An isotropic radiator is a theoretical point source of electromagnetic waves which radiates the same intensity of radiation in all directions. It has no preferred direction of radiation. It radiates uniformly in all directions over a sphere centred on the source.

Isotropic radiators are used as reference radiators with which other sources are compared.

Source: Wikipedia - Isotropic Radiator

One Word Key "theoretical"

Last edited by sapient-prickly-0y. Register to edit

Tags: arrl chapter 9 arrl module 9a

Isotropic antennas are ideal (theoretical) antennas that have equal power in all directions. They are used as references for antenna gain.

The word "isotropic" means "uniform in all orientations/directions".

Last edited by kk4uwv. Register to edit

Tags: arrl chapter 9 arrl module 9a

When the feed point impedance matches the impedance of the feedline and transmitter, the Standing Wave Ratio (SWR) will be 1:1, the minimum SWR.

If we know, for example, that the feedline and antenna impedance is 50 Ohms, we would need the antenna's feed point impedance to also be 50 Ohms.

Coaxial cable feedlines are often designed to exhibit an impedance of 50 Ohms. Transmitters are also often designed to have an output impedance of 50 Ohms.

Last edited by k6yxh. Register to edit

Tags: arrl chapter 9 arrl module 9e

The feed-point impedance of an antenna can be influenced by near-by conductive objects, including proximity to the ground.

Antenna height is the only answer that has anything to do with the antenna and its surrounding environment. The other answers do not affect the antenna.

Another point of view - the question is about what effects the feed point of the antenna. The transmission line length, settings of an antenna tuner, and input power are not part of the impedance at (and after) the feed point of the antenna.

Last edited by liewo. Register to edit

Tags: arrl chapter 9 arrl module 9b

Radiation resistance is that part of an antenna's feedpoint resistance that is caused by the radiation of electromagnetic waves from the antenna, as opposed to loss resistance (also called ohmic resistance) which is caused by ordinary electrical resistance in the antenna, or energy lost to nearby objects, such as the earth, which dissipate RF energy as heat.

The radiation resistance is determined by the geometry of the antenna, whereas the ohmic resistance is primarily determined by the materials of which it is made and its distance from and alignment with other nearby conductors or semi-conductors, and what those are made of.

Both radiation and ohmic resistance depend on the distribution of current in the antenna.

Eliminate the answer referring to "transmission line" because that is not what they are referring to as an "antenna system".

Last edited by bytescream. Register to edit

Tags: arrl chapter 9 arrl module 9b

The beamwidth of an antenna is the width of the radiation of the main lobe from the antenna. As gain increases, it can be expressed as making the main lobe thinner and longer or decreasing its beam's width. see: http://en.wikipedia.org/wiki/Beamwidth

-deanwj

Last edited by ke0ipr. Register to edit

Tags: arrl chapter 9 arrl module 9a

Antenna gain is usually defined as the ratio of the power produced by the antenna from a far-field source on the antenna's beam axis to the power produced by a hypothetical lossless isotropic reference antenna, which is equally sensitive to signals from all directions. Usually this ratio is expressed in decibels, and these units are referred to as "decibels-isotropic" (dBi). An alternative definition compares the antenna to the power received by a lossless half-wave dipole reference antenna, in which case the units are written as dBd. Since a lossless dipole antenna has a gain of 2.15 dBi, the relation between these units is: gain in dBd = gain in dBi − 2.15 dB.

https://en.wikipedia.org/wiki/Antenna_gain

Last edited by w3mit. Register to edit

Tags: arrl chapter 9 arrl module 9a

Bandwidth is a measurement of frequencies, as in the width of a range of frequencies.

An antenna's bandwidth is the range of frequencies it works best on, though it's impossible to give an exact formula without the exact performance requirement.

For example the performance requirement might be "SWR less than 1.7" in which case the antenna could be modeled or tested and the exact range of frequencies for that requirement specified.

Hint: A Band "performs". The answer contains the word "performance".

Last edited by nielsenj. Register to edit

Tags: arrl chapter 9 arrl module 9b

Distractor Effective radiated power divided by transmitter output is sneakily close.
It's wrong because not quite all of the transmitter output energy is delivered to the antenna. A small amount is dissipated in the transmission line and connectors.

Antenna efficiency is a term that relates what portion of the power delivered to the antenna actually gets radiated.

The total resistance is the resistance that is seen at the terminals of the antenna and is composed of two parts:

1. Radiation resistance
2. Loss resistance

Loss resistance turns the power into heat instead of radiating it into the air, while radiation resistance is the part that turns the energy into useful radio waves.

Example An antenna has an input resistance of \(500\) Ohms and a radiation resistance of \(25\) Ohms.

Antenna efficiency would be: \begin{align} \text{antenna efficiency} &= \frac{\text{radiation resistance}}{\text{total resistance}} \times 100\%\\ &=\frac{25 \:\Omega}{500 \:\Omega} \times 100\%\\ &=5\% \end{align}

So \(5\%\) of the power delivered to the antenna gets radiated, the other \(95\%\) is wasted as heat.

Silly hint: Think "part over whole" or "percent over 100" to remember "resistance divided by total resistance"

Last edited by nicenjehr. Register to edit

Tags: arrl chapter 9 arrl module 9b

One of the ways you can improve an antenna's efficiency is to reduce losses from resistance. Anything that dissipates the energy of your RF increases total resistance and soil resistance can be significant.

When you install a system of radial wires at the base of a ground-mounted vertical antenna current flows into the radials instead of into the soil, thus reducing losses.

Last edited by valeria. Register to edit

Tags: arrl chapter 9 arrl module 9b

High soil conductivity improves ground reflections and ground wave propagation. This effect is also the reason why waves are able to travel very far over large bodies of water.

Hint: Ground \(=\) soil

Last edited by zekeabrams116. Register to edit

Tags: arrl chapter 9 arrl module 9b

A 1/2-wave dipole antenna has approximately \(2.15 \text{ dB}\) of gain over an isotropic antenna. So \[6 \text{ dB} - 2.15 \text{ dB} = 3.85 \text{ dB}\]

(The directive gain of a half-wave dipole is 1.64. It has a 2.15 gain over an isotropic antenna or \(10\log_{10}(1.64)\approx2.15\text{ dBi}\))

The distractor of 8.15dB relies on you remembering the 2.15dB difference between isotopic and dipole antennas, so pay careful attention to whether you're adding or subtracting 2.15dB!

Silly way to remember: The 1/2 way number is in the answer, which is a sum of the ends: 3.85... 3+5=8

Last edited by patronics. Register to edit

Tags: arrl chapter 9 arrl module 9a

Given: It's well-known that a half-wave dipole has \(2.15 \text{ dB}\) gain over an ideal isotropic radiator.

Let \(H =\) the gain of an ideal isotropic radiator

Let \(W =\) the gain of a half-wave dipole

Therefore \(W = H + 2.15 \text{ dB}\), or solving for \(H\), \(H = W - 2.15 \text{ dB}\)

Now, let \(X =\) the gain of the antenna in question

If that antenna in question exhibits \(12\text{ dB}\) over an isotropic antenna, then

\(X = H + 12 \text{ dB}\)

Substituting, we get

\(X = W - 2.15\text{ dB} + 12\text{ dB}\) \(= W + 9.85 \text{ dB}\)

Therefore, the antenna in question (\(X\)) exhibits a gain of \(9.85\text{ dB}\) over that (\(W\)) of a half-wave dipole.

Remember that a half-wavength dipole has 2.15 dB gain over an ideal isotropic radiator. Then take the difference between 12 dB and 2.15 dB to arrive at the answer of 9.85 dB.

Last edited by wileyj2956. Register to edit

Tags: arrl chapter 9 arrl module 9a

The electrical resistance of an antenna is composed of its ohmic resistance plus its radiation resistance. The energy lost due to radiation resistance is the energy that is converted to electromagnetic radiation.

It isn't the combined losses of antenna elements and feed line because radiation resistance is not related to feed line losses.

It isn't the specific impedance of the antenna because the radiation resistance is only a portion of the antenna's impedance.

It isn't the resistance in the atmosphere that an antenna must overcome because the radiation resistance is not a property of the atmosphere. It is determined by the geometry of the antenna.

The correct answer is therefore the value of a resistance that would dissipate the same amount of power as that radiated from an antenna

Last edited by cactus121. Register to edit

Tags: arrl chapter 9 arrl module 9b

The power gain G in dB is given by the following equation: \[\text{Gain } G= 10\log\left({P_2 \over P_1}\right)\] where:

• \(P_2\) is the output power in Watts
• \(P_1\) is the power in Watts applied to the input

We may algebraically solve for \(P_2\):

\[P_2 = P_1\left(10^{G/10}\right)\] We now have a simple formula for other similar problems.

The input power is \(P_1 =150 \text{ W}\). The net gain is \(G=7-4.2 = 2.8 \text{ dB}\). Applying the formula: \begin{align} P_2 &= 150\cdot10^{2.8/10}\\ &= 150\cdot10^{0.28}=285.82\\ &\approx286\text{ W} \end{align}

Here's yet another way...

When I was studying for this exam, I came across a much easier way to calculate ERP. This method converts the transmitter power to dB Watts so that you can easily add and subtract gains and losses. This technique requires a calculator that has a log function, which is allowed at the exam.

Using this question as an example: \[10\log(150)=21.8\]

Now you have apples and apples so you can add and subtract gains and losses: \[21.8 -2 -2.2 +7 = 24.56091259\]

Next you will need to divide by 10: \[24.56091259/10 = 2.456091259\]

TIP: You may want to store this temporarily in memory before proceeding.

Finally, apply the inverse log to return the gain (or loss) in Watts): \begin{align} InvLog(2.456091259)&=10^{2.456091259}\\ &\approx285.8\approx286\text{ W} \end{align}

Or use the "cheat": The net gain is \(7 - 4.2 = 2.8 \text{ dB}\). We know that \(3 \text{ dB}\) gain is double, so \(2.8\) is just under that. Double would be \(300\) watts so \(286\) is the only answer that is close.

Last edited by wileyj2956. Register to edit

Tags: arrl chapter 9 arrl module 9e

Sum the losses in the various stages:

\(4 \text{ dB} + 3.2 \text{ dB}+ 0.8\text{ dB}\) adds up to \(8\text{ dB}\) of loss in the total feed system. However, the antenna system gives us \(10\text{ dB}\) of gain (relative to a dipole). Fortunately, the question asks for the effective radiated power (ERP) relative to a dipole so no change to the antenna gain figure is needed.

\(10\text{ dBd}\) antenna gain minus \(8\text{ dB}\) feed system loss gives us an overall gain of \(2\text{ dB}\).

\(\text{Gain }G = 10\log\left(\frac{P_2}{P_1}\right)\), and we need to solve for \(P_2\), the ERP:

\[2\text{ dB} = 10\log\left(\frac{P_2}{200}\right)\]

Divide both sides by \(10\) giving:

\[0.2\text{ dB} = \log\left(\frac{P_2}{200}\right)\]

Take the inverse log of both sides:

\[10^{0.2} = \frac{P_2}{200}\]

evaluate:

\[1.585 = \frac{P_2}{200}\]

Multiply both sides by \(200\):

\[(1.585)(200) = P_2\]

Solve:

\[P_2 = 316.98\]

Alternatively, we may algebraically solve for \(P_2\): \begin{align} P_2 &= P_1 \left(10^{G/10}\right)\\ &= 200\times10^{2/10} = 200\times10^{0.2}\\ &=316.98\\ &\approx317 \text{ W} \end{align}

We now have a simple formula for other similar problems.

Silly Hint: add 4dB and 3.2 dB for 7, only answer with 7 in it

Another silly hint: There are an odd number of inputs to calculate this value. The correct answer is also the only one that's an odd number!

Last edited by michael.lubrect.droid. Register to edit

Tags: arrl chapter 9 arrl module 9e

In this example, the net gain after subtracting the total losses is equal to: \[7\text{ dB} – (2 \text{ dB} + 2.8 \text{ dB} + 1.2 \text{ dB}) = 1 \text{ dB} \] That’s equivalent to a ratio of \(1.26:1\), so the effective radiated power (ERP) is $200 \text{ W} \times 1.26 = 252 \text{ W} $.

Or

\[P_2 = 200 \times 10^{0.1} = 251.8 \text{ W} \]

Where does 0.1 come from? \begin{align} \mathrm{ERP} &= \mathrm{TPO} \times \log^{-1} \left(\frac{\text{system gain}}{10}\right) \\ &= \mathrm{TPO} \times \log^{-1} \left(\frac{1}{10}\right) \\ &= \mathrm{TPO} \times \log^{-1} (0.1) \\ &= 200 \times 10^{0.1} \end{align} Inverse log = \(b^y\) where \(\mathrm{base} = 10\) and \(y = 0.1\)

transmitter power output (TPO)

Where does the ratio \(1.26:1\) come from?

Actually it can be reverse calculated after you find the ERP BY THE 2ND METHOD

Easy Way: If you remember from previous license exams that 3dB is double, you can derive the correct answer from there. 1dB net gain means that the correct answer will be more than the 200W output, but less than double. Only the correct answer choice is between 200 and 400W.

For additional information: https://www.kb6nu.com/extra-class-question-of-the-day-effective-radiated-power/

Last edited by samdypaulson. Register to edit

Tags: arrl chapter 9 arrl module 9e

Effective radiated power (ERP) is used to describe the highest concentration of RF energy that is radiated in a particular direction. As such, it includes all of the gains and losses of the antenna system, including focusing the radiated power. Yagi antennas, and other designs, have "gain" over a reference dipole, and must be considered when calculating effective radiated power. If the antenna has gain in a particular direction, that gain has to be calculated as part of the ERP. If an antenna design results in 3dBd gain (3dB greater than a dipole), then the ERP is twice what it would be with a dipole.

Last edited by k6yxh. Register to edit

Tags: arrl chapter 9 arrl module 9e