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

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

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

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

The term “ground gain” in the context of antennas refers to an increase in signal strength from ground reflections in the environment of the antenna. This means that the correct answer is "An increase in signal strength from ground reflections in the environment of the antenna."

Ground gain is closely related to the environment the antenna is installed in and has a strong dependence on the installation height of said antenna. That is to say, the same antenna can have different gain values when installed at varying heights. For example, an antenna mounted at a height of 1/4 of the wavelength above the ground will have a different gain than if it were mounted 1/2 of the wavelength above ground.

It’s important to note that no new energy is created; it is simply redirected or given directivity from the ground.

Last edited by snippy515831. Register to edit

Tags: none

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

The Fresnel zone is an area around the line-of-sight path between a transmitter and receiver in which obstacles can cause interference. The size of the Fresnel zone depends on the frequency of the signal and the distance between the transmitter and receiver.

In simple terms, higher frequencies have smaller wavelengths. The Fresnel zone radius, which indicates the size of the area where obstacles can interfere, is smaller for higher frequencies.

5.8 GHz is a higher frequency than the others listed. Because 5.8 GHz is higher, it has a smaller first Fresnel zone. This means that the area where obstacles can interfere with the signal is smaller for 5.8 GHz than it would be for lower frequencies -- in a sense, it can punch through a smaller "hole" in trees, etc.

So, the reason 5.8 GHz has the smallest first Fresnel zone is that it operates at a higher frequency, which reduces the size of the Fresnel zone around the line-of-sight path.

Silly hint: first Fresnel and five point eight are alliterative.

Last edited by wstarkwe. Register to edit

Tags: none

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