---
title: "Primary constants from Z0, α, β"
subject: "Electromagnetism"
chapter: 5
key: "semester_3/electromagnetism:primary-constants-from-measurements-1khz"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/primary-constants-from-measurements-1khz
site: BCT Engineering
---

# Primary constants from Z0, α, β

Electromagnetism · Chapter 5: Transmission lines

- Asked in: 2073 Shrawan, 8 marks
- Video: [A Tx line operating at 500 MHz has Zo = 80 Ω, α = 0.04 Np/m, β = 1.5 rad/m. Find R,L, G and C](https://www.youtube.com/watch?v=vxZP3A6UnkU) — ECE by MVK
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/primary-constants-from-measurements-1khz
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-5

## Question

Determine the primary constants (R, L, C and G) on the transmission line when the measurement on the line at 1 kHz gave the following results: Z0 = 710∠−16°, α = 0.01 neper/m and β = 0.035 rad/m.

## Short answer

Using Z = γZ₀ and Y = γ/Z₀ gives R ≈ 13.7 Ω/m, L ≈ 3.49 mH/m, C ≈ 8.16 nF/m, and G ≈ 0.

## Answer

Given at f = 1 kHz: Z₀ = 710∠−16° Ω, α = 0.01 Np/m, β = 0.035 rad/m.

```text
ω = 2π × 1000 = 6283.2 rad/s
γ = α + jβ = 0.01 + j0.035 = 0.03640 ∠74.05° m⁻¹
Z₀ = 710∠−16° = 682.50 − j195.70 Ω
```

Relations: Z₀ = √(Z/Y), γ = √(ZY), so

```text
Z = R + jωL = γ Z₀
Y = G + jωC = γ / Z₀
```

#### Series impedance

```text
Z = 0.03640∠74.05° × 710∠−16°
  = 25.84 ∠58.05°
  = 13.67 + j21.93 Ω/m
```

So R = 13.67 Ω/m and ωL = 21.93 Ω/m:

```text
L = 21.93 / 6283.2 = 3.49×10⁻³ H/m
```

#### Shunt admittance

```text
Y = 0.03640∠74.05° / 710∠−16°
  = 5.127×10⁻⁵ ∠90.05°
  = −4.9×10⁻⁸ + j5.127×10⁻⁵ S/m
```

The real part is essentially zero (G ≈ 0, a numerical rounding result), and

```text
C = 5.127×10⁻⁵ / 6283.2 = 8.16×10⁻⁹ F/m
```

**Answer: R ≈ 13.7 Ω/m, L ≈ 3.49 mH/m, C ≈ 8.16 nF/m, G ≈ 0.**

## Check yourself

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

1. Given Z0 and gamma, how are Z = R + jwL and Y = G + jwC found?
   A. Z = Z0/gamma and Y = gamma Z0
   B. Z = gamma/Z0 and Y = gamma Z0
   C. Z = Z0^2 and Y = gamma^2
   D. Z = gamma Z0 and Y = gamma/Z0
   Answer: D. gamma = sqrt(ZY) and Z0 = sqrt(Z/Y) give Z = gamma Z0 and Y = gamma/Z0. (Key point: Method)

2. What are R and L?
   A. R = 21.93 ohm/m, L = 13.67 mH/m
   B. R = 25.84 ohm/m, L = 3.49 mH/m
   C. R = 13.67 ohm/m, L = 3.49 mH/m
   D. R = 13.67 ohm/m, L = 21.93 H/m
   Answer: C. Z = 25.84 angle 58.05 = 13.67 + j21.93, and L = 21.93/6283 = 3.49 mH/m. (Key point: R and L)

3. What is C?
   A. About 5.13 uF/m
   B. About 8.16 nF/m, from imaginary Y = 5.13x10^-5 S/m divided by w
   C. About 8.16 pF/m
   D. About 0.82 uF/m
   Answer: B. C = 5.127x10^-5/6283 = 8.16x10^-9 F/m. (Key point: C)

4. What is G?
   A. About 0, since the real part of Y is negligible
   B. 13.67 S/m
   C. 8.16 nF/m
   D. 0.0364 S/m
   Answer: A. The real part of Y is -4.9x10^-8 S/m, rounding noise, so G is about 0. (Key point: G)

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