The KCRC TechNet Puzzles For 2019

(Click Here For TechNet Puzzles for 2017)

(Click Here For TechNet Puzzles for 2018)

 

The KCRC TechNet Puzzle for April 8, 2020

How can a sensitive moving coil meter be used to measure a high d.c. voltage?

A) By fixing a resistor in series with it.
B) By fixing a shunt resistor across its terminals
C) By fixing a large capacitor across its terminals
D) By fixing a rectifier in series with it

 

The KCRC TechNet Puzzle for March 25, 2020

Jackie suspects that the poor signal reports she is getting are due to dampness in the coaxial cable feeding her antenna.

She feeds 20 watts into the cable and gets 10 watts into a dummy load at the other end.
So, the loss in her coax is

A) 2 dB
B) 3 dB
C) 4 dB
D) 6 dB

And now for the answer to our March 25, 2020 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

A decibel is a ratio – In the above example the ratio is 10 watts:20 watts, or ½

dB = 10log ½
or
ratio = 1/2 = 10 ^ dB/10

Most people don’t memorize logarithmic tables, and slide rules are pretty much a thing of the past, but there are simple numbers to remember:

Doubling or halving is +/- 3dB
A tenfold increase/decrease is +/-10dB

When you multiply the ratios, you add its decibels
When you divide the ratios, you subtract its decibels
In other words, you can look at an increase by a thousand as:

10 x 10 x 10 = 1,000

In decibels that would be

10dB + 10dB +10dB = 30dB

Just by memorizing 3dB = 2x, and 10dB = 10x you can quickly figure in your head that:

3dB + 3dB = 6dB = 2 x 2 = 4x ratio
And
10 dB – 3dB = 7dB = 10 /2 = 5x ratio

Getting back to the question at hand, when you halve the original signal, you are experiencing 3dB of loss (you can consider it also at -3dB of gain, which would be like a double negative in English).

The correct answer to this question, is: B) 3 dB

 

The KCRC TechNet Puzzle for March 11, 2020

The value on a dBW scale is given in relation to:

A) 1 mW
B) 1 W
C) 10 W
D) 100 W

And now for the answer to our March 11, 2020 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

Decibels are tenth of a ‘Bel’. Bels, named after Alexander Graham Bel, are in themselves unitless ratios – +1 Bel is an increase by 10 times, while -1 Bel is a decrease to 1/10th times. Bels were deemed too large to be useful, so a tenth of a Bel is generally used – the decibel, it just means that 10 decibels = 1 Bel = an increase by 10 times.

Common practice is to add a unit designation after the ‘dB’ term to define what the scale is in relation to – such as dBx, where the unit would be whatever ‘x’ stood for. As an example, the increase or decrease in voltage is designated dBV – a value of 0 dBV is equal to 1 volt, 10 dbV is equal to 10 volts, 20 dBV is equal to 100 volts.

In the case of the unit of power, Watts, a value of 0 dBW is one Watt…

So, the answer to the question above is ‘B’ 1 W.

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The KCRC TechNet Puzzle for February 26, 2020

The modulation envelope of an AM signal carrying a sine wave is displayed on an oscilloscope. Three complete cycles of the sine wave are shown across the screen. The frequency of the time-base is:

A) One third that of the carrier frequency
B) Three times that of the carrier frequency
C) One third that of the modulating sine wave.
D) Three times that of the modulating sine wave

And now for the answer to our February 26, 2020 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

Well, since the oscilloscope is described as taking up three full cycles of the audio modulating waveform, the time base can only be compared to the modulating wave.

An oscilloscope plots voltage against increasing time. The voltage amplitude is displayed in the Y axis and the time is displayed in the X axis – a wave that completes an entire cycle on the oscilloscope’s display is said to have the same cycle period as the oscilloscope’s time base. If the time base is adjusted to half the time period two complete waves will be displayed on the screen…

So, the correct answer to the above question is: ‘C’, one third that of the modulating sine wave.

 

The KCRC TechNet Puzzle for February 12, 2020

A two-tone oscillator is used in:

A) Measuring the SWR on a transmission line
B) The calibration of an oscilloscope time-base
C) Morse practice transmissions on FM
D) The measurement of the output of an SSB transmitter.

And now for the answer to our February 12, 2020 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

A single frequency oscillator will let you test the accuracy of your oscillator’s time base, without a second frequency needed.

Morse code keys your transmitter’s RF oscillator, you select the audio frequency you hear by adjusting the frequency offset of your receiver – no two-tone signals required.

Two tone testing consists of applying two clean non-harmonically related sine waves of approximately equal amplitude to the audio input of an SSB transmitter. The sine waves are typically around 600 – 700Hz and 1800 – 2000Hz i.e. about 300Hz from either end of the audio pass band. The result, in a properly adjusted transmitter, is an RF output that varies from zero to maximum at a rate determined by the difference in frequency between the two audio inputs. Consequently, overdrive (which causes splatter), non-linearity, instability and a host of other problems are easily visible on an oscilloscope. This thorough testing of the transmitting system becomes even more important when the transmitter is followed by a linear amplifier, so as not to compound the problem with more errant RF energy.

So, the correct answer is ‘D’ the measurement of the output of an SSB transmitter.

 

The KCRC TechNet Puzzle for December 11, 2019

An alternating voltage can be measured with a bridge rectifier and a moving coil meter.

The limit to the sensitivity of such an arrangement is determined by the:

A) voltage drop across the rectifier.
B) momentum of the meter
C) capacitance of the range switch on the meter
D) resistance of the probes

And now for the answer to our December 11, 2019 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

Sensitivity is the minimum input magnitude that can be detected and measured. The momentum of the meter might introduce some initial error in the meter reading but is self-corrected in a small interval of time. Capacitance of the range setting rotary switches should not affect the meter’s sensitivity. The resistance of the voltage probes would introduce errors in accuracy of the measured values but would not only limit the device’s sensitivity.

All solid-state rectifiers possess a characteristic forward voltage drop, when forward biased into conducting electricity. Diodes made of different substrates possess different voltage drops – silicon-based diodes have a 0.7V forward voltage drop, while Germanium-based diodes have a 0.3 V forward voltage drop. These voltage drops are independent of the current flowing through the diode.

If the voltage to be measured by the bridge rectifier is below its threshold forward voltage drop it will not rectify ANY current and the meter will be useless. It is this factor in this question, that limits the sensitivity of this measuring device.

The correct answer to this problem is ‘A’ – voltage drop across the rectifier

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The KCRC TechNet Puzzle for November 27, 2019

Bob is testing a home-brew transmitter. He notices that the voltage from his power supply drops when he keys the transmitter. Bob should test especially for:

A. Key clicks
B. Chirp.
C. Over- modulation
D. Parasitic oscillations

And now for the answer to our November 27, 2019 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

For this question you would need to have an understanding about CW transmissions and what leads to degradation of a readable signal

You would think that the best CW transmission would be pure “square waves”:

But when you try this you actually end up with:

Those are heard as ‘key clicks’!
What you REALLY want to transmit are sweeping gradually increasing and decreasing waves:

Over-modulating a CW signal is not easy to do, since it alternates between 0% power and 100%. In theory you can overmodulate it by transmitting ‘sorta CW’ by transmitting AM bursts at the CW tone frequency. So called Modulated CW (MCW) isn’t legal as far as the FCC is concerned, but if you did use it and over-modulated it you would generate harmonics outside of your signal’s normal bandwidth.

Parasitic Oscillations are unwanted oscillations on frequencies other than the one that you are trying to communicate on. They are usually caused by stray resonant circuits in the amplifier chain of your transmitter, not related to your voltage supply’s voltage regulation.

A ‘Chirp’ is a signal whose frequency increases or decreases with time. There are three common causes for this problem:

1) Voltage supply instability due to poor voltage regulation
2) ‘Pulling’ – where the frequency of your transmitter’s other stages change when you key your transmitter.
3) RF Feedback in your transmitter.

To sum it all up, the correct answer is (B) Chirp should be considered if your power supply’s voltage regulation is less than optimum!

 

The KCRC TechNet Puzzle for November 13, 2019

Which one of these circuits is a T-network?

  1. Circuit A.
  2. Circuit B.
  3. Circuit C.
  4. Circuit D.

 

And now for the answer to our November 13, 2019 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

This is a relatively easy question IF you know the naming convention. If you draw horizontal lines for series reactance components, and vertical lines for parallel reactance components you can make stick figure similar to a ‘T’ or a ‘L’ or a Pi (π). Here are some illustrations to demonstrate this:

Circuit ‘D’ is an L network, ‘C’ is a Pi network, and ‘B’ is a balanced L network.

Circuit ‘A’ is a T type network.

So, the correct answer is ‘A’!

 

The KCRC TechNet Puzzle for September 25, 2019

In which category of transmission is the carrier suppressed?

A. FM
B. SSB
C. AM
D. CW

And now for the answer to our September 25, 2019 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

I guess that you just have to know how Frequency Modulation, Single Side Band, Amplitude Modulation, and Continuous Wave transmission works.

Suffice it to say, Single Side Band works by suppressing the carrier and either the upper or lower sideband…

So, the answer is “C” SSB!

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The KCRC TechNet Puzzle for September 11, 2019

The accuracy of a frequency counter is specified as 0.5 parts per million.

If the display reads 2.000000 kHz, the true frequency could be anywhere between:

A. 1.999981 and 2.000019 kHz
B. 1.999987 and 2.000013 kHz
C. 1.999993 and 2.000007 kHz
D. 1.999999 and 2.000001 kHz.

And now for the answer to our September 11, 2019 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

First, let’s discuss ‘Percent error”, how many percents ‘off’ an observed value can be from the true value, due to the measurings device’s accuracy in measuring this value.

It is usually expressed as:
Percent Error = [{(Observed Value) – (True Value)} / True Value] x 100 (percent)

It is simple to see that if we want an error rate expressed as Parts Per Million, rather than Parts Per Hundred (percent), we substitute the 100 in the equation with 1,000,000:

Parts Per Million Error = [{(Observed Value) – (True Value)} / True Value] x 1,000,000 (percent)

For the above question, for the highest observed value, with a 0.5 ppm error, the equation can be substituted as:

0.5 = (X-2)/2 * 1,000,000
0.5 * 2/1,000,000 = X-2
0.000 001 = X -2
2.000 001 = X

That would be answer “D” 1.999 999 and 2.000 001 kHz.

 

The KCRC TechNet Puzzle for August 28, 2019

For a circuit to resonate, an inductor and a capacitor are needed. If the capacitor is replaced by one with four times the value, the frequency of resonance will:

A. Decrease to one quarter of its previous value
B. Be halved.
C. Double
D. Increase by 4 times

And now for the answer to our August 28, 2019 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

Remember the puzzle from June 12, 2019????????

Well, if “C” were replaced with “4C” (four times its original value, then its resonant frequency would be 4’s inverse square root, or one half.

So the answer would be “B” Be halved!

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The KCRC TechNet Puzzle for August 14, 2019

In a purely resistive AC circuit:

A. The current in the resistance leads the applied voltage by 6o°
B. The voltage across the resistance and the current through it are in phase
C. The R.M.S. value of the applied voltage lags its peak value by 18o°
D. the current through the resistance lags the voltage across it by 90°

And now for the answer to our August 14, 2019 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

The reactive part of complex impedance, due to inductance or capacitance will cause a phase difference between an alternating current and its voltage. In a purely resistive circuit, free of any net inductive or capacitive reactance, the current and the voltage of the alternating current remains in phase with each other.

So, the correct answer is ‘B’ that the voltage and current are in phase.

 

The KCRC TechNet Puzzle for July 24, 2019

One of the main causes of fading is:

A. The time difference between the transmitting and receiving stations
B. Instability in the transmitted frequency of the signal
C. The same signal reaching the receiver by two different paths
D. Changes in the beam heading at the transmitter

And now for the answer to our July 24, 2019 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

Well, your local time and the local time of your contact are immaterial to your reception, and gone are the very bad old days of excessive frequency drift of local oscillators, and directional beams tend not to rotate like the second hand of clocks, so just by the process of elimination we are left with answer “C”. Radio waves don’t all travel in the same path and this multipath can introduce phase changes that be additive or subtractive – if two different paths produce two signals 180 degrees out of phase with the other one, they cancel out and you hear it as a “fade”.

So, the best answer is “C”

 

 

The KCRC TechNet Puzzle for July 10, 2019

In calculating SINAD, most of the ‘noise’ comes from:

A. The ionosphere
B. Outer space
C. Nearby electrical interference
D. Within the receiver

And now for the answer to our June 26, 2019 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

SINAD is a receiver audio quality measurement that is typically used for mobile stations operating on an analog system. It is the ratio of Signal+Noise+Distortion divided by Noise+Distortion, expressed in dB.

It takes into account a receiver’s inherent internal noise.
So, answer “D” is the correct one.

 

The KCRC TechNet Puzzle for June 26, 2019

The collector current of an amplifier biased in Class A mode is illustrated in:

A. Graph P
B. Graph Q
C. Graph R
D. Graph S

And now for the answer to our June 12, 2019 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

For this question you need to understand what different “class” amplifiers mean – in the case of “A”, “B”, and “C” it describes the portion of the alternating signal that is linearly amplified. The most linear, distortion-free mode, Class “A” amplifies through the entire cycle, from zero, to positive maximum, back to zero, to negative maximum and back to zero! It has the least nonlinearity BUT is the least efficient mode.

Class “B” amplifies only one half of each cycle – from zero to positive maximum and back to zero.

Class “AB” is somewhere between the amount of each cycle that classes “A” and “B” are amplified.

Class “C” amplifies even less of each cycle than in class “B”.

With all that under your belt, a quick look at the illustrations show that only one of them appear to have both halves of their sinusoidal curves demonstrated – Graph “R”.

So, the answer is “C” Graph R

 

The KCRC TechNet Puzzle for June 12, 2019

Capacitance and inductance make up a parallel resonant circuit.
If the inductance is decreased to one quarter of its original value, what will happen to the resonant frequency?

A. It will decrease to a quarter of its original value.
B. It will halve.
C. It will double.
D. It will increase by 4 times.

And now for the answer to our June 12, 2019 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

I’m sorry, but sometimes you just have to understand a bit of the mathematics of electronics. The frequency of a tuned circuit is calculated by the equation:

So, L is decreased to L/4 inside the square root function of the equation, so the denominator would change by ½, doubling the previous final calculation.

Which is a long way of saying that by quartering the inductance, doubles the resonant frequency!

So, the answer is “C” Double!

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The KCRC TechNet Puzzle for May 8, 2019

The graph below shows the voltage and current in a:

A. Capacitor
B. Resistor
C. Inductor
D. Diode

And now for the answer to our May 8, 2019 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

Is it “current lags voltage” or “current leads voltage”? Just think about it a bit:

If you take a capacitor and apply a voltage potential, the capacitor starts quickly charging at its maximum rate. As the charge develops the current grows less and less and stops when the plate is fully charged – current leads voltage.

If you take a resistor and apply a voltage potential, the current produced remains the same. With an alternating current source, the current and the voltage are perfectly in phase – neither “lags” nor “leads”.

If you take an inductor and apply a voltage potential, the inductor starts building a magnetic field, limiting the current until it builds up completely and then passes current uninhibited. With an alternating current source, the current lags the voltage – the opposite of the illustration above.

A diode will conduct only during one half of the alternating cycle, and that half-wave’s current/voltage waves would be in phase.

So, the answer is “A” Capacitor!

 

The KCRC TechNet Puzzle for April 24, 2019

The power reflected back from an antenna can be measured with:

A. An SWR. meter.
B. A digital frequency counter
C. An oscilloscope
D. A dummy load

And now for the answer to our April 24, 2019 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

“B”, a digital frequency counter shows the frequency of the signal inputted, “C”, an oscilloscope shows signal amplitude vs time, “D”, a dummy load supplies 50 ohms of pure resistance impedance to your transmitter’s output.

Which leaves us with “A”, an S.W.R. meter, which measures forward and reverse power running through the transmission line. With those values it can calculate the Voltage Standing Wave Ratio, so the answer is:

(A) An Standing Wave Ratio meter (S.W.R.)

 

 

The KCRC TechNet Puzzle for April 10, 2019

If your monitor on your computer distorts when you transmit on 80 meters, to remove this, you should:

A. Reduce the resolution of his monitor
B. Re-boot the computer
C. Change to a screened mains lead
D. Fit ferrite rings on the cable between monitor and computer

And now for the answer to our April 10, 2019 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

You’re getting RF into your monitor, that it doesn’t want and can’t deal well with.

Short of locking it up in a Faraday Cage, placing ferrite RF chokes on any cable going into or out of your monitor will help diminish routes for RF to sneak into your monitor, by increasing the impedance that any common code currents will have to deal with.

The correct answer is ‘D’.

 

The KCRC TechNet Puzzle for March 27, 2019

A radio signal in the ionosphere ‘will have the greatest range in a single ‘bounce’ when it is returned to earth from which layer?

A. D region
B. E layer
C. F1 layer
D. F2 layer.

And now for the answer to our March 27, 2019 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

The easiest way to think about this question is that the higher the ionospheric layer, the farther the RF’s ‘bounce’. There’s a good reason how ionospheric layers get named – the ‘D’ layer is closest to Earth’s surface.
Further out you will find the ‘E’ layer, then the ‘F1’ layer and finally the ‘F2’ layer being the farthest away from the surface of the Earth.

The bigger the distance from the Earth the bigger the ‘bounce’ – so answer ‘D’ “the F2 layer” is the correct answer!

 

The KCRC TechNet Puzzle for March 13, 2019

Why are the radial elements of a quarter wave ground plane antenna often made to droop?

A.   To match 50-ohm coaxial cable.
B.   To increase the angle of maximum radiation.
C.   To avoid RF flowing on the outer of the feeder.
D.   To prevent static build up on the vertical element.

And now for the answer to our March 13, 2019 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

It helps to know how antennas work. A ¼ wave ground plane vertical antenna is equivalent to one half of a half wave dipole. Resonant half-wave dipoles in free space possess a nominal impedance of 73 ohms, a quarter-wave ground plane vertical is half of that, 36 ohms in free space, BUT if you droop those radials downward you can increase its impedance at resonance all the way to 50 ohms! You can also position the ground planes to a specific height with regard to Earth ground, to obtain a 50-ohm impedance, but ‘A’ is the best answer for this question.

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The KCRC TechNet Puzzle for February 27, 2019

What frequency would you tune your radio to check that you were not radiating any spurious harmonics from your transmissions on 29.0 MHz?

A. 29.0 MHz
B. 43.5 MHz
C. 52,5 MHz
D. 58.0 MHz

And now for the answer to our February 27, 2019 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

You just have to know how ‘harmonics’ works. The ‘first harmonic’ of 29.0 MHz is 29.0 MHz, where your signal belongs. Its ‘second harmonic’ is twice its primary frequency, or 58.0 MHz in this case.

So, the correct answer is ‘D’, 58.0 MHz.

 

 

The KCRC TechNet Puzzle for February 13, 2019

“ALC” stands for:

A. Automatic Level Control
B. Audio Low-pass Coil
C. Alternating Local Current
D. Antenna Loading Choke

And now for the answer to our February 13, 2019 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

Sometimes ya just gotta know what an abbreviation stands for. In this case it’s just an abbreviation for Automatic Level Control, choice ‘A’.

ALC is a safety feature used with linear amplifiers, so that they are not overdriven beyond their linear range of amplification. Originally, they were designed to feed an increasing negative grid current into the final amplifier tube of a tube-based transmitter, when the linear amplifier detected that the driver power was too high. Although most transmitters are solid state, the ALC remains a negative voltage potential increasing in voltage as the transmitter’s driving power goes higher and higher beyond the linear amplifier’s design specification for an input drive for linear operation.

Technically, it operates much like a receiver’s Automatic Gain Control, except for transmitter power, instead of receiver gain, and is more of a safety feature, than a way to limit your hearing loss from too much receiver gain blowing out your eardrums. Often the ALC uses the same drive limiting circuitry that allows your transmitter to “fold back” transmitter power if your SWR is detected to be too high.

 

 

The KCRC TechNet Puzzle for January 23, 2019

To receive CW, a suitable bandwidth for the IF filter is:

A.   25 Hz
B.  250 Hz
C.  2.5 kHz
D.  25 kHz

And now for the answer to our January 23, 2019 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

The simplest way of thinking about this question, is to ask what bandwidth would be required to receive Morse code. I don’t know about you, but 25 Hz is subsonic and inaudible to me, so ‘A’ isn’t the correct answer. A bandwidth of 2.5 kHz would be fine for SSB phone transmissions, but it’s wasteful for CW, and 25 kHz would be closer to the bandwidth of an FM signal, but even more wasteful for CW.

So, the best answer in this case would be 250 Hz, answer ‘B’.

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The KCRC TechNet Puzzle for January 9, 2019

The block diagram of the sideband generator of an SSB transmitter is drawn below.
What will a probe at point ‘S’ reveal?

A. Carrier plus one sideband
B. Carrier only.
C. Carrier plus two sidebands
D. Two sidebands

.And now for the answer to our January 9, 2019 TechNet Puzzle

(It is in “Invisotext” and will be visible if you highlight the area below!)

Well, if you just look at the labels on the illustration, it seems self-evident that a probe at point ‘S’ would contain only a carrier, but let’s discuss how an SSB transmitter works.

We start with an audio frequency microphone signal – your voice (seen by a probe at point ‘P’). Then we need to generate the RF signal that will propagate through the ionosphere to that distant receiver. That is what the carrier oscillator is designed to do. If we only used a carrier oscillator, all we would have is a CW transmitter, but we are building an SSB transmitter. When alternating currents with different frequencies are ‘mixed’ together in a nonlinear device, this interaction produces additional signals that are the sum and the difference of the two original frequency signals. A ’Balanced Modulator’ is a special kind of mixer – it mixes the carrier frequency with the audio frequency, generating ‘sideband’ frequencies: the sum of the carrier wave frequency and the audio frequency for the upper sideband and the difference between the carrier wave frequency and the audio frequency for the lower sideband. The Balanced modulator also filters out the original carrier wave frequency from the newly created two sidebands (seen by a probe at point ‘Q’).

At this point, you have a double sideband transmission – but all you need is the upper or the lower sideband to get your transmission to be readable.

That is what the filter is for – to remove one of the sidebands, giving you a single sideband signal at probe point ‘R’ – ready for further linear amplification before it goes to the antenna to be radiated to the person that you are communicating with!

So, the correct answer is that a probe at point ‘S’ will show only the carrier signal, choice ‘B’!