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Unformatted text preview: ECE 329 Fall 2010 Homework 10 - Solution Due: Nov. 16, 2010 1. a) We can read the injection coe cient as τ g = 0 . 5 (0 . 5 / 1) directly from the plot. On the other hand, we know τ g = Z R g + Z . Thus, the characteristic impedance of the T.L. Z = R g = 100 Ohm . b) The plot has a jump at t = 2 ns , and we can infer that it takes the wave 2 ns to start from the source, then get re ected by the defect, nally go back to the source. Based on this, we have 2 d v = 2 ns . Therefore, d = 0 . 2 m . c) Again, from the plot we know the re ection coe cient at the defect is Γ d = 1 ((1- . 5) / . 5) . On the other hand, we have Γ d = ( R d + Z )- Z ( R d + Z ) + Z = R d R d + 2 Z . where R d denotes the e ective series resistance of the defect. We see R d should be very/in niely large. So the defect is actually an open circuit. d) Since all the incident wave get re ected at the defect, there will be no voltage response at the load. 2. The voltage and current phasors on the T.L. can be assumed as ( ˜ V ( z ) = ˜ V + e- jβz + ˜ V- e jβz ˜ I ( z ) = ˜ V + Z e- jβz- ˜ V- Z e jβz , where ˜ V + and ˜ V- correspond to the waves travelling along...
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- Spring '08
- Impedance, Trigraph, Transmission line, Impedance matching, Vin, resonant frequencies, resonant angular frequencies