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 S₁₁ would be a measurement to Port 1, from Port 1. (Perhaps an antenna measurement, or other reflection measurement).

S₂₁ 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).

Silly hint: Submarines go to ports Silly hint: Think "Sports" as in football, basebasell, etc. for S Ports

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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.

Hint: When something increases there is more of it.

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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 2-port S (scattering) voltage parameters for a linear electrical network are defined as

S₁₁ = input reflection coefficient
S₁₂ = reverse gain
S₂₁ = forward gain
S₂₂ = output reflection coefficient

Therefore, S₂₁ is the forward gain.

Hint: S_ab is the outgoing wave at Port a when there is an incoming wave at Port B and there is no incoming wave at the other port.

See https://en.wikipedia.org/wiki/Scattering_parameters for a summary of the technical explanation

Mnemonic:
21: Big element followed by a small element is a Yagi with forward gain
12: Small element followed by a big element is a Yagi pointing the other way giving reverse gain.
11: The 1 sees a reflection and 1 looks like I for input.
22: The 2 sees a reflection and 2 does not look like an I.

Poor man's hint: In the United States, you look forward to turning 21 years old.

Another stupid hint: the greater number of 2 in S21 is before (forward) the 1. The reverse is true with reverse gain (pun unintended).

One more stupid hint: Watch out, because there are two very similar questions like this. The way you can remember is this question is shorter, so choose the bigger number. The other question is longer, so choose the smaller number. If that makes sense.

Silly Hint: The shifter on a car goes Park, Reverse, Forward, Overdrive... it's very similar to how the S parameters are structured:

• S11 Input (In park)
• S12 Reverse
• S21 Forward
• S22 Output (Overdrive)

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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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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. SWR can be directly calculated from S11 by a simple formula.

See https://en.wikipedia.org/wiki/Scattering_parameters for a summary of the technical explanation

Memory aid: picture the device facing toward the right, with left side numbered as 1 and right as 2. Then 1-2 and 2-1 would be traveling from one end to the other. Only 1-1 both goes in and (back) out on the left side.

Silly memory tip: The straight lines of "11" look like the tips of an electrical plug you "input" into a electrical socket.

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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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