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Unformatted text preview: ECE 3025A Problem Set # 7 Solutions 11.4. A lossless transmission line having Z = 120 is operating at = 5 10 8 rad/s. If the velocity on the line is 2 . 4 10 8 m/s, find: a) L : With Z = ! L/C and v = 1 / LC , we find L = Z /v = 120 / 2 . 4 10 8 = 0 . 50 H / m . b) C : Use Z v = ! L/C/ LC C = 1 / ( Z v ) = [120(2 . 4 10 8 )] 1 = 35 pF / m . c) Let Z L be represented by an inductance of 0 . 6 H in series with a 100 resistance. Find and s : The inductive impedance is jL = j (5 10 8 )(0 . 6 10 6 ) = j 300. So the load impedance is Z L = 100 + j 300 . Now = Z L Z Z L + Z = 100 + j 300 120 100 + j 300 + 120 = 0 . 62 + j . 52 = 0 . 808 " 40 Then s = 1 +   1   = 1 + 0 . 808 1 . 808 = 9 . 4 11.6. The propagation constant of a lossy transmission line is 1 + j 2 m 1 , and its characteristic impedance is 20 + j 0 at = 1 Mrad/s. Find L , C , R , and G for the line: Begin with Z = " R + jL G + jL = 20 R + jL = 400( G + jC ) (1) Then 2 = ( R + jL )( G + jC ) = (1 + j 2) 2 400( G + jC ) 2 = (1 + j 2) 2 (2) where (1) has been used. Eq. 2 now becomes G + jC = (1 + j 2) / 20. Equating real and imaginary parts leads to G = . 05 S / m and C = 1 / (10 ) = 10 7 = 0 . 1 F / m . Now, (1) becomes 20 = " R + jL 1 + j 2 20 20 = R + jL 1 + j 2 20 + j 40 = R + jL Again, equating real and imaginary parts leads to R = 20 / m and L = 40 / = 40 H / m ....
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This note was uploaded on 10/28/2008 for the course ECE 3025 taught by Professor Citrin during the Spring '08 term at Georgia Institute of Technology.
 Spring '08
 CITRIN

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