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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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!

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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

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/

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in general terms: Yagi antennas commonly have a single strong beam in one direction. Wire-loop antennas commonly have two equal front and back direction patterns. Vertical antennas are mostly omni-directional.

Using a wire loop gives you a 50/50 chance of locating the direction during a "foxhunt" or RDF activity (radio direction finding). Other names are hidden transmitter hunting, radio-orienteering, ARDF, and fox-tailing, A handheld beam antenna, such as a hand-held yagi works best. For a picture, see a foxhunt in progress

at HomingIn.com.

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You need three or more receiving sites distributed at some distance from each other and in different locations around the approximate search area. Each directional receiving antenna is rotated for maximum RSL and the direction azimuths are then plotted on a map. Where the lines cross is the location of the transmitter.

One receiver could drive around the search area, repeating the direction-finding process. However, this approach takes more time.

Hint: The question has the word Triangulation in it. Tri= 3. Thus it would usually require 3 or more separate antenna/stations to determine the location.

Hint 0.0: Triangulation. Triangle. Takes 3 points. 2 listening, one sending.

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A second dipole or vertical antenna known as a sense antenna can be electrically combined with a loop or a loopstick antenna. Switching the second antenna in obtains a net cardioid radiation pattern from which the general direction of the transmitter can be determined. Then switching the sense antenna out returns the sharp nulls in the loop antenna pattern, allowing a precise bearing to be determined.

-cfadams

Silly Memory Hint: Your Sensei will help One find Direction

Another Silly Memory Hint: Null sense antenna

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Loop antenna or Radio antenna is made of a loop of wire or some electrical conductor with a circumference equal to the wavelength. There are two basic sizes, small (magnetic) and large (resonant) type. The ends of the loop are connected to a balanced transmission line.

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As the cardio part of the name implies, cardioid shape means heart shaped.

A transmitted cardioid radiation pattern has much of the radiated signal focused forward, in the direction that the antenna is pointing, with less and less signal energy as you go around the sides and up to almost behind antenna, then there is a distinct null (an area of no radiated energy) directly behind the antenna.

In the case of direction finding, if you rotate the cardioid patterned antenna connected to your receiver, until the incoming signal is lost, then it would result in the antenna pointing directly away from the signal source.

Note that it is the radiation pattern from the antenna that is heart shaped, not the antenna itself. There are a number of different antenna designs of various shapes that make a cardioid radiation pattern.

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Direction finding is an activity that’s both fun and useful. One of the ways that it’s useful is to hunt down noise sources. It can also be used to hunt down stations causing harmful interference.

A variety of directional antennas are used in direction finding, including the shielded loop antenna.

An advantage of using a shielded loop antenna for direction finding is that it is electro-statically balanced against ground, giving better nulls.

The main drawback of a wire-loop antenna for direction finding is that it has a bidirectional pattern. (E9H05)

Source: KB6NU.com - Extra Class question of the day: Direction finding

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