A frequency counter has an internal frequency reference it uses to compare against the signal being measured. "Time base" is another term for "frequency reference" (frequency being the reciprocal of time, therefore a time reference is also a frequency reference). The more accurate the time base, the more accurate the frequency counter.
The other 3 factors mentioned are insignificant -- an attenuator changes the amplitude of the signal but not the frequency; and the logic components (including a decade divider) have no cumulative effect on accuracy since they are referenced to the time base.
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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: \[\pm 0.0000001 \times 146,520,000 \text{ Hz} = \pm 14.652 \text{ Hz}\]
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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There could be as much as 10 Hz error for every million Hz in frequency.
So to calculate the maximum possible error - or the maximum 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.
Because the ppm is 10 in this problem, you can also simply move decimal point five places to the left. +/- .00001 \(\times\) 146,520,000 = 1465.2 Hz.
\[1 \text{ million} = 10^{6}\] \[10 \text{ ppm} = \frac{10}{10^6}=10^{-5}\]
Move decimal point five places to the left, or: \[\pm 0.00001 \times 146,520,000 \text{ Hz} = \pm 1465.20 \text{ Hz}\]
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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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S parameters are a way of measuring the frequency response of devices and can refer to as many ports as are on the device.
For example, an antenna has 1 port, a filter may have 2 ports, a power divider may have 3 or more ports.
The subscripts tell us from which ports the measurements were made and are in the order of "To -> From" or S_{<out><in>}
So S_{11} would be a measurement to Port 1, from Port 1. (Perhaps an antenna measurement, or other reflection measurement).
S_{21} would be a measurement made at port 2 with the signal being delivered from port 1. (Such as measuring a filter to see what frequencies are blocked or passed through).
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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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The magnitude of a complex impedance is always higher than its resistive component (see Pythagorean Theorem). This means that a resistive load with a reactive component will always draw less current than the same load with no reactive component.
When an antenna is tuned to resonance its inductive and capacitive reactances add to zero, leaving only the resistive component of the antenna's impedance. As the antenna's reactance is reduced, more current flows into the antenna. Current is maximum when the antenna's reactance is zero.
Increased current means more power is being delivered to the antenna.
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A simple elimination leads to the correct answer. We use audio to modulate the carrier in SSB mode so the answer saying use radio frequencies is clearly wrong.
The question is asking about measuring the resulting inter-modulated signal. A logic analyzer is clearly not applicable and a peak reading wattmeter is not used to measure distortion.
This leads us to the correct answer:
Modulate the transmitter with two non-harmonically related audio frequencies and observe the RF output with a spectrum analyzer
Still, read the answers slowly as it's very easy to mistakenly choose the answer that says radio and not audio (ericthughes)
Hint: only the correct answer contains both "audio" and "spectrum analyzer" (KG5KOU)
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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
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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. (KG5KOU)
Hint: The word "voltmeter" appears in the question and in the correct answer.
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The 2-port S (scattering) voltage parameters for a linear electrical network are defined as
S_{11} = input reflection coefficient
S_{12} = reverse gain
S_{21} = forward gain
S_{22} = output reflection coefficient
Therefore, S_{21} is the forward gain
See https://en.wikipedia.org/wiki/Scattering_parameters for a summary of the technical explanation
Poor man's hint: In the United States, you look forward to turning 21 years old.
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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)
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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The 2-port \(S\) (scattering) voltage parameters for a linear electrical network are defined as
\[\begin{align} S_{11} &= \text{input reflection coefficient} \\ S_{12} &= \text{reverse gain} \\ S_{21} &= \text{forward gain} \\ S_{22} &= \text{output reflection coefficient} \\ \end{align}\]
SWR and signal return loss are both calculated by using the input reflection coefficient. Therefore, \(S_{11}\) represents return loss or SWR.
See https://en.wikipedia.org/wiki/Scattering_parameters for a summary of the technical explanation
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\(50\Omega\) ohms is the most common impedance used in RF power systems, and amateur transmitters also use this standard. Therefore it is a useful point to calibrate to. After that, short-circuit and open-circuit are two "boundary cases" that ensure the analyzer behaves correctly at the edges of its range.
Calibrating to \(75\Omega\) or \(90\Omega\) ohms might be somewhat helpful, but after covering the full range with the 3 correct answers there is limited value in any further calibration. The other answers are not simple loads or make no sense at all.
Funny reminder: The movie "short circuit" and the robot Johnny 5. Only 1 answer has "short circuit" and 5 (\(50\Omega\)) in it.
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