chapter11_7th_solution-vectorist

# chapter11_7th_solution-vectorist - .1 The parameters of a...

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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 µ H / m, C = 40 pF / m, G = 80 µ S / m, and R = 20 Ω / m. a) Find γ , α , β , λ , and Z : We use γ = √ ZY = p ( R + jωL )( G + jωC ) = 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 = α + jβ Therefore, α = 0 . 10 Np / m , β = 2 . 4 rad / m , and λ = 2 π/β = 2 . 6 m . Finally, Z = r Z Y = s R + jωL G + jωC = s 20 + j 2 . 4 × 10 2 80 × 10 − 6 + j 2 . 4 × 10 − 2 = 100 − j 4 . 0 Ω 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 − (0 . 10)(20) = 0 . 13 or 13 percent Then the phase shift is given by βL , which in degrees becomes φ = βL µ 360 2 π ¶ = (2 . 4)(20) µ 360 2 π ¶ = 2 . 7 × 10 3 degrees 11.2. A lossless transmission line with Z = 60 Ω 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( βl ) + jZ 2 sin( βl ) Z 2 cos( βl ) + jZ 3 sin( βl ) ¸ = j 60tan( 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 ±nd Z in = j 60tan(2 π/ 5) = j 184 . 6 Ω . b) z = − 2 m: Z in = j 60tan(4 π/ 5) = − j 43 . 6 Ω c) z = − 2 . 5 m: Z in = j 60tan(5 π/ 5) = 0 d) z = − 1 . 25 m: Z in = j 60tan( π/ 2) = j ∞ Ω (open circuit) 11.3. The characteristic impedance of a certain lossless transmission line is 72 Ω. If L = 0 . 5 µ 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) 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 Ω. Find Γ and s : Γ = 60 − 72 60 + 72 = − . 09 s = 1 + | Γ | 1 − | Γ | = 1 + . 09 1 − . 09 = 1 . 2 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, ±nd: a) L : With Z = p L/C and v = 1 / √ LC , we ±nd L = Z /v = 120 / 2 . 4 × 10 8 = 0 . 50 µ 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 µ H in series with a 100-Ω 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 6 40 ◦ Then s = 1 + | Γ | 1 − | Γ | = 1 + 0 . 808...
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chapter11_7th_solution-vectorist - .1 The parameters of a...

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