or
Subelement E9
ANTENNAS AND TRANSMISSION LINES
Section E9A
What is an isotropic antenna?
• A grounded antenna used to measure Earth conductivity
• A horizontally polarized antenna used to compare Yagi antennas
A theoretical, omnidirectional antenna used as a reference for antenna gain
• A spacecraft antenna used to direct signals toward Earth

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.

One Word Key "theoretical" -KM6PNZ

Last edited by sacamentakid. Register to edit

Tags: arrl chapter 9 arrl module 9a

What is the effective radiated power relative to a dipole of a repeater station with 150 watts transmitter power output, 2 dB feed line loss, 2.2 dB duplexer loss, and 7 dBd antenna gain?
• 1977 watts
• 78.7 watts
• 420 watts
286 watts

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

What is the radiation resistance of an antenna?
• The combined losses of the antenna elements and feed line
• The specific impedance of the antenna
The value of a resistance that would dissipate the same amount of power as that radiated from an antenna
• The resistance in the atmosphere that an antenna must overcome to be able to radiate a signal

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

Which of the following factors affect the feed point impedance of an antenna?
• Transmission line length
Antenna height
• The settings of an antenna tuner at the transmitter
• The input power level

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.

Last edited by n6sjd. Register to edit

Tags: arrl chapter 9 arrl module 9b

What is included in the total resistance of an antenna system?
• Radiation resistance plus space impedance
• Radiation resistance plus transmission resistance
• Transmission-line resistance plus radiation resistance

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

What is the effective radiated power relative to a dipole of a repeater station with 200 watts transmitter power output, 4 dB feed line loss, 3.2 dB duplexer loss, 0.8 dB circulator loss, and 10 dBd antenna gain?
317 watts
• 2000 watts
• 126 watts
• 300 watts

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.

Last edited by wileyj2956. Register to edit

Tags: arrl chapter 9 arrl module 9e

What is the effective isotropic radiated power of a repeater station with 200 watts transmitter power output, 2 dB feed line loss, 2.8 dB duplexer loss, 1.2 dB circulator loss, and 7 dBi antenna gain?
• 159 watts
252 watts
• 632 watts
• 63.2 watts

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

Last edited by wileyj2956. Register to edit

Tags: arrl chapter 9 arrl module 9e

What is antenna bandwidth?
• Antenna length divided by the number of elements
The frequency range over which an antenna satisfies a performance requirement
• The angle between the half-power radiation points
• The angle formed between two imaginary lines drawn through the element ends

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.

Last edited by kv0a. Register to edit

Tags: arrl chapter 9 arrl module 9b

What is antenna efficiency?
• Radiation resistance divided by transmission resistance
Radiation resistance divided by total resistance
• Total resistance divided by radiation resistance
• Effective radiated power divided by transmitter output

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:

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.

Last edited by wileyj2956. Register to edit

Tags: arrl chapter 9 arrl module 9b

Which of the following improves the efficiency of a ground-mounted quarter-wave vertical antenna?
• Isolating the coax shield from ground
• All these choices are correct

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

Which of the following factors determines ground losses for a ground-mounted vertical antenna operating in the 3 MHz to 30 MHz range?
• The standing wave ratio
• Distance from the transmitter
Soil conductivity
• Take-off angle

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

Hint: Ground $=$ soil

Last edited by philcanete1. Register to edit

Tags: arrl chapter 9 arrl module 9b

How much gain does an antenna have compared to a 1/2-wavelength dipole when it has 6 dB gain over an isotropic antenna?
3.85 dB
• 6.0 dB
• 8.15 dB
• 2.79 dB

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 $10\log_{10}(1.64)\approx2.15\text{ dBi}$)

Last edited by wileyj2956. Register to edit

Tags: arrl chapter 9 arrl module 9a

What term describes station output, taking into account all gains and losses?
• Power factor
• Half-power bandwidth