Lab 7 - Solutions

Lab 7 - Solutions - EE 330 Lab#7 Solution Spring 2007 The...

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EE 330 Lab #7 Spring 2007 Solution The Big Game on Cable TV 1. How far down the line can you expect to find the problem? 8 8 10 49 . 2 10 3 83 . 0 % 83 × = × × = = c v m/s 6 10 1 . 12 × = τ s This is the travel time to and back from the point on the line where the problem is, thus the problem is expected to has occurred at about Ft 4 . 4942 m 45 . 1506 2 = = v from the distribution box. The accuracy, with which the TDR device measures distances of the this order, is . Hence, the problem is m 35 . 3 Ft 11 = m 35 . 3 45 . 1506 ± from the distribution box. 2. How many dB of loss would you expect to see in kilometer of cable? At 50 MHz, the loss is dB/km 52 dB/100m 2 . 5 = ; at 100 MHz, it is . Assuming linear dependence and interpolating, the loss at 67.25 MHz is expected to be dB/km 72 dB/100m 2 . 7 = () dB/km 9 . 58 50 25 . 67 50 100 52 72 52 = + 3. How far back from the slice (toward the generator) you need to place the stub? m 7 . 3 = = f v λ The normalized impedance at the point of measurement is Ω + = + = = 4 2 75 300 150 j j Z Z z c L L , marked with a blue dot on the Smith chart. The corresponding admittance is 1 2 . 0 1 . 0 4 2 1 Ω = + = j j y L , marked with a red dot. The stub must be connected to the TL at a point where the admittance, y d , has a real part equal to one. There are two green points on the Smith chart, which correspond to the two options, 9 . 2 1 1 j jb y d ± = ± = The point representing the admittance with the positive imaginary part is closer to the measured admittance, y L , a normalized distance of 234 . 0 202 . 0 468 . 0 5 . 0 = + = d back from the slice. This normalized distance is denoted by the red ark on the Smith chart. The actual distance is cm 6 . 86 m 866 . 0 234 . 0 = = = d .
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Lab 7 - Solutions - EE 330 Lab#7 Solution Spring 2007 The...

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