- 2076 Chaitra · 3+3+2 marks
Line at 120 MHz, γ v Z0
Short answer
γ = √(ZY) = 0.29 + j3.28 m⁻¹. v = ω/β = 2.3×10⁸ m/s. Z₀ = √(Z/Y) ≈ 69 Ω.
A transmission line operating at 120 MHz has R = 20 Ω/m, L = 0.3 μH/m, C = 63 pF/m and G = 4.2 mS/m. Find (a) propagation coefficient (γ) (b) velocity of wave propagation on the line (v) (c) characteristic impedance (Z0).
Given: f = 120 MHz, R = 20 Ω/m, L = 0.3 μH/m, C = 63 pF/m, G = 4.2 mS/m.
ω = 2π × 120×10⁶ = 7.5398×10⁸ rad/s
ωL = 7.5398×10⁸ × 0.3×10⁻⁶ = 226.19 Ω/m
ωC = 7.5398×10⁸ × 63×10⁻¹² = 0.047501 S/m
Z = R + jωL = 20 + j226.19 Ω/m (|Z| = 227.07, ∠84.95°)
Y = G + jωC = 0.0042 + j0.047501 S/m (|Y| = 0.047687, ∠84.94°)
(a) Propagation coefficient
γ = √(ZY) = √(227.07 × 0.047687 ∠(84.95° + 84.94°)) = √(10.829 ∠169.89°) = 3.2907 ∠84.95°
γ = 0.290 + j3.278 per metre
α = 0.290 Np/m, β = 3.278 rad/m
(b) Velocity of propagation
v = ω/β = 7.5398×10⁸ / 3.278 = 2.30×10⁸ m/s
(c) Characteristic impedance
Z₀ = √(Z/Y) = √(227.07/0.047687 ∠(84.95° − 84.94°)) = √(4761.8 ∠0.01°)
Z₀ ≈ 69.0 + j0 Ω ≈ 69 Ω
(The line is almost exactly distortionless: R/L = 6.7×10⁷ and G/C = 6.7×10⁷, so Z₀ is purely real.)
Answer: γ = 0.29 + j3.28 m⁻¹; v = 2.3×10⁸ m/s; Z₀ ≈ 69 Ω.
✓ Answer checked by Claude Opus 5.5 (high effort)
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