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Coaxial cable: loss and electrical length
50 Ω, attenuation and velocity factor describe different properties of a feed line.
Practice2 min readReviewed 15/10/2026
HB3YAQ · Ajdin HadzicLearning to take awayhttps://hb3yaq.ch/en/knowledge/feed-lines
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WHAT YOU WILL LEARN
- Select relevant data-sheet values.
- Calculate loss and phase length separately.
50 Ω is not DC resistance
A line’s characteristic ImpedanceImpedance is complex opposition to alternating current: it includes resistance and a frequency-dependent reactance. Antenna input impedance is therefore more than a DC resistance reading.Example & full explanation → describes voltage divided by current for a travelling wave. It arises from distributed geometry and materials. An ohmmeter at an open cable end therefore does not simply measure the printed 50 Ω. DC resistance, characteristic impedance and input impedance are different quantities.
With a matched termination, an ideal line’s input impedance remains the same regardless of length. With mismatch, electrical length changes the impedance seen at the input. A Coaxial cableCoaxial cable carries a signal using an inner conductor and a concentric outer conductor. The dielectric and construction determine characteristic impedance, losses and propagation speed.Example & full explanation → is therefore a connection, a lossy component and potentially an impedance transformer at the same time.
What reaches the end of the cable?
- A = 10 dB/100 m × 30 m = 3 dB
- Pout = 100 W × 10^(−3/10) ≈ 50.12 W
- Use the attenuation at the actual frequency. Mismatch is not included in this example.
Quick check
Change this cable example →ARRL · Feed Lines ↗What remains at twice the length with the same assumptions?
Reveal explanation
60 m gives 6 dB: about 25.12 W from 100 W. Attenuation in dB doubles; remaining power approximately halves again.
Attenuation at the right frequency
Cable data sheets usually specify attenuation at several frequencies and a reference length. Use the value for your operating frequency. An example 8.5 dB per 100 m gives 1.7 dB over 20 m. From 100 W, about 67.6 W reaches the end in the matched model; connectors and additional mismatch loss are excluded.
Attenuation also affects reception. A lossy cable before the first amplifier can degrade weak signals. More GainGain describes the ratio of output to input level. In a receiver it can occur before or after digitisation and is often expressed in dB.Example & full explanation → afterwards does not simply restore information already lost. Frequency, length and the whole receiving chain matter, rather than a thick outer jacket alone.
↑ Back to contentsVelocity factor and matching lines
Velocity factor describes propagation speed in the cable relative to c. A quarter-wave line has physical length c·VF/(4f). At 145 MHz with VF = 0.66, that is about 34.1 cm. Connectors and attachments affect the reference plane, so a narrowband matching line should be measured and tuned after assembly.
↑ Back to contentsExplore these relationships in the signal-path lab →
From reading to experimenting
Electrical cable length
145 MHz and VF = 0.66 give a quarter-wave line of approximately 34.1 cm.
Open example with starting values →Frequency: 145 MHz · Velocity / shortening factor: 0.66Cable loss
8.5 dB/100 m and 20 m give 1.7 dB. About 67.61 W remains from 100 W.
Open example with starting values →Attenuation at your frequency: 8.5 dB/100 m · Cable length: 20 m · Input power: 100 WThen change exactly one value. Compare the result and use the formula to explain why it changes. The model limits below the calculator also apply here.
Need a term explained?
These explanations help you keep reading.
Got the idea?
Can an antenna tuner at the radio generally eliminate cable loss from a poorly matched feed line?
Show answer & explanation
No. It can present an appropriate load to the radio; conditions and losses on the following line remain a separate issue.