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Unformatted text preview: ECE 3025A Problem Set # 7 Solutions 11.4. A lossless transmission line having Z = 120Ω is operating at ω = 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 μ 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 μ H in series with a 100Ω resistance. Find Γ and s : The inductive impedance is jωL = j (5 × 10 8 )(0 . 6 × 10 − 6 ) = j 300. So the load impedance is Z L = 100 + j 300 Ω. Now Γ = Z L − Z Z L + Z = 100 + j 300 − 120 100 + j 300 + 120 = 0 . 62 + j . 52 = 0 . 808 " 40 ◦ Then s = 1 +  Γ  1 −  Γ  = 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 Ω at ω = 1 Mrad/s. Find L , C , R , and G for the line: Begin with Z = " R + jωL G + jωL = 20 ⇒ R + jωL = 400( G + jωC ) (1) Then γ 2 = ( R + jωL )( G + jωC ) = (1 + j 2) 2 ⇒ 400( G + jωC ) 2 = (1 + j 2) 2 (2) where (1) has been used. Eq. 2 now becomes G + jωC = (1 + j 2) / 20. Equating real and imaginary parts leads to G = . 05 S / m and C = 1 / (10 ω ) = 10 − 7 = 0 . 1 μ F / m . Now, (1) becomes 20 = " R + jωL 1 + j 2 √ 20 ⇒ 20 = R + jωL 1 + j 2 ⇒ 20 + j 40 = R + jωL Again, equating real and imaginary parts leads to R = 20 Ω / m and L = 40 /ω = 40 μ H / m ....
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 Spring '08
 CITRIN
 #, 2J, Transmission line, Impedance matching, 6 W, 1MW

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