chap13 - .1 The parameters of a certain transmission line...

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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 µ H / m, C = 40 pF / m, G = 80 mS / m, and R = 20 /m . a) Find γ , α , β , λ , and Z : We use γ = √ ZY = p (R + jωL)(G + jωC) = 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 = α + jβ Therefore, α = 2 . 8 Np / m , β = 3 . 5 rad / m , and λ = 2 π/β = 1 . 8 m . Finally, Z = r Z Y = s R + jωL G + jωC = 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 − αL = e − ( 2 . 8 )( 20 ) = 4 . 8 × 10 − 25 or 4 . 8 × 10 − 23 percent! Then the phase shift is given by βL , which in degrees becomes φ = βL ± 360 2 π ² = ( 3 . 5 )( 20 ) ± 360 2 π 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 short-circuited 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 ³ Z 3 cos (βl) + jZ 2 sin (βl) Z 2 cos (βl) + jZ 3 sin (βl) ´ = j 60 tan (βl) where l = − z , and where the phase constant is β = 2 πc/f = 2 π( 3 × 10 8 )/( 6 × 10 7 ) = ( 2 / 5 )π rad / m. Now, with z = − 1 ( l = 1), we find Z in = j 60 tan ( 2 π/ 5 ) = j 184 . 6 . b) z = − 2 m: Z in = j 60 tan ( 4 π/ 5 ) = − j 43 . 6 c) z = − 2 . 5 m: Z in = j 60 tan ( 5 π/ 5 ) = d) z = − 1 . 25 m: Z in = j 60 tan (π/ 2 ) = j ∞ (open circuit) 13.3. The characteristic impedance of a certain lossless transmission line is 72 . If L = . 5 µ 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) β if f = 80 MHz: β = ω √ LC = 2 π × 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 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 = . 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 100-resistance. Find and s : The inductive impedance is jωL = 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 γ = + j . 2 π m − 1 at f = 60 MHz. a) Find L and C for the line: We have β = . 2 π = ω √ LC and Z = 50 = √ L/C . Thus β Z = ωC ⇒ C = β ωZ = . 2 π ( 2 π × 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 (βl) Z + jZ L tan (βl) ² We can then normalize the impedances with respect to Z and write z in = Z in Z = ± (Z L /Z ) + j tan (βl) 1 + j (Z L /Z ) tan (βl) ² = ± z L + j tan (βl) 1 + jz L tan (βl) ² where...
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chap13 - .1 The parameters of a certain transmission line...

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