University of Massachusetts at Amherst

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ECE 585 Microwave Engineering II

NTU: EM 531

 

 

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Homework Help and Worked Problem Examples

Solutions to the HWs will be posted about a week after they are due. In the meantime, I am posting answers to some of the problems for the first few problem sets. This will help you know if you're on the right track. Last updated: January 15, 2005.

HW Set Problem Answer, comments
1 7.2 RL = 26 dB, C = 20 dB, D = 6 dB, I = 26 dB
  7.6 Check your formulas for the Tee network with these specific cases: for Zo = 50 ohms, if alpha = 3 dB (.708), then R1 = 8.6 ohms, R2 = 141.9 ohms
  7.7 Design: Zo = 100 ohms, R = 33.3 ohms; ratio of P3(mismatched) to P3 (matched) = 1.326 (1.2 dB)
2 7.9 Zo3 = 131.7 ohms, Zo2 = 43.9 ohms, R = 115.5 ohms; output impedances are R2 = 28.9 ohms, R3 = 86.7 ohms
  7.13 Bethe hole coupler; ro = 1.90 mm
  7.14 5-hole coupler; ro = r4 = 1.87 mm; r1 = r3 = 2.97 mm; r2 = 3.42 mm
  7.15 ro = r4 = 2.19 mm; r1 = r3 = 2.93 mm; r2 = 2.54 mm
3

7.18

S = 0.24 mm; W = 2.1 mm
  7.19 Zoe = 61 ohms; Zoo = 35 ohms
  7.21 S = 1.15 mm; W = 1.92 mm; line lengths = 6.32 mm
  7.22 S = 0.029 mm (really thin!) ; W = 1.09 mm (not a practical design...)
  7.29 V1- = 0; V2- = (-j / sqrt[2]) (V1 - V4); V3- = (-j / sqrt[2] ) (V1 + V4 ); V4- = 0
  7.33 Here's a helpful hint on solving Prob. 7.33 for the Bailey power divider.
5 8.1 Here are two points from the graph: (1) when z = 0.75 lambda, voltage magnitude = 18.8 V; (2) when z = 2.75 lambda, voltage magnitude = 6.25 V.
  8.13 After using Richards' transforms and first Kuroda identity, transform circuit to one having three series stubs (25 ohms, 100 ohms, 25 ohms), and two main-line sections in the middle, each with 25 ohms Zo
  8.17 from left, electrical lengths are 14.7deg, 35.2 deg, 21.8 deg, 35.2 deg, 14.7 deg; characteristic impedances are (from left) 15 ohms, 200 ohms, 15 ohms, 200 ohms, 15 ohms