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"

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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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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.

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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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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

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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

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

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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.

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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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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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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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