CHAPTER 10. Hence, the circuit in the frequency domain is as shown below. 4 Ω V 1 V 2. 3Vx 10 = + 2 Ω. j4 Ω. V x. At node 1, (1) At node 2, where V
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1 February 5, 006 CHAPTER 0 P.P.0. 0 in(t 0 0, ω H jωl j4 0. F -j.5 jωc Hence, e circuit in e frequency dmain i a hwn belw. -j.5 Ω 4 Ω 0 0 A Ω x j4 Ω x At nde, At nde, 0 - j.5 00 (5 j4 j ( 4 x where x j4 - j j.5 j4(.5( 0 -(7.5 j4 (.5 j.5 ( Put ( and ( in matrix frm. 5 j4 - j (7.5 j4.5 j.5 0 where Δ ( 5 j4(.5 j.5 (-j4(-(7.5 j4.5 j j.5 j4 7.5 j4 5 j j j (00 ( j j (00 ( j
2 n e time dmain, v (t. in(t 60.0 v (t.0 in(t 57. P.P.0. The nly nn-reference nde i a upernde. 5 4 j4 - j 5 -j j4 5 ( j ( j4 ( The upernde give e cntraint f 0 60 ( Subtituting ( int ( give 5 ( j(0 60 ( j 5 ( j( j (-.7 j0.87 (0 j j8.54 Therefre, , P.P.0. Cnider e circuit belw. 0 A -j Ω 6 Ω 8 Ω j4 Ω 0 0 Fr meh, ( 8 j j4 j4 0 ( 8 j j 4 (
3 Fr meh, 6 j4 j Fr meh, - ( Thu, e equatin fr meh becme 6 j4 j4-0 0 ( ( 8 j Frm (, (0.5 j ( j4 Subtituting ( int (, ( 6 j4(0.5 j j4-0 0 ( j4 -(0.66 j5 - (0.66 j5 j4 Hence, 0.66 j j A P.P.0.4 Mehe and frm a upermeh a hwn in e circuit belw. 0 Ω -j4 Ω j8 Ω Ω -j6 Ω Fr meh, 50 (5 j4 ( j j4 j ( ( Fr e upermeh, j8 j4 (5 j6 (5 j4 0 ( ( Al, (
4 Eliminating frm ( and ( ( 5 j4 (-5 j4 60 (4-5 j4 (5 j -0 j (5 ( Frm (4 and (5, 5 j4-5 j4-5 j4 5 - j 60-0 j 5 j4-5 j4 Δ 58 j j4 5 - j 60-5 j4 Δ 98 j j 5 - j Δ Thu, A Δ ' ' P.P.0.5 Let, where and are due t e vltage urce and current urce repectively. Fr ' cnider e circuit in Fig. (a. ' -j Ω 6 Ω 8 Ω j4 Ω 0 0 (a Fr meh, ( 8 j j4 0 ( 0.5 j ( Fr meh, 6 j4 j ( ( Subtituting ( int (, ( 6 j4(0.5 j j4 0 0 ' j0.556 j4
5 Fr cnider e circuit in Fig. (b. 0 A -j Ω 6 Ω 8 Ω j4 Ω j4 Let 8 j Ω, 6 j4.846 j. 769 Ω 6 j4 ((.846 j.769 ( j j0.77 ' Therefre, j A ' ' P.P.0.6 Let v v v, where v i due t e vltage urce and i due t e current urce. Fr ' v, we remve e current urce. 0 in(5t 0 0, ω 5 0. F -j jωc j(5(0. H jωl j(5( The circuit in e frequency dmain i hwn in Fig. (a. 8 Ω (b j5 v 0 0 ' -j Ω j5 Ω (a
6 Nte at - j j5 -j.5 By vltage diviin, - j.5 ' ( j.5 Thu, v ' 4.6in(5t 8. Fr v, we remve e vltage urce. c(0t 0, ω 0 0. F -j0.5 jωc j(0(0. H jωl j(0( The crrepnding circuit in e frequency dmain i hwn in Fig (b. 8 Ω j0 Ω -j0.5 Ω 0 Let -j0.5 j80, 8 j0 8 j j. 9 By current diviin, ( - j(4.877 j.9 (-j0.5 ((-j j Thu, v.05c(0t 86.4 ' Therefre, v v v v 4.6 in(5t c(0t 86.4 (b j0
7 P.P.0.7 f we tranfrm e current urce t a vltage urce, we btain e circuit hwn in Fig. (a. 4 Ω -j Ω Ω j Ω S j5 Ω Ω -j Ω (a (j4(4 j j6 We tranfrm e vltage urce t a current urce a hwn in Fig. (b. j6 Let 4 j j 6 j. Then,.5 j. 6 j S 6 Ω Ω j5 Ω -j Ω -j Ω (b (6 j(j5 0 Nte at j5 ( j. 6 j By current diviin, 0 ( j (.5 j 0 ( j ( j 0 j j A
8 P.P.0.8 When e vltage urce i et equal t zer, 0 (-j4 (6 j (-j4(6 j j 0.4 j..4 j. Ω By vltage diviin, - j4 (0 0 6 j j4 (4-90 ( (-j4(0 0 6 j P.P.0.9 T find, cnider e circuit in Fig. (a. 8 j4 8 j4 5 0 a S a 4 j 0. 4 j 0. 0 (a b (b b At nde, j 8 j4 - ( j 50 ( j0.5( 50 ( j0.5 ( j0.5 ( At nde, , where. 8 j4 Hence, e equatin fr nde becme
9 5 0.( 0 8 j4 50 ( j0.5 Subtituting ( int (, j ( j0.5 ( j0.5 (50 j ( j (5 j j j T find, we remve e independent urce and inert a - vltage urce between terminal a-b, a hwn in Fig. (b. At nde a, j4 4 j But, and 8 j4 8 j4 4 j S, 8 j4.6 j0.8 (0. j j j and j j Ω
10 P.P.0.0 T find N, cnider e circuit in Fig. (a. 4 Ω j Ω 4 Ω j Ω 8 Ω Ω -j Ω a 8 Ω Ω -j Ω a N 0 0 -j4 Ω N (a b (b b N N (4 j (9 j.76 j0.706 Ω (4 j(9 j j T find N, hrt-circuit terminal a-b a hwn in Fig. (b. Ntice at mehe and frm a upermeh. Fr e upermeh, 0 8 ( j (9 j 0 ( - Al, j4 ( Fr meh, j 8 ( j 0 ( ( Slving fr, we btain 50 j N 9 j A N Uing e Nrtn equivalent, we can find a in Fig. (c. N N 0 j5 Ω (c
11 By current diviin, N.76 j0.706 N ( N 0 j5.76 j4.94 (.54.5 ( A P.P.0. 0 nf -j0 kω -9 jωc j(5 0 (0 0 0 nf -9 -j0 kω jωc j(5 0 (0 0 Cnider e circuit in e frequency dmain a hwn belw. -j0 kω 0 0 kω 0 kω -j0 kω A a vltage fllwer, At nde, 0 - j0 0 4 ( j ( j ( At nde, j0 ( j ( Subtituting ( int ( give 4 j6 r - 90
12 Hence, Nw, But frm ( Hence, v (t c(5000t 90 (t v - j0k in(5000t -j (t i (t i j66.66 μa - j0k c(5000t 90 μa in(5000t μa P.P.0. Let R R jωc jωrc R R The lp gain i / G R R R R R jωrc -6 where ωrc (000(0 0 ( 0 0 jωrc jωrc j / G j G
13 P.P.0. The chematic i hwn belw. Since ω πf 000 rad / f Hz. Setup/Analyi/AC Sweep a Linear fr pint tarting and ending at a frequency f Hz. When e chematic i aved and run, e utput file include Frequency M(_PRNT P(_PRNT 4.775E E E0 Frequency M($N_0005 P($N_ E0.68E E0 Frm e utput file, we btain and ma Therefre, (t 0.68 c(000t 54.6 v (t i c(000t 55. ma
14 P.P.0.4 The chematic i hwn belw. We elect ω rad/ and f Hz. We ue i t btain e value f capacitance, where C ωx c, and inductance, where L X L ω. Nte at AC de nt allw fr an AC PHASE cmpnent; u, we have ued AC in cnjunctin wi G t create an AC current urce wi a magnitude and a phae. T btain e deired utput ue Setup/Analyi/AC Sweep a Linear fr pint tarting and ending at a frequency f Hz. When e chematic i aved and run, e utput file include Frequency M(_PRNT P(_PRNT.59E-0.584E00.580E0 Frequency M($N_0004 P($N_ E E E0 Frm e utput file, we btain and A x 6 R 0 0 P.P.0.5 ( -9 C eq C μf R 0 0 P.P.0.6 f R R R.5 kω and C C C nf f khz πrc (π(.5 0 ( 0 x
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