The KCRC TechNet Puzzles For 2021

(Click Here For TechNet Puzzles for 2017)

(Click Here For TechNet Puzzles for 2018)

(Click Here For TechNet Puzzles for 2019-2020)

 

The KCRC TechNet Puzzle for June 23, 2021

Which of the following are found in the output from a balanced modulator of a Single Sideband transmitter?

…..Carrier    audio       lower           upper
………………….   ….sideband     sideband

A)    yes        yes           no               no
B)     no         yes          yes              no
C)     no          no          yes              yes
D)    yes         no           no               yes

 

The KCRC TechNet Puzzle for June 9, 2021

The left-hand graph shows the electric field of a radio wave. The associated magnetic field is drawn in diagram:
A) a
B) b
C) c
D) d

 

And now for the answer to our June 9, 2021 TechNet Puzzle

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

I guess this one you just have to know. The magnetic field is ‘temporally in phase’ with the electric field – that means that when the electric field is maximum, so is the magnetic field, and when the electric field is minimum, so is the magnetic field. That knocks out answers A, B and D as possible true answers.

Another thing (that you don’t need to know to answer this particular question) is that the magnetic wave is rotated 90 degrees from the plane of the electric wave – they are said to be ‘orthogonal’ with each other. As you may see, curve c is orthogonal to the initial curve.

Answer ‘C’ is the right answer!

 

 

The KCRC TechNet Puzzle for May 26, 2021

A 3-element trap Yagi antenna contains a total of 6 traps.

How many amateur bands will it resonate?

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

And now for the answer to our May 26, 2021 TechNet Puzzle

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

The driven element for any Yagi is simply a dipole half-wave resonant dipole. The simplest trapped dipole, with on trap on either side, is only resonant for two bands. Since it is a Yagi, it must include a slighter longer parasitic element behind the driven element that acts as a ‘reflector’ and a parasitic element slighter shorter in front of the driven element, acting as a ‘director’. In order for the reflector and the director to work on both of the driven element’s resonant bands, they must also have a pair of traps on each of them!

2 + 2 + 2 = 6

That means you need six traps to build a two band trap Yagi!
So, the correct answer is ‘A’ 2.

(click to see illustration)

 

The KCRC TechNet Puzzle for May 12, 2021

When the base current of a transistor is 3 microamps, its collector current is 1.5 milliamps.

The current gain β for such a transistor would be:

A) 200
B) 250
C) 400
D) 500

And now for the answer to our May 12, 2021 TechNet Puzzle

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

The beta value for a given transistor (β) is defined as the ratio of the collector current over the base current, whin in this case would be:

β = 1,500 µA / 3 µA

β = 500

Answer ‘D’ 500 is the correct answer!

 

The KCRC TechNet Puzzle for April 28, 2021

The above circuit shows a:

A) Peak R.F. Indicator
B) Key click filter
C) Antenna Matching Unit
D) Class A amplifier

And now for the answer to our April 21, 2021 TechNet Puzzle
(It is in “Invisotext” and will be visible if you highlight the area below!)

Well, there are some obvious giveaways to what this circuit is decidedly not!

There is no meter or indicator of any kind to be seen, so it couldn’t possibly be a ‘peak R.F. indicator.

There is no vacuum tube, nor a transistor, so the circuit can’t amplify any signals, so answer ‘D’ Class A amplifier is wrong.

Antenna matching circuits are impedance matching devices. There are many, many types – Pi networks, L networks, T networks. The illustration above does not look like any of them. Probably because it cannot match impedances and isn’t an antenna tuner circuit.

This leaves us with answer ‘B’ – Key click filter. The fact that the illustration clearly shows a labeled ‘key’ should be a dead giveaway that this is the right answer. When a morse code key ‘unkeys’ the signal drops precipitously – that sharp drop is rich in harmonics and produces an annoying click sound on the received signal.

The difference between an unfiltered key switch with a click, and a filter key switch without an annoying click can be seen in the following illustration:

Answer ‘B’ Key click filter is the correct answer.

 

 

The KCRC TechNet Puzzle for April 14, 2021

Another Ham is heard by you, with a signal strength of S-2, two S units. He tells you that he is operating with an output power of 100 Watts, and that he will turn on his linear amplifier and increase his signal output power to 1,000 Watts.

Assuming no other complicating factors, if your HF radio’s S Meter is properly calibrated, what would you expect your contacts signal strength to now be, based upon your S meter:

  1. A) 3.7 ‘S‘ units
  2. B) 5.2 ‘S‘ units
  3. C) 7.4 ‘S‘ units
  4. D) 9.1 ‘S‘ units

 

And now for the answer to our April 14, 2021 TechNet Puzzle

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

As we’ve discussed during past TechNet, a 10-fold increase in transmitter power is the same as a 10dB increase in power. One ‘S’ unit on HF is defined as a 6dB increase in received power.

In ‘S’ units, a 10dB increase would translate as a 10/6 ‘S’ unit increase, or 1.6666…

If your original signal were ‘S’ of 2, then the new signal strength with the linear amplifier turned on would be 2 + 1.7 = 3.7

So, answer ‘A’ S of 3.7 units is the correct answer!

.

The KCRC TechNet Puzzle for March 24, 2021

A trap dipole antenna will resonate on two different bands.

When resonant on its higher frequency band, the current distribution on one side of the antenna is as shown in which of the following graphs?

A) Graph A

B) Graph B

C) Graph C

D) Graph D

 

And now for the answer to our March 24, 2021 TechNet Puzzle

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

 

The way that a trap dipole works is that it contains LC traps that block RF at high frequencies. This means that at higher and higher frequencies the mechanical length of a given trap dipole is shorter than it would be at lower frequencies, when the traps would allow RF current to flow passed the traps.

At a high frequency, the trap would block RF current, and the antenna would appear as a mechanically shorted antenna resonant for the higher frequency.

Current on one leg of a centerfed dipole is highest at the feedpoint, so Graph ‘A’ is wrong from the start. Graph ‘D’ shows what the current would be at high frequencies, when the RF current could flow through the trap without any impedance, so that is wrong.

Graph ‘C’ shows current flowing past the point of the high frequency trap, so that is wrong.

Graph ‘B’ shows the proper current distribution of a trap dipole when the antenna is fed at or higher than the highest frequency band that it was designed to be resonant at.

Answer ‘B’ is the correct answer.  

 

 

The KCRC TechNet Puzzle for March 10, 2021

Recently, a local repeater was received with a signal strength of -77dBm on a local Ham’s radio. After a failure of the repeater, its transmission signal was received at -94dBm by the same Ham on the same radio. What fraction of the original signal was being radiated after the repeater failure:

A) 30%
B) 7%
C) 2%
D) 0.1%

And now for the answer to our March 10, 2021 TechNet Puzzle

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

 

The difference between the signal strength before the failure and after the failure:

(-94dbm) – (-77dBm) = -17dBm

Sure, you can take the antilog of 17/10, but let’s see if he can do this ourselves…

We learned earlier to recall just a few numbers related to decibels:

-10dB = 1/10 of the original power

And

-3dB = ½ of the original factor

We also (hopefully) recall that:

dB = Log(A)/10

Log(A)/10 / Log(B)/10 = Log(A-B)/10

Or

7dB loss is equal to 10db loss, divided by 3dB loss (Log0.7 = Log 1.0 – Log0.3)

-7dB = 1/10 / ½

-7dB = 1/5

Then, remembering that 10dB loss reduces power to 1/10

-17dB = -10dB – 7dB

-17dB = 1/10 / 1/5

-17dB = 1/50 = 2%

So, the correct answer is “C” 2% of the pre-failure output was being transmitted after the repeater failure.

 

The KCRC TechNet Puzzle for February 24, 2021

Which one of the following is not a factor affecting accuracy when measuring frequency using the station receiver and a crystal oscillator?

A) The stability of the transmitter under test.
B) The automatic gain control (AGC) of the receiver.
C) The crystal specification.
D) Aural error in determining the point of zero beat.

And now for the answer to our February 24, 2021 TechNet Puzzle

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

First, keep in mind that this is one of those sneaky problems, that isn’t asking for the single best answer, but the single worse one!

Answer ‘A’ is the wrong answer, you would need the transmitter’s frequency stability in order to measure your receiver’s frequency accuracy – it would be like shooting at a target while being blindfolded.

Answer ‘C’ is the wrong answer, since the receiver’s crystal’s specification in the crystal oscillator will have a direct effect in the accuracy of measuring the receiver’s frequency accuracy.

Answer ‘D’ is also the wrong answer, since you would need to ‘zero beat’ the receiver to the crystal oscillator’s signal in order to get to the closest approximation of the crystal oscillator’s frequency. If you erred by 10 or 20 Hz on either side, it would introduce an additional 10 or 20 Hz error to a more accurate frequency measurement.

So, the answer is ‘B’ The automatic gain control (AGC) of the receiver. Changing the AGC should not affect the frequency of the receiver, and therefore it would not change the accuracy of the receiver’s frequency measurement.

 

The KCRC TechNet Puzzle for February 10, 2021

The measurement of forward and reflected power on a transmission line is carried out with a:

A) digital multimeter
B) signal strength meter
C) grid dip meter
D) standing wave ratio meter.

And now for the answer to our February 10, 2021 TechNet Puzzle

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

What this question is really asking is: Which of these devices has a directional coupler, as part of its design?

A directional coupler can measure currents traveling in opposite directions on the same transmission line. If your transmitter is perfectly impedance matched to your antenna system, all of the RF power is transferred to the radiating antenna, and nothing is reflected back.

A directional coupler would detect all the power going in one direction, and nothing traveling in the other direction. If you have an impedance mismatch, i.e. a SWR >1:1, then part of the power traveling through the transmission line to the antenna will be reflected down the transmission line in the opposite direction. As your SWR gets higher and higher the impedance mismatch gets higher and higher and the percentage of reflected power gets higher and higher and can be seen as a reverse current using a directional coupler set to detect that directional current.

What I have just described is an SWR Meter!

So, answer “D” is correct! Multimeters, signal strength meters and grid dip meters are not designed to detect directional currents, so they are all the wrong answer for this question.

 

The KCRC TechNet Puzzle for January 27, 2021

This is the block diagram of a

A) phase locked loop
B) transverter
C) voltmeter
D) frequency meter

And now for the answer to our January 27, 2021 TechNet Puzzle

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

Well, I guess you should know what each choice is, and how they work, to answer this one:

A phase lock loop is a control circuit, whose output signal is referenced by the phase of its input signal. A common use for a phase lock loop in our radios is variable frequency oscillator, a phase detector, and a feedback loop. The development of this circuit made it possible for us to throw away all those old crystals that plugged into the front of old transmitters! The diagram for a phase lock loop doesn’t look anything like the one in our question, so ‘A’ isn’t right.

A transverter is used to up ‘up convert’ and ‘down convert’ a radio signal, so that a transceiver can operate beyond its designed frequency range. Its block diagram shows a local oscillator, a mixer and band filters. That doesn’t look anything like the block diagram seen in this question, so answer ‘B’ is wrong.

There are earlier TechNet Puzzles that go into how voltmeters operate. They are galvanometers, with voltage drop serial resistors, so our answer isn’t ‘C’ either.

That leaves us with ‘D’ frequency meter, which shapes the incoming sinewave signal into square waves, use a calibration crystal to reset a counter circuit to an accurate time interval and a display to output the counted square waves during the accurate time interval. That looks exactly like our block diagram above! That’s because ‘D’ is the right answer for this question!

 

The KCRC TechNet Puzzle for January 13, 2021

Which control on an oscilloscope determines the speed at which the trace moves from left to right?

A) X shift
B) Timebase.
C) Y gain
D) Brilliance

And now for the answer to our January 13, 2021 TechNet Puzzle

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

You have to keep in mind what an oscilloscope is: a display of amplitude (in the Y-axis), over time (the X-axis). The X-shift and Y-shift knobs only move the display to another part of the display – they do not alter any amplitude or time-based signals.

The Brilliance knob does not alter the time-based signals or the amplitude – it just makes the display brighter or dimmer, depending upon its setting.

The Y-gain alters the deflection of the image for a given amplitude. The higher the Y-gain, the smaller a change will be visible on the display.

The X-shift, the Y-gain, and the brilliance controls do not affect the speed at which the trace moves from left to right.

The only remaining answer, ‘B’ Time base changes the rate that the trace sweeps the X-axis as a function of time.

So, ‘B’ Time base is the correct answer!