chapter11_7th_solution-vectorist

chapter11_7th_solution-vectorist - CHAPTER 11 11.1. The...

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Unformatted text preview: CHAPTER 11 11.1. The parameters of a certain transmission line operating at 6 10 8 rad/s are L = 0 . 4 H / m, C = 40 pF / m, G = 80 S / m, and R = 20 / m. a) Find , , , , and Z : We use = ZY = p ( R + jL )( G + jC ) = p [20 + j (6 10 8 )(0 . 4 10 6 )][80 10 6 + j (6 10 8 )(40 10 12 )] = 0 . 10 + j 2 . 4 m 1 = + j Therefore, = 0 . 10 Np / m , = 2 . 4 rad / m , and = 2 / = 2 . 6 m . Finally, Z = r Z Y = s R + jL G + jC = s 20 + j 2 . 4 10 2 80 10 6 + j 2 . 4 10 2 = 100 j 4 . 0 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 (0 . 10)(20) = 0 . 13 or 13 percent Then the phase shift is given by L , which in degrees becomes = L 360 2 = (2 . 4)(20) 360 2 = 2 . 7 10 3 degrees 11.2. 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 short-circuited at z = 0, nd 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 60tan( 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 nd Z in = j 60tan(2 / 5) = j 184 . 6 . b) z = 2 m: Z in = j 60tan(4 / 5) = j 43 . 6 c) z = 2 . 5 m: Z in = j 60tan(5 / 5) = 0 d) z = 1 . 25 m: Z in = j 60tan( / 2) = j (open circuit) 11.3. The characteristic impedance of a certain lossless transmission line is 72 . If L = 0 . 5 H / m, nd: a) C : Use Z = p L/C , or C = L Z 2 = 5 10 7 (72) 2 = 9 . 6 10 11 F / m = 96pF / m 1 11.3b) 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 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, nd: a) L : With Z = p L/C and v = 1 / LC , we nd L = Z /v = 120 / 2 . 4 10 8 = 0 . 50 H / m . b) C : Use Z v = p 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 6 40 Then s = 1 + | | 1 | | = 1 + 0 . 808...
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This note was uploaded on 02/07/2010 for the course EE FMW taught by Professor Kaplan during the Spring '10 term at Johns Hopkins.

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chapter11_7th_solution-vectorist - CHAPTER 11 11.1. The...

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