A frequency counter counts the oscillations of the signal for a given period of time. For the result to be accurate, this period of time must be measured accurately.

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A bridge circuit uses an adjustable known reference impedance connected to the unknown impedance. The reference impedance is adjusted until a signal null is achieved. At that point, the reference impedance is equal in value to the unknown impedance. The reference impedance can then be measured.

Memory aid: In folktales, a troll (sounds like null) lives under a bridge.

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There could be as much as 1 Hz error for every million Hz in frequency.

So to calculate the maximum possible error - or the max difference between read frequency and the actual frequency.

Divide the frequency (in Hz) by 1,000,000 and multiply by the “parts per million” (also in Hz) to get the answer.

146,520,000 / 1,000,000 x 1.0 gives us 146.52 Hz

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\[1 \text{ million} = 10^{6}\] \[0.1 \text{ ppm} = \frac{0.1}{10^6}=\frac{10^{-1}}{10^{6}}=10^{-7}\]

Move decimal point seven places to the left, or: \[\pm 0.0000001 \times 146,520,000 \text{ Hz} = \pm 14.652 \text{ Hz}\]

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Better done: divide the frequency by 1,000,000 and multiply by the “parts per” to get the answer in Hz.

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ppm means “parts per million” divide the frequency by 1,000,000 and multiply by the “parts per” to get the answer in Hz.

(146,520,000 / 1,000,000) x 10 = 1465.20 Hz

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Where \(P\) is power:

\begin{align} \text{(load absorption)} &= P_\text{forward} - P_\text{reflected} &=100-25=75\:\text{Watts} \end{align}

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There are multiple potential issues with having a longer ground connection:

1. The scope is measuring the voltage differential between the ground and whatever you connect the probe to, so by keeping the ground short you minimize any voltage drop there may be.

2. The probe tip and ground of an oscilloscope acts like a sensitive antenna loop. The bigger the loop the more undesired signals or noise it will pick up. Also, the inductance of the ground wire increases with length, which can distort high-frequency signals. So, keep it short as possible.

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A voltmeter should be an infinite impedance attachment to the circuit of interest so that it has no effect in the circuit. In practice, it becomes part of the circuit and affects the signal being measured. Keeping the impedance as high as possible minimizes this effect.

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Caution, this one is easy to misread.

Both choices that Modulate the transmitter using two RF signals are distractors because we use audio frequencies (here abbreviated AF) to modulate the carrier in SSB mode.

You can rule out the distractors mentioning logic analyzer, which only handles digital logic signals, and peak reading wattmeter, because it just outputs a watt number - you'd have no way to know if that number represents your signal peak or a intermodulation distortion peak.
Instead you want a spectrum analyser, which shows many frequencies at once, and their current power level, letting you spot intermodulation distortion at nearby frequencies.

That leaves the correct answer Modulate the transmitter with two non-harmonically related audio frequencies and observe the RF output with a spectrum analyzer.

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A portable antenna analyzer is a device that is used to analyze the characteristics of an antenna and often the feed line. You can see some pictures of analyzers on the Wikipedia page, but generally it has at minimum a connector to attach an antenna / feed line to, a readout and dial for selecting the frequency, and a readout for the SWR of the antenna/feed line system at that frequency.

There is no need for a dummy load with an antenna analyzer, and the analyzer is designed to connect to an antenna so it generally connects the same way a transceiver would -- by connecting the feedline directly to the analyzer.

-kd7bbc

Hint: Connected, connector.

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\begin{align} \frac{\text{Ohms}}{\text{volt}} \times \text{full scale volts} &= \text{full scale impedance} \\ &=\text{input impedance} \end{align} (drichmond60)

Hint: all the "When used as..." answers are incorrect.

Hint: The word "voltmeter" appears in the question and in the correct answer.

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The probe is adjusted until the horizontal flats of a square wave are as flat as possible.

This adjustment is equivalent to making the probe have uniform attenuation over the range of frequencies to be measured. In passive probes this adjustment is usually a small adjustable capacitor.

Silly trick: Cubes are three dimensional, but squares are as flat as possible!

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Remember that a dip meter is an instrument used to check the circuit without direct connection to the circuit under test. This way it does not cause harmonics, cross modulation or intermodulation distortion to occur. What results is the readings are less accurate.

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Quick silly attempt at remembering the right answer:

The right answer has "frequency response" in it. So, Every Question (Q) deserves a response

(you're welcome)

Q stands for quality. The frequency response is a measurement of quality.

Definition Of \(Q\) Factor: In the context of resonators, \(Q\) is defined in terms of the ratio of the energy stored in the resonator to the energy supplied by a generator, per cycle, to keep signal amplitude constant, at a frequency \(f_r\) (the resonant frequency), where the stored energy is constant with time:

\[\begin{align} Q &= 2π \times \left( \frac{\text{Energy}_{\text{stored}}}{\text{Energy}_{\text{dissipated per cycle}}} \right) \\ &= 2π\times f_r \times \left( \frac{\text{Energy}_{\text{stored}}}{\text{Power}_{\text{loss}}} \right) \end{align}\]

http://en.wikipedia.org/wiki/Q_factor#Explanation

There are a few ways to define \(Q\). With regard to this question, the bandwidth is the width of the range of frequencies for which the energy is at least half its peak value. The higher the \(Q\), the narrower the bandwidth. That is,

\[\text{bandwidth} = \frac{f_r}{Q}\]

-wileyj2956

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