This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 shortcircuited 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...
View
Full
Document
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.
 Spring '10
 Kaplan

Click to edit the document details