University of Massachusetts at Amherst
ECE 585 Microwave Engineering II
Homework Assignments and Solutions
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 | |