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Unformatted text preview: CHAPTER 13 13.1. The parameters of a certain transmission line operating at 6 10 8 rad/s are L = . 4 H / m, C = 40 pF / m, G = 80 mS / m, and R = 20 /m . a) Find , , , , and Z : We use = ZY = p (R + jL)(G + jC) = q [20 + j ( 6 10 8 )( . 4 10 6 ) ][80 10 3 + j ( 6 10 8 )( 40 10 12 ) ] = 2 . 8 + j 3 . 5 m 1 = + j Therefore, = 2 . 8 Np / m , = 3 . 5 rad / m , and = 2 / = 1 . 8 m . Finally, Z = r Z Y = s R + jL G + jC = s 20 + j 2 . 4 10 2 80 10 3 + j 2 . 4 10 2 = 44 + j 30 b) If a voltage wave travels 20 m down the line, what percentage of the original amplitude remains, and by how many degrees is it phase shifted? First, V 20 V = e L = e ( 2 . 8 )( 20 ) = 4 . 8 10 25 or 4 . 8 10 23 percent! Then the phase shift is given by L , which in degrees becomes = L 360 2 = ( 3 . 5 )( 20 ) 360 2 = 4 . 10 3 degrees 13.2. A lossless transmission line with Z = 60 is being operated at 60 MHz. The velocity on the line is 3 10 8 m/s. If the line is shortcircuited at z = 0, find Z in at: a) z = 1m: We use the expression for input impedance (Eq. 12), under the conditions Z 2 = 60 and Z 3 = 0: Z in = Z 2 Z 3 cos (l) + jZ 2 sin (l) Z 2 cos (l) + jZ 3 sin (l) = j 60 tan (l) where l = z , and where the phase constant is = 2 c/f = 2 ( 3 10 8 )/( 6 10 7 ) = ( 2 / 5 ) rad / m. Now, with z = 1 ( l = 1), we find Z in = j 60 tan ( 2 / 5 ) = j 184 . 6 . b) z = 2 m: Z in = j 60 tan ( 4 / 5 ) = j 43 . 6 c) z = 2 . 5 m: Z in = j 60 tan ( 5 / 5 ) = d) z = 1 . 25 m: Z in = j 60 tan (/ 2 ) = j (open circuit) 13.3. The characteristic impedance of a certain lossless transmission line is 72 . If L = . 5 H / m, find: a) C : Use Z = L/C , or C = L Z 2 = 5 10 7 ( 72 ) 2 = 9 . 6 10 11 F / m = 96 pF / m 213 13.3b) v p : v p = 1 LC = 1 p ( 5 10 7 )( 9 . 6 10 11 ) = 1 . 44 10 8 m / s c) if f = 80 MHz: = LC = 2 80 10 6 1 . 44 10 8 = 3 . 5 rad / m d) The line is terminated with a load of 60 . Find and s : = 60 72 60 + 72 = . 09 s = 1 +   1   = 1 + . 09 1 . 09 = 1 . 2 13.4. 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 = . 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 100resistance. Find and s : The inductive impedance is jL = j ( 5 10 8 )( . 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 = . 62 + j . 52 = . 808 6 40 Then s = 1 +   1   = 1 + . 808 1 . 808 = 9 . 4 13.5. Two characteristics of a certain lossless transmission line are Z = 50 and = + j . 2 m 1 at f = 60 MHz. a) Find L and C for the line: We have = . 2 = LC and Z = 50 = L/C . Thus Z = C C = Z = . 2 ( 2 60 10 6 )( 50 ) = 1 3 10 10 = 33 . 3 pF / m Then L = CZ 2 = ( 33 . 3 10 12 )( 50 ) 2 = 8 . 33 10 8 H / m = 83 . 3 nH / m . b) A load, Z L = 60 + j 80 is located at z = 0. What is the shortest distance from the load to a point at which Z in = R in + j 0? I will do this using two different methods: The Hard Way : We use the general expression Z in = Z Z L + jZ tan (l) Z + jZ L tan (l) We can then normalize the impedances with respect to Z and write z in = Z in Z = (Z L /Z ) + j tan (l) 1 + j (Z L /Z ) tan (l) = z L + j tan (l) 1 + jz L tan (l) where...
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This note was uploaded on 06/10/2008 for the course EE 325 taught by Professor Brown during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Brown
 Electromagnet

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