Answers to Selected Supplementary Problems

Chapter 1

  1. Q1.1   Z = (51 + j25) Ω
  2. Q1.2   Distance = 0.02λ
  3. Q1.3   ρ = 0.72 ∠ −37° VSWR = 6
  4. Q1.4   ρ = 0.36 ∠ 25°
  5. Q1.5   ZL = (232.5 − j157.5) Ω
  6. Q1.6   YL = (4 − j1.5) mS distance = 0.191λ
  7. Q1.7   ZL = (7.5 − j4) Ω
  8. Q1.8   Distance from load to stub = 0.035 m

    Length of stub = 0.038 m

  9. Q1.9   Distance from load to stub = 0.035 m

    Length of stub = 0.138 m

  10. Q1.10   Distance from load to stub = 0.048 m

    Length of stub = 0.034 m

  11. Q1.11   Length = 16.5 mm
  12. Q1.12   Length = 70.27 mm
  13. Q1.13   Zin = (6.5 + j18) Ω
  14. Q1.14   Zin = (18 + j23.5) Ω
  15. Q1.15   Zin = (24 + j2.5) Ω
  16. Q1.16   
    Solution 1: Cseries = 64.30 pF Lshunt = 6.89 nH
    Solution 2: Lseries = 9.98 nH Cshunt = 12.15 pF
  17. Q1.17   
     (i) Solution 1: Cshunt = 0.42 pF Lseries = 3.18 nH
    Solution 2: Lshunt = 4.57 nH Cseries = 0.88 pF
    (ii) Solution 1: ρin = 0.14 ∠ −43°
    Solution 2: ρin = 0.20 ∠ −124°
  18. Q1.18   Lshunt (nearest to load) = 2.63 nH

    Lseries (nearest to source) = 2.39 nH

Chapter 2

  1. Q2.2   λs = 30.78 mm
  2. Q2.3   
    Zo (Ω) w (μm) λs (mm) Vp (m/s)
    25 2159.0 10.95 1.10 × 108
    50 571.5 11.68 1.17 × 108
    75 190.5 12.05 1.20 × 108
    100 63.5 12.27 1.23 × 108
  3. Q2.5   17.98%
  4. Q2.6   Minimum thickness at 1 GHz = 10.55 μm

    Minimum thickness at 10 GHz = 3.35 μm

    Minimum thickness at 100 GHz = 1.05 μm

  5. Q2.7   Percentage error in 50 Ω line = 6.0%

    Percentage error in 70 Ω line = 11.4%

  6. Q2.8   
    (a) V2 = 0.707 ∠ −90°  V3 = 0  V4 = 0.707 ∠ −270°
    (c) Ring: w = 175 μm  mean diameter ...

Get RF and Microwave Circuit Design now with the O’Reilly learning platform.

O’Reilly members experience books, live events, courses curated by job role, and more from O’Reilly and nearly 200 top publishers.

Start your free trial