =t -e ) ( i ) 0 ( i ) ( i ) t ( i= ) t ( i A e 2-8t 8t -e ) 2 )( 8 - (21dtdiL ) t ( v |.|
\|= == ) t ( v V e 8 --8t Chapter 7, Solution 64. Let i be the inductor current. For t < 0, the inductor acts like a short circuit and the 3 resistor is short-circuited so that the equivalent circuit is shown in Fig. (a). v 6 (b)i 3 2 + 10 (a) i 3 6 + io10 A 667 . 1610) 0 ( i i = = =For t > 0, = + = 4 6 || 3 2 Rth, 144RLth= = = To find, consider the circuit in Fig. (b).) ( i 610v2v3v6v 10= + = 652v) ( i i = = =| | + =t -e ) ( i ) 0 ( i ) ( i ) t ( i( )A e 165e6561065) t ( it - t - = |.|
\| + = ovis the voltage across the 4 H inductor and the 2 resistor t - t - t -oe610610e -1) (65) 4 ( e610610dtdiL i 2 ) t ( v = |.|
\|+ + = + == ) t ( vo( ) V e 1 667 . 1-t Chapter 7, Solution 65. Since | |) 1 t ( u ) t ( u 10 vs = , this is the same as saying that a 10 V source is turned on at t = 0 and a -10 V source is turned on later at t = 1.This is shown in the figure below. For 0 < t < 1,,0 ) 0 ( i = 2510) ( i = = vs -10 10 1 t 4 20 || 5 Rth= = , 2142RLth= = = | | + =t -e ) ( i ) 0 ( i ) ( i ) t ( i( )A e 1 2 ) t ( i-2t =( )729 . 1 e 1 2 ) 1 ( i-2= = For t > 1,since0 v 0 ) ( i = s = =) 1 t ( -e ) 1 ( i ) t ( iA e 729 . 1 ) t ( i) 1 t ( -2 = Thus, = ) t ( i( )> < < 1 t A e 729 . 11 t 0 A e 1 2) 1 t ( 2 -2t - Chapter 7, Solution 66. Following Practice Problem 7.14, =t -Te V ) t ( v-4 ) 0 ( v VT= = , 501) 10 2 )( 10 10 ( C R6 - 3f= = = -50te -4 ) t ( v =0 t , e 4 ) t ( v - ) t ( v-50to> = = == =50t -3oooe10 104R) t ( v) t ( i 0 t , mA e 4 . 0-50t> Chapter 7, Solution 67. The op amp is a voltage follower so thatv vo =as shown below. R vovo++voC v1 R R At node 1, o 1o 1 1 1 ov32vRv vR0 vRv v= += At the noninverting terminal, 0Rv vdtdvC1 o o=+o o o 1 oov31v32v v vdtdvRC = = = RC 3vdtdvo o =3RC t -T oe V ) t ( v =V 5 ) 0 ( v Vo T= = , 1003) 10 1 )( 10 10 )( 3 ( RC 36 - 3= = = = ) t ( voV e 53 100t - Chapter 7, Solution 68. This is a very interesting problem and has both an important ideal solution as well as an important practical solution.Let us look at the ideal solution first.Just before the switch closes, the value of the voltage across the capacitor is zero which means that the voltage at both terminals input of the op amp are each zero.As soon as the switch closes, the output tries to go to a voltage such that the input to the op amp both go to 4 volts.The ideal op amp puts out whatever current is necessary to reach this condition.An infinite (impulse) current is necessary if the voltage across the capacitor is to go to 8 volts in zero time (8 volts across the capacitor will result in 4 volts appearing at the negative terminal of the op amp).So vo will be equal to 8 volts for all t > 0. What happens in a realcircuit?Essentially, the output of the amplifier portion of the op amp goes to whatever its maximum value can be.Then this maximum voltage appears across the output resistance of the op amp and the capacitor that is in series with it.This results in an exponential rise in the capacitor voltage to the steady-state value of 8 volts. vC(t)=Vop amp max(1 e-t/(RoutC)) volts, for all values of vC less than 8 V, =8 Vwhen t is large enough so that the 8 V is reached. Chapter 7, Solution 69. Letvbe the capacitor voltage. xFor t < 0,0 ) 0 ( vx= For t > 0, the 20 k and 100 k resistors are in series since no current enters the op amp terminals.As t , the capacitor acts like an open circuit so that1348) 4 (10 100 20100 20) ( vx=+ + += = + = k 120 100 20 Rth,3000 ) 10 25 )( 10 120 ( C R3 - 3th= = = | | + =t -x x x xe ) ( v ) 0 ( v ) ( v ) t ( v( )3000 t -xe 11348) t ( v = = = ) t ( v120100) t ( vx o( ) V e 113403000 t - Chapter 7, Solution 70. Let v = capacitor voltage. For t < 0, the switch is open and0 ) 0 ( v = . For t > 0, the switch is closed and the circuit becomes as shown below. 2 1 C R + v + vo+vS s 2 1v v v = = (1) dtdvCRv 0s= (2) where(3)v v v v v vs o o s = = From (1), 0RCvdtdvs= = = + =RCv t -) 0 ( v dt vRC1 -vss Since v is constant, 1 . 0 ) 10 5 )( 10 20 ( RC-6 3= =mV t -200 mV0.1t 20 -v = = From (3), t 200 20 v v vs o+ = ==ov mV ) t 10 1 ( 20 + Chapter 7, Solution 71. Let v = voltage across the capacitor. Letv = voltage across the 8 k resistor. o For t < 2, so that0 v = 0 ) 2 ( v = . For t > 2, we have the circuit shown below. +vo 10 k + 20 k+v 100 mF10 k+io 4 V 8 k Since no current enters the op amp, the input circuit forms an RC circuit. 1000 ) 10 100 )( 10 10 ( RC3 - 3= = = | | + =) 2 t ( -e ) ( v ) 2 ( v ) ( v ) t ( v( )1000 ) 2 t ( -e 1 4 ) t ( v = As an inverter, ( )1 e 2 vk 20k 10 -v1000 ) 2 t ( -o = = = =8vioo( ) A 1 e 25 . 01000 ) 2 t ( - Chapter 7, Solution 72. The op amp acts as an emitter follower so that the Thevenin equivalent circuit is shown below. C + v + io3u(t)R Hence, [ ] + =t -e ) ( v ) 0 ( v ) ( v ) t ( vV 2 - ) 0 ( v = ,V 3 ) ( v = ,1 . 0 ) 10 10 )( 10 10 ( RC-6 3= = = -10t -10te 5 3 e 3) - -2 ( 3 ) t ( v = + = 10t - 6 -oe ) 10 - )( 5 - )( 10 10 (dtdvC i = ==oi 0 t , mA e 5 . 0-10t> Chapter 7, Solution 73. Consider the circuit below. Rf +vo + v + R1 +C v2v1v3v1 At node 2, dtdvCRv v12 1=(1) At node 3, fo 3Rv vdtdvC= (2) But and0 v3 =2 3 2v v v v = = .Hence, (1) becomes dtdvCRv v11= dtdvC R v v1 1= or C RvC Rvdtdv111= +which is similar to Eq. (7.42).Hence, ( )> + ==ov V ) t ( u e 6 --5t Chapter 7, Solution 74. Let v = capacitor voltage. Rf +vo + v + R1 +C v2v1v3v1 For t < 0,0 ) 0 ( v =For t > 0,.Consider the circuit below.A 10 is = RvdtdvC is+ =(1) [ ] + =t -e ) ( v ) 0 ( v ) ( v ) t ( v (2) It is evident from the circuit that 1 . 0 ) 10 50 )( 10 2 ( RC3 6= = = is+vo C Rf +R is At steady state, the capacitor acts like an open circuit so thatipasses through R.Hence, sV 5 . 0 ) 10 50 )( 10 10 ( R i ) ( v3 6s= = = Then, ( )V e 1 5 . 0 ) t ( v-10t = (3) But f s ofosR i - vRv 0i = = (4) Combining (1), (3), and (4), we obtain dtdvC R vRR -vffo =dtdv) 10 2 )( 10 10 ( v51 -v6 - 3o =( )-10t -2 -10toe 10 - ) 5 . 0 )( 10 2 ( e 0.1 -0.1 v + =1 . 0 e 2 . 0 v-10to ==ov ( ) V 1 e 2 1 . 0-10t Chapter 7, Solution 75. Letv= voltage at the noninverting terminal. 1Let 2v= voltage at the inverting terminal. For t > 0,4 v v vs 2 1= = =o1siRv 0=, = k 20 R1 R i - vo o = (1) Also, dtdvCRvi2o+ = , = k 10 R2,F 2 C =i.e. dtdvCRvRv -2 1s+ = (2) This is a step response. [ ] + =t -e ) ( v ) 0 ( v ) ( v ) t ( v ,1 ) 0 ( v =where 501) 10 2 )( 10 10 ( C R6 - 32= = = At steady state, the capacitor acts like an open circuit so thatipasses through .Hence, as o2R t2o1sR) ( viRv - = =i.e.-2 ) 4 (2010 -vRR -) ( vs12= = = -50te 2) (1 -2 ) t ( v + + =-50te 3 -2 ) t ( v + = But o sv v v =or t 50 -s oe 3 2 4 v v v + = ==ov V e 3 6t 50 - mA -0.220k4 -Rv -i1so= = =or= + =dtdvCRvi2omA 0.2 - Chapter 7, Solution 76. The schematic is shown below.For the pulse, we use IPWL and enter the corresponding values as attributes as shown.By selecting Analysis/Setup/Transient, we let Print Step = 25 ms and Final Step = 2 s since the width of the input pulse is 1 s.After saving and simulating the circuit, we select Trace/Add and display V(C1:2).The plot of V(t) is shown below. Chapter 7, Solution 77. The schematic is shown below.We click Marker and insert Mark Voltage Differential at the terminals of the capacitor to display V after simulation.The plot of V is shown below.Note from the plot that V(0) = 12 V and V() = -24 V which are correct. Chapter 7, Solution 78. (a)When the switch is in position (a), the schematic is shown below.We insert IPROBE to display i.After simulation, we obtain, i(0)=7.714 A from the display of IPROBE. (b)When the switch is in position (b), the schematic is as shown below.For inductor I1, we let IC = 7.714.By clicking Analysis/Setup/Transient, we let Print Step = 25 ms and Final Step = 2 s.After Simulation, we click Trace/Add in the probe menu and display I(L1) as shown below.Note that i() = 12A, which is correct. Chapter 7, Solution 79. When the switch is in position 1,io(0) = 12/3 = 4A.When the switch is in position 2, 3 / 80 , 3 / 8 4 // ) 5 3 ( A, 5 . 03 54) ( = = = + = =+ = LRR iThTh o A 5 . 4 5 . 0 )] ( ) 0 ( [ ) ( ) (80 / 3 / t to o o oe e i i i t i + = + = Chapter 7, Solution 80. (a)When the switch is in position A, the 5-ohm and 6-ohm resistors are short-circuited so that 0 ) 0 ( ) 0 ( ) 0 (2 1= = =ov i i but the current through the 4-H inductor is iL(0) =30/10 = 3A. (b) When the switch is in position B, 5 . 0 4 / 2 , 2 6 // 3 = = = = =LRRThTh A 3 3 0 )] ( ) 0 ( [ ) ( ) (2 5 . 0 / / t t tL L L Le e e i i i t i = + = + = (c) A 0 ) (93) ( , A 25 1030) (2 1= = =+= Li i i V 0 ) ( ) ( = =oLovdtdiL t v Chapter 7, Solution 81. The schematic is shown below.We use VPWL for the pulse and specify the attributes as shown.In the Analysis/Setup/Transient menu, we select Print Step = 25 ms and final Step = 3 S.By inserting a current marker at one termial of LI, we automatically obtain the plot of i after simulation as shown below. hapter 7, Solution 82. C === = 6 --310 10010 3CR RC 30 Chapter 7, Solution 83. s x x x RC v v 510 10 15 10 34 , 0 ) 0 ( , 120 ) (6 6= = = = = ) 1 ( 120 6 . 85 )] ( ) 0 ( [ ) ( ) (510 / / t te e v v v t v = + = Solving for t gives s t 16 . 637 488 . 3 ln 510 = = speed = 4000m/637.16s = 6.278m/s Chapter 7, Solution 84. LetIo be the final value of the current.Then 50 / 1 8 / 16 . 0 / ), 1 ( ) (/= = = =L R e I t ito .ms 33 . 184 . 01ln501) 1 ( 6 . 050= = =t e I Ito o Chapter 7, Solution 85. (a)s 24 ) 10 6 )( 10 4 ( RC-6 6= = = Since | | + =t -e ) ( v ) 0 ( v ) ( v ) t ( v| | = 1t -1e ) ( v ) 0 ( v ) ( v ) t ( v (1) | | = 2t -2e ) ( v ) 0 ( v ) ( v ) t ( v (2) Dividing (1) by (2), = ) t t (211 2e) ( v ) t ( v) ( v ) t ( v |.|
\| = =) ( v ) t ( v) ( v ) t ( vln t t t211 2 0 = = |.|
\|= ) 2 ( ln 24120 30120 75ln 24 t0s 63 . 16(b)Since, the light flashes repeatedly everyt t0 0, we have an RL circuit 1 . 0400 10050RL=+= = ,0 ) ( i = | | + =t -e ) ( i ) 0 ( i ) ( i ) t ( i-10te 2 . 1 ) t ( i = At t = 100 ms = 0.1 s, = =-1e 2 . 1 ) 1 . 0 ( i A 441 . 0which is the same as the current through the resistor. Chapter 7, Solution 88. (a)s 60 ) 10 200 )( 10 300 ( RC-12 3u = = = As a differentiator, ms 6 . 0 s 600 10 T = u = >i.e.=minT ms 6 . 0(b)s 60 RC u = = As an integrator, s 6 1 . 0 T u = = = = = = = = + = + == = = = = ( )t 268 t 32 . 372 12 1 2 1t 32 . 372t 68 . 212o2e e 928 . 6 ) t ( iA 928 . 6 A to leads This240 A 32 . 37 A 68 . 2dt) 0 ( di, A A 0 ) 0 ( ie A e A ) t ( i32 . 37 , 68 . 2 3 10 20 300 20 s = = = = = + = =+ = = = = = get we , 60 dt ) t ( iC1) t ( v , Sincet0+= v(t) = (60 + 64.53e-2.68t 4.6412e-37.32t) V Chapter 8, Solution 18. When the switch is off, we have a source-free parallel RLC circuit. 5 . 021, 21 25 . 01 1= = = = =RCx LCo 936 . 1 25 . 0 4 cased underdampe2 2d= = = < o o Io(0) = i(0) = initial inductor current = 20/5 = 4A Vo(0) = v(0) = initial capacitor voltage = 0 V ) 936 . 1 sin 936 . 1 cos ( ) sin cos ( ) (2 15 . 02 1t A t A e t A t A e t vtd dt+ = + = v(0) =0 = A1 ) 936 . 1 cos 936 . 1 936 . 1 sin 936 . 1 ( ) 936 . 1 sin 936 . 1 cos )( 5 . 0 (2 15 . 02 15 . 0t A t A e t A t A edtdvt t+ + + = 066 . 2 936 . 1 5 . 0 41) 4 0 ( ) ( ) 0 (2 2 1 = + = =+ =+ = A A ARCRI Vdtdvo o Thus, t e t vt936 . 1 sin 066 . 2 ) (5 . 0 = Chapter 8, Solution 19. Fort0,we have a series RLC circuit as shown in Figure (b)with R=0=. o= LC1=41=0.5=d i(t)=[Acos0.5t + Bsin0.5t],i(0)=12=A v=-Ldi/dt,and-v/L=di/dt=0.5[-12sin0.5t + Bcos0.5t], which leads to-v(0)/L=0=B Hence, i(t)=12cos0.5t Aandv=0.5 However,v=-Ldi/dt=-4(0.5)[-12sin0.5t]=24sin0.5tV Chapter 8, Solution 20. Fort0,we have a series RLC circuit. =R/(2L)=2/(2x0.5)=2 o = 1/ 2 2 4 1 x 5 . 0 / 1 LC = = Since is less than o, we have an under-damped response. 2 4 82 2o d= = = i(t) =(Acos2t + Bsin2t)e-2t i(0) = 6 = A di/dt =-2(6cos2t + Bsin2t)e-2t + (-2x6sin2t + 2Bcos2t)e-t di(0)/dt=-12 + 2B=-(1/L)[Ri(0) + vC(0)]=-2[12 12]=0 Thus, B=6andi(t) =(6cos2t + 6sin2t)e-2t A Chapter 8, Solution 21. By combining some resistors, the circuit is equivalent to that shown below. 60||(15 + 25)=24 ohms. 12 + + v t = 0 i 24 6 3 H 24V (1/27)F Att=0-,i(0)=0,v(0)=24x24/36=16V Fort>0,we have a series RLC circuit.R=30 ohms,L=3 H,C=(1/27) F =R/(2L)=30/6=5 27 / 1 x 3 / 1 LC / 1o= = =3,clearly > o(overdamped response) s1,2=2 2 2o23 5 5 = =-9, -1 v(t)=[Ae-t+ Be-9t],v(0)=16=A + B(1) i=Cdv/dt=C[-Ae-t- 9Be-9t] i(0)=0=C[-A 9B]orA=-9B(2) From (1) and (2),B=-2andA=18. Hence, v(t)=(18e-t 2e-9t) V Chapter 8, Solution 22. =20=1/(2RC)orRC=1/40(1) 2 2o d50 = = which leads to 2500 + 400=o2=1/(LC) Thus,LC1/2900(2) In a parallel circuit,vC=vL=vR But,iC=CdvC/dtoriC/C=dvC/dt =-80e-20tcos50t 200e-20tsin50t + 200e-20tsin50t 500e-20tcos50t =-580e-20tcos50t iC(0)/C=-580whichleads to C=-6.5x10-3/(-580)=11.21 F R=1/(40C)=106/(2900x11.21)=2.23 kohms L=1/(2900x11.21)=30.76 H Chapter 8, Solution 23. Let Co=C + 0.01.For a parallel RLC circuit, =1/(2RCo),o=1/oLC =1=1/(2RCo),we then have Co=1/(2R)=1/20 =50 mF o=1/ 5 . 0 x 5 . 0 =6.32>(underdamped) Co=C + 10 mF=50 mFor40 mF Chapter 8, Solution 24. For t < 0,u(-t)1,namely, the switch is on. v(0)=0,i(0)=25/5=5A For t > 0,the voltage source is off and we have a source-free parallel RLC circuit. =1/(2RC)=1/(2x5x10-3)=100 o=1/310 x 1 . 0 / 1 LC= =100 o=(critically damped) v(t)=[(A1 + A2t)e-100t] v(0)=0=A1 dv(0)/dt=-[v(0) + Ri(0)]/(RC)=-[0 + 5x5]/(5x10-3)=-5000 But, dv/dt=[(A2 + (-100)A2t)e-100t] Therefore, dv(0)/dt=-5000=A2 0 v(t)=-5000te-100tV Chapter 8, Solution 25. In the circuit in Fig. 8.76, calculate io(t) and vo(t) for t>0. (1/4)F + 8 2 t=0, note this is a make before break switch so the inductor current is not interrupted. 1 Hio(t) + vo(t) 30V Figure 8.78For Problem 8.25. At t = 0-, vo(0) = (8/(2 + 8)(30) = 24 For t > 0, we have a source-free parallel RLC circuit. =1/(2RC)= o = 1/ 2 4 1 x 1 / 1 LC = = Since is less than o, we have an under-damped response. 9843 . 1 ) 16 / 1 ( 42 2o d= = = vo(t) =(A1cosdt + A2sindt)e-t vo(0) = 24 = A1 and io(t) = C(dvo/dt) = 0 when t = 0. dvo/dt =-(A1cosdt + A2sindt)e-t + (-dA1sindt + dA2cosdt)e-t at t = 0, we get dvo(0)/dt = 0 = -A1 + dA2 Thus, A2 = (/d)A1 =(1/4)(24)/1.9843=3.024 vo(t) =(24cosdt + 3.024sindt)e-t/4 volts Chapter 8, Solution 26. s2 + 2s + 5=0,which leads tos1,2=220 4 2 =-1j4 i(t)=Is + [(A1cos4t + A2sin4t)e-t],Is=10/5=2 i(0)=2==2 + A1,orA1=0 di/dt=[(A2cos4t)e-t] + [(-A2sin4t)e-t]=4=4A2,or A2=1 i(t)=2 + sin4te-t A Chapter 8, Solution 27. s2 + 4s + 8=0 leads to s= 2 j 2232 16 4 = v(t)=Vs + (A1cos2t + A2sin2t)e-2t 8Vs = 24 means that Vs = 3 v(0) = 0 = 3 + A1 leads to A1 = -3 dv/dt = -2(A1cos2t + A2sin2t)e-2t + (-2A1sin2t + 2A2cos2t)e-2t 0 = dv(0)/dt = -2A1 +2A2 or A2 = A1 = -3 v(t) = [3 3(cos2t + sin2t)e-2t] volts Chapter 8, Solution 28. The characteristic equation is s2 + 6s + 8with roots2 , 4232 36 62 , 1 = = sHence, t tsBe Ae I t i4 2) ( + + = 5 . 1 12 8 = =s sI I B A i + + = = 5 . 1 0 0 ) 0 ( (1) t tBe Aedtdi4 24 2 =B A B Adtdi2 1 0 4 2 2) 0 (+ + = = = (2) Solving (1) and (2) leads to A=-2 and B=0.5. t te e t i4 25 . 0 2 5 . 1 ) ( + =A Chapter 8, Solution 29. (a)s2 + 4=0which leads tos1,2=j2(an undamped circuit) v(t)=Vs+Acos2t + Bsin2t 4Vs=12orVs=3 v(0)=0=3 + AorA=-3 dv/dt=-2Asin2t + 2Bcos2t dv(0)/dt=2=2BorB=1,therefore v(t)=(3 3cos2t + sin2t) V (b)s2 + 5s + 4=0which leads tos1,2=-1, -4 i(t)=(Is + Ae-t + Be-4t) 4Is=8orIs=2 i(0)=-1=2 + A + B,orA + B=-3(1) di/dt=-Ae-t-4Be-4t di(0)/dt=0=-A 4B,orB=-A/4(2) From (1) and (2)we getA=-4and B=1 i(t)=(2 4e-t + e-4t) A (c)s2 + 2s + 1=0,s1,2=-1, -1 v(t)=[Vs + (A + Bt)e-t],Vs=3. v(0)=5=3 + Aor A=2 dv/dt=[-(A + Bt)e-t] + [Be-t] dv(0)/dt=-A + B=1or B=2 + 1=3 v(t)=[3 + (2 + 3t)e-t] V Chapter 8, Solution 30. 2 222 21800 , 500o os s = = + = = LRs s2650 2 13002 1= = = = + Hence, mH 8 . 153650 22002= = =xRL LCs so o145 . 623 2 3002 22 1= = = = F 25 . 16) 45 . 632 (12 = =LC Chapter 8, Solution 31. For t=0-,we havethe equivalent circuit in Figure (a).For t=0+, the equivalent circuit is shown in Figure (b).By KVL, v(0+)=v(0-)=40,i(0+)=i(0-)=1 ByKCL,2=i(0+) + i1=1 + i1which leads toi1=1.By KVL,-vL + 40i1 + v(0+)=0which leads tovL(0+)=40x1 + 40=80 vL(0+)=80 V, vC(0+)=40 V 40 10 i10.5H +v 50V + +vL 40 10 i + v 50V + (a)(b) Chapter 8, Solution 32. For t=0-, the equivalent circuit is shownbelow. 2 A i 6 + v i(0-)=0,v(0-)=-2x6=-12V For t > 0,we have a series RLC circuit with a step input. =R/(2L)=6/2=3,o=1/ 04 . 0 / 1 LC = s= 4 j 3 25 9 3 = Thus, v(t)=Vf + [(Acos4t + Bsin4t)e-3t] whereVf=final capacitor voltage=50 V v(t)=50 + [(Acos4t + Bsin4t)e-3t] v(0)=-12=50 + Awhich gives A=-62 i(0)=0=Cdv(0)/dt dv/dt=[-3(Acos4t + Bsin4t)e-3t] + [4(-Asin4t + Bcos4t)e-3t] 0=dv(0)/dt=-3A + 4BorB=(3/4)A=-46.5 v(t)={50 + [(-62cos4t 46.5sin4t)e-3t]} V Chapter 8, Solution 33. We may transform the current sources to voltage sources.Fort=0-, the equivalent circuit is shown in Figure (a). 1 H i + 30V +v 4F i + 5 10 + v 10 30V (a)(b) i(0)=30/15=2 A,v(0)=5x30/15=10 V For t > 0, we have a series RLC circuit. =R/(2L)=5/2=2.5 4 / 1 LC / 1o= = =0.25,clearly > o(overdamped response) s1,2= 25 . 0 25 . 6 5 . 22o2 = =-4.95, -0.05 v(t)=Vs + [A1e-4.95t+ A2e-0.05t],v=20. v(0)=10=20 + A1 + A2(1) i(0)=Cdv(0)/dtordv(0)/dt=2/4=1/2 Hence, =-4.95A1 0.05A2(2) From (1) and (2),A1=0,A2=-10. v(t)={20 10e-0.05t} V Chapter 8, Solution 34. Before t = 0, the capacitor acts like an open circuit while the inductor behaves like a short circuit. i(0) = 0, v(0) = 20 V For t > 0, the LC circuit is disconnected from the voltage source as shown below. + Vx (1/16)F() H i This is a lossless, source-free, series RLC circuit. =R/(2L)=0, o = 1/ LC=1/41161+=8,s =j8
Since is less than o, we have an underdamped response.Therefore, i(t) = A1cos8t + A2sin8twhere i(0) = 0 = A1 di(0)/dt = (1/L)vL(0) = -(1/L)v(0) = -4x20 = -80 However,di/dt = 8A2cos8t, thus, di(0)/dt = -80 = 8A2 which leads to A2 = -10 Now we havei(t)=-10sin8t A Chapter 8, Solution 35. At t=0-,iL(0)=0,v(0)=vC(0)=8 V For t > 0, we have a series RLC circuit with a step input. =R/(2L)=2/2=1, o = 1/ LC=1/ 5 / 1= 5 s1,2= 2 j 12o2 = v(t)=Vs + [(Acos2t + Bsin2t)e-t],Vs=12. v(0)=8=12 + Aor A=-4,i(0)=Cdv(0)/dt=0. But dv/dt=[-(Acos2t + Bsin2t)e-t] + [2(-Asin2t + Bcos2t)e-t] 0=dv(0)/dt=-A + 2Bor2B=A=-4 and B=-2 v(t)={12 (4cos2t+ 2sin2t)e-t V. Chapter 8, Solution 36. For t=0-, 3u(t)=0.Thus, i(0)=0,and v(0)=20 V. For t > 0, we have the series RLC circuit shown below. 20 V 2 0.2 F i 10 + + 5 H10 + v 15V =R/(2L)=(2 + 5 + 1)/(2x5)=0.8 o = 1/ LC=1/ 2 . 0 x 5=1 s1,2= 6 . 0 j 8 . 02o2 = v(t)=Vs + [(Acos0.6t + Bsin0.6t)e-0.8t] Vs=15 + 20=35Vandv(0)=20=35 + Aor A=-15 i(0)=Cdv(0)/dt=0 But dv/dt=[-0.8(Acos0.6t + Bsin0.6t)e-0.8t] + [0.6(-Asin0.6t + Bcos0.6t)e-0.8t] 0=dv(0)/dt=-0.8A + 0.6Bwhich leads to B=0.8x(-15)/0.6=-20 v(t)={35 [(15cos0.6t + 20sin0.6t)e-0.8t]}V i=Cdv/dt=0.2{[0.8(15cos0.6t + 20sin0.6t)e-0.8t] + [0.6(15sin0.6t 20cos0.6t)e-0.8t]} i(t)=[(5sin0.6t)e-0.8t] A Chapter 8, Solution 37. Fort=0-,the equivalent circuit is shown below. 6 + 10V + i2i1+ v(0) 6 6 30V 18i2 6i1=0ori1=3i2(1) -30 + 6(i1 i2) + 10=0ori1 i2 =10/3(2) From(1)and(2).i1=5,i2=5/3 i(0)=i1=5A -10 6i2 + v(0)=0 v(0)=10 + 6x5/3=20 Fort > 0,we have a series RLC circuit. R=6||12=4 o = 1/ LC=1/ ) 8 / 1 )( 2 / 1 (=4 =R/(2L)=(4)/(2x(1/2))=4 =o,therefore the circuit is critically damped v(t)=Vs +[(A + Bt)e-4t],and Vs=10
v(0)=20=10 + A,orA=10 i=Cdv/dt=-4C[(A + Bt)e-4t] + C[(B)e-4t] i(0)=5=C(-4A + B)which leads to40=-40 + BorB=80 i(t)=[-(1/2)(10 + 80t)e-4t] + [(10)e-4t] i(t)=[(5 40t)e-4t] A Chapter 8, Solution 38. Att=0-, the equivalent circuit is as shown. 2 A 10 i i15 + v 10 i(0)=2A,i1(0)=10(2)/(10 + 15)=0.8 A v(0)=5i1(0)=4V Fort>0,we have a source-free series RLC circuit. R=5||(10 + 10)=4 ohms o = 1/ LC=1/ ) 4 / 3 )( 3 / 1 (=2 =R/(2L)=(4)/(2x(3/4))=8/3 s1,2= = 2o2-4.431,-0.903 i(t)=[Ae-4.431t + Be-0.903t] i(0)=A + B=2(1) di(0)/dt=(1/L)[-Ri(0) + v(0)]=(4/3)(-4x2 + 4)=-16/3=-5.333 Hence,-5.333=-4.431A 0.903B(2) From (1) and (2),A=1and B=1. i(t) =[e-4.431t + e-0.903t] A Chapter 8, Solution 39. Fort=0-, the equivalent circuit is shown in Figure (a).Where 60u(-t)=60and30u(t)=0. + 30V 20 + + v 20 30 0.5F0.25H30 60V (a)(b) v(0)=(20/50)(60)=24and i(0)=0 Fort>0,the circuit is shown in Figure (b). R=20||30=12 ohms o = 1/ LC=1/ ) 4 / 1 )( 2 / 1 (= 8 =R/(2L)=(12)/(0.5)=24 Since >o, we have an overdamped response. s1,2= = 2o2-47.833,-0.167 Thus,v(t)=Vs + [Ae-47.833t + Be-0.167t], Vs=30 v(0)=24=30 + A + Bor-6=A + B(1) i(0)= Cdv(0)/dt=0 But,dv(0)/dt=-47.833A 0.167B=0 B=-286.43A(2) From (1) and (2),A=0.021andB=-6.021 v(t)=30 + [0.021e-47.833t 6.021e-0.167t] V Chapter 8, Solution 40. Att=0-,vC(0)=0andiL(0)=i(0)=(6/(6 + 2))4=3A Fort>0, we have a series RLC circuit with a step input as shown below. + v 6 + + 12V24V i 14 0.02 F 2 H o = 1/ LC=1/ 02 . 0 x 2=5 =R/(2L)=(6 + 14)/(2x2)=5 Since =o, we have a critically damped response. v(t)=Vs + [(A + Bt)e-5t], Vs=24 12=12V v(0)=0=12 + AorA=-12 i=Cdv/dt=C{[Be-5t] + [-5(A + Bt)e-5t]} i(0)=3=C[-5A + B]=0.02[60 + B]or B=90 Thus,i(t)=0.02{[90e-5t] + [-5(-12 + 90t)e-5t]} i(t)={(3 9t)e-5t} A Chapter 8, Solution 41. At t=0-, the switch is open.i(0)=0,and v(0)=5x100/(20 + 5 + 5)=50/3 For t>0, we have a series RLC circuit shown in Figure (a).After source transformation, it becomes that shown in Figure (b). 1 H + 20V+ v i 4 20 10 F10 H 5A (a) 5 0.04F (b) o = 1/ LC=1/ 25 / 1 x 1=5 =R/(2L)=(4)/(2x1)=2 s1,2= = 2o2-2 j4.583 Thus,v(t)=Vs + [(Acosdt + Bsindt)e-2t], whered=4.583and Vs=20 v(0)=50/3=20 + AorA=-10/3 i(t)=Cdv/dt=C(-2) [(Acosdt + Bsindt)e-2t] + Cd[(-Asindt + Bcosdt)e-2t] i(0)=0=-2A+ dB B=2A/d=-20/(3x4.583)=-1.455 i(t)=C{[(0cosdt + (-2B - dA)sindt)]e-2t} =(1/25){[(2.91 + 15.2767) sindt)]e-2t} i(t)= {0.7275sin(4.583t)e-2t} A Chapter 8, Solution 42. Fort=0-, we have the equivalent circuit as shown in Figure (a). i(0)=i(0)=0,andv(0)=4 12=-8V + v 6 + 12V1 H i 4V + + + v(0) 12V 1 5 0.04F (a)(b)Fort>0, the circuit becomes that shown in Figure (b) after source transformation. o = 1/ LC=1/ 25 / 1 x 1=5 =R/(2L)=(6)/(2)=3 s1,2= = 2o2-3 j4 Thus,v(t)=Vs + [(Acos4t + Bsin4t)e-3t], Vs=-12 v(0)= -8=-12 + AorA=4 i=Cdv/dt,ori/C=dv/dt=[-3(Acos4t + Bsin4t)e-3t] + [4(-Asin4t + Bcos4t)e-3t] i(0)=-3A + 4BorB=3 v(t)={-12 + [(4cos4t + 3sin4t)e-3t]} A Chapter 8, Solution 43. Fort>0, we have a source-free series RLC circuit. = = = = 8 5 . 0 8 2 22x x L RLR 836 64 900 302 2= = = =o o d mF 392 . 25 . 0 8361 1 12= = = =x LCLC oo Chapter 8, Solution 44. 491010 1001 1, 5001 210002= = = = = =xLCx LRo > ounderdamped. Chapter 8, Solution 45. o = 1/ LC=1/ 5 . 0 x 1= 2 =R/(2L)=(1)/(2x2x0.5)=0.5 Since 0,we have a parallel RLC circuit with a step input, as shown below. 5F8mH +v i 2 k 6mA =1/(2RC)=(1)/(2x2x103 x5x10-6)=50 o = 1/ LC=1/6 310 x 5 x 10 x 8 =5,000 Since 0,the 20-ohm resistor is short-circuited and we have a parallel RLC circuit with a step input. =1/(2RC)=(1)/(2x5x0.01)=10 o = 1/ LC=1/ 01 . 0 x 1=10 Since =o, we have a critically damped response. s1,2=-10 Thus,i(t)=Is + [(A + Bt)e-10t], Is=3 i(0)=1=3 + AorA=-2 vo=Ldi/dt= [Be-10t] + [-10(A + Bt)e-10t] vo(0)=0=B 10AorB=-20 Thus,vo(t)= (200te-10t) V Chapter 8, Solution 48. Fort=0-,we obtaini(0)=-6/(1 + 2)=-2andv(0)=2x1=2. Fort>0,the voltage is short-circuited and we have a source-free parallel RLC circuit. =1/(2RC)=(1)/(2x1x0.25)=2 o = 1/ LC=1/ 25 . 0 x 1=2 Since =o, we have a critically damped response. s1,2=-2 Thus,i(t)=[(A + Bt)e-2t], i(0)=-2=A v=Ldi/dt= [Be-2t] + [-2(-2 + Bt)e-2t] vo(0)=2=B + 4orB=-2 Thus,i(t)= [(-2 - 2t)e-2t] A andv(t)=[(2 + 4t)e-2t] V Chapter 8, Solution 49. Fort=0-,i(0)=3 + 12/4=6andv(0)=0. Fort > 0,we have a parallel RLC circuit with a step input. =1/(2RC)=(1)/(2x5x0.05)=2 o = 1/ LC=1/ 05 . 0 x 5=2 Since =o, we have a critically damped response. s1,2=-2 Thus,i(t)=Is + [(A + Bt)e-2t], Is=3 i(0)=6=3 + AorA=3 v=Ldi/dtorv/L=di/dt= [Be-2t] + [-2(A + Bt)e-2t] v(0)/L= 0=di(0)/dt=B 2x3orB=6 Thus,i(t)= {3 + [(3 + 6t)e-2t]} A Chapter 8, Solution 50. Fort=0-,4u(t)=0,v(0)=0,and i(0)=30/10=3A. Fort>0,we have a parallel RLC circuit. i 40 6A3A 10 mF+v 10 10 H Is=3 + 6=9AandR=10||40=8 ohms =1/(2RC)=(1)/(2x8x0.01)=25/4=6.25 o = 1/ LC=1/ 01 . 0 x 4=5 Since >o, we have a overdamped response. s1,2= = 2o2-10, -2.5 Thus,i(t)=Is + [Ae-10t] + [Be-2.5t], Is=9 i(0)=3=9 + A + BorA + B=-6 di/dt=[-10Ae-10t] + [-2.5Be-2.5t], v(0)= 0=Ldi(0)/dtordi(0)/dt=0=-10A 2.5BorB=-4A Thus,A=2andB=-8 Clearly,i(t)= { 9 + [2e-10t] + [-8e-2.5t]} A Chapter 8, Solution 51. Leti=inductor current and v=capacitor voltage. Att=0,v(0)=0andi(0)=io. Fort>0,we have a parallel, source-free LC circuit(R=). =1/(2RC)=0ando=1/ LC which leads to s1,2= jo v=Acosot + Bsinot,v(0)=0A iC=Cdv/dt=-i dv/dt=oBsinot=-i/C dv(0)/dt=oB=-io/CthereforeB=io/(oC) v(t)=-(io/(oC))sinot Vwhereo = LC Chapter 8, Solution 52. RC 21300 = = (1) LCo o d1575 . 264 300 400 4002 2 2 2= = = = = (2) From (2), F 71 . 28510 50 ) 575 . 264 (13 2 = =x xCFrom (1), = = = 833 . 5 ) 3500 (300 2121x CR Chapter 8, Solution 53. C1R2 + v1 i2i1 + C2+ vo R1 vS i2=C2dvo/dt(1) i1=C1dv1/dt(2) 0=R2i2 + R1(i2 i1) +vo(3) Substituting (1) and (2) into (3) we get, 0=R2C2dvo/dt + R1(C2dvo/dt C1dv1/dt)(4) Applying KVL to the outer loop produces, vs=v1 + i2R2 + vo=v1 + R2C2dvo/dt + vo,which leads to v1=vs vo R2C2dvo/dt(5) Substituting (5) into (4) leads to, 0=R1C2dvo/dt + R1C2dvo/dt R1C1(dvs/dt dvo/dt R2C2d2vo/dt2) Hence,(R1C1R2C2)(d2vo/dt2) + (R1C1 + R2C2 +R1C2)(dvo/dt)=R1C1(dvs/dt) Chapter 8, Solution 54. Let i be the inductor current. dtdv vi 5 . 04 + = (1) dtdii + = 2 v (2) Substituting (1) into (2) gives 0 3 5 . 2214122222= + + + + + = vdtdvdtv ddtv ddtdvdtdv vv 199 . 1 25 . 1 0 3 5 . 22j s s s = = + + t Be t Ae vt t199 . 1 sin 199 . 1 cos25 . 1 25 . 1 + = v(0) = 2=A.Letw=1.199 ) cos sin ( ) sin cos ( 25 . 125 . 1 25 . 1 25 . 1 25 . 1wt Be wt Ae w wt Be wt Aedtdvt t t t + + + = 085 . 2199 . 12 25 . 125 . 1 0) 0 (= = + = =XB Bw Adtdv V 199 . 1 sin 085 . 2 199 . 1 cos 225 . 1 25 . 1t e t e vt t + = Chapter 8, Solution 55. At the top node, writing a KCL equation produces, i/4 +i=C1dv/dt,C1=0.1 5i/4= C1dv/dt= 0.1dv/dt i=0.08dv/dt(1) But,v= ) idt ) C / 1 ( i 2 (2 + ,C2=0.5 or-dv/dt=2di/dt + 2i(2) Substituting (1) into (2) gives, -dv/dt=0.16d2v/dt2 + 0.16dv/dt 0.16d2v/dt2 + 0.16dv/dt+ dv/dt=0,ord2v/dt2 + 7.25dv/dt=0 Which leads to s2 + 7.25s=0=s(s + 7.25)ors1,2=0, -7.25 v(t)=A + Be-7.25t(3) v(0)=4=A + B(4) From (1),i(0)=2=0.08dv(0+)/dtordv(0+)/dt=25 But,dv/dt=-7.25Be-7.25t, which leads to, dv(0)/dt=-7.25B=25orB=-3.448andA=4 B=4 + 3.448=7.448 Thus,v(t)={7.45 3.45e-7.25t} V Chapter 8, Solution 56. Fort0,the circuit is as shown below. 4 ioi 6 + 0.04Fi 20 0.25H Applying KVL to the larger loop, -20 +6io +0.25dio/dt + 25+ dt ) i i (o=0 Taking the derivative, 6dio/dt + 0.25d2io/dt2 + 25(io + i)=0(1) For the smaller loop,4 + 25+ dt ) i i (o=0 Taking the derivative, 25(i + io)=0ori=-io(2) From (1) and (2)6dio/dt + 0.25d2io/dt2=0 This leads to,0.25s2 + 6s=0ors1,2=0, -24 io(t)=(A + Be-24t)and io(0)=0=A + BorB=-A As t approaches infinity, io()=20/10=2=A,thereforeB=-2 Thus,io(t)=(2 - 2e-24t)=-i(t)ori(t)=(-2 + 2e-24t) A Chapter 8, Solution 57. (a)Letv=capacitor voltage and i=inductor current.Att=0-, the switch is closed and the circuit has reached steady-state. v(0-)=16Vandi(0-)= 16/8=2A Att=0+, the switch is open but,v(0+)=16andi(0+)=2. We now have a source-free RLC circuit. R8 + 12=20 ohms,L=1H,C =4mF. =R/(2L)=(20)/(2x1)=10 o = 1/ LC=1/ ) 36 / 1 ( x 1=6 Since >o, we have a overdamped response. s1,2= = 2o2-18, -2 Thus, the characteristic equation is (s + 2)(s + 18)=0ors2 + 20s +36=0. (b)i(t)=[Ae-2t + Be-18t]andi(0)=2=A + B(1) To get di(0)/dt, consider the circuit below att=0+. +vL 12 (1/36)F+ v i 8 1 H -v(0) + 20i(0) + vL(0)=0, which leads to, -16 + 20x2 + vL(0)=0orvL(0)=-24 Ldi(0)/dt=vL(0)which givesdi(0)/dt=vL(0)/L=-24/1=-24 A/s Hence-24=-2A 18Bor12=A + 9B(2) From (1) and (2),B=1.25and A=0.75 i(t)=[0.75e-2t + 1.25e-18t]=-ix(t)orix(t)=[-0.75e-2t - 1.25e-18t] A v(t)=8i(t)=[6e-2t + 10e-18t] A Chapter 8, Solution 58. (a)Let i =inductor current,v = capacitor voltage i(0) =0,v(0) = 4 V/s 85 . 0) 0 4 ( )] 0 ( ) 0 ( [ ) 0 ( =+ =+ =RCRi vdtdv (b) For, the circuit is a source-free RLC parallel circuit.0 t 21 25 . 01 1, 11 5 . 0 2121= = = = = =x LCx x RCo 732 . 1 1 42 2= = = od Thus, ) 732 . 1 sin 732 . 1 cos ( ) (2 1t A t A e t vt+ = v(0) = 4 = A1 t A e t A e t A e t A edtdvt t t t732 . 1 cos 732 . 1 732 . 1 sin 732 . 1 sin 732 . 1 732 . 1 cos2 2 1 1 + =309 . 2 732 . 1 8) 0 (2 2 1 = + = = A A Adtdv ) 732 . 1 sin 309 . 2 732 . 1 cos 4 ( ) ( t t e t vt =V Chapter 8, Solution 59. Leti = inductor current andv = capacitor voltage v(0) = 0,i(0) = 40/(4+16) = 2A For t>0, the circuit becomes a source-free series RLC with 2 , 216 / 1 41 1, 24 2162= = = = = = = =o ox LCx LR t tBte Ae t i2 2) ( + =i(0) = 2 = A t t tBte Be Aedtdi2 2 22 2 + =4 ), 0 32 (412 )] 0 ( ) 0 ( [12) 0 ( = + = + + = + = B B A v RiLB Adtdi t tte e t i2 24 2 ) ( =ttttt ttttt e e dt te dt e v idtCv02002 2020) 1 2 (46416 64 32 ) 0 (1 = = + = tte v232=V Chapter 8, Solution 60. Att=0-,4u(t)=0so thati1(0)=0=i2(0)(1) Applying nodal analysis, 4=0.5di1/dt + i1 + i2 (2) Also,i2=[1di1/dt 1di2/dt]/3or3i2=di1/dt di2/dt(3) Taking the derivative of (2),0=d2i1/dt2 + 2di1/dt + 2di2/dt(4) From (2) and (3),di2/dt=di1/dt 3i2=di1/dt 3(4 i1 0.5di1/dt) =di1/dt 12 + 3i1 + 1.5di1/dt Substituting this into (4), d2i1/dt2 + 7di1/dt + 6i1=24which givess2 + 7s + 6=0=(s + 1)(s + 6) Thus,i1(t)=Is + [Ae-t + Be-6t],6Is=24orIs=4 i1(t)=4 + [Ae-t + Be-6t]and i1(0)=4 + [A + B](5) i2=4 i1 0.5di1/dt= i1(t)=4 + -4 - [Ae-t + Be-6t] [-Ae-t - 6Be-6t] =[-0.5Ae-t + 2Be-6t]andi2(0)=0=-0.5A + 2B (6) From (5) and (6),A=-3.2andB=-0.8 i1(t)={4 + [-3.2e-t 0.8e-6t]} A i2(t)=[1.6e-t 1.6e-6t] A Chapter 8, Solution 61. Fort>0,we obtain the natural response by considering the circuit below. 0.25F+vC aiL4 1 H 6 At node a,vC/4 + 0.25dvC/dt + iL=0(1) But,vC=1diL/dt + 6iL(2) Combining (1) and (2), (1/4)diL/dt + (6/4)iL + 0.25d2iL/dt2 + (6/4)diL/dt + iL=0 d2iL/dt2 + 7diL/dt + 10iL=0 s2 + 7s + 10=0=(s + 2)(s + 5)ors1,2=-2, -5 Thus,iL(t)= iL()+ [Ae-2t + Be-5t], where iL() represents the final inductor current=4(4)/(4 + 6)=1.6 iL(t)=1.6 + [Ae-2t + Be-5t]and iL(0)=1.6 + [A+B]or-1.6=A+B(3) diL/dt=[-2Ae-2t - 5Be-5t] and diL(0)/dt=0=-2A 5BorA=-2.5B(4) From (3) and (4),A=-8/3andB=16/15 iL(t)=1.6 + [-(8/3)e-2t + (16/15)e-5t] v(t)=6iL(t)={9.6 + [-16e-2t + 6.4e-5t]} V vC=1diL/dt + 6iL=[ (16/3)e-2t - (16/3)e-5t] + {9.6 + [-16e-2t + 6.4e-5t]} vC={9.6 + [-(32/3)e-2t + 1.0667e-5t]} i(t)=vC/4={2.4 + [-2.667e-2t + 0.2667e-5t]} A Chapter 8, Solution 62. This is a parallel RLC circuit as evident when the voltage source is turned off. =1/(2RC)=(1)/(2x3x(1/18))=3 o = 1/ LC=1/ 18 / 1 x 2=3 Since =o, we have a critically damped response. s1,2=-3 Let v(t)=capacitor voltage Thus,v(t)=Vs + [(A + Bt)e-3t]where Vs=0 But-10 + vR + v=0orvR=10 v Therefore vR=10 [(A + Bt)e-3t]where A and B are determined from initial conditions. Chapter 8, Solution 63. -R R v1 +vo vs v2 CC At node 1, dtdvCRv vs 1 1=(1) At node 2, dtdvCRv vo o=2(2) As a voltage follower, v v v = =2 1.Hence (2) becomes dtdvRC v voo + = (3) and (1) becomes dtdvRC v vs+ = (4) Substituting (3) into (4) gives 222 2dtv dC RdtdvRCdtdvRC v vo o oo s+ + + =or s oo ov vdtdvRCdtv dC R = + + 2222 2
Chapter 8, Solution 64. C2vs R1 21 v1 R2+C1 vo At node 1,(vs v1)/R1=C1 d(v1 0)/dtorvs=v1 + R1C1dv1/dt(1) At node 2,C1dv1/dt=(0 vo)/R2 + C2d(0 vo)/dt orR2C1dv1/dt= vo + C2dvo/dt(2) From (1) and (2),(vs v1)/R1=C1 dv1/dt=-(1/R2)(vo + C2dvo/dt) orv1 = vs + (R1/R2)(vo + C2dvo/dt)(3) Substituting (3) into (1) produces, vs= vs + (R1/R2)(vo + C2dvo/dt) + R1C1d{vs + (R1/R2)(vo + C2dvo/dt)}/dt = vs + (R1/R2)(vo)+ (R1C2/R2) dvo/dt) + R1C1dvs/dt + (R1R1C1/R2)dvo/dt+ (R12 C1C2/R2)[d2vo/dt2] Simplifying we get, d2vo/dt2+ [(1/ R1C1) + (1/ C2)]dvo/dt + [1/(R1C1C2)](vo)=- [R2/(R1C2)]dvs/dt Chapter 8, Solution 65. At the input of the first op amp, (vo 0)/R = Cd(v1 0)(1) At the input of the second op amp, (-v1 0)/R=Cdv2/dt (2) Let us now examine our constraints.Since the input terminals are essentially at ground, then we have the following, vo=-v2orv2=-vo (3) Combining (1), (2), and (3), eliminating v1andv2we get,
0 v 100dtv dvC R1dtv do2o2o2 2 2o2= = Which leads tos2 100=0 Clearly this produces roots of 10 and +10. And, we obtain, vo(t)=(Ae+10t + Be-10t)V Att=0,vo(0+)= v2(0+)=0=A + B,thusB=A This leads to vo(t)=(Ae+10t Ae-10t)V.Now we can use v1(0+)=2V. From (2),v1=RCdv2/dt=0.1dvo/dt=0.1(10Ae+10t + 10Ae-10t) v1(0+)=2=0.1(20A)=2AorA=1 Thus,vo(t)=(e+10t e-10t)V It should be noted that this circuit is unstable (clearly one of the poles lies in the right-half-plane). Chapter 8, Solution 66. C2 R4R2 +C1vS 1 2 R3R1 vo Note that the voltage across C1 isv2=[R3/(R3 + R4)]vo This is the only difference between this problem and Example 8.11,i.e.v=kv,wherek=[R3/(R3 + R4)]. At node 1, (vs v1)/R1=C2[d(v1 vo)/dt] + (v1 v2)/R2 vs/R1=(v1/R1) + C2[d(v1)/dt] C2[d(vo)/dt] + (v1 kvo)/R2(1) At node 2, (v1 kvo)/R2= C1[d(kvo)/dt] orv1=kvo + kR2C1[d(vo)/dt](2) Substituting (2) into (1), vs/R1=(kvo/R1) + (kR2C1/R1)[d(vo)/dt] + kC2[d(vo)/dt] + kR2C1C2[d2(vo)/dt2] (kvo/R2) + kC1[d(vo)/dt] (kvo/R2) + C2[d(vo)/dt] We now rearrange the terms. [d2(vo)/dt2] + [(1/C2R1) + (1/ R2C2) + (1/R2C1) (1/ kR2C1)][d(vo)/dt] + [vo/(R1R2C1C2)]= vs/(kR1R2C1C2) If R1=R210 kohms,C1=C2=100 F,R3=20 kohms,andR4=60 kohms, k=[R3/(R3 + R4)]=1/3 R1R2C1C2=104 x104 x10-4 x10-4=1 (1/C2R1) + (1/ R2C2) + (1/R2C1) (1/ kR2C1)=1 + 1 + 1 3=3 3=0 Hence, [d2(vo)/dt2] + vo=3vs=6,t>0,and s2 + 1=0,ors1,2=j vo(t)=Vs + [Acost + B sint],Vs=6 vo(0)=0=6 + AorA=6 dvo/dt=Asint + Bcost,but dvo(0)/dt=0=B Hence, vo(t)=6(1 cost)u(t) volts. Chapter 8, Solution 67. At node 1, dt) 0 v ( dCdt) v v ( dCRv v12o 1111 in+=(1) At node 2, 2o 12Rv 0dt) 0 v ( dC=,or2 2o 1R Cvdtdv = (2) From (1) and (2), 2o1o1 1o2 21 11 inRvRdtdvC RdtdvR CC Rv v = 2o1o1 1o2 21 1in 1RvRdtdvC RdtdvR CC Rv v + + + = (3) C1 From (2) and (3), 0Vvin C2R2+v11 2 R1 vo dtdvRRdtv dC RdtdvR CC RdtdvdtdvR Cvo212o21 1o2 21 1 in 12 2o+ + + = = dtdvC R1R R C CvdtdvC1C1R1dtv din1 1 1 2 2 1o o2 1 22o2 = ++ + But C1C2R1R2 =10-4 x10-4 x104 x104=1 210 x 102C R2C1C1R14 41 2 2 1 2= = =+ dtdvvdtdv2dtv dinoo2o2 = + + Which leads to s2 + 2s + 1=0or (s + 1)2=0ands=1, 1 Therefore,vo(t)=[(A + Bt)e-t] + Vf As t approaches infinity, the capacitor acts like an open circuit so that Vf=vo()=0 vin=10u(t) mVand the fact that the initial voltages across each capacitor is 0 means that vo(0)=0which leads to A=0. vo(t)=[Bte-t] dtdvo=[(B Bt)e-t](4) From (2),0R C) 0 ( vdt) 0 ( dv2 2o o=+ =+ From (1) at t=0+, dt) 0 ( dvCR0 1o11+ =which leads to1R C1dt) 0 ( dv1 1o = =+ Substituting this into (4) gives B=1 Thus,v(t)=te-tu(t) V Chapter 8, Solution 68. The schematic is as shown below.The unit step is modeled by VPWL as shown. We insert a voltage marker to display V after simulation.We set Print Step=25 ms and final step =6s in the transient box.The output plot is shown below. Chapter 8, Solution 69. The schematic is shown below.The initial values are set as attributes of L1 and C1.We set Print Step to 25 ms and the Final Time to 20s in the transient box.A current marker is inserted at the terminal of L1 to automatically display i(t) after simulation.The result is shown below. Chapter 8, Solution 70. The schematic is shown below. After the circuit is saved and simulated, we obtain the capacitor voltage v(t) as shown below. Chapter 8, Solution 71. The schematic is shown below.We use VPWL and IPWL to model the 39 u(t) V and 13 u(t) A respectively.We set Print Stepto25 ms and Final Step to 4s in the Transient box.A voltage marker is inserted at the terminal of R2 to automatically produce the plot of v(t) after simulation.The result is shown below. Chapter 8, Solution 72. When the switch is in position 1, we obtain IC=10 for the capacitor and IC=0 for the inductor. When the switch is in position 2, the schematic of the circuit is shown below. When the circuit is simulated, we obtain i(t) as shown below. Chapter 8, Solution 73. (a)Fort0, we have the schematic shown below.To display i(t) and v(t), we insert current and voltage markers as shown.The initial inductor current and capacitor voltage are also incorporated.In the Transient box, we set Print Step=25 ms and the Final Time to 4s.After simulation, we automatically have io(t) and vo(t) displayed as shown below. Chapter 8, Solution 74. 10 5 + 20 V 2F 4H - Hence the dual circuit is shown below. 2H 4F0.2 20A 0.1 Chapter 8, Solution 75. The dual circuit is connected as shown in Figure (a).It is redrawn in Figure (b). 0.5 H 2 F 12A 24A 0.25 0.1 10 F12A + 24V10 H 10 0.25 24A + 12V 0.5 F 10 H (a)4 0.1 (b) Chapter 8, Solution 76. The dual is obtained from the original circuit as shown in Figure (a).It is redrawn in Figure (b). 120 A 2 V + 2 A 30 1/3 10 0.1 120 V+ 60 V+ 60 A 4F 1 F 4 H1 H 20 0.05 (a) 0.05 60 A 1 H 120 A 1/4 F 0.1 + 2V 1/30 (b) Chapter 8, Solution 77. The dual is constructed in Figure (a) and redrawn in Figure (b). + 5 V1/4 F 1 H 2 1 1/3 12 A 1 12 A 1/3 3 1/2 5 V + 5 A + 12V1/4 F 1 F 1/4 H 1 H (a)1 2 (b) Chapter 8, Solution 78. The voltage across the igniter is vR=vCsince the circuit is a parallel RLC type. vC(0)=12,andiL(0)=0. =1/(2RC)=1/(2x3x1/30)=5 o 30 / 1 x 10 x 60 / 1 LC / 13 = = =22.36 0,the circuit is as shown below. + vo C1+v a R2R1 C2 At node a, (vo v/R1=(v/R2) + C2dv/dt vo=v(1 + R1/R2) + R1C2 dv/dt 60=(1 + 5/2.5) + (5x106 x5x10-6)dv/dt 60=3v + 25dv/dt v(t)=Vs + [Ae-3t/25] where3Vs=60yieldsVs=20 v(0)=0=20 + AorA= 20 v(t)=20(1 e-3t/25)V Chapter 8, Solution 83. i=iD + Cdv/dt(1) vs + iR + Ldi/dt + v=0(2) Substituting (1) into (2), vs=RiD + RCdv/dt + Ldi/dt + LCd2v/dt2 + v=0 LCd2v/dt2 + RCdv/dt + RiD + Ldi/dt=vs d2v/dt2 + (R/L)dv/dt + (R/LC)iD + (1/C)di/dt=vs/LC Chapter 9, Solution 1. (a)angular frequency=103 rad/s (b)frequencyf= 2 =159.2 Hz (c)periodT= f =16.283 ms (d)Since sin(A)=cos(A 90), vs=12 sin(103t + 24)=12 cos(103t + 24 90) vs in cosine form isvs=12 cos(103t 66) V (e)vs(2.5 ms)= 12 ) 24 ) 10 5 . 2 )( 10 sin((3 - 3 + =12 sin(2.5 + 24)=12 sin(143.24 + 24) =2.65 V Chapter 9, Solution 2. (a)amplitude=8 A (b)=500=1570.8 rad/s (c)f= 2 =250 Hz (d)Is=8-25 A Is(2 ms)=) 25 ) 10 2 )( 500 cos(( 83 - =8 cos( 25)=8 cos(155) =-7.25 A Chapter 9, Solution 3. (a)4 sin(t 30)=4 cos(t 30 90)=4 cos(t 120) (b)-2 sin(6t)=2 cos(6t + 90) (c)-10 sin(t + 20)=10 cos(t + 20 + 90)=10 cos(t + 110) Chapter 9, Solution 4. (a)v=8 cos(7t + 15)=8 sin(7t + 15 + 90)=8 sin(7t + 105) (b)i=-10 sin(3t 85)=10 cos(3t 85 + 90)=10 cos(3t + 5) Chapter 9, Solution 5. v1=20 sin(t + 60)=20 cos(t + 60 90)=20 cos(t 30) v2=60 cos(t 10) This indicates that the phase angle between the two signals is 20 and that v1 lags v2. Chapter 9, Solution 6. (a)v(t)=10 cos(4t 60) i(t)=4 sin(4t + 50)=4 cos(4t + 50 90)=4 cos(4t 40) Thus, i(t) leads v(t) by 20. (b)v1(t)=4 cos(377t + 10) v2(t)=-20 cos(377t)=20 cos(377t + 180) Thus, v2(t) leads v1(t) by 170. (c)x(t)=13 cos(2t) + 5 sin(2t)=13 cos(2t) + 5 cos(2t 90) X=130 + 5-90=13 j5=13.928-21.04 x(t)=13.928 cos(2t 21.04) y(t)=15 cos(2t 11.8) phase difference=-11.8 + 21.04=9.24 Thus, y(t) leads x(t) by 9.24. Chapter 9, Solution 7. If f()=cos + j sin, ) ( f j ) sin j (cos j cos j -sinddf = + = + = = d jfdf Integrating both sides ln f=j + ln A f=Aej=cos + j sin f(0)=A=1 i.e.f()=ej=cos + j sin Chapter 9, Solution 8. (a) 4 j 345 15 + j2= 53.13 - 545 15+ j2 =398.13 + j2 =-0.4245 + j2.97 + j2 =-0.4243 + j4.97 (b)(2 + j)(3 j4)=6 j8 + j3 + 4=10 j5=11.18-26.57 j4) - j)(3 (220 - 8+ +j12 5 -10+ = 26.57 - 11.1820 - 8+144 25) 10 )( 12 j 5 - (+ =0.71566.57 0.2958 j0.71 =0.7109 + j0.08188 0.2958 j0.71 = 0.4151 j0.6281 (c)10 + (850)(13-68.38)=10+104-17.38 =109.25 j31.07 Chapter 9, Solution 9. (a)2 +8 j 54 j 3+ =2 +64 25) 8 j 5 )( 4 j 3 (+ + + =2 +8932 20 j 24 j 15 + + =1.809 + j0.4944 (b)4-10 + 6 32 j 1 =4-10 + 6 363.43 - 236 . 2 =4-10 + 0.7453-69.43 = 3.939 j0.6946 + 0.2619 j0.6978 =4.201 j1.392 (c) + 50 4 80 920 - 6 10 8 = 064 . 3 j 571 . 2 863 . 8 j 5628 . 1052 . 2 j 638 . 5 3892 . 1 j 879 . 7 + + + = 799 . 5 j 0083 . 16629 . 0 j 517 . 13+ = 86 . 99 886 . 581 . 2 - 533 . 13 =2.299-102.67 =-0.5043 j2.243 Chapter 9, Solution 10. (a)z 9282 . 6 4 zand , 5 66 . 8 z, 8 63 2 1j j j = = =93 . 19 66 . 103 2 1j z z z = + + (b) 499 . 7 999 . 932 1jzz z+ = Chapter 9, Solution 11. (a) =(-3 + j4)(12 + j5) 2 1z z =-36 j15 + j48 20 =-56 + j33 (b) 21zz = 5 j 124 j 3 -+ = 25 144) 5 j 12 )( 4 j 3 (-++ + =-0.3314 + j0.1953 (c) =(-3 + j4) + (12 + j5)=9 + j9 2 1z z +2 1z z =(-3 + j4) (12 + j5)=-15 j 2 12 1z zz z+ = ) j 15 ( -) j 1 ( 9++ = 2 21 15j) - 15 )( j 1 ( 9 -+ = 226) 14 j 16 ( 9 - + =-0.6372 j0.5575 Chapter 9, Solution 12. (a) =(-3 + j4)(12 + j5) 2 1z z =-36 j15 + j48 20 =-56 + j33 (b) 21zz = 5 j 124 j 3 -+ = 25 144) 5 j 12 )( 4 j 3 (-++ + =-0.3314 + j0.1953 (c) =(-3 + j4) + (12 + j5)=9 + j9 2 1z z +2 1z z =(-3 + j4) (12 + j5)=-15 j 2 12 1z zz z+ = ) j 15 ( -) j 1 ( 9++ = 2 21 15j) - 15 )( j 1 ( 9 -+ = 226) 14 j 16 ( 9 - + =-0.6372 j0.5575 Chapter 9, Solution 13. (a) 1520 . 0 2749 . 1 ) 2534 . 0 8425 . 0 ( ) 4054 . 0 4324 . 0 j j j ( + = + + (b) 0833 . 2150 2430 50 = oo (c) (2+j3)(8-j5) (-4) = 35 +j14 Chapter 9, Solution 14. (a) 5116 . 0 5751 . 011 1514 3jjj+ =+ (b) 55 . 11 922 . 17 . 2134 06 . 2469 . 6944 24186) 5983 . 10 96 . 16 )( 84 67 () 80 60 )( 80 56 . 138 82 . 231 116 . 62 (jj j jj j j =+=+ + + + (c) ( ) 89 . 200 4 . 256 ) 120 260 ( 4 22j j j = + Chapter 9, Solution 15. (a) j 1 - 5 -3 j 2 6 j 10+ + =-10 j6 + j10 6 + 10 j15 =-6 j11 (b) 45 3 0 1610 - 4 - 30 20 =6015 + 64-10 =57.96 + j15.529 + 63.03 j11.114 =120.99 j4.415 (c) j 1 j0 j j 1j 1 j 1j 1 j0 j j 1 + = 1 ) j 1 ( j ) j 1 ( j 0 1 0 12 2+ + + + += 1 ) j 1 j 1 ( 1 + + = 1 2=-1 Chapter 9, Solution 16. (a)-10 cos(4t + 75)=10 cos(4t + 75 180) =10 cos(4t 105) The phasor form is 10-105 (b)5 sin(20t 10)=5 cos(20t 10 90) =5 cos(20t 100) The phasor form is 5-100 (c)4 cos(2t) + 3 sin(2t)=4 cos(2t) + 3 cos(2t 90) The phasor form is 40 + 3-90=4 j3=5-36.87 Chapter 9, Solution 17. (a)Let A=8-30 + 60 =12.928 j4 =13.533-17.19 a(t)=13.533 cos(5t + 342.81) (b)We know that -sin=cos( + 90). Let B=2045 + 30(20 + 90) =14.142 + j14.142 10.261 + j28.19 =3.881 + j42.33 =42.5184.76 b(t)=42.51 cos(120t + 84.76) (c)Let C=4-90 + 3(-10 90) =-j4 0.5209 j2.954 =6.974265.72 c(t)=6.974 cos(8t + 265.72) Chapter 9, Solution 18. (a) = ) t ( v160 cos(t + 15) (b) =6 + j8=1053.13 2V) t ( v2 =10 cos(40t + 53.13) (c) = ) t ( i12.8 cos(377t /3) (d) =-0.5 j1.2=1.3247.4 2I) t ( i2 =1.3 cos(103t + 247.4) Chapter 9, Solution 19. (a)310 5-30=2.954 + j0.5209 4.33 + j2.5 =-1.376 + j3.021 =3.32114.49 Therefore,3 cos(20t + 10) 5 cos(20t 30)=3.32 cos(20t + 114.49) (b)4-90 + 3-45=-j40 + 21.21 j21.21 =21.21 j61.21 =64.78-70.89 Therefore,40 sin(50t) + 30 cos(50t 45)=64.78 cos(50t 70.89) (c)Using sin=cos( 90), 20-90 + 1060 5-110=-j20 + 5 + j8.66 + 1.7101 + j4.699 =6.7101 j6.641 =9.44-44.7 Therefore,20 sin(400t) + 10 cos(400t + 60) 5 sin(400t 20) =9.44 cos(400t 44.7) Chapter 9, Solution 20. (a)o o o oj j 399 . 4 966 . 8 2139 . 3 83 . 3 2 464 . 3 40 5 90 60 4 = = = V Hence, ) 399 . 4 377 cos( 966 . 8ot v = (b)5 , 90 20 8 0 10 = + = o o oj I ,i.e.o oI 04 . 16 51 . 49 20 40 10 = + = ) 04 . 16 5 cos( 51 . 49ot i + = Chapter 9, Solution 21. (a)o o o oj F 86 . 34 3236 . 8 758 . 4 8296 . 6 90 30 4 15 5 = + = = ) 86 . 34 30 cos( 324 . 8 ) (ot t f + = (b) Go o oj 49 . 62 565 . 5 9358 . 4 571 . 2 50 4 90 8 = = + = ) 49 . 62 cos( 565 . 5 ) (ot t g = (c) ( ) 40 , 90 5 0 101= + = o ojHi.e.o o oj H 6 . 116 2795 . 0 125 . 0 25 . 0 180 125 . 0 90 25 . 0 = = + = ) 6 . 116 40 cos( 2795 . 0 ) (ot t h = Chapter 9, Solution 22. Letf(t) = +tdt t vdtdvt v ) ( 2 4 ) ( 10oVjVV j V F 30 20 , 5 ,24 10 = = + = oj j V j V j V F 97 . 92 1 . 440 ) 10 32 . 17 )( 6 . 19 10 ( 4 . 0 20 10 = = + = ) 97 . 92 5 cos( 1 . 440 ) (ot t f =Chapter 9, Solution 23. (a)v(t)=40 cos(t 60) (b)V=-3010 + 5060 =-4.54 + j38.09 =38.3696.8 v(t)=38.36 cos(t + 96.8) (c)I=j6-10=6(90 10)=680 i(t)=6 cos(t + 80) (d)I= j2+ 10-45=-j2 + 7.071 j7.071 =11.5-52.06 i(t)=11.5 cos(t 52.06) Chapter 9, Solution 24. (a) 1 , 0 10j= =+VV10 ) j 1 ( = V = + == 45 071 . 7 5 j 5j 110VTherefore,v(t)=7.071 cos(t + 45) (b) 4 ), 90 10 ( 20j45 j = =+ + VV V =+ + 80 - 204 j45 4 j V =+ = 96 . 110 - 43 . 33 j 580 - 20VTherefore,v(t)=3.43 cos(4t 110.96) Chapter 9, Solution 25. (a) 2 , 45 - 4 3 2j = = + I I = + 45 - 4 ) 4 j 3 ( I = =+ = 98.13 - 8 . 013 . 53 545 - 4j4 345 - 4ITherefore,i(t)=0.8 cos(2t 98.13) (b)5 , 22 5 6 jj10 = = + +I II = + + 22 5 ) 6 5 j 2 j - ( I = =+ = 56 . 4 - 745 . 056 . 26 708 . 622 53 j 622 5ITherefore,i(t)=0.745 cos(5t 4.56) Chapter 9, Solution 26. 2 , 0 1j2 j = =+ + II I12 j12 2 j =+ + I =+= 87 . 36 - 4 . 05 . 1 j 21ITherefore,i(t)=0.4 cos(2t 36.87) Chapter 9, Solution 27. 377 , 10 - 110j100 50 j = =+ + VV V = + 10 - 110377100 j50 377 j V = 10 - 110 ) 45 . 82 6 . 380 ( V = 45 . 92 - 289 . 0 V Therefore, v(t)=0.289 cos(377t 92.45). Chapter 9, Solution 28. = = =8) t 377 cos( 110R) t ( v) t ( is13.75 cos(377t) A. Chapter 9, Solution 29. 5 . 0 j -) 10 2 )( 10 ( j1C j16 - 6=== Z = = = 65 - 2 ) 90 - 5 . 0 )( 25 4 ( IZ V Therefore v(t)=2 sin(106t 65) V. Chapter 9, Solution 30. Z 2 j ) 10 4 )( 500 ( j L j3 -= = = = = = 155 - 3090 265 - 60ZVITherefore,i(t)=30 cos(500t 155) A. Chapter 9, Solution 31. i(t)=10 sin(t + 30)=10 cos(t + 30 90)=10 cos(t 60) Thus,I=10-60 v(t)=-65 cos(t + 120)=65 cos(t + 120 180)=65 cos(t 60) Thus,V=65-60 = = = 5 . 660 - 1060 - 65IVZ Since V and I are in phase, the element is a resistor with R=6.5 . Chapter 9, Solution 32. V=18010,I=12-30, =2 + = = = = 642 . 9 j 49 . 11 0 4 1530 - 120 1 180IVZ One element is a resistor with R=11.49 . The other element is an inductor with L=9.642orL=4.821 H. Chapter 9, Solution 33. 2L2Rv v 110 + =2R2Lv 110 v == =2 2L85 110 v69.82 V Chapter 9, Solution 34. if0 vo =LC1C1L = = = = ) 10 2 )( 10 5 (13 3 100 rad/s Chapter 9, Solution 35. = 0 5sV2 j ) 1 )( 2 ( j L j = = 2 j -) 25 . 0 )( 2 ( j1C j1= = = = =+ = ) 0 5 )( 90 1 ( 0 522 j2 j 2 j 22 js oV V590 Thus, 5 cos(2t + 90)= = ) t ( vo-5 sin(2t) V Chapter 9, Solution 36. LetZ be the input impedance at the source. 20 10 100 200 mH 1003j x x j L j = = 500200 10 101 1F 106jx x j C j = = 1000//-j500 = 200 j400 1000//(j20 + 200 j400) = 242.62 j239.84 oj Z 104 . 6 2255 84 . 239 62 . 2242 = = mA 896 . 3 61 . 26104 . 6 225510 60oooI = = ) 896 . 3 200 cos( 1 . 266ot i = Chapter 9, Solution 37. 5 j ) 1 )( 5 ( j L j = = j -) 2 . 0 )( 5 ( j1C j1= = LetZ ,j -1 =5 j 210 j5 j 2) 5 j )( 2 (5 j || 22+=+= = Z Then, s2 12xIZ ZZI+= ,where = 0s2 I =+=+++= 32 12 . 28 j 520 j) 2 (5 j 210 jj -5 j 210 jxI Therefore,= ) t ( ix2.12 sin(5t + 32) A Chapter 9, Solution 38. (a)2 j -) 6 / 1 )( 3 ( j1C j1F61= = = = 43 . 18 - 472 . 4 ) 45 10 (2 j 42 j -IHence, i(t)=4.472 cos(3t 18.43) A = = = 43 . 18 - 89 . 17 ) 43 . 18 - 472 . 4 )( 4 ( 4I VHence, v(t)=17.89 cos(3t 18.43) V (b)3 j -) 12 / 1 )( 4 ( j1C j1F121= = 12 j ) 3 )( 4 ( j L j H 3 = = = = = 87 . 36 10j3 40 50ZVIHence, i(t)=10 cos(4t + 36.87) A = += 69 . 33 6 . 41 ) 0 50 (j12 812 jVHence, v(t)=41.6 cos(4t + 33.69) V Chapter 9, Solution 39. 10 j 810 j 5 j) 10 j - )( 5 j (8 ) 10 j - ( || 5 j 8 + =+ = + = Z = =+ = = 34 . 51 - 124 . 334 . 51 403 . 620j10 80 40ZVI = == 34 . 51 - 248 . 6 210 j 5 j10 j -1I I I = = = 66 . 128 3.124 -5 j -5 j2I I I Therefore,= ) t ( i16.248 cos(120t 51.34) A = ) t ( i2 3.124 cos(120t + 128.66) A Chapter 9, Solution 40. (a)For ,1 =j ) 1 )( 1 ( j L j H 1 = = 20 j -) 05 . 0 )( 1 ( j1C j1F 05 . 0 = = 802 . 0 j 98 . 120 j 240 j -j ) 20 j - ( || 2 j + =+ = + = Z = =+ = = 05 . 22 - 872 . 105 . 22 136 . 20 4j0.802 1.980 4oZVIHence, i = ) t (o1.872 cos(t 22.05) A (b)For ,5 =5 j ) 1 )( 5 ( j L j H 1 = = 4 j -) 05 . 0 )( 5 ( j1C j1F 05 . 0 = = 2 . 4 j 6 . 12 j 14 j -5 j ) 4 j - ( || 2 5 j + =+ = + = Z = =+ = = 14 . 69 - 89 . 014 . 69 494 . 40 4j4 1.60 4oZVIHence, i = ) t (o0.89 cos(5t 69.14) A (c)For ,10 =10 j ) 1 )( 10 ( j L j H 1 = = 2 j -) 05 . 0 )( 10 ( j1C j1F 05 . 0 = = 9 j 12 j 24 j -10 j ) 2 j - ( || 2 10 j + =+ = + = Z = =+ = = 66 . 3 8 - 4417 . 066 . 83 9.0550 49 j 10 4oZVIHence, i = ) t (o0.4417 cos(10t 83.66) A Chapter 9, Solution 41. ,1 = j ) 1 )( 1 ( j L j H 1 = = j -) 1 )( 1 ( j1C j1F 1 = = j 211 j -1 ) j - ( || ) j 1 ( 1 =++ = + + = Z j 210s= =ZVI ,I I ) j 1 (c+ = == = + = 18.43 - 325 . 6j 2) 10 )( j 1 () j 1 ( ) j 1 )( j - ( I I V Thus,v(t)=6.325 cos(t 18.43) V Chapter 9, Solution 42. 200 =100 j -) 10 50 )( 200 ( j1C j1F 506 -== 20 j ) 1 . 0 )( 200 ( j L j H 1 . 0 = = 20 j 40j2 - 1j100 -j100 50) (50)(-j100-j100 || 50 = == = = + += 90 14 . 17 ) 0 60 (7020 j) 0 60 (20 j 40 30 20 j20 joV Thus,= ) t ( vo17.14 sin(200t + 90) V or= ) t ( vo17.14 cos(200t) V Chapter 9, Solution 43. 2 =2 j ) 1 )( 2 ( j L j H 1 = = 5 . 0 j -) 1 )( 2 ( j1C j1F 1 = = = +=+ = 69 . 33 328 . 3 0 45 . 1 j 15 . 1 j1 5 . 0 j 2 j5 . 0 j 2 joI I Thus,= ) t ( io3.328 cos(2t + 33.69) A Chapter 9, Solution 44. 200 =2 j ) 10 10 )( 200 ( j L j mH 10-3= = j -) 10 5 )( 200 ( j1C j1mF 53 -== 4 . 0 j 55 . 010j 35 . 0 j 25 . 0j 312 j141 =++ =+ + = Y 865 . 0 j 1892 . 14 . 0 j 55 . 01 1+ == =YZ =+ =+ = 7.956 - 96 . 0865 . 0 j 1892 . 60 650 6ZI Thus,i(t)=0.96 cos(200t 7.956) A Chapter 9, Solution 45. We obtainIby applying the principle of current division twice. oI I2 I2IoZ1 -j2 Z2 2 (a)(b) 2 j -1 = Z ,3 j 1j2 - 2j4 -j4 2 || -j2) ( 4 j2+ = + = + = Z j 1j10 -) 0 5 (3 j 1 2 j -2 j -2 112+= + +=+= IZ ZZI =+=+= =1 110 -j 1j10 -j - 1j -j2 - 2j2 -2 oI I-5 A Chapter 9, Solution 46. = + = 40 5 ) 40 t 10 cos( 5 is sI j -) 1 . 0 )( 10 ( j1C j1F 1 . 0 = = 2 j ) 2 . 0 )( 10 ( j L j H 2 . 0 = = Let6 . 1 j 8 . 02 j 48 j2 j || 41+ =+= = Z ,j 32 = Z ) 40 5 (6 . 0 j 8 . 3j1.6 0.8s2 11o ++=+= IZ ZZI = = 46 . 94 325 . 297 . 8 847 . 3) 40 5 )( 43 . 63 789 . 1 (oI Thus,= ) t ( io2.325 cos(10t + 94.46) A Chapter 9, Solution 47. First, we convert the circuit into the frequency domain. -j10 Ix + 2 j4 5020 = = +=+ + + += 63 . 52 4607 . 063 . 52 854 . 105626 . 8 j 588 . 4 254 j 20 10 j) 4 j 20 ( 10 j25Ix is(t) = 0.4607cos(2000t +52.63) A Chapter 9, Solution 48. Converting the circuit to the frequency domain, we get: 10 30 V1 + j20 Ix -j20 20-40 We can solve this using nodal analysis. A ) 4 . 9 t 100 sin( 4338 . 0 i4 . 9 4338 . 020 j 3029 . 24 643 . 15I29 . 24 643 . 1503462 . 0 j 12307 . 040 2V40 2 ) 01538 . 0 j 02307 . 0 05 . 0 j 1 . 0 ( V020 j 300 V20 j0 V1040 20 Vxx111 1 1 + = = = = = = + + =++ Chapter 9, Solution 49. 4j 1) j 1 )( 2 j (2 ) j 1 ( || 2 j 2T=++ = + = Z1 I Ixj2 -j I I Ij 12 jj 1 2 j2 jx+= += , where 210 5 .x= 0 = I4 jj 12 jj 1x+=+= I I = =+=+= = 45 - 414 . 1 j 1jj 1) 4 (4 jj 1T sZ I V= ) t ( vs 1.414 sin(200t 45) V Chapter 9, Solution 50. Since = 100, the inductor = j100x0.1 = j10 and the capacitor = 1/(j100x10-3)= -j10. j10 Ix -j10 + vx 20 540 Using the current dividing rule: V ) 50 t 100 cos( 50 v50 50 I 20 V50 5 . 2 40 5 . 2 j 40 510 j 20 10 j10 jIxx xx = = = = = + + = Chapter 9, Solution 51. 5 j -) 1 . 0 )( 2 ( j1C j1F 1 . 0 = = j ) 5 . 0 )( 2 ( j L j H 5 . 0 = = The current I through the 2- resistor is 4 j 3 2 j 5 j 11ss=+ + =II I ,where = 0 10 I = = 13 . 53 - 50 ) 4 j 3 )( 10 (sI Therefore, = ) t ( is 50 cos(2t 53.13) A Chapter 9, Solution 52. 5 . 2 j 5 . 2j 15 j5 j 525 j5 j || 5 + =+=+= 101 = Z ,5 . 2 j 5 . 2 5 . 2 j 5 . 2 5 j -2 = + + = ZI2Z1Z2IS s s s2 112j 545 . 2 j 5 . 1210I I IZ ZZI==+= ) 5 . 2 j 5 . 2 ( + =2 oI V s sj 5) j 1 ( 10) j 1 )( 5 . 2 (j 5430 8 I I+= += =+ =) j 1 ( 10) j 5 )( 30 8 (sI2.884-26.31 A Chapter 9, Solution 53. Convert the delta to wye subnetwork as shown below. Z1Z2 Io2 Z3 + 10 60V 8o30 - Z , 3077 . 2 4615 . 02 104 6, 7692 . 0 1532 . 02 104 22 1jjx jZ jjx jZ + == == 2308 . 0 1538 . 12 10123jjZ + == 6062 . 0 726 . 4 ) 3077 . 2 5385 . 9 //( ) 2308 . 0 1538 . 9 ( ) 10 //( ) 8 (2 3j j j Z Z + = + + = + + 163 . 0 878 . 6 6062 . 0 726 . 4 21j j Z Z = + + + = A 64 . 28 721 . 83575 . 1 88 . 630 60Z30 60Iooo oo = = =Chapter 9, Solution 54. Since the left portion of the circuit is twice as large as the right portion, the equivalent circuit is shown below. Vs V1++ V2 + Z2 Z ) j 1 ( 2 ) j 1 (o 1 = = I V) j 1 ( 4 21 2 = = V V) j 1 ( 62 1 s = + = V V V=sV8.485-45 V Chapter 9, Solution 55. -j4 I I1+Vo I2+ Z12 -j20 Vj8 -j0.58 j48 jo1= = =VI j8 j4 -) 8 j ( (-j0.5)j4 -) 8 j (12+ =+=+=Z Z Z II 5 . 0 j8j8-j0.52 1+ = + + = + =Z ZI I I ) 8 j ( 12 j20 -1+ + = Z I I ) 8 j (2j -2j812 j20 - + + + = ZZ =21j23j26 - 4 - Z = == 279.68 64 . 1618.43 - 5811 . 125 . 261 31 . 2621j23j26 - 4 -Z Z=2.798 j16.403 Chapter 9, Solution 56. 30 H 3 j L j = 30 /13F jC j = 15 /11.5F jC j = 06681 . 015301530) 15 / //( 30 jjjjx jj j == = + = = m 333 606681 . 0 2 033 . 0) 06681 . 0 2 ( 033 . 0) 06681 . 0 2 //(30jj jj jjjZ Chapter 9, Solution 57. 2 H 2 j L j = jC j = 11F 2 . 1 j 6 . 2j 2 2 j) j 2 ( 2 j1 ) j 2 //( 2 j 1 Z + = + + = + = S 1463 . 0 j 3171 . 0Z1Y = =Chapter 9, Solution 58. (a)2 j -) 10 10 )( 50 ( j1C j1mF 103 -== 5 . 0 j ) 10 10 )( 50 ( j L j mH 10-3= = ) 2 j 1 ( || 1 5 . 0 jin + = Z2 j 22 j 15 . 0 jin+ = Z) j 3 ( 25 . 0 5 . 0 jin + = Z=inZ0.75 + j0.25 (b)20 j ) 4 . 0 )( 50 ( j L j H 4 . 0 = = 10 j ) 2 . 0 )( 50 ( j L j H 2 . 0 = = 20 j -) 10 1 )( 50 ( j1C j1mF 13 -== For the parallel elements, 20 j -110 j1201 1p+ + =Z 10 j 10p+ = ZThen, =inZ10 + j20 +=pZ 20 + j30 Chapter 9, Solution 59. ) 4 j 2 ( || ) 2 j 1 ( 6eq+ + = Z ) 4 j 2 ( ) 2 j 1 () 4 j 2 )( 2 j 1 (6eq+ + + + = Z 5385 . 1 j 308 . 2 6eq + = Z =eqZ8.308 j1.5385 Chapter 9, Solution 60. + = + + = + + + = 878 . 9 1 . 51 122 . 5 097 . 26 15 25 ) 10 30 //( ) 50 20 ( ) 15 25 ( j j j j j j ZChapter 9, Solution 61. All of the impedances are in parallel. 3 j 115 j12 j 11j 11 1eq++ +++=Z 4 . 0 j 8 . 0 ) 3 . 0 j 1 . 0 ( ) 2 . 0 j - ( ) 4 . 0 j 2 . 0 ( ) 5 . 0 j 5 . 0 (1eq = + + + + =Z ==4 . 0 j 8 . 01eqZ1 + j0.5 Chapter 9, Solution 62. 2 20 j ) 10 2 )( 10 10 ( j L j mH-3 3= = 100 j -) 10 1 )( 10 10 ( j1C j1F 16 - 3= = 50 j20 + + V+ Vin10 A2V -j100 50 ) 50 )( 0 1 ( = = V ) 50 )( 2 ( ) 100 j 20 j 50 )( 0 1 (in+ + = V80 j 150 100 80 j 50in = + = V = =0 1ininVZ150 j80 Chapter 9, Solution 63. First, replace the wye composed of the 20-ohm, 10-ohm, and j15-ohm impedances with the corresponding delta. 5 . 22 j 1020450 j 200z , 333 . 13 j 3015 j450 j 200z45 j 2010300 j 150 j 200z3 21+ =+= =+=+ =+ += j12 j16 8 z2z3j16 z1 10 ZT 10 Now all we need to do is to combine impedances. = + + = = = = 93 . 6 j 69 . 34 ) 821 . 3 j 7 . 21 938 . 8 j 721 . 8 ( z 12 j 8 Z821 . 3 j 70 . 21 ) 16 j 10 ( z938 . 8 j 721 . 833 . 29 j 40) 16 j 10 )( 333 . 13 j 30 () 16 j 10 ( z1 T32 Chapter 9, Solution 64. A 7 . 104 527 . 1 4767 . 1 j 3866 . 0Z90 30I5 j 192 j 6) 8 j 6 ( 10 j4 ZTT = + = = = + + = Chapter 9, Solution 65. ) 4 j 3 ( || ) 6 j 4 ( 2T+ + = Z 2 j 7) 4 j 3 )( 6 j 4 (2T+ + = Z =TZ6.83 + j1.094 =6.9179.1 = = =1 . 9 917 . 610 120TZVI17.350.9 A Chapter 9, Solution 66. ) j 12 (1451705 j 60) 10 j 40 )( 5 j 20 () 10 j 40 ( || ) 5 j 20 (T =++ = + = Z =TZ14.069 j1.172 =14.118-4.76 = = = 76 . 94 25 . 476 . 4 - 118 . 1490 60TZVII I120 +VabI2j10 I I Ij 122 j 85 j 6010 j 401++=++= I I Ij 12j 45 j 605 j 202+=+= 2 1 ab10 j 20 - I I V + = I I Vj 1240 j 10j 12) 40 j (160 -ab+++++= I I V145j)(150) -12 (j 12150 -ab+=+= ) 76 . 97 25 . 4 )( 24 . 175 457 . 12 (ab = V =abV52.94273 V Chapter 9, Solution 67. (a)20 j ) 10 20 )( 10 ( j L j mH 20-3 3= = 80 j -) 10 5 . 12 )( 10 ( j1C j1F 5 . 126 - 3== ) 80 j 60 ( || 20 j 60in + = Z60 j 60) 80 j 60 )( 20 j (60in + = Z = + = 22 . 20 494 . 67 33 . 23 j 33 . 63inZ = =inin1ZY0.0148-20.22 S (b)10 j ) 10 10 )( 10 ( j L j mH 10-3 3= = 50 j -) 10 20 )( 10 ( j1C j1F 206 - 3== 20 60 || 30 = ) 10 j 40 ( || 20 50 j -in+ + = Z10 j 60) 10 j 40 )( 20 (50 j -in+ ++ = Z = = 56 . 74 - 75 . 50 92 . 48 j 5 . 13inZ = =inin1ZY0.019774.56 S=5.24 + j18.99 mS Chapter 9, Solution 68. 4 j -1j 312 j 51eq+++= Y ) 25 . 0 j ( ) 1 . 0 j 3 . 0 ( ) 069 . 0 j 1724 . 0 (eq+ + + = Y =eqY0.4724 + j0.219 S Chapter 9, Solution 69. ) 2 j 1 (412 j -141 1o+ = + =Y 6 . 1 j 8 . 05) 2 j 1 )( 4 (2 j 14o ==+= Y 6 . 0 j 8 . 0 jo = + Y ) 6 . 0 j 8 . 0 ( ) 333 . 0 j ( ) 1 (6 . 0 j 8 . 013 j -111 1o+ + + =+ + =Y = + =41 . 27 028 . 2 933 . 0 j 8 . 11oY 2271 . 0 j 4378 . 0 41 . 27 - 4932 . 0o = =Y 773 . 4 j 4378 . 0 5 jo+ = +Y 97 . 22773 . 4 j 4378 . 05 . 0773 . 4 j 4378 . 0121 1eq+ =++ =Y 2078 . 0 j 5191 . 01eq =Y ==3126 . 02078 . 0 j 5191 . 0eqY1.661 + j0.6647 S Chapter 9, Solution 70. Make a delta-to-wye transformation as shown in the figure below. c bnaZanZcnZbn2 Zeq 8 -j5 9 j 75 j 15) 10 j 15 )( 10 (15 j 10 10 j 5) 15 j 10 )( 10 j - (an =+=+ + += Z 5 . 3 j 5 . 45 j 15) 15 j 10 )( 5 (bn+ =++= Z 3 j 1 -5 j 15) 10 j - )( 5 (cn =+= Z ) 5 j 8 ( || ) 2 (cn bn an eq + + + = Z Z Z Z ) 8 j 7 ( || ) 5 . 3 j 5 . 6 ( 9 j 7eq + + = Z 5 . 4 j 5 . 13) 8 j 7 )( 5 . 3 j 5 . 6 (9 j 7eq ++ = Z 2 . 0 j 511 . 5 9 j 7eq + = Z = = 2 . 9 j 51 . 12eqZ15.53-36.33 Chapter 9, Solution 71. We apply a wye-to-delta transformation. j4 Zeq-j2 ZbcZab Zac a bc 1 j 12 j2 j 22 j4 j 2 j 2ab =+=+ = Z j 122 j 2ac+ =+= Z j2 -2j -2 j 2bc+ =+= Z 8 . 0 j 6 . 13 j 1) j 1 )( 4 j () j 1 ( || 4 j || 4 jab =+ = = Z 2 . 0 j 6 . 0j 2) j 1 )( 1 () j 1 ( || 1 || 1ac+ =++= + = Z 6 . 0 j 2 . 2 || 1 || 4 jac ab = + Z Z 6 . 0 j 2 . 212 j 2 -12 j -1 1eq+++ =Z 1154 . 0 j 4231 . 0 25 . 0 j 25 . 0 5 . 0 j + + = = + = 66 . 64 4043 . 0 3654 . 0 j 173 . 0 =eqZ2.473-64.66 =1.058 j2.235 Chapter 9, Solution 72. Transform the delta connections to wye connections as shown below. a R3R2R1-j18 -j9 j2 j2 j2 b 6 j - 18 j - || 9 j - = , =+ += 810 20 20) 20 )( 20 (R1, = = 450) 10 )( 20 (R2, = = 450) 10 )( 20 (3R 4 4) j6 (j2 || ) 8 2 j ( j2ab+ + + + = Z j4) (4 || ) 2 j 8 ( j2 4ab + + + = Z j2 - 12) 4 j j2)(4 (8j2 4ab ++ + = Z 4054 . 1 j 567 . 3 j2 4ab + + = Z =abZ7.567 + j0.5946 Chapter 9, Solution 73. Transform the delta connection to a wye connection as in Fig. (a) and then transform the wye connection to a delta connection as in Fig. (b). a R3R2R1-j18 -j9 j2 j2 j2 b 8 . 4 j -10 j486 j 8 j 8 j) 6 j - )( 8 j (1= = += Z-j4.81 2= = Z Z4 . 6 jj1064 -10 j) 8 j )( 8 j (3= = = Z = + + + + + + ) )( 2 ( ) )( 4 ( ) 4 )( 2 (3 1 3 2 2 1Z Z Z Z Z Z6 . 9 j 4 . 46 ) 4 . 6 j )( 8 . 4 j 2 ( ) 4 . 6 j )( 8 . 4 j 4 ( ) 8 . 4 j 4 )( 8 . 4 j 2 ( + = + + 25 . 7 j 5 . 14 . 6 j6 . 9 j 4 . 46a =+= Z688 . 6 j 574 . 38 . 4 j 46 . 9 j 4 . 46b+ =+= Z945 . 8 j 727 . 18 . 4 j 26 . 9 j 4 . 46c+ =+= Z 3716 . 3 j 07407688 . 12 j 574 . 3) 88 . 61 583 . 7 )( 90 6 (|| 6 jb+ =+ = Z602 . 2 j 186 . 025 . 11 j 5 . 1j7.25) -j4)(1.5 (|| 4 j -a = = Z1693 . 5 j 5634 . 0945 . 20 j 727 . 1) 07 . 79 11 . 9 )( 90 12 (|| 12 jc+ =+ = Z ) || 12 j || 4 j - ( || ) || 6 j (c a b eqZ Z Z Z + =) 5673 . 2 j 7494 . 0 ( || ) 3716 . 3 j 7407 . 0 (eq+ + = Z=eqZ1.50875.42 =0.3796 + j1.46 Chapter 9, Solution 74. One such RL circuit is shown below. Z+ Vi = 10 j20 20 20 Vj20 + Vo We now want to show that this circuit will produce a 90 phase shift. ) 3 j 1 ( 42 j 120 j 20 -40 j 20) 20 j 20 )( 20 j () 20 j 20 ( || 20 j + =++=+ += + = Z ) j 1 (313 j 63 j 1) 0 1 (12 j 2412 j 420i+ =++= ++=+= VZZV = = ++=+= 90 3333 . 03j) j 1 (31j 1j20 j 2020 joV V This shows that the output leads the input by 90. Chapter 9, Solution 75. Since, we need a phase shift circuit that will cause the output to lead the input by 90.) 90 t sin( ) t cos( + = This is achieved by the RL circuit shown below, as explained in the previous problem. 10 10 + Vi j10 j10 + Vo This can also be obtained by an RC circuit. Chapter 9, Solution 76. LetZ= R jX,where fC 21C1X== 394 . 95 66 116 R | Z | X X R | Z |2 2 2 2 2 2= = = = + = F 81 . 27394 . 95 x 60 x 21fX 21C === Chapter 9, Solution 77. (a) iccojX RjX -V V=where979 . 3) 10 20 )( 10 2 )( 2 (1C1X9 - 6 c= == )) 5 3.979 ( tan -90 (979 . 3 5979 . 3j3.979 - 5j3.979 -1 -2 2io+ += =VV ) 51 . 38 -90 (83 . 15 25979 . 3io +=VV = 51.49 - 6227 . 0ioVV Therefore, the phase shift is 51.49 lagging (b)) R X ( tan -90 -45c-1+ = = C1X R ) R X ( tan 45c c1 -= = = RC1f 2 = = = ==) 10 20 )( 5 )( 2 (1RC 21f9 - 1.5915 MHz Chapter 9, Solution 78. 8+j6 R Z -jX [ ] 5) 6 ( 8)] 6 ( 8 [) 6 ( 8 // = + + += + =X j RX j RX j R Z i.e8R + j6R jXR = 5R + 40 + j30 j5X Equating real and imaginary parts: 8R = 5R + 40 whichleads to R=13.33 6R-XR =30-5 which leadstoX=4.125. Chapter 9, Solution 79. (a)Consider the circuit as shown. 40 Z1V1V2j30 + Vi j60 + Vo j10 30 20 Z2 21 j 390 j 30) 60 j 30 )( 30 j () 60 j 30 ( || 30 j1+ =+ += + = Z = + =+ += + = 21 . 80 028 . 9 896 . 8 j 535 . 131 j 43) 21 j 43 )( 10 j () 40 ( || 10 j1 2Z Z LetV . = 0 1i 896 . 8 j 535 . 21) 0 1 )( 21 . 80 028 . 9 (20i222+ =+= VZZV = 77 . 57 3875 . 02V =++=+=03 . 26 85 . 47) 77 . 57 3875 . 0 )( 87 . 81 213 . 21 (21 j 4321 j 3402 2111V VZZV = 61 . 113 1718 . 01V 1 1 1 o) j 2 (522 j 12 j60 j 3060 jV V V V + =+=+=) 6 . 113 1718 . 0 )( 56 . 26 8944 . 0 (o = V = 2 . 140 1536 . 0oV Therefore, the phase shift is 140.2 (b)The phase shift is leading. (c)If, thenV 120i = V = = 2 . 140 43 . 18 ) 2 . 140 1536 . 0 )( 120 (oV V and the magnitude is 18.43 V. Chapter 9, Solution 80. = = 4 . 75 j ) 10 200 )( 60 )( 2 ( j L j mH 2003 - ) 0 120 (4 . 75 j 50 R4 . 75 j4 . 75 j 50 R4 . 75 ji o + +=+ += V V (a)When =100 R , = +=69 . 26 88 . 167) 0 120 )( 90 4 . 75 () 0 120 (4 . 75 j 1504 . 75 joV=oV53.8963.31 V (b)When = 0 R , = +=45 . 56 47 . 90) 0 120 )( 90 4 . 75 () 0 120 (4 . 75 j 504 . 75 joV=oV10033.55 V (c)To produce a phase shift of 45, the phase of =90 + 0 =45. oVHence, =phase of (R + 50 + j75.4)=45. For to be 45,R + 50=75.4 Therefore,R=25.4 Chapter 9, Solution 81. LetZ , 1 1R =22 2C j1R+ = Z , 3 3R = Z , and xx xC j1R+ = Z . 213xZZZZ=+ =+2213xxC j1RRRC j1R = = = ) 600 (4001200RRRR213x 1.8 k = = = = ) 10 3 . 0 (1200400CRRCC1RRC16 -231x2 13x 0.1 F Chapter 9, Solution 82. = = = ) 10 40 (2000100CRRC6 -s21x 2 F Chapter 9, Solution 83. = = = ) 10 250 (1200500LRRL3 -s12x 104.17 mH Chapter 9, Solution 84. Let s1 1C j1|| R= Z , 2 2R = Z , 3 3R = Z , and. x x xL j R + = Z1 C R jRC j1RC jRs 11s1s11+ =+= Z Since 213xZZZZ= , ) C R j 1 (RR RR1 C R jR R L j Rs 113 21s 13 2 x x + =+ = + Equating the real and imaginary components, 13 2xRR RR= ) C R (RR RLs 113 2x = implies that s 3 2 xC R R L = Given that, = k 40 R1 = k 6 . 1 R2, = k 4 R3, andF 45 . 0 Cs = = = = = k 16 . 0 k40) 4 )( 6 . 1 (RR RR13 2x 160 = = = ) 45 . 0 )( 4 )( 6 . 1 ( C R R Ls 3 2 x 2.88 H Chapter 9, Solution 85. Let, 1 1R = Z22 2C j1R+ = Z , 3 3R = Z , and 44 4C j1|| R= Z . j C RR j -1 C R jR4 444 444 =+ = Z Since3 2 4 1 2134Z Z Z Z ZZZZ = = , = 22 34 41 4CjR Rj C RR R j - 232 3 242424 4 1 4CjRR R1 C R) j C R ( R R j - =+ + Equating the real and imaginary components, 3 2 242424 1R R1 C RR R=+ (1) 2324242424 1CR1 C RC R R=+ (2) Dividing (1) by (2), 2 24 4C RC R1 = 4 4 2 22C R C R1= 4 4 2 2C R C R1f 2 = = 4 2 4 2C C R R 21f= Chapter 9, Solution 86. 84 j -195 j12401+ + = Y 0119 . 0 j 01053 . 0 j 10 1667 . 43 -+ = Y =+= =2 . 18 3861 . 4100037 . 1 j 1667 . 41000 1YZ Z=228-18.2 Chapter 9, Solution 87. ) 10 2 )( 10 2 )( 2 (j -50C j1506 - 3 1 + =+ = Z 79 . 39 j 501 = Z ) 10 10 )( 10 2 )( 2 ( j 80 L j 80-3 32 + = + = Z 66 . 125 j 802+ = Z 1003 = Z 3 2 11 1 1 1Z Z Z Z+ + = 66 . 125 j 80179 . 39 j 5011001 1+++ =Z ) 663 . 5 j 605 . 3 745 . 9 j 24 . 12 10 ( 1013 - + + + =Z
3 -10 ) 082 . 4 j 85 . 25 ( + = = 97 . 8 10 17 . 26-3 Z=38.21-8.97 Chapter 9, Solution 88. (a)20 j 120 30 j 20 j - + + = ZZ=120 j10 (b)If the frequency were halved, C f 21C1=L f 2 L would cause the capacitive impedance to double, while = would cause the inductive impedance to halve.Thus, 40 j 120 15 j 40 j - + + = ZZ=120 j65 Chapter 9, Solution 89. + =C j1R || L jinZ + +=+ ++ =C1L j RR L jCLC j1L j RC j1R L jinZ 22inC1L RC1L j R R L jCL + += Z To have a resistive impedance,0 ) Im(in = Z .Hence, 0C1LCLR L2= C1L C R2 = 1 LC C R2 2 2 2 = C1 C RL22 2 2+ =(1) Ignoring the +1 in the numerator in (1), = = = ) 10 50 ( ) 200 ( C R L9 - 2 2 2 mH Chapter 9, Solution 90. Let, = 0 145sV L 377 j L ) 60 )( 2 ( j L j X = = = jX R 800 145jX R 80s+ + =+ +=VI jX R 80) 145 )( 80 (801+ += = I V jX R 80) 145 )( 80 (50+ += (1) jX R 80) 0 145 )( jX R () jX R (o+ + += + = I V jX R 80) 145 )( jX R (110+ ++= (2) From (1) and (2), jX R8011050+= = +511) 80 ( jX R 30976 X R2 2= + (3) From (1), 23250) 145 )( 80 (jX R 80 = = + + 53824 X R R 160 64002 2= + + + 47424 X R R 1602 2= + + (4) Subtracting (3) from (4), = = R 16448 R 160102.8 From (3), 20408 10568 30976 X2= = = = = L L 377 86 . 142 X0.3789 H Chapter 9, Solution 91. L j || RC j1in += Z L j RLR jCj -in ++= Z 2 2 22 2 2L RLR j R LCj - + + += To have a resistive impedance,0 ) Im(in = Z . Hence, 0L RLRC1 -2 2 22= ++ 2 2 22L RLRC1 += 2 22 2 2LRL RC += where 710 2 f 2 = = ) 10 9 )( 10 20 )( 10 4 () 10 400 )( 10 4 ( 10 9C4 6 14 212 14 2 4 + = nF7216 9C22 += C=235 pF Chapter 9, Solution 92. (a) = = =oooox YZZ 5 . 13 4 . 47110 48 45075 1006 (b)o o ox x ZY 5 . 61 2121 . 0 10 48 450 75 1006 = = = Chapter 9, Solution 93. L s2 Z Z Z Z + + =A ) 9 . 18 6 . 0 5 . 0 ( j ) 2 . 23 8 . 0 1 ( + + + + + = Z20 j 25 + = Z = =66 . 38 02 . 320 115SLZVI =LI3.592-38.66 A Chapter 10, Solution 1. 1 = 45 - 10 ) 45 t cos( 10 + 60 - 5 ) 30 t sin( 5j L j H 1 = j -C j1F 1 = The circuit becomes as shown below. Vo3 + 10-45 V + 2 Io j 5-60 V Applying nodal analysis, j - j) 60 - 5 (3) 45 - 10 (o o oV V V= + oj 60 - 15 45 - 10 j V = + = + = 9 . 247 15.73 150 - 15 45 - 10oVTherefore,= ) t ( vo 15.73 cos(t + 247.9) V Chapter 10, Solution 2. 10 = 45 - 4 ) 4 t 10 cos( 4 + 150 - 20 ) 3 t 10 sin( 2010 j L j H 1 = 5 j -2 . 0 j1C j1F 02 . 0 = = The circuit becomes that shown below. Io Vo10 j10 -j5 20-150 V+ 4-45 A Applying nodal analysis, 5 j - 10 j45 - 410) 150 - 20 (o o oV V V+ = + o) j 1 ( 1 . 0 45 - 4 150 - 20 V + = + =+ + = = 98 . 150 816 . 2) j 1 ( j45 - 4 150 - 210 jooVITherefore,= ) t ( io 2.816 cos(10t + 150.98) A Chapter 10, Solution 3. 4 = 0 2 ) t 4 cos( 2-j16 90 - 16 ) t 4 sin( 16 = 8 j L j H 2 = 3 j -) 12 1 )( 4 ( j1C j1F 12 1 = = The circuit is shown below. Vo+ 20 A -j16 V 4 -j3 6 1 j8 Applying nodal analysis, 8 j 6 123 j 416 j -o o o++ = + V V V o8 j 613 j 411 23 j 416 j -V|.|
\|+++ = + = =+= 02 . 35 - 835 . 388 . 1 2207 . 115 . 33 - 682 . 404 . 0 j 22 . 156 . 2 j 92 . 3oV Therefore,= ) t ( vo3.835 cos(4t 35.02) V Chapter 10, Solution 4. 16 4 , 10 - 16 ) 10 t 4 sin( = 4 j L j H 1 = j -) 4 1 )( 4 ( j1C j1F 25 . 0 = = j4 1 Ix -j 16-10 V + V10.5 Ix+ Vo j 1 214 j) 10 - 16 (1x1= + VIV But 4 j) 10 - 16 (1xVI = So, j 1 8 j) ) 10 - 16 (( 31 1= V V 4 j 1 -10 - 481+ = V Using voltage division, =+ == 04 . 69 - 232 . 8) 4 j 1 j)(- - (110 - 48j 111 oV V Therefore,= ) t ( vo 8.232 sin(4t 69.04) V Chapter 10, Solution 5. Let the voltage across the capacitor and the inductor be Vx and we get: 03 jV2 jV430 10 I 5 . 0 Vx x x x= ++ xxx x xV 5 . 0 j2 jVI but 30 30 I 5 . 1 V ) 4 j 6 j 3 ( == = +Combining these equations we get: A 38 . 97 615 . 425 . 1 j 330 305 . 0 j I25 . 1 j 330 30V or 30 30 V ) 75 . 0 j 2 j 3 (xx x =+ =+ = = + Chapter 10, Solution 6. Let Vo be the voltage across the current source.Using nodal analysis we get: 010 j 20V320V 4 Vo x o=++ where o xV10 j 2020V+= Combining these we get: 30 j 60 V ) 3 5 . 0 j 1 ( 010 j 20V310 j 20V 420Voo o o+ = + =++ + =+ =+ +=5 . 0 j 2) 3 ( 20V or5 . 0 j 230 j 60Vx o29.11166 V. Chapter 10, Solution 7. At the main node, ||.|
\|+ ++= + ++ =+ 50130j20 j 401V3 j 196 . 520 j 40058 . 31 j 91 . 11550V30 jV30 620 j 40V 15 120oo V 154 08 . 1240233 . 0 j 04 . 07805 . 4 j 1885 . 3Vo =+ = Chapter 10, Solution 8. , 200 = 20 j 1 . 0 x 200 j L j mH 100 = = 100 j10 x 50 x 200 j1C j1F 506 = = u The frequency-domain version of the circuit is shown below. 0.1 Vo 40 V1Io V2 + -j100 6o15 20 Vo j20 - At node 1, 40V V100 jV20VV 1 . 0 15 62 1 1 11o++ = + or 2 1025 . 0 ) 01 . 0 025 . 0 ( 5529 . 1 7955 . 5 V V j j + = +(1) At node 2, 2 1212 1V ) 2 j 1 ( V 3 020 jVV 1 . 040V V + = + =(2) From (1) and (2), B AV or0) 5529 . 1 j 7955 . 5 (VV) 2 j 1 ( 3025 . 0 ) 01 . 0 j 025 . 0 (21=||.|
\| +=||.|
\|
+ Using MATLAB, V = inv(A)*B leads to V 09 . 161 3 . 110 , 23 . 127 63 . 702 1j V j + = = o 2 1o17 . 82 276 . 740V VI ==Thus, A ) 17 . 82 t 200 cos( 276 . 7 ) t ( ioo = Chapter 10, Solution 9. 103 310 , 0 10 ) t 10 cos( = 10 j L j mH 10 = 20 j -) 10 50 )( 10 ( j1C j1F 506 - 3== u Consider the circuit shown below. V120 V2-j20 20 Io30 + Vo 4 Io+ 100 V j10 At node 1, 20 j - 20 20102 1 1 1V V V V + = 2 1j ) j 2 ( 10 V V + = (1) At node 2, 10 j 30 20) 4 (20 j -2 1 2 1++ = V V V V,where 201oVI=has been substituted. 2 1) 8 . 0 j 6 . 0 ( ) j 4 - ( V V + = + 2 1j 4 -8 . 0 j 6 . 0V V++= (2) Substituting (2) into (1) 2 2jj 4 -) 8 . 0 j 6 . 0 )( j 2 (10 V V + + += or 2 . 26 j 6 . 01702= V =+=+= 26 . 70 154 . 62 . 26 j 6 . 0170j 3310 j 30302 oV V Therefore,= ) t ( vo6.154 cos(103 t + 70.26) V Chapter 10, Solution 10. 2000 , 100 j 10 x 50 x 2000 j L j mH 503= = = 250 j10 x 2 x 2000 j1C j1F 26 = = u
Consider the frequency-domain equivalent circuit below. V1 -j250V2
36=0 t , e0 t , eett) t ( + = = 110110t j t t j t t jdt e e dt e e dt e ) t ( y ) ( Y 10t ) j 1 (01t ) j 1 () j 1 (ej 1e + + = +
+ + + +=j 1sin j cosj 1sin j cose1212 Y()= | | ) sin (cos e 11212 + Chapter 18, Solution 11. f(t) = sin t [u(t) - u(t - 2)] ( ) = = 2020t j t j t j t jdt e e ej 21dt e t sin ) ( F
+ + + +20t ) ( j t ) ( jdt ) e e (j 21=
+ + + 20t ) ( j20t ) ( j) ( jee) ( j1j 21= ||.|
\| + + 2 j 2 je 1 e 121= ( ) + 2 j2 2e 2 2) ( 21= F()= ( ) 1 e2 j2 2 Chapter 18, Solution 12. (a)F = dt e dt e e ) (020t ) j 1 ( t j t = = = 20t ) j 1 (ej 11 j 11 e2 j 2 (b) + = 0110t j t jdt e ) 1 ( dt e ) ( H ( ) ( ) ) cos 2 2 (j11 ej1e 1j1j j + = + = = =j2 / sin 4222 /2 / sinj |.|
\| Chapter 18, Solution 13. (a)We know that )] a ( ) a ( [ ] at [cos + + = F . Using the time shifting property, ) a ( e ) a ( e )] a ( ) a ( [ e )] a 3 / t ( a [cos3 / j 3 / j a 3 / j+ + = + + = F (b) sin t sin sin t cos cos t sin ) 1 t ( = + = + g(t) = -u(t+1) sin (t+1) Letx(t) = u(t)sin t,then 2 2111 ) j (1) ( X =+ = Using the time shifting property, 1ee11) ( G2jj2 = = (c )Lety(t) = 1 + Asin at, then Y )] a ( ) a ( [ A j ) ( 2 ) ( + + = h(t) = y(t) cos bt Using the modulation property, )] b ( Y ) b ( Y [21) ( H + + = | | | | ) b a ( ) b a ( ) b a ( ) b a (2A j) b ( ) b ( ) ( H + + + + + + + + = (d)) 1 4 j (eje 1) 1 t j (ejedt e ) t 1 ( ) (24 j 4 j2402t j40t jt j+ = = = I Chapter 18, Solution 14. (a)) t 3 cos( ) 0 ( t 3 sin ) 1 ( t 3 cos sin t 3 sin cos t 3 cos ) t 3 cos( = = = +( f ) t ( u t 3 cos e ) tt = F()=( )( ) 9 j 1j 12+ + + (b) | | ) 1 t ( u ) 1 t ( u t cos ) t ( ' g =g(t) t1 -1 -1 1 --11 g(t)t ) 1 t ( ) 1 t ( ) t ( g ) t ( " g2 + + = + = j j 2 2e e ) ( G ) ( G( ) = = sin j 2 ) e e ( ) ( Gj j 2 2 G()=2 2sin j 2 Alternatively, we compare this with Prob. 17.7 f(t) = g(t - 1) F() = G()e-j () ( ) = = j j2 2je e e ) ( F G 2 2sin 2 j = G()=2 2sin j 2 (c)t cos ) 0 ( t sin ) 1 ( t cos sin t sin cos t cos ) 1 t ( cos = + = + = Lete x = ) t ( h e ) 1 t ( u ) 1 t ( cos ) t (2 ) 1 t ( 2 = and ) t ( u ) t cos( e ) t ( yt 2 = 2 2) j 2 (j 2) ( Y + + += ) 1 t ( x ) t ( y = = je ) ( X ) ( Y ( )( )2 2jj 2e j 2) ( X + + += ) ( H e ) ( X2 = ) ( X e ) ( H2 = =( )( )222 jj 2e j 2 + + + (d)Letx ) t ( y ) t ( u ) t 4 sin( e ) t (t 2 = =) t ( x ) t ( p =where) t ( u t 4 sin e ) t ( yt 2= ( )2 24 j 2j 2) ( Y+ + += ( ) 16 j 2j 2) ( Y ) ( X2+ = = = = ) ( X ) ( p( ) 16 2 j2 j2+ (e) 2 j 2 jej1) ( 2 3 ej8) ( Q ||.|
\|+ += Q()=2 j 2 je ) ( 2 3 ej6 + Chapter 18, Solution 15. (a)F = = 3 j 3 je e ) ( 3 sin j 2 (b)Letg = =je 2 ) ( G ), 1 t ( 2 ) t ( = ) ( FF\||.| tdt ) t ( g ) ( ) 0 ( Fj) ( G += ) ( ) 1 ( 2je 2j += = jej2 (c)F| | 121) t 2 ( = = = j21131) ( F2j31 Chapter 18, Solution 16. (a) Using duality properly 22t 2t22 or 4t42 F()=F=|.|\2t4
| 4 (b) t ae
2 2aa 2 + 2 2t aa 2+ ae 2 2 2t a8+
2e 4 G()= F=|.|\+2t 48
| 2e 4 Chapter 18, Solution 17. (a) Since H()= F( ) ( ) ( | |0 0 0F F21) t ( f t cos + + = ) where F()= F( ) | | () 2 ,j1t u0 = + = () ( )( )( )( )
+ ++ + + = 2 j122 j1221H ( ) ( ) | |( )( )
+ + + + + =2 22 22j2 22 H()= ( ) ( ) | |4j2 222 + + (b)G()= F| | ( ) ( | |0 0 0F F2j) t ( f t sin + = ) where F()= F( ) | | ()+ =j1t u () ( )( )( )( )
+ + + = 10 j11010 j1102jG ( ) ( ) | |
+ + + =10j10j2j10 102j = ( ) ( ) | |1001010 1022j + Chapter 18, Solution 18. Letf ( ) ( ) t u e tt = () += j j1F ( ) t cos t f ( ) ( | | 1 F 1 F21+ + ) Hence ()( ) ( )
+ ++ += 1 j 111 j 1121Y ( ) | | ( ) | |
+ + + + + + +=1 j 1 1 j 1j j 1 j j 121 1 j j j j 1j 12+ + + + += =2 j 2j 12+ + Chapter 18, Solution 19. () ( ) + = = dt e e e21dt e ) t ( f Ft j10t 2 j t 2 j t j ()( ) ( )| | + + = 10t 2 j t 2 jdt e e21F ( )( )( )( )10t 2 j t 2 je2 j1e2 j121
+ + = + ( )( )( )( )
+ + = + 2 j1 e2 j1 e212 j 2 j But = = + =2 j 2 je 1 2 sin j 2 cos e ()|.|
\| + + ||.|
\| = 2121j1 e21Fj = ( ) 1 e4jj2 2 Chapter 18, Solution 20. (a)F (cn) = cn() F( ) ( )o nt jnnn c e co = F|= |.|\ =nt jnnoe c ( ) = no nn c (b) = 2 T1T2o== ( ) |.|
\|+ = = T0 0jnt t jnn0 dt e 121dt e t fT1co ( ) 1 en 2jejn121jn0jnt=||.|
\|= Buten jn) 1 ( n cos n sin j n cos = = + = ( ) | |
= == =even n , 00 n , odd n ,njnn1 1n 2jc for n = 0 ==0n21dt 121c Hence = = =odd n0 nnjntenj21) t ( f F()= ( )= = odd n0 nnnnj21 Chapter 18, Solution 21. Using Parsevals theorem, = d | ) ( F |21dt ) t ( f2 2 Iff(t) = u(t+a) u(t+a),then |.|
\|= = = daa sina 421a 2 dt ) 1 ( dt ) t ( f22aa2 2 or aa 4a 4daa sin22== |.|
\| as required. Chapter 18, Solution 22. F| |( ) = dt ej 2e e) t ( f t sin ) t ( ft jt j t joo o ( ) ( )
= + dt e dt e ) t ( fj 21t j t jo o = ( ) ( | |o oF Fj 21 + ) Chapter 18, Solution 23. (a)f(3t)leads to( )( ) ( )( ) + += + +j 15 j 6303 / j 5 3 / j 21031 F| | ( ) = t 3 f( )( ) j 15 j 630 (b)f(2t)( )( ) ( )( ) + += + +j 10 j 4202 / j 15 2 / j 21021 f(2t-1) = f [2(t-1/2)]( )( ) + + j 10 j 4e 202 / j (c)f(t) cos 2t( ) ( ) 2 F212 F21+ + + =| | ( ) ( ) | | ( ) ( ) | | | | 2 j 5 2 j 252 j 5 2 j 2 + +++ + + +5 (d) F| | ( ) ()( )( ) + += =j 5 j 210 jF j t ' f (e)( )f tdt t()()( ) () +0 FjF ( )( )()5 x 210 xj 5 j 2 j + + + 10= =( )( )() + + + j 5 j 2 j10 Chapter 18, Solution 24. (a)() () = F X+ F [3] = () ( ) 1 ej+ j6 (b)( ) ( ) 2 t f t y = () () = = F e Y2 j( ) 1 ejej2 j (c)Ifh(t) = f '(t) () () ( ) = = = 1 ejj F j Hj je 1 (d)( )|.|
\| + |.|
\| = |.|
\|.
\|=53F53x 1023F23x 4 ) ( G , t35f3f t + ||10 t24 g ( ) ( ) 1 e53j 61 e23j65 / 3 j 2 / 3 j+ = = ( ) ( ) 1 e10 j1 e4 j5 / 3 j 2 / 3 j+ Chapter 18, Solution 25. (a)( )( ) =++ =+= j s ,2 sBs 2 s ssA 10F 5210B , 5210A == = = ()2 j5j5F+ = f(t)= ( ) ( ) t u e 5 t2t 2 sgn5 (b)()( )( ) 2 jB1 jA2 j 1 j4 jF+ ++ =+ + = ( )( )( ) =+++=+ + = j s ,2 sB1 sA2 s 1 s4 ss F A = 5,B = 6 () ++ += j 26j 15F f(t)= ( ) ( ) t u e 6 e 5t 2 t + Chapter 18, Solution 26. (a) ) t ( u e ) t () 2 t ( = f (b) ) t ( u te ) t (t 4 = h (c)If = + =sin2 ) ( X ) 1 t ( u ) 1 t ( u ) t ( x By using duality property, tt sin 2) t ( g ) 1 ( u 2 ) 1 ( u 2 ) ( G= + = Chapter 18, Solution 27. (a)Let( )( ) =++ =+= j s ,10 sBs 10 s ssA 100F 1010100B , 1010100A == = = ()10 j10j10F+ = f(t)= ( ) ( ) t u e 10 t sgnt 10 5 (b)( )( )( ) =++=+ = j s ,3 sBs 2As 3 s 2s 10s G 6530B , 4520A == = = ()3 j62 j4G+ + == g(t)= ( ) ( ) t u e 6 t u et 3 t 2 4 (c)()( ) ( ) 900 20 j601300 40 j j60H2 2+ + =+ + = h(t)= ( ) t u ) t 30 sin( et 20 2 ( )( )( )( ) = =+ + =21211 j j 2d e21t yt j41 Chapter 18, Solution 28. (a) + + = = d) j 2 )( j 5 (e ) (21d e ) ( F21) t ( ft jt j = = =201) 2 )( 5 (121 0.05 (b) + = + + =) 1 2 j )( 2 j (e210d e) 1 j ( j) 2 ( 1021) t ( ft 2 jt j = =2 j 1e25 jt 2 j + 2e ) j 2 (t 2 j (c) + + = + + =) j 3 )( j 2 (e220d) 5 3 )( j 2 (e ) 1 ( 2021) t ( fjt t j =+ =) j 5 5 ( 2e 20jt
jte ) j 1 ( (d)Let) ( F ) ( F) j 5 ( j5) j 5 () ( 5) (2 1 + = + + +F = = = + = 5 . 05125d ej 5) ( 521) t ( ft j1 1 B , 1 A ,5 sBsA) s 5 ( s5) s (2 = =++ =+= F 5 j1j1) ( F2+ = t 5 t 52e ) t ( u21e ) t sgn(21) t ( + = =f = + = ) t ( f ) t ( f ) t ( f2 1ut 5e ) t ( Chapter 18, Solution 29. (a)f(t) = F -1[ + )] (F -1 | | ) 3 ( 4 ) 3 ( 4 + + +=t 3 cos 421 =( ) t 3 cos 8 121+ (b)If) 2 t ( u ) 2 t ( u ) t ( h + = = 2 sin 2) ( H ) ( H 4 ) ( G = tt 2 sin 821) t ( = g g(t)=tt 2 sin 4 (c)Sincecos(at) ) a ( ) a ( + + Using the reversal property, ) 2 t ( ) 2 t ( 2 cos 2 + + or F -1 | | = 2 cos 6) 2 t ( 3 ) 2 t ( 3 + + Chapter 18, Solution 30. (a) += = =j a1) ( X ,j2) ( Y ) t sgn( ) t ( y )] t ( u ) t ( u [ a ) t ( 2 ) t ( hja 22j) j a ( 2) ( X) ( Y) ( H + = + = +== (b) += += j 21) ( Y ,j 11) ( X ) t ( u e ) t ( ) t ( hj 211j 2j 1) ( Ht 2 = + = + += (c )In this case, by definition, ) t ( u bt sin e ) t ( y ) t ( hat = =Chapter 18, Solution 31. (a) += += j a1) ( H ,) j a (1) ( Y2 ) t ( u e ) t ( xj a1) ( H) ( Y) ( Xat = +== (b)By definition, ) 1 t ( u ) 1 t ( u ) t ( y ) t ( x + = = (c ) = += j2) ( H ,) j a (1) ( Y ) t ( u e2a) t (21) t ( x) j a ( 2a21) j a ( 2j) ( H) ( Y) ( Xat = + = +== Chapter 18, Solution 32. (a)Since 1 jj+ e ( )) 1 t ( u e1 t andF f(-t)( ) ()1 jej1+ = F( )( ) ( ) 1 t u e t f1 t1 = f1(t)= e( ) ( ) 1 t u1 t + (b)From Section 17.3, 1 t22+
e 2 If () , e 2 F2 = then f2(t)=( ) 1 t22+ (b)By partial fractions ()( ) ( ) ( )( )( )1 j411 j411 j411 j411 j 1 j1F2 2 2 23 ++ ++ = + = Hence( ) ( ) ( ) t u e te e te41tt t t t3 + + = f = ( ) ( ) ( ) (t u e 1 t41t u e 1 t4t t + +)1 (d)( ) ( )() + = =2 j 1e21d e F21t ft jt j1 4 = 21 Chapter 18, Solution 33. (a)Let( ) ( ) ( ) | | 1 t u 1 t u t sin 2 t x + = From Problem 17.9(b), ()2 2sin j 4X = Applying duality property, ( ) ( )( )2 2tt sin j 2t X21t f = = f(t)=2 2tt sin j 2 (b)() ( ) ( ) = sin j cosj2 sin j 2 cosjF ( )= jejee ej2 j jj 2 j= ( ) ( ) ( ) 2 t sgn211 t sgn21t f = Butsgn( 1 ) t ( u 2 ) t = ( ) ( ) ( )212 t u211 t u t f + = = ( ) u ( 2 t u 1 t ) Chapter 18, Solution 34. First, we find G() for g(t) shown below. ( ) ( ) ( ) | | ( ) ( ) | | 1 t u 1 t u 10 2 t u 2 t u 10 t g + + + =( ) ( ) ( ) | | ( ) ( ) | | 1 t 1 t 10 2 t 2 t 10 t ' g + + + = The Fourier transform of each term gives 20tg(t)100 21 1 2 10(t+2)10(t+1) 1 21 2tg (t) 0 10(t-2)10(t-1) () ( ) ( ) + = j j 2 j 2 je e 10 e e 10 G j + = sin j 20 2 sin j 20 () = + = sin 20 2 sin 20G 40 sinc(2) + 20 sinc() Note that G() = G(-). () ( ) = G 2 F ( ) ( ) t G21t f= =(20/)sinc(2t) + (10/)sinc(t) Chapter 18, Solution 35. (a)x(t) = f[3(t-1/3)].Using the scaling and time shifting properties, ) j 6 (ee3 / j 2131) ( X3 / j3 / j += += (b)Using the modulation property,
++ =
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