Paul seems to explain *return loss* like this: You throw a glass of
water straight up into the air. Return loss is the difference between how much
water you tossed up there and how much lands back on you... or something like that!

Introduction:

Return loss is the difference in dB between forward and reflected power measured at any given point in an RF system. Like SWR, return loss does not vary with the power level at which it is measured.

Computing Return Loss:

As stated above, Return Loss is *the difference (in dB) between
the forward and reflected power, measured at any given point
in an RF system.* Typically, but not
always, this point of measurement is the input of a "component
exhibiting a frequency response" used in an RF system.

Note that this subtraction uses logarithm-based quantities (dB). This is radically different from a subtraction of Watts or milliWatts. Remember "0 dBm" means "1 mW", while 0 milliWatts means "no power" (none, zero, zip, nada). Running some example calculations through the power level addition (use this calculator) might help with understanding this principle.

A "component exhibiting a frequency response" is anything with characteristics that vary with frequency. An example with which most hams are familiar is a single-band, high-gain yagi antenna. It is generally understood that such an antenna is GENERALLY designed to work over a limited range of frequencies, or a bandwidth, within a specific ham band.

Implications of SWR:

Within this useful frequency range (bandwidth), the SWR (Standing Wave Ratio) is relatively low, typically 1.5:1 or less, 1.0:1 being the best possible SWR. As one varies the frequency outside the useful bandwidth, the SWR increases. Depending on several variables, discussion of which is beyond the scope of this article, it is generally accepted that an antenna is useful only within the bandwidth at which the SWR is 2.0:1 or lower. 1.5:1 is often cited as the maximum acceptable SWR.

Note that the SWR of a yagi antenna is a mathmatical relationship between the forward power and the reflected power measured at the driven element of the antenna. Here, again, the relationship is linear, not logarithmic, and the less reflected power (the better situation), the lower the value of SWR. With return loss, less reflected power means smaller number of dBm (or dBW; see below) to be subtracted off, resulting in larger return loss for the better situations.

Using our antenna example and the definition of return loss given above, return loss is the difference between forward and reflected power, in dB, generally measured at the input to the coaxial cable connected to the antenna. Ideally, one would want to measure the antenna input directly, but since said antenna is generally at the top of a tower, this is not practical.

The accepted practice is to make the measurement as mentioned, at the input to the coaxial cable going to the antenna. See the caveat below regarding the pitfalls of this practice.

Calculations:

Since return loss is in dB,
one must convert power levels to dB units, generally either dBm
(decibels referenced to one milliWatt**;** 1 mW = 0dBm) or dBW
(decibels referenced to one Watt**;** 1W = 0dBW = 30dBm).

To calculate return loss:

- Convert forward and reflected power to dB (as above).
- Subtract reflected power in dB from forward power in dB.

- dBm = 10logP where p is in milliWatts

- dBW = 10logP where p is in Watts.

*Note:* Subtract dBm from dBm and dBW from dBW, do NOT mix dBm and dBW!!

Example - The best case: Perfect match (SWR=1

Fwd pwr = 1500 Watts (61**.**76 dBm); Ref'd pwr = 0 Watts

This example appears to break down mathematically because it is not possible to calculate 10log(0). It does have a solution, however (see math refresher, below). We can say 10log(0) is, to all practical purposes, equivalent to negative infinity.

- Return Loss = 61

Remember your arithmetic: Subtracting a negative is like adding a positive ("two negatives make a positive!).

Also, remember that any finite number added to infinity doesn't increase infinity enough to make a difference, so 61.76 + infinity = infinity, as shown.

Example - The outstanding case: Perfect match (SWR=1.05:1)

Fwd pwr = 1500W (31**.**76 dBW); Ref'd pwr = 1W (0 dBW)

- Return Loss = 31

Example - The not-so-great case: SWR=1.92:1

Forward power = 100W (20 dBW); Reflected Power = 10W (10 dBW)

- Return Loss = 20 dBW - 10 dBW = 10 dB

Example - The worst case: SWR=

Fwd pwr = Ref'd pwr = 100W (50 dBm)

This situation illustrates a dead short or complete open, with 100% of the forward power reflected back to the source.

- Return Loss = 50 dBm - 50 dBm = 0 dB

The relationship between practical values of SWR and the associated return loss is shown below. Examples of forward and reflected powers are given and return loss calculated by subtracting reflected power in dB from forward power in dB, per the examples above.

*Note:* The difference between all forward and reflected
powers **in dB** at each SWR is identical. Note also that as in the
four examples above, return loss increases as SWR decreases, i.e.
high return loss is good and low SWR is good.

Here is a table of SWR forward/reflected relationships showing return loss:

SWR | Forward Power W / dBm / dBW |
Reflected Power W / dBm / dBW |
Return Loss dB |
---|---|---|---|

2:1 |
10 / 40 / 10 | 1.1 / 30.5 / 0.5 |
9.5 |

100 / 50 / 20 | 11 / 40.5 / 10.5 | ||

1000 / 60 / 30 | 111 / 50.5 / 20.5 | ||

1500 / 61.8 / 31.8 |
166.7 / 52.3 / 22.3 | ||

1.92:1 |
10 / 40 / 10 | 1 / 30 / 0 | 10 |

100 / 50 / 20 | 10 / 40 / 10 | ||

1000 / 60 / 30 | 100 / 50 / 20 | ||

1500 / 61.8 / 31.8 |
150 / 51.8 / 21.8 | ||

1.67:1 |
10 / 40 / 10 | 0.6 / 28 / -2 |
12 |

100 / 50 / 20 | 6.3 / 38 / 8 | ||

1000 / 60 / 30 | 63.1 / 48 / 18 | ||

1500 / 61.8 / 31.8 |
94.7 / 49.8 / 29.8 | ||

1.5:1 |
10 / 40 / 10 | 0.4 / 26 / -4 |
14 |

100 / 50 / 20 | 4 / 36 / 6 | ||

1000 / 60 / 30 | 39.8 / 46 / 16 | ||

1500 / 61.8 / 31.8 |
59.7 / 46.8 / 26.8 | ||

1.38:1 |
10 / 40 / 10 | 0.25/ 24 / -6 |
16 |

100 / 50 / 20 | 2.5/ 34 / 4 | ||

1000 / 60 / 30 | 25.1 / 44 / 14 | ||

1500 / 61.8 / 31.8 |
37.7 / 45.8 / 25.8 | ||

1.29:1 |
10 / 40 / 10 | 0.16 / 22 / -8 |
18 |

100 / 50 / 20 | 1.6 / 32 / 2 | ||

1000 / 60 / 30 | 15.9 / 42 / 12 | ||

1500 / 61.8 / 31.8 |
23.8 / 45.8 / 25.8 | ||

1.22:1 |
10 / 40 / 10 | 0.1 / 20 / -10 |
20 |

100 / 50 / 20 | 1 / 30 / 0 | ||

1000 / 60 / 30 | 10 / 40 / 10 | ||

1500 / 61.8 / 31.8 |
15 / 41.8 / 11.8 | ||

1.1:1 |
10 / 40 / 10 | 0.025 / 14 / -16 |
26 |

100 / 50 / 20 | 0.25 / 24 / -6 | ||

1000 / 60 / 30 | 2.51/ 34 / 4 | ||

1500 / 61.8 / 31.8 |
3.77 / 35.8 / 5.8 |

Values are rounded off, e.g., 1.76 to 1.8, 40.46 to 40.5, etc.

**Caveat:** Making
antenna measurements at the input to the transmission line (coax)
can result in an artifically good reading of antenna SWR. The higher
the loss (attenuation), the worse the disparity. Run some sample
calculations, varying transmission line attenuation, at Antenna
System Paramters to see how coax loss affects the actual
value of SWR at an antenna versus that measured at the input of the
associated transmission line!

**Math:**
One can do myriad example calculations of 10 log(p) with
powers of ever decreasing magnitude, and it is clear that as power
gets smaller and smaller, approaching 0, 10log(p) becomes more and
more negative (the negative number becomes larger and larger),
approaching negative infinity.

Mathematically, anything "close" to infinity is so "huge" that it cannot be differentiated FROM infinity, so we can say 10log(0) is, to all practical purposes, equivalent to negative infinity.

The valuation of return loss in the "zero reflected power" situation can be verified as well by a quick refresher on the exponential basis for logarithms:

X^{4} **/** X^{2} = X^{4 - 2} = X^{2}

Notice the division of X^{4} by
X^{2} is a subtraction of exponents. Logarithms are nothing
more than exponents. For the illustrations below, recall that
anything raised to an exponent of "zero" is equal to one, and
anything not showing an exponent actually has an exponent of one:

X = X^{1}

Also note that "log(p)" means "log_{10}(p)".

log(10^{0}) = log(1) = 0

log(10^{1}) = log(10) = 1

log(10^{2}) = log(100) = 2

log(10^{3}) = log(1000) = 3

etc.!

The discussion above should remind the reader that subtraction of logaritms is the same as division, i.e.

log(X^{4}) - log(X^{2}) = log(X^{4}**/**X^{2})

So in the best case example above with "zero reflected power:"

- 10log(p

We know division by zero is undefined. Recall from the discussion above about "approaching zero" and "approaching infinity", that in division, as the divisor gets smaller and smaller (approaching zero), the quotient gets larger and larger (approaching infinity).

- Return Loss = 10log(1500/0)

Q.E.D. *Paul*

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