329fall10hw9sol

# 329fall10hw9sol - ECE 329 Fall 2010 Homework 9 - Solution...

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Unformatted text preview: ECE 329 Fall 2010 Homework 9 - Solution Due: Nov. 9, 2010 1. a) Since the re ected power can be expressed in terms of the incident power as P r = | | 2 P i , | | 2 = 1 4 and | | = 1 2 = | 2- | | 2 + | Since the material is an imperfect dielectric, we can approximate the impedance as 2 q r and since r &gt; 1 , 2 &lt; = q . Thus, we must take the negative branch of the square root in the numerator of the above equation for . Thus we have | | = 1 2 =- ( 2- ) 2 + 1 2 =- q r- q q r + q r 1 r + 1 = 2- 2 r 1 r r = 9 b) The transmitted power decays as P t ( z ) = P t (0) e- 2 z where 2 2 = 5 employing the imperfect dielectric approximation. We can also relate P t (0) = 1- | | 2 P i . So our nal equation is P t ( z ) P i = 1- | | 2 e- 2 z 1 20 = 3 4 e- 2 5 z z =- 5 2 ln 1 15 = 2 . 15 m The question could be interpreted to mean the power in the second material is 1/20 of the transmitted value (rather than the incident value). Because of the ambiguity, this answer will also be accepted. If the power lost in transmission at the interface is not included, then P t ( z ) P t = e- 2 z 1 20 = e- 2 5 z z =- 5 2 ln 1 20 = 2 . 38 m 1 ECE 329 Fall 2010 t ( s) z (m) 1 2 3 4 300 1/2 1/6 V t ( s) z (m) 1 2 3 4 300 1/100-1/300 I Figure 1: Bounce diagrams for 3b. 2. a) The wavefront travels down the line, re ects o the load resistor, and then returns to the input in 20 ns (the change in voltage at this time is due to the re ected wave). Thus, the wave velocity on the line is...
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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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329fall10hw9sol - ECE 329 Fall 2010 Homework 9 - Solution...

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