---
title: "Line at 120 MHz, γ v Z0"
subject: "Electromagnetism"
chapter: 5
key: "semester_3/electromagnetism:line-120mhz-primary-constants"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/line-120mhz-primary-constants
site: BCT Engineering
---

# Line at 120 MHz, γ v Z0

Electromagnetism · Chapter 5: Transmission lines

- Asked in: 2076 Chaitra, 3+3+2 marks
- Video: [Transmission Line Example on Characteristics Impedance, Attenuation Constant & Phase Constant](https://www.youtube.com/watch?v=3LP1Lu7MwUs) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/line-120mhz-primary-constants
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-5

## Question

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).

## Short answer

γ = √(ZY) = 0.29 + j3.28 m⁻¹. v = ω/β = 2.3×10⁸ m/s. Z₀ = √(Z/Y) ≈ 69 Ω.

## Answer

Given: f = 120 MHz, R = 20 Ω/m, L = 0.3 μH/m, C = 63 pF/m, G = 4.2 mS/m.

```text
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°

```text
γ = 0.290 + j3.278 per metre
α = 0.290 Np/m,   β = 3.278 rad/m
```

#### (b) Velocity of propagation

```text
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°)

```text
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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is gamma found from R, L, G, C?
   A. gamma = sqrt(R G) + j sqrt(L C)
   B. gamma = (R + jwL)(G + jwC)
   C. gamma = sqrt((R + jwL)(G + jwC))
   D. gamma = sqrt((R + jwL)/(G + jwC))
   Answer: C. gamma = sqrt(ZY), giving 0.29 + j3.28 per metre. (Key point: Propagation coefficient)

2. What is the wave velocity on the line?
   A. v = 3x10^8 m/s
   B. v = w/beta = 7.54x10^8/3.278 = 2.3x10^8 m/s
   C. v = 1/(alpha) = 3.4 m/s
   D. v = w/alpha = 2.6x10^9 m/s
   Answer: B. v = w/beta, not alpha. (Key point: Velocity)

3. How is Z0 found and what is it?
   A. Z0 = sqrt(Z/Y) = about 69 ohm, almost real
   B. Z0 = sqrt(Z Y) = 3.29 ohm
   C. Z0 = Z/Y = 4762 ohm
   D. Z0 = Z + Y
   Answer: A. Z0 = sqrt(Z/Y), with Z and Y at nearly equal angles. (Key point: Characteristic impedance)

4. Why is Z0 purely real here?
   A. Because R is much larger than G
   B. Because the line is lossless
   C. Because frequency is 120 MHz
   D. R/L = G/C (both 6.7x10^7), so the line is distortionless
   Answer: D. When R/L = G/C the angles of Z and Y match and Z0 = sqrt(L/C) is real. (Key point: Why Z0 is real)

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