329fall10hw11sol - ECE 329 Fall 2010 Homework 11 Solution...

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Unformatted text preview: ECE 329 Fall 2010 Homework 11 - Solution Due: Nov. 30, 2010 1. This problem is solved by performing some simpli cations based on our knowledge of λ/ 2 and λ/ 4 transformers. Since the line connected to the generator is one wavelength long (i.e. two λ/ 2 lines combined), the voltage and current at the junction of the three transmission lines are equal to the voltage and current at the generator. Then, we see that the rst of the two lines connected in parallel is a λ/ 2 line. Therefore we know that V L 1 =- V g . The power dissipated in resistor R L 1 is P L 1 = 1 2 Re ( | V L 1 | 2 R * L 1 ) = 1 2 Re ( |- 100 | 2 50 ) . = 100 W Then examining the second line, we know that the λ/ 4 transformer has a load current I L 2 =- j V g Z 02 . The expression for the time average dissipated power in resistor R L 2 is P L 2 = 1 2 Re n | I L 2 | 2 R L 2 o = 1 2 Re (- j 100 100 2 (50) ) . = 25 W 2. a) We are given the voltage at the load and the load resistance so the power dissipated in the load is easily calculated P L = 1 2 Re ( | V L 1 | 2 R * L 1 ) = 1 2 Re ( | j 15 | 2 150 ) ....
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This note was uploaded on 11/06/2011 for the course ECE 329 taught by Professor Kim during the Spring '08 term at University of Illinois, Urbana Champaign.

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329fall10hw11sol - ECE 329 Fall 2010 Homework 11 Solution...

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