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Amateur Extra Class (2016-2020)
Subelement E9
ANTENNAS AND TRANSMISSION LINES
Section E9A
Basic Antenna parameters: radiation resistance, gain, beamwidth, efficiency, beamwidth; effective radiated power, polarization
What describes an isotropic antenna?
• A grounded antenna used to measure earth conductivity
• A horizontally polarized antenna used to compare Yagi antennas
A theoretical antenna used as a reference for antenna gain
• A spacecraft antenna used to direct signals toward the 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

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What antenna has no gain in any direction?
• Quarter-wave vertical
• Yagi
• Half-wave dipole
Isotropic antenna

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".

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Why would one need to know the feed point impedance of an antenna?
To match impedances in order to minimize standing wave ratio on the transmission line
• To measure the near-field radiation density from a transmitting antenna
• To calculate the front-to-side ratio of the antenna
• To calculate the front-to-back ratio of the antenna

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.

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Which of the following factors may affect the feed point impedance of an antenna?
• Transmission-line length
Antenna height, conductor length/diameter ratio and location of nearby conductive objects
• The settings of an antenna tuner at the transmitter
• Sunspot activity and time of day

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.

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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 plus ohmic 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".

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How does the beamwidth of an antenna vary as the gain is increased?
• It increases geometrically
• It increases arithmetically
• It is essentially unaffected
It decreases

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

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What is meant by antenna gain?
The ratio of the radiated signal strength of an antenna in the direction of maximum radiation to that of a reference antenna
• The ratio of the signal in the forward direction to that in the opposite direction
• The ratio of the amount of power radiated by an antenna compared to the transmitter output power
• The final amplifier gain minus the transmission line losses

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

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What is meant by 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.

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How is antenna efficiency calculated?
• (radiation resistance / transmission resistance) x 100 percent
(radiation resistance / total resistance) x 100 percent
• (total resistance / radiation resistance) x 100 percent
• (effective radiated power / transmitter output) x 100 percent

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.

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Which of the following choices is a way to improve the efficiency of a ground-mounted quarter-wave vertical antenna?
Install a good radial system
• Isolate the coax shield from ground
• Shorten the radiating element
• Reduce the diameter of the radiating element

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.

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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 effect is also the reason why waves are able to travel very far over large bodies of water.

Hint: Ground $=$ soil

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

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

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How much gain does an antenna have compared to a 1/2-wavelength dipole when it has 12 dB gain over an isotropic antenna?
• 6.17 dB
9.85 dB
• 12.5 dB
• 14.15 dB

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.

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What is meant by 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

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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.

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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.

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What is the effective 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

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