VSWR and Return Loss Converter

VSWR, Return Loss and Reflection Coefficient Converter

Enter any one of VSWR, return loss, reflection coefficient or mismatch loss and get the other three, with the percentage of power reflected and delivered and the two load resistances that would produce that standing wave ratio on your line.

VSWR and return loss

Any one → all four
Fill in that box; the other three are locked and show the equivalent figures.
1 is a perfect match. It can never be less than 1.
A bigger number is a better match. 20 dB means a hundredth of the power comes back.
0 is a perfect match, 1 is a short or an open.
How much of the forward power never reaches the load because it is reflected.
50 Ω for radio and test gear, 75 Ω for television and video.
What the transmitter is pushing into the line. Used only for the wattage figures.
A source driving a load through a line of the impedance you named. The dots show the forward wave; the reflected wave travels back along the same line and is given as a number rather than drawn, because two sets of dots on one wire would be unreadable. The load shown is the higher of the two resistances that produce this standing wave ratio — the lower one gives exactly the same reading.
1.8000: 1Example

A VSWR of 1.8 : 1 on a 50 Ω line carrying 100 W

Advertisement

One measurement, four names

Γ = (VSWR − 1) ÷ (VSWR + 1)     VSWR = (1 + |Γ|) ÷ (1 − |Γ|)
return loss = −20 log10|Γ| dB     |Γ| = 10−RL/20
power reflected = |Γ|²     mismatch loss = −10 log10(1 − |Γ|²) dB
for a purely resistive load:   Γ = (R − Z₀) ÷ (R + Z₀), so R = Z₀ × VSWR  or  R = Z₀ ÷ VSWR
Gamma
the reflection coefficient: the fraction of the VOLTAGE wave that comes back from the load
VSWR
voltage standing wave ratio — biggest to smallest voltage along the line. Always 1 or more
return loss
|Gamma| in decibels, quoted positive. Bigger is better
mismatch loss
the power that never reaches the load, in decibels. Even a 2:1 VSWR costs only half a decibel
Z0
the line’s characteristic impedance — 50 ohms for radio, 75 for video

Worked example

A VSWR of 1.8 : 1 on a 50 Ω line carrying 100 W
Γ = (1.8 − 1) ÷ (1.8 + 1) = 0.8 ÷ 2.8 = 0.28571
Return loss = −20 log₁₀(0.28571) = 10.88 dB
Power reflected is Γ² = 8.16%, so 8.163 W comes back and 91.84 W gets through
Mismatch loss = −10 log₁₀(1 − Γ²) = 0.3698 dB — less than half a decibel, which nobody on the other end could detect
On a 50 Ω line, a resistive load of either 90.0 Ω or 27.78 Ω gives exactly this VSWR: a standing wave ratio is a magnitude and cannot tell you which side of the line impedance the load is on

VSWR against return loss and the power you lose

VSWRReturn lossReflection coefficientPower reflectedMismatch loss
1.00 : 1∞0.00000.00%0.0000 dB
1.10 : 126.44 dB0.04760.23%0.0099 dB
1.20 : 120.83 dB0.09090.83%0.0360 dB
1.30 : 117.69 dB0.13041.70%0.0745 dB
1.50 : 113.98 dB0.20004.00%0.1773 dB
1.70 : 111.73 dB0.25936.72%0.3022 dB
2.00 : 19.54 dB0.333311.11%0.5115 dB
2.50 : 17.36 dB0.428618.37%0.8814 dB
3.00 : 16.02 dB0.500025.00%1.2494 dB
The column that surprises people is the last one. A 2 : 1 VSWR throws 11% of the power back, which sounds terrible, and costs half a decibel, which is nothing. The reason to chase VSWR is the transmitter’s protection circuit and the voltage on the line, not the signal at the far end.

What VSWR is actually telling you

When a transmission line meets a load that is not its own characteristic impedance, part of the wave comes back. The forward and reflected waves add along the line, in phase at some points and out of phase at others, which makes a standing pattern of voltage maxima and minima. The ratio of the biggest to the smallest is the voltage standing wave ratio, and it is 1 when nothing is reflected and infinite when everything is. Every other number on this page is a rearrangement of that one fraction.

Why there are two load resistances. For a purely resistive load, Γ = (R − Z₀)/(R + Z₀), and VSWR uses only the MAGNITUDE of Γ. A load of 90 Ω on a 50 Ω line and a load of 27.8 Ω give the same magnitude with opposite signs, so both read 1.8 : 1 on a meter. A VSWR meter cannot tell you which, and it cannot tell you anything about reactance either — a load of 50 Ω in series with a large reactance also shows a high VSWR. That is the whole reason an antenna analyser or a vector network analyser, which measure phase as well, are worth having.

Mismatch loss is smaller than everyone thinks. At 2 : 1, 11.1% of the forward power is reflected — and the mismatch loss is 0.51 dB. Half a decibel is a difference nobody has ever heard on a radio signal. Even 3 : 1 costs only 1.25 dB. What high VSWR actually does is three other things: it raises the voltage and current maxima on the line, which stresses the cable and the connectors; it makes the transmitter see an impedance it was not designed for, and a solid-state power amplifier will fold its output back or shut down to protect the output devices; and, on a lossy feeder, the reflected wave travels the cable twice, so the cable’s own loss is paid twice over. A long run of poor coax turns a bad match into real loss; a short run of good coax does not.

Sign conventions. Return loss is quoted as a positive number here and on every antenna analyser: 20 dB of return loss is better than 10 dB. A vector network analyser plots the same quantity as S₁₁, in negative decibels, so −20 dB on the screen is 20 dB of return loss. Both are correct; the sign is a convention about which direction you are counting, not a disagreement about physics. If a specification says “return loss better than 14 dB” it means a VSWR under 1.50 : 1.

To fix a mismatch rather than measure it, the L-network matching calculator designs the two-component network that does it. The decibels themselves are the decibel calculator, the forward power in dBm is the dBm to watts calculator, and the line that carries it on a circuit board is the microstrip impedance calculator. Antenna dimensions are on the antenna length calculator.

Advertisement

Frequently asked questions

What is a good VSWR?

Under 1.5 : 1 is excellent and under 2 : 1 is fine for almost any purpose — at 2 : 1 you are losing about half a decibel, which is undetectable. Above 3 : 1 the problem is not the lost power but the transmitter, which will start protecting itself.

How do I convert VSWR to return loss?

Work out Γ = (VSWR − 1)/(VSWR + 1), then return loss = −20 log₁₀|Γ| dB. A VSWR of 2 : 1 is 9.54 dB of return loss; 1.5 : 1 is 13.98 dB; 3 : 1 is 6.02 dB.

How much power does a 2:1 SWR lose?

11.11% of the forward power is reflected, which is a mismatch loss of 0.51 dB. In practice a transmitter with a matching network re-reflects most of that back towards the antenna, so the real loss is smaller still. The reflected power is not burned up — it goes back down the line.

Why does my VSWR meter give the same reading for two different antennas?

Because VSWR is a magnitude and throws away the sign and the phase. On a 50 Ω line, 100 Ω and 25 Ω both read 2 : 1, and a 50 Ω load with reactance added reads high too. Use an antenna analyser or a vector network analyser if you need to know which way to move.

Is return loss positive or negative?

By convention it is quoted positive, and bigger is better. The same quantity appears on a network analyser as S₁₁ in negative decibels. −20 dB of S₁₁ and 20 dB of return loss are the same match.

Does a high SWR damage the transmitter?

It can. A solid-state power amplifier’s output devices see a load impedance that swings with the mismatch, and both the voltage and the current can exceed their ratings. Modern radios detect high SWR and fold the power back; older or cheaper ones may not. Valve transmitters with a tuned output stage are much more tolerant.

Related calculators

References

  1. Pozar DM. Microwave Engineering, 4th ed. Wiley, 2012. §3.8 on microstrip (the quasi-TEM model, effective permittivity and the design equations) and §2.3 on the terminated lossless line — reflection coefficient, standing wave ratio and return loss.
  2. American Radio Relay League. The ARRL Antenna Book, and the ARRL Frequency Allocations chart for the US amateur bands used in the table below. The half-wave dipole’s traditional 468/f(MHz) feet is 492/f (a free-space half wave) times an end-effect factor of about 0.95.
  3. International Electrotechnical Commission. IEC 60027-3:2002, Letter symbols to be used in electrical technology — Part 3: Logarithmic and related quantities, and their units. The standard that defines the decibel and the neper, and the rule that a power quantity takes 10 log10 while a root-power (field) quantity takes 20 log10.
  4. Hewlett-Packard / Keysight. Application note 154, S-parameter design — the relation between S₁₁, reflection coefficient, return loss and standing wave ratio, and the sign convention each uses.