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Chapter 14, Solution 1. Vo R jωRC = = Vi R + 1 jωC 1 + jωRC jω ω 0 , 1 + jω ω 0 H (ω) = H (ω) = where ω 0 = 1 RC H = H (ω) = ω ω0 1 + (ω ω0 ) 2 φ = ∠H (ω) = ω π − tan -1 2 ω0 This is a highpass filter. The frequency response is the same as that for P.P.14.1 except that ω0 = 1 RC . Thus, the sketches of H and φ are shown below. H 1 0.7071 0 ω0 = 1/RC ω φ 90° 45° 0 ω0 = 1/RC ω .Chapter 14, Solution. | Chapter 14 Solution 1. H V R j RC O X R 1 j C 1 joRC H o j .0 where 0 v 1 j 0 0 RC o o0 z x n o H H o . z 0 b ZH - tan 1 I 1 0 o - 2 0 This is a highpass filter. The frequency response is the same as that for P.P.14.1 except that 0 1 RC. Thus the sketches of H and are shown below. Chapter 14 Solution 2. H o R joL 1 joL R 1 jo o0 where o0 R L H H o 71 o 2 ZH o - tan-1 The frequency response is identical to the response in Example 14.1 except that o0 R L. Hence the response is shown below. Chapter 14 Solution 3. a The Thevenin impedance across the second capacitor where Vo is taken is Z R R 1 sC R Th 1 sRC V Th V -V-R 1 sC 1 1 sRC ZTh 1 sC o V -----L V. ------- --------- o Z Th 1 sC Th 1 sRC 1 sCZ Th . .V__1_1__ H s V 1 sCZ Th 1 sRC 1 sRC 1 sRC sRC 1 sRC _1_ H s s2R2C2 3sRC 1 b RC 40 X 103 2 X10-6 80 X10-3 0.08 There are no zeros and the poles are at -0.383 R s 787. -2.617 s 2 RC -32.712 Chapter 14 Solution 4. 1R a R - j C 1 j RC R H Vl 1 j RC R o Vi R R j L 1 j RC J 1 j RC R H v 7 - 2RLC R j L b H o R j L j C R j L R j L 1 joC 1 j C R j L - 2LC j RC H --7 v 1 - 2LC j .
