Chapter 14, Solution 1. RCj1 RCj Cj1R R )( i o ω+ ω = ω+ ==ω V V H =ω)(H 0 0 j1 j ωω+ ωω , where RC 1 0 =ω 2 0 0 )(1 )(H ωω+ ωω =ω= H ω ω − π =ω∠=φ 0 1- tan 2 )(H This is a highpass filter. The frequency response is the same as that for P.P.14.1 except that RC1 0 =ω . Thus, the sketches of H and φ are shown below. H ω ω 0 = 1/RC 1 0.7071 0 0 90 ° φ ω ω 0 = 1/RC 45 ° Chapter 14, Solution 2. = ω+ = ω+ =ω RLj1 1 LjR R )(H 0 j1 1 ωω+ , where L R 0 =ω 2 0 )(1 1 )(H ωω+ =ω= H ω ω =ω∠=φ 0 1- tan-)( H The frequency response is identical to the response in Example 14.1 except that LR 0 =ω . Hence the response is shown below. φ H ω ω 0 = R/L 0.7071 1 0 ω ω 0 = R/L -45° -90° 0° Chapter 14, Solution 3. (a) The Thevenin impedance across the second capacitor where V is taken is o sRC1 R RsC1||RR Th + +=+=Z sRC1sC1R sC1 i iTh + = + = V VV Z Th sC 1 + V o − + − V Th )sC1)(sRC1(sC1 sC1 Th i Th Th o Z V V Z V ++ =⋅ + = ))sRC1(sRCsRC1)(sRC1( 1 )sRC1)(sC1( 1 )s Thi o ++++ = ++ == ZV V H ( =)s(H 1sRC3CRs 1 222 ++ (b) 08.01080)102)(1040(RC -3-63 =×=××= There are no zeros and the poles are at == RC 0.383- s 1 4.787- == RC 2.617- s 2 32.712- Chapter 14, Solution 4. (a) RCj1 R Cj 1 ||R ω+ = ω )RCj1(LjR R RCj1 R Lj RCj1 R )( i o ω+ω+ = ω+ +ω ω+ ==ω V V H =ω)(H LjRRLC- R 2 ω++ω (b) )LjR(Cj1 )LjR(Cj Cj1LjR LjR )( ω+ω+ ω+ω = ω+ω+ ω+ =ωH =ω)(H RCjLC1 RCjLC- 2 2 ω+ω− ω+ω Chapter 14, Solution 5. (a) Cj1LjR Cj1 )( i o ω+ω+ ω ==ω V V H =ω)(H LCRCj1 1 2 ω−ω+ (b) RCj1 R Cj 1 ||R ω+ = ω )RCj1(LjR )RCj1(Lj )RCj1(RLj Lj )( i o ω+ω+ ω+ω = ω++ω ω ==ω V V H =ω)(H RLCLjR RLCLj 2 2 ω−ω+ ω−ω Chapter 14, Solution 6. (a) Using current division, Cj1LjR R )( i o ω+ω+ ==ω I I H )25.0)(10()25.0)(20(j1 )25.0)(20(j LCRCj1 RCj )( 22 ω−ω+ ω = ω−ω+ ω =ωH =ω)(H 2 5.25j1 5j ω−ω+ ω (b) We apply nodal analysis to the circuit below. 1/jωC + − I o V x 0.5 V x R I s jωL Cj1Lj 5.0 R xxx s ω+ω − += VVV I But )Cj1Lj(2 Cj1Lj 5.0 ox x o ω+ω=→ ω+ω = IV V I Cj1Lj 5.0 R 1 x s ω+ω += V I )Cj1Lj(2 1 R 1 )Cj1Lj(2 o s ω+ω += ω+ω I I 1 R )Cj1Lj(2 o s + ω+ω = I I )LC1(2RCj RCj R)Cj1Lj(21 1 )( 2 s o ω−+ω ω = ω+ω+ ==ω I I H )25.01(2j j )( 2 ω−+ω ω =ωH =ω)(H 2 5.0j2 j ω−ω+ ω Chapter 14, Solution 7. (a) Hlog2005.0 10 = Hlog105.2 10 -3 =× == × -3 105.2 10H 005773.1 (b) Hlog206.2- 10 = Hlog0.31- 10 = == -0.31 10H 4898.0 (c) Hlog207.104 10 = Hlog235.5 10 = == 235.5 10H 5 10718.1 × Chapter 14, Solution 8. (a) 05.0H = == 05.0log20H 10dB 26.02- , =φ °0 (b) 125H = == 125log20H 10dB 41.94 , =φ °0 (c) °∠= + = 43.63472.4 j2 10j )1(H == 472.4log20H 10dB 01.13 , =φ °43.63 (d) °∠=−= + + + = 23.55-254.47.1j9.3 j2 6 j1 3 )1(H == 254.4log20H 10dB 577.12 , =φ °23.55- Chapter 14, Solution 9. )10j1)(j1( 1 )( ω+ω+ =ω H 10/j1log20j1log20-H 1010dB ω+−ω+= )10/(tan)(tan- -1-1 ω−ω=φ The magnitude and phase plots are shown below . H dB 0.1 -40 10/j1 1 log20 10 ω+ ω+ j1 1 log20 10 -20 10 1 100 ω 1 j1 1 arg φ -135° 10/j1 arg ω+ ω+ -90° -180° 0.1 10 1 100 ω -45° Chapter 14, Solution 10. ω +ω = ω+ω =ω 5 j 1j1 10 )j5(j 50 )j(H ωj 1 log20 ω + 5 j 1 1 log20 20 log1 -40 0.1 10 1 100 ω 20 -20 H dB 40 φ -135° 5/j1 1 arg ω+ ωj 1 arg -90° -180° 0.1 10 1 100 ω -45° Chapter 14, Solution 11. )2j1(j )10j1(5 )( ω+ω ω+ =ω H 2j1log20jlog2010j1log205log20H 10101010dB ω+−ω−ω++= 2tan10tan90- -1-1 ω−ω+°=φ The magnitude and phase plots are shown below . -40 0.1 10 1 100 ω 20 14 -20 H dB 40 34 φ -45° 45° ω 1001 10 0.1 -90° 90° Chapter 14, Solution 12. 201.0log20, )10/1( )1(1.0 )( −= + + = ωω ω jj j wT The plots are shown below. |T| (db) 20 0 ω 0.1 1 10 100 -20 -40 arg T 90 o 0 ω 0.1 1 10 100 -90 o Chapter 14, Solution 13. )10j1()j( )j1)(101( )j10()j( j1 )( 22 ω+ω ω+ = ω+ω ω+ =ωG 10j1log20jlog40j1log2020-G 101010dB ω+−ω−ω++= 10tantan-180 -1-1 ω−ω+°=φ The magnitude and phase plots are shown below . G dB 40 20 0.1 1 10 100 ω -20 -40 φ 90° 0.1 1 10 100 ω -90° -180°