continuousWave --> Whaler --> Reference

### by James W. Hebert

A common situation in boating is the use of VHF Marine Band radios for communication between two recreational vessels. This article analyzes the radio circuit using typical radios, antennas, and distances encountered in the VHF Marine Radio service. The signal levels are analyzed in technical but simplified terms. Antenna height will be found to be the primary determinant of communication range between two recreational vessels.

### Typical Marine VHF Radio Circuit (156 MHz) of Ten Miles

Let us look at a common situation: two recreational vessels with 25-watt radios attempting to communication with each other over a distance of ten miles of open water. The signal level available at the receiver will determine the quality of the communication circuit. Received signal level is a function of:

• Transmitter Power
• Transmission Line Loss to Antenna
• Transmit Antenna Gain
• Path Loss
• Transmission Line Loss to Receiver

Once the received signal level has been calculated, it can then be assessed by comparison to the receiver's rated sensitivity to determine how well the communication link will work. The math involved is not particularly difficult.

A general formula to determine the signal power Pr available at the receiver input:

```
Pr = Pt - Lp + Gt + Gr - Lt - Lr

where

Pt = transmitter power output
(dBm or dBW, same units as Pr)

Lp = free space path loss between isotropic antennas (dB)

Gt = transmit antenna gain (dBi)

Gr = receive antenna gain (dBi)

Lt = transmission line loss between transmitter
and transmit antenna (dB)

Lr = transmission line loss between receive antenna
```

The individual terms can be readily calculated or determined from specifications. Let us investigate each one and find an appropriate value for them in this situation.

#### Pt Transmitter Power Output Relative to One Milliwatt

Expressing Pt of 25 watts in dBm, we have:

```Pt  = 25 watts
= 10log(25) dBw
= 14 dBw
= 14 + 30 (to convert to reference to 1 milliwatt)
Pt  = 44 dBm
```

#### Lp Free Space Path Loss Between Isotropic Antennas in dB

Path loss may be calculated by this formula:

```Lp = 36.6 + 20log(f) + 20log(d)  (f in MHz, d in miles)

= 36.6 + 20log(156) + 20log(10)

= 36.6 + 43.9 + 20

= 100.5 dB
```

It is interesting to note that path loss increases 6 dB with each doubling of the distance. The path loss for a 20-mile circuit will be only 106.5 dB. Keep this in mind when we calculate the signal margin in our circuit below .

#### GtTransmit Antenna Gain in dB Relative to Isotropic

The transmit antenna is assumed to be a typical marine whip with gain of about 3-dB. This is a reasonable expectation from most marine whip antennas.

```Gt = 3 dB (typical 4-foot marine antenna)
```

#### Gr Receive Antenna Gain In dBi

The receive antenna is assumed to be a typical marine whip with gain of about 3-dB. This is a reasonable expectation from most marine whip antennas.

```Gr = 3 dB (typical 4-foot marine antenna)
```

#### Lt Transmission Line Loss Between Transmitter and Transmit Antenna

Here we will assume the typical 20-feet of RG-58/U coax cable. From the specifications of a high-quality cable manufacturer a figure of 1 dB loss is obtained. Thus:

```Lt = 1 dB
```

The same feed line type and length as used in the transmitter are assumed to be used at the receiver. Thus

```Lr = 1 dB
```

#### Pr Power At Receiver in dBm

Now that we have evaluated all the terms, we just have to add them all to obtain the power at the receiver:

```Rr = Pt - Lp + Gt + Gr - Lt - Lr

Rr = 44 - 100.5 + 3 + 3 - 1 - 1

Rr = - 52.5 dBm
```

Thus in the case of two VHF Marine Radios separated by ten miles and transmitting with 25-watts of power, at the input of the distant receiver each should produce a signal of

```- 52.5 dBm
```

How strong is a signal of -52.5 dBm? It is less than one-ten-thousandths of a milliwatt of energy. However, in terms of radio signals, it is really quite a strong signal. Will this be enough signal for good communications? What will be the signal-to-noise ratio of the recovered audio from the receiver? The answers to these questions depend on the receiver and its sensitivity.

Receiver sensitivity is often specified in terms of the signal level at the antenna input necessary to produce audio output with a particular signal-to-noise ratio. Because when making measurements of this type it is difficult to distinguish between noise and distortion, the more correct term for the measurement is signal-to-noise-and-distortion, abbreviated SINAD.

Let us assume we have a typical VHF Marine Radio transceiver like the ICOM M-401 model which has a rated sensitivity of 0.25 uV for a SINAD of 12 dB. To make the math easier, let's de-rate that to 1.0 uV, or 0.000001 volt. This will also account for some degradation of the receiver sensitivity from local noise sources, such as the ignition noise from the outboard motor or noise emitted by other electronic devices in close proximity like a fish finder. (If you have a c.1980 outboard motor without resistor spark plugs and a very hot spark voltage, the receiver sensitivity can be severely degraded, to the point were even the strongest signals will be interfered with by the ignition sparking noise.) First, we need to convert this voltage level into power in dBm at the receiver input terminal assuming a 50-ohm antenna.

```   P = IE and
I = E/R thus
P = E2/R
If
E = 1 uV = 10-6 Volt
and
R = 50 Ohms

then

P =  (10-6)2 / 50

=  10-12/50

=  2 X 10-14 watts

= 10log(2 X 10-14) dBw (reference to 1 watt)

= 10 [log(2) + log(10-14)]

= 10 (0.3 - 14)

= 10 (-13.7)

= -137 dBw + 30 (convert to dBm or reference to 1 milliwatt)

= -107 dBm
```

A generic solution for converting micro-volts to dBm is shown in a companion article.

This is the minimum or threshold signal level necessary to produce communication quality recovered audio. Now we compare this to the actual received signal level calculated above to see how they compare. In this case, the calculated received signal level is quite a bit stronger than the threshold level. Thus the receiver will have a very ample signal above threshold:

```Margin = -52.5 - -107
=  54.5 dB above minimum sensitivity
```

A SINAD of 12dB is not particularly excellent reception but usually is sufficient for communication use. This would be the minimum useful for communication quality reception. As the received signal level increases, so will the SINAD. When using frequency modulation techniques, the SINAD will improve very rapidly with increasing signal level. As the SINAD improves, the apparent noise in the circuit decreases. This is often referred to as "quieting" of the receiver. At some point there is sufficient signal to reduce the noise to the residual noise level of the receiver itself. This is called "full quieting" and often produces recovered audio with a SINAD of greater than 60 dB. Here we are far above the signal level threshold for 12 dB SINAD, and we should have excellent signal with very little noise, i.e., a "full quieting" signal.

So far our analysis has made several very big assumptions: that the path is a Line of Sight (LOS) path, that the propagation medium is free space, and that there is no variation in the path due to atmospheric effects or motion in the antennas. In actual use, of course, the two antennas may not be entirely in line of sight, the propagation is through the atmosphere and over the irregular surface of the earth, and the antennas are polarized and directional while also in motion.

### Line Of Sight Path

Radio waves propagating at VHF Marine Band frequencies (156 MHz) tend to travel without much atmospheric refraction. If the antennas are not in line of sight, the path loss will be very much greater. Calculating the precise amount of additional path loss is beyond the scope of this analysis. Therefore, the assumption is made that beyond the LOS distance the radio communication circuit will very rapidly degrade.

For the 10-mile path to be Line Of Sight (LOS), each boat must have sufficient antenna height for a range of 5 miles. The approximate radio horizon is given by

```d = 1.42 X h0.5  where d is in miles and h is in feet.

Solving for d=5, (i.e., each antenna has a range to a 5-mile horizon)

h = (d/1.42)2

h = 12.4 feet
```

If we want a line of sight path between two boats, their antennas must both be 12.4 feet above the water. If using the typical 8-foot whip antenna, and considering the radiation to come from the center of the whip, this means the base of the antenna must be 8.4-feet above the water. This is a typical mounting height on a small boat with a radar arch or hard top. Such a configuration is about the tallest that can be easily achieved on a small boat. On boats with the antenna mounted directly to the boat's hull, the antenna height will probably be much lower, and the corresponding line of sight range will be shorter. Note that range changes as a function of the square-root of the antenna height, so to double the range you have to raise the antenna four times higher. If an antenna is located 6-feet above the water, its LOS range will be about 3.5 miles. Thus two boats with antennas of this height will have a range of about 7 miles.

As shown above, the path loss over 10 miles was not great enough to affect the received signal level, so it follows that over a shorter path the path loss in line of sight range will be even more favorable. Even doubling the path to twenty miles will only increase the path loss by 6-dB. There is ample margin for this in the circuit. Communication range will be most affected by antenna height. Propagation beyond line of sight distances will have much higher path loss. Within the line of sight distance, use of 25-watt transmitter power produces a very reliable circuit with typically 50-dB of fade margin.

So far we have only considered free space propagation. In the real world there are other propagation effects, including refraction, diffraction, and reflections which can affect path loss. It is not uncommon for these factors to increase the path loss by 20 dB. If we wish to allow for these, our circuit margin is now reduced by a like amount to

```54.5 - 20 = 34.5 dB margin
```

In the case of two boats in communications, the antennas of both boats may be in random motion due to wave conditions. The gain which the antennas have is created by narrowing their radiation from an isotropic pattern. As the antennas move (due to boat movement from waves) their orientation changes, the best signal may no longer be directed in the desired axis or orientation, and the off-axis gain will be lower. The more the on-axis gain is greater, the more the off-axis gain is lower. This is part of the disadvantage of using very high gain antennas on small boats. Boat movement is greater on small boats, so there will be more problems from the antenna orientation being off-axis due to pitching and rolling of the boat. It is not unusual for the gain to decrease 10 dB or more. If both antennas happen to move so that each decreases its gain toward the other by 10 dB, the circuit margin is reduced by 20 dB. Thus the margin is now down to

```34.5 - 20 = 14.5 dB
```

Let us assume for a moment that each radio was to decrease transmit power to 1-watt. This is a reduction of

```dB = 10 log(1/25)

= -14 dB
```

Now the margin is down to

```14.5 - 14 = 0.5 dB
```

or at minimum communication SINAD at the receiver. Thus, the circuit may experience deep fades or drop outs due to variables in propagation and antenna orientation. Propagation variables and antenna movement have reduced the circuit to marginal levels, and now the 25-watts of power provides an important margin.

### Other Factors Affecting Radio Communication

There are many other factors which can affect radio communication between two vessels. Foremost among these are interfering signals. Interference can come in many forms.

Local noise sources in the vicinity of the receiver can have a significant impact on receiver sensitivity. It is very common that an outboard motor will generate significant radio noise from its ignition system, primarily from the spark plugs. The typical outboard motor is not well shielded, and usually has a plastic cowling. The spark plug leads are typically not shielded. Many older motors do not use spark plugs designed to suppress radio frequency interference, usually accomplished with "resistor" plugs. Very strong radio interference can be generated from such a motor, and it will make reception of weak signals impossible. A quick test to determine if your engine is interfering with reception is to shut off the motor while receiving a weak signal. If the interference disappears, you have found the source.

Other sources of local radio frequency noise interference are electronic devices in the boat. GPS receivers, SONAR devices, chart plotters, RADAR, etc., are all possible sources of radio frequency noise which could interfere with radio communications. In most cases, however, these sources are not as significant as the engine spark noise.

Interference from remote sources is also very likely in marine radio communications. Other signals being received can affect reception of the desired signal. Interfering signals can be either on-channel of off-channel. If the interfering signal is on-channel and stronger than the desired signal there is little remedy possible except to wait for the interfering signal to stop or to change to another channel.

If the desired signal itself is very strong, it is possible that it can overload the receiver and produce garbled output. This is a very common situation when two recreational boats attempt communication with 25-watt power levels while in sight (visually) of each other. Many VHF Marine radios will be overloaded in this situation.

A very typical situation observed in VHF marine radio communication is an exchange like this: "Your radio sounds good on 1-watt but it is distorted on 25-watt." The real problem is usually in the receiver. The strong signal is overloading the receiver and producing distortion.

The remedy is to reduce power to 1-watt. I also suggest getting a higher quality radio. There are many quality VHF marine radios available which can tolerate strong local signals without distortion.

Another possible source of difficulty in radio communication is the transmitter. Two common problems are poor battery voltage and poor modulation.

Transmitter power output is proportional to battery voltage. The rated power output (25-watts) will only be produced when the supply voltage is at the rated level, typically 13.2-volts. At lower supply voltages the power output of the radio will be lower. If the supply voltage has poor regulation or contains audio-frequency noise, this can be superimposed on the transmitted signal.

The combination of a poor battery and strong charging current will often induce an audio-frequency tone on the modulation of a transmitter. This tone will vary in frequency in proportion to engine speed. Interference of this type is commonly called "alternator whine."

Modulation of the radio frequency carrier is necessary for communication. Many factors can affect the quality of the modulation. Speaking too loudly or too softly into the microphone will produce over-modulation or under-modulation. It is very common on a small boat with a noisy outboard motor that the radio operator will yell into the microphone. This usually produces garbled and distorted modulation.

Many operators do not use proper microphone technique when transmitting. This can result in very low modulation, drastically reducing the effectiveness of the communication, particularly when received signals are weak. To learn better microphone technique I recommend monitoring your own transmission with a nearby radio. In this way you can hear how your transmission actually sounds at a remote receiver. This will help you learn effective microphone technique.

If the transmitting antenna is located very close to the radio, the transmitted signal may interfere with the transmitter microphone, creating a howling sound known as "RF feedback". The remedy for this is to locate the antenna farther from the transmitter and microphone. Also, this is usually not a problem in better quality radios. Distortion in transmitted audio due to RF feedback usually only occurs at the 25-watt level. Test by reducing power to 1-watt. Do not confuse true transmitted distortions with apparent distortion due to overload at the receiver used for testing.

Here is the derivation of the decibel loss form of the path loss calculation from the free space path loss calculation. It has always bothered me that this is seldom shown in reference literature. Thus:

#### Converting Path Loss to Decibel Equation

```
It has been demonstrated that path loss in free space can be described as being

Power Loss Ratio = (4π / c)2 x f2 x d2

Rearranging algebraicly

Power Loss Ratio = (4πdf/c)2

where d = distance in meters
f = frequency in hertz, i.e., 1/sec
c = speed of light in same units, thus c = 3 x 108 meters/sec

We now change units to more convenient forms, as

where d = distance in kilometers, i.e., d x 103
f = frequency in Megahertz, i.e., f x 106/sec
c = speed of light in meters, thus c = 3 x 108 meters/sec

and also where
d = distance in miles
f = frequency in Megahertz, i.e., f x 106/sec
c = speed of light in miles, thus c = 1.86 x 105 miles/sec

Now we have new formulas:

For Megahertz and Kilometers we have

Power Loss Ratio = (4π d x 103 x  f x 106 / 3 x 108)2

For Megahertz and Miles we have

Power Loss Ratio = (4π  d x f x 106 / 1.86 x 105)2

Converting these to decibels according to db = 10 log (Power Loss Ratio)

For Megahertz and Kilometers

dB loss = 10 log [ (4π d x 103 f x 106 / 3 x 108)2]
dB loss = 20 log (4π x 109 / 3 x 108) + 20 log (f) + 20 log (d)
dB loss = 20 log (4π x 10 / 3)        + 20 log (f) + 20 log (d)
dB loss = 20 log (41.88) + 20 log (f) + 20 log (d)
dB loss = 32.44          + 20 log (f) + 20 log (d)

and since log(a) + log(b) = log(a x b)

db loss = 32.44 + 20 log ( f x d )

For Megahertz and Miles
dB loss = 10 log [ (4π x d x f x 106 / 1.86 x 105)2 ]
dB loss = 20 log (4π x 106 / 1.86 x 105) + 20 log (f) + 20 log (d)
dB loss = 20 log (4π x 10 / 1.86)       + 20 log (f) + 20 log (d)
dB loss = 20 log (67.56) + 20 log (f) + 20 log (d)
dB loss = 36.59          + 20 log (f) + 20 log (d)

and since log(a) + log(b) = log(a x b)

db loss = 36.59 + 20 log ( f x d )
```