Sun Coast Amateur
Radio Club Society - hambasics:sections

b001

Under Construction: VA7FI is editing this section, please do not edit it until this notice is taken down.

Regulations

To prevent interference between different users, the generation and transmission of radio waves is strictly:

regulated by national laws and
coordinated by an international body called the International Telecommunication Union (ITU).¹⁾

A Few Canadian Rules

ISED regulates the amateur radio service as defined in the Radiocommunications Regulations derived from the Radiocommunication Act.
Although the Amateur Certificate is free and valid for life, you must notify ISED of any change of address.
The certificate must be retained at the station (at the address provided) and an operator may be asked to produce it to an inspector within 48 hours. However the holder of an Amateur Radio Certificate can operate an amateur radio anywhere in Canada.
There is no age limit to hold an Amateur Radio Certificate with Basics Qualification, but you must have a valid Canadian address.
Payments for third-party message is also not allowed.
Offensive language, business-related conversations, music, deceptive signals, encrypted, coded, or secret messages, harmful interference, are not allowed on ham bands. In other words, only conversations of a personal or technical nature are allowed.
Broadcasts to the general public, or communications with non-ham are also not allowed on ham bands except if it is required for the immediate safety of life of individuals or the immediate protection of property. The only non-emergency one-way transmissions allowed are from beacons.
In emergency situations, a ham operator can use any means of radio communications necessary, including transmitting outside the ham bands and at any power output.

If you have an Amateur Radio Certificate with Basics Qualification:²⁾

You are allowed to let your friend who doesn't have a license talk on your radio as long as you are there to operate and control the station.
But if a foreign station breaks in to talk to your friend, have your friend wait until you determine from the foreign station if their administration permits third-party traffic. Canada doesn't expressly forbids other countries to pass third party traffic to us, but the other country might.
The following organizations are not considered third-party:
- Canadian Forces Affiliate Radio Service (CFARS)
- United States Military Auxiliary Radio System (MARS)
Your friend is allowed to listen without you present, but not transmit.
You are not allowed to install, place in operation, modify, repair, maintain, or permit the operation of the radio apparatus on behalf of your friend.

If you have an Amateur Radio Certificate and are using another amateur's stations,

both of you are responsible for the proper operation of the station.
you can only operate your friend's station to the lesser of the two certifications.

‒

Required Reading

Read the following from the ISED Website. Hopefully the sections above will help put these in more context.

Questions

B-001-001-001 → B-001-002-007
B-001-003-004 → B-001-004-002
B-001-004-007 → B-001-006-002
B-001-007-001 → B-001-008-003
B-001-009-001 → B-001-010-002
B-001-010-005 → B-001-010-006
B-001-010-009
B-001-011-001 → B-001-012-004
B-001-014-001 → B-001-014-009
B-001-014-011 → B-001-015-002
B-001-020-001 → B-001-022-005

SKIPPED B-001-016-009 B-001-023-001

¹⁾

See: Radio Spectrum

²⁾

See RBR-4, Sec 6, p.2, and RIC-3, Sec 5.3, p.13 and RIC-3, Sec 6, p.15

codes

The Standard International Phonetic Alphabet

To make your call sign better understood when using voice transmission, you should use the Standard International Phonetics for each letter:

Alpha	Foxtrot	Kilo	Papa	Uniform	Zulu
Bravo	Golf	Lima	Quebec	Victor
Charlie	Hotel	November	Romeo	Whisky
Delta	India	Mike	Sierra	X-Ray
Echo	Juliet	Oscar	Tango	Yanky

Q-Code

“The Q code is a standardized collection of three-letter codes all of which start with the letter “Q”. It is a brevity code initially developed for commercial radiotelegraph communication and later adopted by other radio services, especially amateur radio.”³⁾

There are a lot of them, but here are the ones you'll need to memorize:

Code	Question	Answer or Statement	Other Usage
QRL	Is this frequency in use?	This frequency is in use. Please do not interfere.
QRM	Do you have interference from other stations?	I have interference form other stations.
QRN	Are you troubled by static?	I am troubled by static.
QRP	Shall I decrease power?	Decrease power.	I'm using low power
QRS	Shall I send more slowly?	Send more slowly
QRT	Shall I cease or suspend operation?	I am suspending operation / shutting off the radio
QRU	Have you anything for me?	I have nothing for you.	No traffic
QRX	Shall I standby / When will you call me again?	Please standby / I will call you again
QRZ	Who is calling me?	You are being called by
QSB	Are my signals fading?	Your signals are fading.
QSL	Can you acknowledge receipt?	I am acknowledging receipt.	Same as “Roger”
QSO	Can you communicate with ... direct or by relay?	I can communicate with ... direct (or by relay through ...).	A conversation
QSY	Shall I change to on another frequency?	Change to on another frequency.
QTH	What is your location?	My location is

Signal Report

A short way to tell another station how well you receive them is to use the “RST” (Readability, Signal Strength, Tone) code. From Wikipedia:

Readability is a qualitative assessment of how easy or difficult it is to correctly copy the information being sent during the transmission. In a Morse code telegraphy transmission, readability refers to how easy or difficult it is to distinguish each of the characters in the text of the message being sent; in a voice transmission, readability refers to how easy or difficult it is for each spoken word to be understood correctly. Readability is measured on a scale of 1 to 5.

1. Unreadable
2. Barely readable, occasional words distinguishable
3. Readable with considerable difficulty
4. Readable with practically no difficulty
5. Perfectly readable

Strength is an assessment of how powerful the received signal is at the receiving location. Although an accurate signal strength meter can determine a quantitative value for signal strength, in practice this portion of the RST code is a qualitative assessment, often made based on the S meter of the radio receiver at the location of signal reception. “Strength” is measured on a scale of 1 to 9.

1. Faint signal, barely perceptible
2. Very weak
3. Weak
4. Fair
5. Fairly good
6. Good
7. Moderately strong
8. Strong
9. Very strong signals

For a quantitative assessment, quality HF receivers are calibrated so that S9 on the S-meter corresponds to a signal of 50 μV at the antenna standard terminal impedance 50 ohms. One “S” difference should correspond to 6 dB at signal strength (2x voltage = 4x power). On VHF and UHF receivers used for weak signal communications, S9 often corresponds to 5 μV at the antenna terminal 50 ohms. Amateur radio (ham) operators may also use a signal strength of “20 to 60 over 9”, or “+20 to +60 over 9.” This is in reference to a signal that exceeds S9 on a signal meter on a HF receiver.

The strength assessment is a bit tricky when using a repeater because every station using the repeater would appear to your S-metre to have the same strength since you're listening to the repeater, not the individual station. In that case, you have to use a qualitative assessment based on readability more than strength since the other station wants to know how well the repeater is receiving them, not how well you receive the repeater. More on repeaters in the next section.

Tone is only used in Morse code and digital transmissions and is therefore omitted during voice operations. With modern transmitter technology, imperfections in the quality of the transmitter modulation that can be detected by humans are rare. Tone is measured on a scale of 1 to 9.

1. Sixty cycle a.c or less, very rough and broad
2. Very rough a.c., very harsh and broad
3. Rough a.c. tone, rectified but not filtered
4. Rough note, some trace of filtering
5. Filtered rectified a.c. but strongly ripple-modulated
6. Filtered tone, definite trace of ripple modulation
7. Near pure tone, trace of ripple modulation
8. Near perfect tone, slight trace of modulation
9. Perfect tone, no trace of ripple or modulation of any kind

Examples

“Your signal is 5 7” = “Your signal is readable and moderately strong”
“Your signal report is 3 3” = “Your signal is readable with considerable difficulty and weak in strength”
“Your are 5 9 plus 20 dB” = “You are perfectly readable with a signal strength of 20 decibels greater than S9”
“Your signal report is 1 1” = “Your signal is unreadable, and barely perceptible”
“RST 579” in Morse Code = “Your signal is perfectly readable, moderately strong, and with perfect tone”
“RST 459” in Morse Code = “Your signal is quite readable, fair strength, and with perfect tone”

On HF radio, each unit on the S-meter represent a four-fold increase in power, which is 6 dB⁴⁾. But after S9, the meter shows a dB scale.

For example: “20 over S9” means

20 dB over S9, which is 10² = 100 times more powerful than a signal at S9, or
26 dB over S8, which is 10^2.6 = 400 times more powerful than a signal at S8, or
32 dB over S7, which is 10^3.2 = 1600 times more powerful than a signal at S7, or

So if a station gives you a 20 over 9 signal report when you transmit with 100W, they would still receive you at S9 if you transmitted at 1W. What a power saving!

Here are a few more examples. How many times stronger is station A than station B if:

Station A is S9 + 40 and Station B is S9 + 10?
- The difference is 30 dB
- Which means station B is 10³ = 1000 times stronger
Station A is S9 and Station B is S6
- Each S unit is 4 times more powerful than the previous S unit
- 3 x 4 = 12 times stronger
Station A is S7 and station B is S9 + 20
- From S9 to S9 + 20 is a difference of 20 dB, which is 10² = 100 times stronger.
- From S7 to S9 is 2 x 4 = 8 times stronger.
- So from S7 to S9 + 20 is 8 x 100 = 800 times stronger.

Questions

B-002-002-001 → B-002-002-011
B-002-006-001 → B-002-007-011

³⁾

See Q code

⁴⁾

For more information about decibels (dB), see the Math Basic page

digitalconcepts

Digital Concepts

Here, we explore concepts of digital signal transmission. We'll see different digital modulation schemes like Frequency-Shift Keying (FSK), Phase-Shift Keying (PSK) and Quadrature Amplitude Modulation (QAM), and explain the difference between bit rate (bit/s) and Baud (Bd).

Morse to FSK

Let's start with Morse code and “transform” it until we can use it as a basic digital system.

Say we (arbitrarily) represented dots by 0s, and dashes by 1s, the letter “A” (•−) would then be represented as “01”.

The first thing we need to fix is that the dashes last longer than the dots. In fact, to speed things up, “in 1910, Reginald Fessenden invented a two-tone method of transmitting Morse code. Dots and dashes were replaced with different tones of equal length.”⁵⁾

For example, here's how “VE7” (•••− • −−•••) could sound like if dots were tones of 600 Hz and dashes tones of 800 Hz (with some dead space in between each note for us to hear the breaks):

afsk_ve7.ogg

The second thing we need to fix is that the number of 0s and 1s (number of bits) that a letter needs is different for different letters. For example: “V” has four bits (0001) where as “E” only has one (0) and “7” has five (11000).

Since the biggest number of bits in Morse Code is 5, it might be tempting to try and fix this problem by adding leading zeros to all the other letters, but this doesn't work because letters with less than 5 bits might map onto existing letters once we add the extra zeros. For example: we can't make “V” 00001, because that's already “4”, and “E” can't be 00000 since that's already “5”, and so on.

So in addition to having every bit be the same length, we'd also prefer an encoding method where every character has the same number of bits. To that end, we could use the Baudot Code, which can encode 32 characters of 5 bits each, or the ASCII Standard, which can encode 128 characters of 7 bits each.

The third thing we need to fix is to make the notes as short as they can possibly be. With morse code, the speed someone can receive is dependent on their skills, but if a computer had “perfect skills”, would there be a theoretical limit to how short a note can be? It turns out there is! The shortest note has to be long enough to include a full cycle of the lowest frequency used. For example, as we'll see below, APRS uses 1200 Hz and 2200 Hz tones for its transmissions. One cycle of 1200 Hz is 1/1200th second (0.000833... s) and 2200 Hz has a cycle of 1/2200th second (0.0004545... s). That means that each “note” has to be at least 1/1200th second long.

The fourth and final thing we need to fix is to eliminate the spaces between each note so that the transmission is one long continuous two-note song. This is easy enough, but it leads to a technical issue that we'll address in a moment: how will the receiver know when a note starts and ends if they're all smooched together?

Before we answer that, let's have a look at a real example⁶⁾ of a digital transmission that uses 1200 Hz and 2200 Hz tones:

AFSK_1200_baud.ogg

If we download the audio file and open it in Audacity, we can see the wave form visually:

The first half of the transmission contains the same pattern of one long wave (1200 Hz) followed by 12 short waves (2200 Hz) repeated 76 times. I separated these in red. Since each of these sections lasts 1/150th of a second and each note lasts 1/1200th of a second, each section contains 8 notes (separated here in yellow).

But why spend so much time sending essentially no information? Remember when we removed all the spaces between notes and wondered how the receiver would be able to know when a note starts and finishes? This part acts as a sort of clock to synchronize the receiver to the transmitter. By sending a series of “ticks”, the receiver now knows where each group of 8 notes starts.

Let's take a look at the beginning of the message that follows the synchronization period:

From here, there's a few ways to decide how we should encode the information: do we want each note to literally represent a 0 and a 1? or do we want the change in note to represent a change in bit? See this short but very instructive page for more details.

Summary

What we have created here is a Frequency-Shift Keying (FSK) system.⁷⁾ The characteristics of (A)FSK are that:

0s and 1s are represented by two different tones.
Each character has the same number of bits.
Each tone is as short as it can be (one cycle of the lowest tone).
All notes are transmitted back to back without spaces in between so a sync method is required.

In a sense, FSK is the digital analog of FM, where as ASK (Amplitude-Shift Keying), which we're going to skip here, would be the analog of AM.

Now, there's a distinction we need to make that's not clear yet (but will be soon). There are two ways to describe the speed of the transmission:

Bit rate (in bit/s, or bps) is the number of 0s and 1s that can be transmitted per second. In our case, it's 1200 bps.
The Baud (Bd) is the number of notes that can be transmitted per second. In our case, that's also 1200 Bd.

So what's the difference? Well, with FSK, the bit rate is equal to the Baud so the distinction is not clear. But imagine there was a system where each note could somehow encode more than one bit. The bit rate would then be greater than the Baud. We'll see this soon...

PSK

Where as with FSK, the information was encoded in the change of frequency (using two different tones to represent 0s and 1s), PSK encodes the information in the phase. This is a little trickier to conceptualize because while the human ear can detect differences in frequency (as different notes) and differences in amplitude (as different volumes), it can't detect differences in phase.⁸⁾

In a nutshell, the phase of a wave describes where it starts. Let's take a look at an example:

Supposing we use the blue wave as a reference wave, the green wave is 90° out of phase, while the red wave is 180° out of phase. Notice that all three waves have the same frequency (the number of cycles per second) and the same amplitude (the height of the wave), but they start at different points. To the human ear, all three notes would sound exactly the same, but a computer can be made to detect the difference, which means we can use two of those waves to represent different bits.

Constellation Diagram

A useful way of representing the different “states” that the waves can be in is using a constellation diagram.⁹⁾ For example, using the Blue and Red waves to encode a message, the constellation diagram would have the two dots on the horizontal axis (one on the right at (1,0) and one on the left at (-1,0)).

The distance between a particular point and the centre of the graph represents the amplitude of the wave. In this case, both dots have the same amplitude. The phase is represented as the angle the point makes with the right side of the horizontal axis. So the blue wave has a phase of 0°, while the red wave has a phase of 180°. Finally, we can assign each wave a bit (0 or 1).

Notice that this diagram doesn't tell us anything about the actual frequency of the waves because they are made to be the same. What changes here is the phase, not the frequency.

Before we continue, let's briefly review the difference between a bit rate and Baud:

Bit rate (in bit/s, or bps) is the number of 0s and 1s that can be transmitted per second.
The Baud (Bd) is the number of notes that can be transmitted per second.

Again, in this case, the bit rate is the same as the Baud since as with FSK, each note encodes a one bit. But this is about to change!

Higher order PSK

If instead of only using two different phases (0° and 180°) we used eight different phases (0°, 45°, 90°, 135°, 180°, 225°, 270°, and 315°) each “note” can now encode three bits instead of just one! This particular scheme is called 8-PSK.¹⁰⁾

The advantage is that while we might only be able to transmit 1200 notes per second (1200 Baud), the data rate is now triple (3600 bps). That is, each note encodes 3 bits.

But why stop at eight states? Well, in reality, the more states we use, the closer together they become, which means that it can get more and more difficult to tell them apart, which leads to more errors.

Quadrature Amplitude Modulation (QAM)

In 8-PSK, the eight states were all on the same circle because their amplitudes were all the same. One way to add more states while keeping them apart as much as possible is to spread them around over the area of a “square” instead of on the perimeter of a circle. This is called Quadrature Amplitude Modulation (QAM).¹¹⁾

QAM is kind of a mix between PSK and ASK (Amplitude Shift Keying). That is, in addition to being able to vary the phase of the waves (where they start), we can also vary their amplitude (how strong the signal is). In the case of 16-QAM, each note can encode 4 bits. So a 1200 Baud signal has a bit rate of 4800 bps.

ADSL technology for copper twisted pairs uses a constellation size of up to 32768-QAM, equivalent to 15 bits per tone.¹²⁾

⁵⁾

Source: Wikipedia: Frequency-Shift Keying

⁶⁾

Audio file source: AFSK 1200 Baud file is from Wikipedia: Wikipedia: Frequency Shift Keying

⁷⁾

Technically, what we just built is an Audio Frequency-Shift Keying (AFSK) system. The difference between the two is that FSK changes the radio frequency directly while AFSK changes the audio frequency being transmitted. This matters if we're using FM to modulate the signal, but on SSB, changing the audio frequency is equivalent to changing the radio frequency. See Wikipedia: FSK for more details

⁸⁾

This is a bit of an over simplification. Two notes out of phase will sound different compared to two notes in phase if they are each played in different ears. That difference is subtle to hear. Some people can't hear the difference, and those who can find it very hard to describe. Phase information between both ears is part of how the brain processes direction (see Wikipedia: Stereophonic sound for more information).

If you listen to this clip with headphones, you'll hear a note (880 Hz tone) being played in phase for 5 seconds, then slowly get out of phase for the next 36 seconds (10° per second), then come back in phase for the last 5 seconds.

It's hard to describe but the out of phase section has a weird 3D depth quality to it. The point is that the human ear can't detect phase nearly as well as frequency (pitch) or amplitude (volume).

This is also the basic idea behind what W8JI calls stereo diversity, which is a super interesting subject in its own right!

⁹⁾

BPSK Constellation diagram was modified from Wikipedia

¹⁰⁾

8-PSK Diagram from Wikipedia: Phase-shift keying

¹¹⁾

16-QAM animation is from Wikipedia: Quadrature Amplitude Modulation

¹²⁾

Source: Wikipedia: QAM

electronics

Under Construction: VA7FI is editing this section, please do not edit it until this notice is taken down.

Electronics

In this section we'll discuss the three basic electronic components:

Name	Property	Unit	Source
Resistor (R)	Resistance	Ohm (Ω)	Resistor
Inductor (L)	Inductance	Henry (H)	Inductor
Capacitor (C)	Capacitance	Farad (F)	Capacitor

Resistor

The easiest component to start with is the resistor.

Resistors have many usage:

❝In electronic circuits, resistors are used to reduce current flow, adjust signal levels, divide voltages, bias active elements, and terminate transmission lines, among other uses. High-power resistors that can dissipate many watts of electrical power as heat [...] or as test loads for generators. Fixed resistors have resistances that only change slightly with temperature, time or operating voltage. Variable resistors can be used to adjust circuit elements (such as a volume control or a lamp dimmer), or as sensing devices for heat, light, humidity, force, or chemical activity.❞ Wikipedia: Resistor

RLC Impedance

Impedance (Ω)	Low Frequency	Medium Frequency	High Frequency
Resistance, R	Doesn't depend on frequency
Inductive Reactance \$X_L = 2\pi f L\$	Low	Medium	High
Capacitive Reactance \$X_C = \frac{1}{2\pi f C}\$	High	Medium	Low

RLC Addition

	Series	Parallel
Resistor, R [Ω]
Resistor, R [Ω]	\$R = R_1 + R_2\$	\$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}\$
Inductor, L [H]
Inductor, L [H]	\$L = L_1 + L_2\$	\$\frac{1}{L} = \frac{1}{L_1} + \frac{1}{L_2}\$
Capacitor, C [F]
Capacitor, C [F]	\$\frac{1}{C} =\frac{1}{C_1} + \frac{1}{C_2}\$	\$C = C_1 + C_2\$

Questions

empos

Emergency Operation and Procedures

Most of the information pertaining to emergency operation procedures is contained in the RIC-22 document. Although this document was originally created for the Restricted Operator's Certificate, which is no longer being issued, the general information provided is still relevant to all radio operators, and continues to be offered for that purpose.

General Information

Below is a summary of the general information relating to emergency communication. More details can be found in the RIC-22 document.

Emergency conditions are classified in accordance with the degree of danger or hazard.¹³⁾ The procedures for each will be explained further down, but here they are in order priority:
1. Distress: A condition of being threatened by grave and/or imminent danger and requiring immediate assistance. Distress calls have priority over all other types of calls.
2. Urgency: A condition concerning the safety of an aircraft or other vehicle, or of someone on board or within sight, but which does not require immediate assistance. Urgent calls have priority over all other types of calls except distress calls.
3. Safety: An indication that the station calling is about to transmit a message concerning the safety of navigation or important meteorological warnings. Safety calls have priority over regular communication but must yield to distress and urgent calls.
The first rule to know is that it's forbidden to transmit any false or fraudulent distress call or message of any kind.¹⁴⁾
Normally, all radio stations need to be operated so as not to interfere with other radiocommunications. However, the only situation under which you may interrupt or interfere with the normal working of another station is when you are required to transmit a higher priority message, for example, distress, urgency or other priority calls or messages.¹⁵⁾
Normally, radio operators and all persons who become acquainted with radiocommunications are required to preserve the privacy of those communications, except as permitted by the addressee of the message. However, these restrictions do not apply to a message of distress, urgency, safety or to messages addressed to “ALL STATIONS”.¹⁶⁾
Normally, the base station has control of communications over mobile stations. However, this does not apply in the cases of distress or urgency communications, where control lies with the station initiating the priority call.¹⁷⁾

Operating Procedures

Distress Communications

Important Disclaimer: The distress communications procedures outlined below should not prevent a station in distress from making use of any means at its disposal to attract attention, to make known its position, and to obtain assistance.

Sending A Distress Call

In radiotelephony (voice), the spoken word for distress is “MAYDAY” spoken three times and pronounced as the French m'aider. In CW (Morse code) it is “SOS”. Since the distress call has absolute priority over all other transmissions, all stations who hear it must immediately cease regular transmission and continue to listen on the frequency.

The distress call should have the following form:

“MAYDAY, MAYDAY, MAYDAY”
“THIS IS”
[call sign of the station in distress] (spoken three times).

To interrupt a lower priority conversation to signal a distress call, break-in immediately following the transmission of the active party with the above sequence.

Responding To A Distress Call

If you hear a distress call:

Immediately stop your contact, acknowledge the emergency call, and determine its location and what assistance may be needed.
If you are unable to render direct assistance, you should contact authorities and then maintain watch until you are certain that assistance will be forthcoming.
Continue to monitor the frequency on which the distress message was received and, if possible, establish a continuous watch on appropriate distress and emergency frequencies.
Notify any station with direction-finding or radar facilities and request assistance unless it is known that this action has been, or will be, taken by the station acknowledging receipt of the distress message.
Cease all transmissions that may interfere with the distress traffic.

To acknowledge the receipt of a distress message, use the following form:

“MAYDAY” (only once)
[call sign of the station in distress] (spoken three times)
“THIS IS”
[your call sign] (spoken three times)
“RECEIVED MAYDAY”.

Urgency Communications

The urgency signal indicates that the station calling has a very urgent message to transmit concerning the safety of a station or a person, but does not require immediate assistance. The urgency signal and message may be addressed to all stations or to a specific station.

Sending an Urgency Signal

The urgency signal is the word “PAN PAN” spoken three times and can be addressed to “ALL STATIONS” or to a specific station.

The urgency call should have the following form:

“PAN PAN, PAN PAN, PAN PAN”
[the name of the station addressed] or the words “ALL STATIONS” (three times)
“THIS IS”
[your call sign]
[the nature of the urgency condition and other useful information]

Responding To An Urgency Call

The urgency signal has priority over all other communications except distress.

Stations that hear the urgency signal shall continue to listen for at least three minutes on the frequency which the signal was heard. After that, if no urgency message has been heard, stations may resume normal service. All stations that hear the urgency signal must take care not to interfere with the urgency message that follows.

Safety Communications

The safety signal is used mainly in the maritime mobile service. It indicates that the station calling is about to transmit a message concerning the safety of navigation or important meteorological warnings.

The safety signal is the word “SECURITE” spoken three times and pronounced as the French sécurité and can be addressed to “ALL STATIONS” or to a specific station.

Sending an Safety Signal

The safety call should have the following form:

“SECURITE, SECURITE, SECURITE”
[the station addressed] or “ALL STATIONS” (repeated three times)
“THIS IS”
[your call sign]
[the nature of the condition]
[your call sign]

Other Emergency Considerations

It's a good idea to have a way to operate your amateur station without using commercial AC power lines so you may provide communications in an emergency.
The most important accessory to have for a hand-held radio in an emergency is several sets of charged batteries.
A portable dipole antenna is a good choice to have as part of a portable HF amateur station that could be set up in case of an emergency.

In a real or simulated emergency, a person operating radio apparatus in the amateur radio service may only communicate with a radio station that is in the amateur radio service in order to transmit a message that relates to the real or simulated emergency on behalf of a person, government or relief organization. ¹⁸⁾

Example

Here's a real MAYDAY call and exchange found on the Mayday Wikipedia page.

Questions

B-002-008-001 → B-002-008-011

¹³⁾

RIC-22, Sec 3.1 and RIC-22, Sec 5.1

¹⁴⁾

¹⁵⁾

¹⁶⁾

¹⁷⁾

¹⁸⁾

See RIC-3, Sec 6, p.18 and SOR-96-484 Sec 48

hfops

Intro to HF

Hopefully you'll get more than 80% on the test and be able to get into HF. These bands are a lot of fun because even in simplex, they can be used to communicate around the world. The radios are also more complex, and the antennas much bigger (more on that later).

HF Band Plan

As we've seen, using LSB or USB has advantages over using AM, but either sideband is as good as the other. As long as both radios are using the same method, they'll be able to communicate. By convention, however, when using SSB, the HF band plan stipulates that for voice operation:

LSB be used on 160m, 80m, 60m, and 40m, and
USB be used on bands 20m and above.

A quick trick is to use lower sideband for the lower bands, and upper sideband for the upper bands.¹⁹⁾

HF Etiquette

Listen before transmitting so as to not interrupt communications already in progress.
Keep transmissions short.
Give enough time between transmissions for others to break in.
Move to a VHF/UHF simplex frequency whenever possible.
Move to a different frequency if the propagation changes and you start hearing others on your frequency that you couldn't hear before.

Nets

Nets are regularly scheduled “meetings” where hams meet. Different nets have different purposes and procedures but generally they have the following in common:

They meet on the same frequency at the same time on a given day.
There's a net controller who is responsible for calling the net and taking checkins.

On HF, if other operators are using the designated net frequency just before the start of the net, the net controller can ask them to relinquish the frequency for the net operations. Although they should move to a different frequency as a courtesy, they are not obliged to. In this case the net controller should conduct the net on a frequency 3 to 5 kHz away from the regular frequency so that regular net participant can easily find the net without causing interference to the others using the regular net frequency.

See our Nets page for more details.

Morse

Although Morse code is not required for the basic licence, it's important to know some of the rules regarding this mode of operation.

The way to initiate a contact in Morse is to send “CQ” (calling any station) three times followed by “DE” (from in French) and your call sign three times.
“CQ” should be sent at the speed which you are able to receive.
You'll also sometimes see “K” which means “Any station please reply”
Or “DX” which means “distant station”
To answer a call, send the other station's call twice, followed by “DE” and your call sign twice.
“73” (not “73s”) means “Best Regards”
Where as voice bandwidth usually is 3 kHz wide, CW bandwidth is much narrower. As such, you should stay between 150 Hz and 500 Hz away from other transmitting stations.

Azimuthal Maps

Suppose you wanted to fly your own plane from Vancouver to the Caspian Sea. Which direction would be the shortest?

Looking at a map, it looks like the answer would be to fly roughly due East along the 49th parallel, over Canada, across the Atlantic Ocean, and over Europe:

Or would it be to go West instead, across the Pacific Ocean and over China?

Well, it turns out that neither is right. The shortest path is over the north pole:

Drawing a straight line on a “conventional” map usually results in an actual path that curves and is longer than it could be. For example, if you flew due East along the 49th parallel, you'd need to be turning left a little bit all the time to stay at the same latitude. This may seem like a weird idea, but supposed that you were right at the North Pole. If you moved off a bit and walked circles around the North Pole flag, you'd be walking on one latitude line (probably like 89.99999º or something). But the point is that unless you're at the equator, following one latitude line due East or West means that you have to be turning (left if going East and right if going west). So you're not going in a straight line²⁰⁾. The converse is also true: a real straight line will have to be drawn like a curve on a “conventional” map unless it's right on the equator or going in the north-south direction.

For example, the path from Vancouver to the Caspian Sea over the North pole is the shortest path between those two points. It's a real straight line. But on a “conventional” map, it would look something like this²¹⁾:

For ham operators, being able to find the shortest straight line between two points on Earth is important because radio waves travel in straight lines. Sure, they bounce, reflect and refract, but they don't just turn left to follow a latitude line. So if you want to know where to point your antenna to reach a specific region, a conventional map won't be much help. Instead, we use what's called an Azimuthal map.

For example, here's one centred on Vancouver showing that the shortest straight line to the Caspian Sea has a bearing of 5° East of North. That's where you'd need to point your antenna if you wanted to talk to someone there:

Here are a few important things to know about Azimuthal maps:

The circles shown are NOT latitudes; they are equidistant lines. That is: every place along a specific circle is the same distance away from the centre, but in different direction. So if you know the scale of your particular map, you can very quickly read off how far a place is as well as which direction.
But more importantly: there is NO one Azimuthal map. Instead, everyone needs their own with their location at the centre. These maps only work from the centre outward. You can't use these maps to find the distance and direction between two off-centre points. This is one case where putting yourself at the centre of the world is necessary.

Here's a really convenient online tool to create your own map.
And here's a few tricks on how to use the tool.

Fun fact: the United Nations flag is an Azimuthal map centred on the North Pole.

Short / Long Path

As we've seen above, in general, there's only one shortest straight line path between two points on the Earth (called the short path) and it can be found using an Azimuthal map. There is, however, one other straight line path (called the long path) in the opposite direction, 180º, from the short path. Pointing your antenna in the long path direction can sometimes allow you to hear better depending on the conditions or the skip zone (more on that later).

QSL Cards and Logbooks

QSL cards are signed post cards listing contact date, time, frequency, mode, and power that amateur operator sometime exchange as written proof of the communication. They are optional and many hams do not use them.

A logbook is also optional but many amateurs keep one in case of interference complaints, for operating awards, or to preserve memories over the years.

Both are always kept in UTC (Universal Time Coordinated) formerly known as GMC (Greenwich Mean Time). Incidentally, to set your station clock accurately to UTC off the air, you can use: CHU (in Ontario), WWV (in Colorado), or WWVH (in Hawaii)

Questions

B-002-003-009 → B-002-003-011
B-002-004-006 → B-002-005-011
B-002-009-001 → B-002-009-011

¹⁹⁾

Digital modes, however, always use USB no matter what band.

²⁰⁾

On a sphere, a “straight line” is called a geodesic and it's always along a great circle

²¹⁾

Turns out that there's now a way to show geodesic lines on Google Maps:

right click at one point and select “Measure Distance”.
Then left click somewhere else.
Once the geodesic is drawn, you can move either end to see how the path changes.

Notice how North-South paths appear straight, but not East-West paths

howtouse

Ham Basics	About The Test	How To Use This Site	References	Study Sections

intro

Intro to Radio Waves

Before we can start discussing Amateur (Ham) Radio, we need to talk a little bit about radio waves. We'll explore this topic in much more detail later on, but for now, let's look at some foundational concepts.

Imagine the radio in your car could not only listen but also transmit on any frequency you like. What would happen as you move up and down the dial?²²⁾

Starting in the FM radio range, let's turn the dial down:

At 88.1 MHz, you'd be transmitting on top of CBC Radio 1 FM (in the Vancouver area).
At 0.690 MHz (or 690 kHz) you'd be transmitting on top of CBC Radio 1 AM (in the Vancouver area).
Around 1 kHz (or 1000 Hz) you'd be interfering with military submarine radio communications.

At this point, you should start thinking about the relationship between MHz, kHz, and Hz. We'll add more to the list below.

There's a lot of stuff in between, but it's pretty much radio waves all the way down. However, turning the dial above the FM radio stations yields some surprises:

At 2.4 GHz (or 2400 MHz) and 5.8 GHz (or 5800 Mhz), you'd be interfering with WiFi signals.
Between 30 and 120 THz (30,000 and 120,000 GHz), you'd be in the mid-infrared range and your antenna would start to feel warm.
At 400 THz, the antenna would start glowing red.²³⁾ By increasing the frequency, you'd go through all the colours of the rainbow until the last purple would vanish around 790 THz.
Between 790 THz and 30 PHz (30,000 THz) you'd create UV rays, which are invisible but could blind you.
Between 30 PHz and 30 EHz (30,000 PHz) you'd create X-rays, which we could be used to take pictures of your bones.
And passed that you'd create gamma rays.

Electromagnetic Spectrum

So radio waves are a small part of what we call the Elecromagnetic Spectrum²⁴⁾, which also contains visible light and a lot of other invisible stuff:

Take a moment to look at the spectrum and see which terms you're not familiar with.

Let's now unpack some of what we just saw...

Hz

A Hertz (Hz) is a measure of how fast something vibrates. For example, the A-string of a guitar vibrates 440 times per second, so we say that it vibrates at 440 Hz. The next A (an octave higher) vibrates twice as fast at 880 Hz. The human ear can hear sounds between roughly 20 Hz and 20,000 Hz.²⁵⁾ With sound, the higher the frequency, the higher the pitch. With light, the higher the frequency, the “colder” the colour.

Electromagnetic (EM) waves and sound waves are completely different things. The only thing they have in common is that “something” oscillates, but many things oscillate so that's not saying much. Just seeing “Hz” doesn't tell you anything about what it is that's oscillating in the same way that seeing “°C” doesn't tell you anything about what it is that has temperature. “Hz” is a unit of measure, not a thing itself.

Now back to radio waves...

Without going into too much detail (yet), radio waves are created by oscillating electric currents. How many times this current oscillates per second is called the frequency, which is measured in Hz (or kHz, MHz, GHz).

The “k” (kilo), “M” (mega), or “G” (giga) that you'll often see in front of Hz is a quick way of multiplying by 1000:

1 kHz	= 1000 Hz		= 10³ Hz (1 followed by 3 zeros)
1 MHz	= 1000 kHz	= 1,000,000 Hz	= 10⁶ Hz (1 followed by 6 zeros)
1 GHz	= 1000 MHz	= 1,000,000,000 Hz	= 10⁹ Hz (1 followed by 9 zeros)

These prefixes are not only used for frequencies. You've seen them in other places before:

1 km (kilometer) = 1000 m (meter)
1 MW (megawatt) = 1,000,000 W (watt)
1 GB (gigabyte) = 1,000,000,000 B (byte)

So no matter what the unit of measure, these prefixes mean:

kilo (k) = a thousand
mega (M) = a million
giga (G) = a billion
tera (T) = a trillion

This might be a good time to mention that we also have prefixes for small units (more on this later):

milli (m) = a thousandth
micro (μ) = a millionth
nano (n) = a billionth
pico (p) = a trillionth

Now, let's take another look at the Electromagnetic Spectrum picture. You should be able to make sense of pretty much all of it:

FM radio and TV broadcasting is between 50 MHz and 1000 MHz (called VHF and UHF bands).
But some radio waves go even lower than 10⁶ Hz (or 1 MHz).
Above radio waves are Microwaves, Infrared, Visible light, UV, Xray, and Gamma-rays.

Next, let's look at where Ham radio frequencies are on that spectrum.

Ham Bands Overview

Ham radio operators are allowed to transmit on very specific slices of the Electromagnetic Spectrum depending on which qualifications we have (“Basic” or “Basic +”):

In green are VHF, UHF, SHF, and EHF bands that require only the Basic qualification.
In orange are LF, MF, and HF bands that require Basic with Honours, Basic with Morse, or Basic with Advanced qualification.
In blue are CB bands, which don't require any qualification but can only be used with unmodified CB radios at relatively low power (for reference).
In red are the AM and FM radio broadcasting bands (for reference).
In maroon are the old VHF (1-13) and UHF (14-50) TV channels.²⁶⁾

Bandwidth

Although the human ear can detect sounds between 20 Hz and 20 kHz, human speech typically uses sounds between 300 Hz and 3000 Hz. Modulating these sounds (more on that later) onto a radio wave means that the radio will actually transmit over a range of frequencies that we call bandwidth. For example, using SSB, the bandwidth would be 2700 Hz (300 Hz to 3000 Hz). So a radio tuned to transmit at 3.800 MHz would actually transmit between 3.7973 MHz and 3.7997 Mhz. Using AM, the bandwidth would be 6 kHz so the transmitted frequencies would be between 3.797 MHz and 3.803 MHz.

We'll explore this in much more detail later, but for now, the important concept is that to transmit a signal, the radio must transmit over a range of frequencies, not just one single frequency. This range is called bandwidth.

In addition to only being allowed to transmit on specific frequencies, ham operators also have to make sure that they don't transmit over a greater bandwidth than allowed for the specific frequencies.

That is, there are restrictions on where we transmit on the spectrum as well as how wide the transmissions are.

This is important because different modes have different bandwidth requirements. From lowest to highest:

Mode	Required Bandwidth
CW	~300 Hz
300 Baud Packet	~600 Hz
SSB Voice	2.7 kHz
Slow Scan TV	3 kHz
AM Voice	6 kHz
FM Voice	20 kHz
Fast Scan TV	6 MHz

We'll look at that picture in more details soon, but for now, let's just point out how the AM signal is twice as wide as the SSB signal.

Certificate Qualifications Overview

There are three different qualifications: Basic, Morse, and Advanced.²⁷⁾

Basic requires 70% to pass, but 80% or greater gives Honours privileges.
Morse requires 5 wpm.
Advanced is a different test that requires more electronics.

Privilege	Basic	Basic with Honours or Basic + Morse 5wpm	Basic + Advanced
Frequencies above 30 MHz	Y	Y	Y
Power up to 250 W	Y	Y	Y
Frequencies below 30 MHz	N	Y	Y
Power up to 1 kW	N	N	Y
Build or Modify Radios	N	N	Y
Manage a repeater	N	N	Y
Remote Control Radios	N	N	Y

The Basic (70% ‒ 79%) certificate gives access to these frequencies:

VHF and UHF bands are local bands (more on that later). Your range could vary between 1km and 100km depending on your setup (more on that later).

The “2m band” (144 ‒ 148 MHz) is the most popular band above 30 MHz. You should make sure your first radio covers at least this band. Radios that can cover both the “2m” and the “70cm” bands are very common.

The Basic with Honours (80% or more) certificate adds access to these frequencies:

HF bands are “long range” bands. Depending on the conditions, you could talk to someone in the next town or halfway around the world.

Full Frequency List

Here is the full frequency list. Highlighted information might be on the test.

The Band name is given in meter or cm. You'll need to know them.
The Maximum Bandwidth is the maximum width of the radio signal. You'll also need to know these.
Under License,
- “B” means Basic, and
- “B+” means Basic with Honours, or Basic with Morse, or Basic with Advanced.
The Notes in the last column are really important since some bands have restrictions you need to be aware of.

	Band	Range (MHz)	Max Bandwidth	License	Notes
LF	2200m	0.1357 ‒ 0.1378	100 Hz	B +	D, 1
LF	630m	0.472 ‒ 0.479	1 kHz	B +	2
MF	160m	1.8 ‒ 2.0	6 kHz	B +
HF	80m	3.5 ‒ 4.0	6 kHz	B +
HF	60m	5.332, 5.348, 5.3515 ‒ 5.3665, 5.373, 5.405	2.8 kHz	B +	3
HF	40m	7.0 ‒ 7.3	6 kHz	B +	4
HF	30m	10.10 ‒ 10.15	1 kHz	B +	D, 5
HF	20m	14.00 ‒ 14.35	6 kHz	B +
HF	17m	18.068 ‒ 18.168	6 kHz	B +
HF	15m	21.00 ‒ 21.45	6 kHz	B +
HF	12m	24.89 ‒ 24.99	6 kHz	B +
HF	10m	28.00 ‒ 29.7	20 kHz	B +
VHF	6m	50 ‒ 54	30 kHz	B
VHF	2m	144 ‒ 148	30 kHz ^¥	B
VHF	135cm	219 ‒ 225	100 kHz	B	6
UHF	70cm	430 ‒ 450	12 MHz	B	☆
UHF	35cm	902 ‒ 928 ^§	12 MHz	B	☆
		Range (GHz)
UHF		1.24 ‒ 1.30		B	☆
UHF		2.30 ‒ 2.45 ^‡		B	☆
SHF		3.3 ‒ 3.5		B	☆
SHF		5.650 ‒ 5.925		B	☆
SHF		10.0 ‒ 10.5		B	☆
SHF		24.00 ‒ 24.05		B
SHF		24.05 ‒ 24.25		B	☆
EHF		47.0 ‒ 47.2		B	☆
EHF		76.0 ‒ 77.5		B	☆
EHF		77.5 ‒ 78.0		B
EHF		78.0 ‒ 81.0		B	☆
EHF		81.0 ‒ 81.5		B	7
EHF		122.25 ‒ 123.00		B	☆
EHF		134.0 ‒ 136.0		B
EHF		136.0 ‒ 141.0		B	☆
EHF		241.0 ‒ 248.0		B	☆
EHF		248.0 ‒ 250.0		B

^¥ Since Fast Scan TV requires 6 MHz of bandwidth, it can't be transmitted below the 70cm band.
^§ The 902 ‒ 928 MHz band may be heavily occupied by licence exempt devices, which are lower power devices that don't require a license but can't be interered with (like cordless phones)
^‡ The 2.30 ‒ 2.45 GHz band is shared with Industrial Scientific Medical (ISM) licence exempt devices.

Non-ham frequencies for comparison:

	Range (MHz)	Details
MF	0.535 ‒ 1.705	AM Radio
HF	26.965 ‒ 27.405	CB (40 channels)
VHF	54 ‒ 88	VHF TV Channels 2 ‒ 6
VHF	88 ‒ 108	FM Radio
VHF	174 ‒ 216	VHF TV Channels 7 ‒ 13
UHF	462.550 ‒ 462.725	FRS (channels 1 ‒ 7 and 15 ‒ 22)
UHF	467.5625 ‒ 467.7125	FRS (channels 8 ‒ 13)
UHF	470 ‒ 692	UHF TV Channels (14 ‒ 50)²⁸⁾

Name of frequency ranges:

Name	Abbreviation	Frequency Range
Very Low Frequency	VLF	3 ‒ 30 kHz
Low Frequency	LF	30 ‒ 300 kHz
Medium Frequency	MF	300 ‒ 3000 kHz
High Frequency	HF	3 ‒ 30 MHz
Very High Frequency	VHF	30 ‒ 300 MHz
Ultra High Frequency	UHF	300 ‒ 3000 MHz
Super High Frequency	SHF	3 ‒ 30 GHz
Extremely High Frequency	EHF	30 ‒ 300 GHz

Important Notes

Information quoted here was taken from ISED's RBR-4.

D	For the 2200m and 30m bands, the maximum bandwidth allowed is too narrow for phone (voice) transmissions. Therefore, only digital or CW modes are allowed.
☆	Secondary User “means that transmissions shall not cause interference nor be protected from interference from stations licensed in other services operating in that band. Operating provisions defined below are excerpts from the Canadian Table of Frequency Allocations, which is amended from time to time.” Basically, this means that the Amateur Radio Service is secondary to some other service and that we must yield the frequency when they need it. We must not cause interference to these other services, and we must accept that they have first priority and may interfere with us at anytime.
1	“Stations in the amateur service using frequencies in the band 135.7 ‒ 137.8 kHz shall not exceed a maximum radiated power of 1 W (e.i.r.p.) and shall not cause harmful interference to stations of the radionavigation service.” This is the 2200m band referred to in Note D. It is mostly used for experimental purposes and transmissions must be at very low power and can't cause interference.
2	“The maximum equivalent isotropically radiated power (e.i.r.p.) of stations in the amateur service using frequencies in the band 472-479 kHz shall not exceed 1 W. Administrations may increase this limit of e.i.r.p. to 5 W in portions of their territory which are at a distance of over 800 km from the borders of Algeria, Saudi Arabia, Azerbaijan, Bahrain, Belarus, China, Comoros, Djibouti, Egypt, United Arab Emirates, the Russian Federation, Iran (Islamic Republic of), Iraq, Jordan, Kazakhstan, Kuwait, Lebanon, Libya, Morocco, Mauritania, Oman, Uzbekistan, Qatar, Syrian Arab Republic, Kyrgyzstan, Somalia, Sudan, Tunisia, Ukraine and Yemen. In this frequency band, stations in the amateur service shall not cause harmful interference to, or claim protection from, stations of the aeronautical radionavigation service. (WRC-12)”
3	“Amateur service operators may transmit in the frequency band 5351.5 ‒ 5366.5 kHz and on the following four centre frequencies: 5332 kHz, 5348 kHz, 5373 kHz and 5405 kHz. Amateur stations are allowed to operate with a maximum effective radiated power of 100 W PEP in each channel and are restricted to the following emission modes and designators: telephony (2K80J3E), data (2K80J2D), RTTY (60H0J2B) and CW (150HA1A). Transmissions in any channel may not occupy a bandwidth of more than 2.8 kHz. Such use is not in accordance with international frequency allocations. Canadian amateur operations shall not cause interference to fixed and mobile operations in Canada or in other countries and, if such interference occurs, the amateur service may be required to cease operations. The amateur service in Canada may not claim protection from interference by the fixed and mobile operations of other countries.” Other amateurs around the world do not have permission to transmit on these frequencies so Canadian amateurs are secondary users, which mean that they must stop transmitting if they interfere with other users.
4	“The use of the band 7.200 ‒ 7.300 MHz in Region 2 (North America) by the amateur service shall not impose constraints on the broadcasting service intended for use within Region 1 and Region 3 (Europe and Asia).” Although Amateurs are allowed to transmit between 7.200 ‒ 7.300 MHz in North America, they'll have to work around foreign broadcasting stations who also use this section of the spectrum.
5	“The use of the band 10.100 ‒ 10.150 MHz by the amateur service in Canada is not in accordance with the international frequency allocations. Canadian amateur operations shall not cause interference to fixed service operations of other administrations and if such interference should occur, the amateur service may be required to cease operations. The amateur service in Canada may not claim protection from interference by the fixed service operations of other administrations.” Other amateurs around the world do not have permission to transmit on these frequencies so Canadian amateurs are secondary users, which mean that they must stop transmitting if they interfere with other users.
6	“In the band 219 ‒ 220 MHz, the amateur service is permitted on a secondary basis. In the band 220 ‒ 222 MHz, the amateur service may be permitted in exceptional circumstances on a secondary basis to assist in disaster relief efforts.”
7	“The 81.0 ‒ 81.5 GHz band is also allocated to the amateur and amateur-satellite services on a secondary basis.”

Questions

B-001-003-001
B-001-004-002
B-001-004-004 → B-001-004-006
B-001-005-002
B-001-006-005 → B-001-006-006
B-001-008-004 → B-001-008-006
B-001-010-003 → B-001-010-004
B-001-010-007 → B-001-010-008
B-001-010-010 → B-001-010-011
B-001-015-001 → B-001-016-011
B-001-018-004
B-001-020-004

²²⁾

Photo by Bryan Costin licensed under CC By-Nc-Sa 2.0

²³⁾

As we'll see later, the “antenna” would also get smaller, the higher the frequency. So you can think of the rods and cones in your eyes as antennas tuned to the colours we can see.

²⁴⁾

Picture of the Electromagnetic Spectrum modified from Electromagnetic_spectrum#Boundaries

²⁵⁾

Cool side note: We can double 20 Hz about 10 times before it gets to 20,000 Hz. That means that we can hear about 10 octaves. But with visible light, if we double 400 THz (red), it goes over 790 THz (Violet), which means that we can't even see a full “octave” of light. What would it feel like to see two colours that are exactly one octave apart?

²⁶⁾

Remember those? (source)

²⁷⁾

See RIC-3, Sec 1, p.2

²⁸⁾

Channel 37 (608 ‒ 614 MHz) is reserved for Radio Astronomy so there are no TV broadcasts there. See: Channel_37

mathbasics

Metric Prefix

A metric prefix is a unit prefix that precedes a basic unit of measure to indicate a multiple or fraction of the unit.

In the previous section, we saw that MHz means a million Hertz. Here's a list of the most common ones:

Name	Symbol	Base 10	Decimal
Tera	T	10¹²	1 000 000 000 000
Giga	G	10⁹	1 000 000 000
Mega	M	10⁶	1 000 000
Kilo	k	10³	1 000
hecto	h	10²	100
deca	da	10¹	10
		10⁰	1
deci	d	10^-1	0.1
centi	c	10^-2	0.01
milli	m	10^-3	0.001
micro	μ	10^-6	0.000 001
nano	n	10^-9	0.000 000 001
pico	p	10^-12	0.000 000 000 001

The decibel

A decibel (dB) is a way of saying how many times bigger (or smaller) something is compared to something else. For example, let's use the faintest sound that the human ear can detect as the basis. A normal conversation would be a million times louder, and a jack hammer would be a hundred billion times louder. To represent these vast differences in numbers, we use a logarithmic scale based on powers of 10.

Here's the idea. Start with representing numbers by exponents of 10:

Number	Base 10	Exponent
1 =	10⁰	0
10 =	10¹	1
100 =	10²	2
1000 =	10³	3
10 000 =	10⁴	4
100 000 =	10⁵	5
1 000 000 =	10⁶	6

If your math is a little rusty, notice how the exponent counts the number of zeros after the one (1 000 000 has 6 zeros so it's equal to 10⁶)

For now, let's call the exponent a “bel” (not decibel yet). For example:

Instead of saying that a sound is 1000 times louder, we could say that it's 3 bels louder, and
Instead of saying that it's 10,000 times louder, we could say that it's 4 bels louder.

So the bel is just the number of zeros after the 1. But what if the number doesn't start with 1. For example: what if it's 5000 times louder?

Since 5000 is between 1000 and 10,000, it might be tempting to say that it would 3.5 bels louder (midway between 3 and 4), but let's check to see if it works: 10^3.5 = 3162, which is between 1000 and 10,000, but not 5000.
By trial and error, you might get pretty close: 10^3.7 = 5011... so 5000 times louder is roughly 3.7 bels.
But a faster way of finding the exponent of a base 10 number is to use the Log button on your calculator: \$\log(5000) \approx 3.699\$²⁹⁾

In the same way that milli means a thousandth, and centi means a hundredth (think millimetre and centimetre), a deci means a tenth. So 1 bel contains 10 decibels (like 1 metre contains 10 decimetres, 100 centimetres, or 1000 millimetres).

So another way of saying 3.7 bels is to say 37 decibels, or 37 dB. So here's our table again with a few common “in between” numbers:

Number	Base 10	Exponent (bels)	decibels (dB)
1 =	10⁰	0	0
2 ≈	10^0.3	0.3	3
4 ≈	10^0.6	0.6	6
8 ≈	10^0.9	0.9	9
10 =	10¹	1	10
100 =	10²	2	20
1000 =	10³	3	30
10 000 =	10⁴	4	40
100 000 =	10⁵	5	50
1 000 000 =	10⁶	6	60

In addition to being more convenient to represent big numbers, decibels also allow us to multiply big (or small) numbers more easily by adding the dB instead. For example, look at:

100 x 1000 = 100,000

Let's write the same thing again in powers of 10:

10² x 10³ = 10⁵

Notice that while the numbers are multiplying (100 x 1000), the exponents are adding (2 + 3). So 100 x 1000 is the same as saying 20 dB + 30 dB = 50 dB. That is, adding the dB representation of numbers, is the same as multiplying those numbers.

Look back at the previous table; if you double the power, three times in a row:

The final power is 2 x 2 x 2 = 8 times more powerful than the original.
In terms of dB, we can instead add 3dB three times: 3dB + 3dB + 3dB = 9dB.

Here's a quick exercise: Using only the table above, what is the dB representation of 4000?

Click to display ⇲

Click to hide ⇱

A 4000 fold increase is the same as a 1000 fold followed by two doubling. That is:

1000 x 2 x 2 = 4000

Which means:

30dB + 3dB + 3dB = 36dB

Quick calculator check: 10^3.6 = 3981 ≈ 4000

So the only thing to memorize with decibels is that:

a bel (10 dB) is the number of zeros after the 1. So that takes care of knowing 10dB, 20dB, 30dB, ...
3dB is one doubling.

With these two things, we can now estimate a bunch of in between numbers like 36dB (30dB + 3dB + 3dB), or 13dB (10dB + 3dB)

Here's Dave explaining this in more details:

Alternative Formulation

If you like formulas, what I did above was define the dB implicitly as:

\$$ \text{ratio} = 10^{\left(\frac{\text{dB}}{10} \right)} \$$

For example, a ratio of 1000 is equivalent to 30 dB because:

\$$ 1000 = 10^{\left(\frac{\text{30}}{10} \right)} \$$

But what you'll often find in books is the following explicit definition:

\$$ \text{dB} = 10\log\left(\text{ratio}\right) \$$

While the two are definitions are mathematically equivalent³⁰⁾, I personally prefer the implicit definition because I find it easier to picture the dB in terms of powers of 10, where as I can't really picture the log function that well. But they are equivalent, so use the one you like most.

dBm

A related measurement is the dBm. While the decibel (dB) is a ratio between two quantities (saying “20 dB” is the same as saying “100 times more”), the dBm is a ratio between one quantity and 1 mW. That is, it's a measure of how much stronger (or weaker) the power of something is compared to 1 milliwatt.

For example, a typical fibre optic light signal to a house for internet has a light level of -16 dBm. Before we go through the math, recall that \$16 \text{ dB} = 10 \text{ dB} + 3 \text{ dB} + 3 \text{ dB}\$. Which means, the ratio is \$10 \times 2 \times 2 = 40\$. Let's now see why -16 dBm is equal to a power of 25 μW (microwatts):

\$$-16 \text{ dBm} = 1 \text{ mW} \div 10^{1.6} \approx 1 \text{ mW} \div 40 = 0.025 \text{ mW} = 25~\mu \text{W} \$$

Notice how the minus sign in front of the dBm means that the power is less (not more) so it divides (not multiply).

Binary Numbers

Base 10

It's kind of weird explaining how to count in base 10 because we're so used to it. But to understand other bases, it helps to understand how base 10 really works first.

So base 10 means we have 10 different symbols to represent numbers, and when we need more, we just add more digits to the number. For example, let's count and pay close attention to what we're really doing:

With one digit, we can count to nine because we have ten symbols: \$$0, 1, 2, 3, 4, 5, 6, 7, 8, 9\$$

To continue past nine, we need to add a new digit in front of the “counting digit”, and then we just do what we did above again: \$$10, 11, 12, 13, 14, 15, 16, 17, 18, 19\$$

And that new digit will increase in the same way until we run out of symbols: \$$20, 21, 22, 23, 24, 25, 26, 27, 28, 29\$$ \$$ 30, 31, 32, 33, 34, 35, 36, 37, 38, 39\$$ \$$\vdots \$$ \$$ 90, 91, 92, 93, 94, 95, 96, 97, 98, 99 \$$

At which point we need to add a third digit (and so on).

It's strange to think about this but the ability to count is completely separate from having a sense of how big the numbers are. What I mean is: if you know how to change the symbols (from 0 to 1, and 1 to 2, and 2 to 3, etc) and how to add new digits (9 to 10, 99 to 100, ...) you can count forever without having a “number sense”. In base 10, it's hard to imagine not knowing how big a number is, but we'll actually do this in base 2 in a moment. You'll see, it's a weird feeling!

But first, here's another weird way to think of numbers. Take 234 for example. Let's “decompose” it into its powers of 10:

\begin{align*} {\color{teal}{234}} &= {\color{teal}2}\times 100 + {\color{teal}3}\times 10 + {\color{teal}4}\times 1 \\ &= {\color{teal}2}\times 10^2 + {\color{teal}3}\times 10^1 + {\color{teal}4}\times 10^0 \end{align*}

This is a very tedious way of expressing a simple number, but I promise it'll be invaluable when we try to convert from base 2 to base 10 in a moment (can you think of why?).

Base 2

Now let's do the same thing again in base 2. Imagine we only have two symbols we can use to count: 0 and 1. Let's see what counting to fifteen looks like starting at zero:

0, 1, 10, 11, 100, 101, 110, 111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111

Do you see the pattern? Here's another way of writing it with leading zeros and keeping track of where we are in base 10:

Base 2	Base 10
0000	0
0001	1
0010	2
0011	3
0100	4
0101	5
0110	6
0111	7
1000	8
1001	9
1010	10
1011	11
1100	12
1101	13
1110	14
1111	15

You see what I mean about being able to count without knowing how big the number is? All we're doing is “flipping bits” from 0 to 1 in the right order. If we wanted, we could count to 1001010011010 and have no idea of how big that number is unless we kept track of it in base 10. That's because over the years, we developed a number sense for base 10 numbers, but we didn't for base 2 numbers since we don't typically use them.

Conversion

Now let's look at how to convert from one base to the other without counting.

Base 2 to Base 10

We'll use our example of 1101 in base 2 since we already know it's 13 in base 10. Before we do this, however, let's use subscripts to denote the base. For example, we'd write:

\$$ 1101_2 = 13_{10} \$$

Now let's see how we convert from base 2 to base 10 without counting. In the same way that each position in base ten is a factor of 10, each position in base 2 is a factor of 2:

\begin{align*} {\color{teal}{1101}}_2 &= {\color{teal}1}\times 2^3 + {\color{teal}1}\times 2^2 + {\color{teal}0}\times 2^1 + {\color{teal}1}\times 2^0 \\ &= {\color{teal}1}\times 8 + {\color{teal}1}\times 4 + {\color{teal}0}\times 2 + {\color{teal}1}\times 1 \\ &= 8 + 4 + 0 + 1 \\ &= 13_{10} \end{align*}

Let's try that again with the crazy number I mentioned earlier: 1001010011010. Skipping the powers with a coefficient of 0 (to save space): \begin{align*} 1001010011010_2 &= 2^{12} + 2^9 + 2^7 + 2^4 + 2^3 + 2^1 \\ &= 4096 + 512 + 128 + 16 + 8 + 2 \\ &= 4762_{10} \end{align*}

Base 10 to Base 2

Converting the other way around is a bit different, but kind of makes sense if you think about it. Here are the steps to convert \$ 13_{10} \$ back to \$1101_2\$ and I'll explain a bit more after:

Division by 2	In Decimal	As fraction	Quotient	Remainder
13÷2	6.5	6 + \$\frac{1}{2}\$	6	1
6÷2	3.0	3 + \$\frac{0}{2}\$	3	0
3÷2	1.5	1 + \$\frac{1}{2}\$	1	1
1÷2	0.5	0 + \$\frac{1}{2}\$	0	1

Now we read the remainders from the bottom up: 1101.

Recall that the remainder is what's left of a division; it's not the decimal portion. For example:

\begin{align*} 22 \div 5 &= 4.4 \\ &= 4 \tfrac{2}{5} \\ &= 4 \text{ remainder 2 (of 5)} \end{align*}

In our case, the remainder is what's left after we divided by 2. So why does this method work?

What we really want to do is decompose 13 into the sum of its powers of 2. That is: \begin{align*} 13_{10} &= \; 8 + 4 + 0 + 1 \\ &= \; 2^3 + 2^2 + 0 + 2^0 \\ &= (1 \quad \ \ 1 \quad \ \ \, 0 \quad \ 1)_2 \end{align*}

When we have a non-zero power of 2, we represent that position with a 1, when we have a zero power of 2, we write 0.

What the table above is doing is systematically dividing by 2 to see if the result is a whole number or not. If it is, it means that that position is 0 because the factor of two can be absorbed further up the chain. For example, the fact that 13 ÷ 2 = 4.5 means that the first digit has to be 1.

Let's look at a trivial example converting 2037 from base 10 to base 10 (trivial indeed!):

Division by 10	In Decimal	As fraction	Quotient	Remainder
2037÷10	203.7	203 + \$\frac{7}{10}\$	203	7
203÷10	20.3	20 + \$\frac{3}{10}\$	20	3
20÷10	2	2 + \$\frac{0}{10}\$	2	0
2÷10	0.2	0 + \$\frac{2}{10}\$	0	2

Other Bases?

There are other useful bases such as hexadecimal (base 16). Here's an example of counting to twenty in hex:

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, 10, 11, 12, 13, 14

The convention to denote a hexadecimal number is to use either \$ 14_{16} \$ or 0x14.

The same logic applies to convert between hexadecimal and base 10:

\begin{align*} \text{F61B}_{16} &= 15 \times 16^3 + 6 \times 16^2 + 1 \times 16^1 + 11 \times 16^0 \\ &= 15 \times 4096 + 6 \times 256 + 1 \times 16 + 11 \\ &= 63003_{10} \end{align*}

One reason why hexadecimal is so useful in computer science is because one hexadecimal “digit” can represent four binary “digits”. For example,

Decimal	Hexadecimal	Binary
1	1	1
2	2	10
3	3	11
4	4	100
...
7	7	111
8	8	1000
...
15	F	1111
16	10	1 0000
...
31	1F	1 1111
32	20	10 0000
...
63	3F	11 1111
64	40	100 0000
...
127	7F	111 1111
128	80	1000 0000
...
255	FF	1111 1111
256	100	1 0000 0000
...
4095	FFF	1111 1111 1111
4096	1000	1 0000 0000 0000

²⁹⁾

Depending on your calculator, you might have to type it in this order 5000Log

³⁰⁾

These two definitions are equivalent because the log function is the inverse of the power function. That is, it helps us find the exponent when we know the other quantities. Mathematically, we say that: \$a = 10^b \quad \Leftrightarrow \quad b = \log_{10}(a) \$

polarization

How To Make A Radio Wave

Back on the Intro Page, we introduced to the idea of frequency and saw that

A Hertz (Hz) is a measure of how fast something vibrates [...]

Just seeing “Hz” doesn't tell you anything about what it is that's oscillating in the same way that seeing “°C” doesn't tell you anything about what it is that has temperature. “Hz” is a unit of measure, not a thing itself.

Radio waves are created by oscillating electric currents. How many times this current oscillates per second is called the frequency, which is measured in Hz.

It's now time to add a few more details to this story. Here is a basic recipe for making a radio wave:

Get a length of conducting wire and lay it in a straight line.
Cut it in half right in the middle and bend both ends at right angle.
Connect the two middle ends to each side of an alternating current generator.

Voila! The antenna we've made is called a dipole. Assuming that its length matches the frequency of the current generator (more on this shortly), and that the antenna is high enough above the ground, you've created a radio wave.³¹⁾ As electrons move up and down the length of the wires, they create varying electric and magnetic fields that couple together according to Maxwell's Equations and propagate outward in a doughnut shape.³²⁾

Horizontal vs Vertical Polarization

Here's the critical part though: In the same way that an alternating current through an antenna creates a radio wave, a radio wave hitting an antenna induces an alternating current through it if the radio wave hitting the antenna is in the same “direction” as the antenna.

This “direction” is called polarization.

Effect on Communication

The polarization of an antenna is determined by the orientation of the electric field relative to the Earth's surface. So a horizontal antenna will produce a radio wave that has an horizontal electric field. The magnetic field part of the wave is always perpendicular to both the electric field and the direction of propagation, so in this case, it would be vertical.

In practice, polarization is more important for VHF and UHF communication because signals travel more or less directly from the transmitting antenna to the receiving one. For skywave HF communications, however, the ionosphere can change the polarization of the signal from moment to moment as the radio wave refracts, reflects, or goes through magnetic fields in the atmosphere. As such polarization of the antennas on HF frequency doesn't matter much because you never know what polarization your signal will end up having once it arrives at its destination.

Wavelength and Antenna Length

As we saw previously, the wavelength (λ) in metres of the wave is dictated by the frequency \$f\$ in MHz and the speed of light:

\$$ \lambda = \frac{300}{f} \qquad \text{or} \qquad f = \frac{300}{\lambda}\$$

This explains the Band name in the table on the intro page. For example, the frequency range of the 2m band is 144 Mhz ‒ 148 Mhz. If we calculate the wavelength of 146 Mhz, we get: 300 ÷ 146 = 2.05 m

Now it turns out that the size of the antenna is very closely related to the wavelength of the signal we wish to transmit or receive. For a dipole, the total length is roughly half of the wavelength. So an antenna for the 2m band should be roughly 1m long, while an antenna for the 160m band would be roughly 80m long!

We say “roughly” here because as we'll see later, the speed of electricity in a conductor is a bit less than the speed of light, so the antenna length end up a being roughly 5% shorter than calculated here.

Cellphone Antenna

LTE band 1 operates roughly at 2100 MHz.³³⁾ How long would a dipole have to be?

The wave length is: 300 ÷ 2100 = 0.14m = 14cm
Half of the wave length would be 7cm, which is small enough to fit in the cellphone.

Questions

³¹⁾

GIF from Wikipedia Dipole Antenna

³²⁾

Picture modified from Wikipedia Dipole Antenna

³³⁾

Wikipedia LTE

practice

Under Construction: VA7FI is editing this section, please do not edit it until this notice is taken down.

Recall

Modulation

Receivers

There are three main characteristics of a receiver: sensitivity, selectivity, and stability.

Sensitivity

A signal is always accompanied by some sort of noise, and very roughly speaking, if the signal is stronger than the noise, then it can be heard. To quantify this, we use a term called Signal-to-Noise Ratio (SNR or S/N):

\$$\text{SNR} = \frac{\text{Signal}}{\text{Noise}}\$$

Since SNR is a ratio:

If SNR > 1, then the signal is stronger than the noise.
If SNR = 1, then the signal and the noise have the same strength.
If SNR < 1, then the noise is stronger than the signal.

Like other ratios, we often express SNR in decibel. Recall that a ratio of 1 = 0 dB, so the above could be stated as:

If SNR > 0 dB, then the signal is stronger than the noise.
If SNR = 0 dB, then the signal and the noise have the same strength.
If SNR < 0 dB, then the noise is stronger than the signal.

Now back to the receiver. The sensitivity of a receiver is its ability to pick out weak signals from the noise. That is, it indicates how faint an input signal can be and still be successfully received by the receiver.

For example, here's the specs sheet from the IC-7300:

For example, a receiver with a sensitivity of -123 dBm can pick out a signal of 0.0000000000005 mW.³⁴⁾

Selectivity

Recall from the intro section that radio signals always take up some bandwidth on the radio spectrum:

CW takes the least amount of “space” because it's essentially just a single note being turned on and off. A 250 Hz filter would work well to isolate the signal and reject neighbouring signals.
RTTY is a digital mode that uses two notes to represent 0s and 1s, thus it takes a little bit more space than CW.
SSB signals usually have a bandwidth between 2 kHz and 3 kHz. A 2.4 kHz filter would work well to isolate the signal and reject neighbouring signals.
FM, needs about 20 kHz, which is why FM radio stations sound better than AM radio stations.

Now back to the receiver. The selectivity of a receiver is its ability to pass only the signal of interest and reject everything else.

Filters

Sometimes a specific noise or interference is mixed in the signal and we'd like to reduce or eliminate it. One way is to use filters. You can think of filters a little bit like the equalizer on a Hi-Fi stereo.

A filter that lets high frequencies through but blocks low frequencies is called a high pass filter.
A filter that lets low frequencies through but blocks high frequencies is called a low pass filter.
Combining a high pass and low pass filter we get a band pass filter, which lets audio between two frequencies. The narrower the bandwidth of the mode we use, the narrower the band pass filter we can use to clean up the audio. For example:
- An SSB voice signal can use a 2.4 kHz filter where as
- A CW signal can use a much narrower 250 Hz filter.
The “inverse” of a band pass filter is called a notch filter. It blocks a narrow band of frequencies in the middle of the audio spectrum. Here's an example:

This picture has three main parts. From top to bottom:

The frequency is 7.100 MHz on LSB.
The scope shows this frequency relative to the entire band (7.000 Mhz to 7.300 Mhz)
At the bottom is the audio spectrum of what we hear (with a 2.4 kHz band pass filter).

You'll notice that there's a very intense line around 1.5 kHz on the audio spectrum. This line, coming from an interfering carrier signal, sounds like a loud note (a little above \$ \text{F}_{6}^{\text{#}} \$ for the musicians out there) mixed with the voice. To remove this tone, we can “notch” it. We now see a dark spot around 1.5 kHz. If we need to, we can make this spot more or less wide.

Stability

This is the ability of a receiver to stay on the right frequency and not drift.

Transmitters

Questions

B-003-010-001 → B-003-011-001

³⁴⁾

-123 dBm = 10^-12.3 mW ≈ 5 × 10^-13 mW

propagation

Propagation

Radio wave propagation describes the way in which radio waves travel from one point to another. As we saw in the previous page, radio waves (like light waves) have polarization and are affected by the phenomena of reflection, refraction, diffraction, and scattering. As we'll see next, these give rise to different ways that the signal can propagate through the atmosphere.

Direct Waves (Line Of Sight)

VHF radio waves (above 50 MHz) travel more or less in a straight line, and so cannot go much beyond the horizon. To increase the distance that an antenna can “see”, we raise our antennas as high as possible. The radio horizon is given roughly by: \$d = 4.12 \sqrt{h} \$ where \$d\$ is in kilometre and \$h\$ is in meters.³⁵⁾

For example, VA7FI's antenna is 20m above the ground, at an elevation of 100m overlooking the water. It means that his antenna can see about 45 km in that direction.

For direct waves to occur, the height of the antenna needs to be many times greater than the wavelength of the radio wave so that the signal doesn't interact with the ground. In this example, the antenna (120m above sea level) is 60 wavelengths high on the 2 meter band, and 170 wavelengths on the 70 cm band.

Another similar station could be reached at about 90 km.

Ground Waves

Ground waves occur when the signal curves with the Earth until it becomes too weak to be detected. This phenomena happens because of diffraction for vertically polarized radio waves when the frequency is below 3 MHz. The radio wave interacts with the ground where it loses some of its energy but also curves toward it. Depending on the frequency, these waves can go beyond the horizon out to about 200 km.

Skywaves

Depending on the frequency and atmospheric conditions, it's possible for radio waves going up to reflect back down to Earth. From our location in British Columbia, we can very easily talk to people in Japan using Skywaves.

This process uses Ionospheric Refraction, which we'll see next.³⁶⁾

Ionosphere

The region of our atmosphere between 50km and 400km altitude is called the ionosphere³⁷⁾, and to radio waves, it can act like:

a mirror that refracts and reflects a signal back to earth,
a clear window that lets a signal escape to space,
or a tinted window that absorbs the signal.

The reason for this complex behaviour is that the ionosphere is composed of electrons and electrically charged atoms and molecules (called ions) caused by the Sun's ultraviolet radiation (solar flux). Gas at higher altitude is more ionized because it is less dense, which makes recombination into neutral molecules more difficult. Depending on the frequency, radio waves travelling into the ionized gas can see an index of refraction that is less than that of the air below, which means that they can refract and reflect the way light does through water. Also, because ionization depends primarily on the Sun's activity, three main cycles dictate the characteristics of the ionosphere:

The Day / Night cycle
The Summer / Winter cycle
11-year Sunspot cycle

Depending on the time of day, the ionosphere separates into 3 or 4 layers (of different gas composition):

D-Layer (50km ‒ 90km)
E-Layer (90km ‒ 150km)
F-Layer (150km ‒ 400km)

The distance radio waves can propagate via ionospheric refraction depends on many factors:

Take Off Angle and Layer Height

A radio signal will reach further when:³⁸⁾

The take off angle is as low towards the horizon as possible, and
The ionosphere layer is as high as possible.

The above animation is a gross oversimplification to illustrate the point that, all else being equal, signals sent near the horizon using the F layer will go further. For example:

The maximum distance using one hop of the F₂ layer is around 4000 km, while
The maximum distance using one hop of the E layer is around 2000 km.

In reality, the ionosphere is a medium with a continuously varying index of refraction rather than a series of discrete “mirrors”. As such, how much signals “curve” also depends on the takeoff angle, and just like what we saw in the previous section, there's a critical angle that must be met for Total Internal Reflection to occur. So the real picture is more like this one:³⁹⁾

Note the following important terms on the above image:

The Skip Distance is the distance between the transmitter and the first hop of the reflected sky-wave.
The Skip Zone is the zone where no signal reaches. It's too far for the ground wave can reach, but too close for the first reflected sky-wave hop.

And finally, real antennas do NOT transmit their signal at a single take off angle but over a range of them, which can vary depending on the antenna type and how high it is over the ground. So in reality, many of these paths are used at the same time and even reflect off the ground and go back for a second or third hop. Communications exceeding 5000 km uses multihop propagation, which looks like this:⁴⁰⁾

Frequency

Whether a layer lets a radio signal through, reflects it back, or absorbs it depends on the frequency. In general:

The D-Layer forms about half an hour after sunrise and disappears half an hour before sunset. It tends to absorb frequencies below 5 MHz and lets others through. At our HF frequencies, it acts either like a tinted window, or a clear window, and so it never really does anything good for us.
The E-Layer can reflect high angle 160m and 80m signals that made it through the D-Layer during daylight hours.
The F-Layer splits into two layers about half an hour before sunrise and recombines into one layer about half an hour after sunset. It refracts higher frequency HF bands (40m, 20m, 10m), but VHF frequencies (above 50 MHz) go straight through it, acting either like a mirror or a clear window.

If the goal is to get our signal to travel as far as possible, there's a kind of Goldilocks range of frequencies that we can use. If the frequency is too high, the signal will travel straight through the ionosphere as if it was a clear window. If the frequency is too low, the signal will get absorbed by the ionosphere as if it was a tinted window. The game is to find the frequency that will get reflected by the ionosphere as if it were a mirror, and not only that but we also would like to use the F layer because the higher the point of reflection, the further the signal will travel.

The Maximum Usable Frequency (MUF) is the maximum frequency that the F layer will reflect before it turns transparent and lets the signal escape into space.
The Lowest Usable Frequency (LUF) is the lowest frequency that the D layer will let through before it turns opaque and absorbs the signal.

In the previous animation, frequencies below 6 MHz don't get through the D layer, and frequencies above 44 Mhz escape into space. Only frequencies between the LUF and MUF get reflected back down to Earth by different layers.

Sometimes, depending on the atmospheric and/or solar conditions, the LUF, is greater than the MUF. In that case, no reflection is possible. Signals either get absorbed by the D layer, or get through all of them and escape into space:

In general, during the day:

The Maximum Usable Frequency (MUF) is around 50 MHz, which means the 6m, 2m, and 70cm bands are all direct waves.
The Lowest Usable Frequency (LUF) is around 15 Mhz, Which means that the 40m, 80m, and 160m bands are all ground waves.
Only frequencies between the 20m and 10m bands will be skywaves.

At night:

The D-Layer recombined into neutral molecules.
The MUF lowers to around 10 MHz, which means that even the 10m and 20m band escape into space.
The LUF also lowers so that the 160m and 80m band can be skywaves.

So normally, higher frequencies are better during the day (up to about 6m), and lower frequencies are better at night.

Here's a map of MUF that's updated regularly. You can think of this as a “weather map” for ham radio.

The Gray Zone

Finally, because the D-Layer disappears before the F-Layer recombines, and reappears after the F-Layer splits, the propagation can be interesting around sunrise and sunset. This is called the gray zone.

Animation

Here's the animation if you want to move the dials and experiment with it yourself.

Download earthpropagation.ggb

Meteor Scattering

When meteors enter the ionosphere, they create intensely ionized columns of air that can scatter radio waves for very short periods of time (from a fraction of a second to a couple seconds per event). This mode can be used on VHF frequencies between 30 MHz and 100 MHz but is most effective on the 6m band (50 MHz).

Auroral backscatter

Auroral activity creates strong ionization of the E-region. HF (and sometimes VHF) radio waves can backscatter and be heard up to 2000 km in the east-west direction. CW is the best mode to use to take advantage of this mode.

Sporadic-E propagation

Sporadic-E propagation (not to be confused with ordinary E-layer propagation) takes advantage of ionization patches in the E-layer that drift westwards at speeds of a few hundreds of kilometres per hour. You can think of it as invisible clouds of ionized gases that move in the E-layer. If your signal is lucky enough to enter one of these clouds, it can bounce between 1000 and 2000 km in a single hop. Sporadic-E is most often observed on the 6m band.

Troposphere

In the previous section, we discussed how the Ionosphere (the region of our atmosphere between 50km and 400km altitude) can, reflect and refract radio waves, let them pass straight through, or absorbed them completely mostly due to the sun's ionization of the gas in these layers.

Here we discuss how the troposphere (the lowest region of our atmosphere below 20km altitude) can also affect radio waves because of variation in temperature, pressure, or water vapour content. Normal VHF tropospheric propagation can have a range of roughly 800 km.

Tropospheric Ducting

The index of refraction of air is lower when the air is warmer. So during a temperature inversion, the air on the ground is colder than the air above, which means that radio waves go from a high to a low index of refraction medium (that is, from a slow to a fast medium). This causes the radio wave to refract back down toward the Earth. Tropospheric ducting is when the radio wave follows the curve of that inversion layer until it exits back to the Earth after travelling several hundreds of kilometres (up to 2000 kilometers).

Unlike Ionospheric refraction, Tropospheric ducting is observed at VHF frequencies as opposed to HF frequencies.

Scattering

At VHF frequencies, small variations in the density of the troposphere (around 10 km) can scatter some of the radio waves back toward the ground to distances of 800 km.

Scattering can also allow HF signals from the skipzone to be heard. Scatter is most likely involved when weak or distorted signals near or above the maximum usable frequency (ie, they should escape into space) are heard over unusual paths.

References

Questions

B-007-001-001 → B-007-004-002
B-007-005-001 → B-007-008-001
B-007-008-006 → B-007-008-011

³⁵⁾

See Wikipedia: Line Of Sight Propagation for more details

³⁶⁾

The previous three images were taken from Milo Carroll's Satellite Time Transfer presentation: http://slideplayer.com/slide/5941869/

³⁷⁾

Picture from Wikipedia: Ionosphere

³⁸⁾

The animations that follow are not to scale.

³⁹⁾

Picture is from Fig. 2-14 at https://www.globalsecurity.org/intell/library/policy/army/fm/24-18/fm24-18_3.htm

⁴⁰⁾

Picture from http://www.ferzkopp.net/Personal/Thesis/node8.html

stationassembly

Block Diagrams

The next few sections use block diagrams to illustrate the configuration of various pieces of equipment. Here's an excellent introduction that should help with what follows:

HF Station

Some of the components shown above may be integrated into one device, and others may be optional. But if all are included, this is how they should be connected.

The transceiver takes the audio from the microphone and creates a modulated radio signal. Typical HF radios can output about 100W of power. Hams with their advanced ticket can feed that into...
The amplifier takes the radio signal from the transceiver and amplifies its power to 1 kW or even 1.5 kW.
From there, the signal may contain higher frequencies (called harmonics) that are not desirable, so it goes through a Low Pass Filter, which passes low frequencies and filters out high ones.
After that, the SWR Bridge measures how much of the signal is reflected back toward the radio from the antenna system. We saw earlier that the length of the antenna needs to match the frequency we use. When the match isn't perfect, some of the radio signal “bounces” at the antenna back to the radio, which isn't good for the equipment. The SWR Bridge measures this.
A trick we use to protect the radio equipment is to add a Tuner. This device uses varying combinations of capacitors and inductors (more on this later) to match the impedance of the antenna system to the radio (more on that later). Although there is still reflection at the antenna back toward the radio system, the tuner will “protect” the radio from it.
An Antenna Switch is a handy piece of equipment to quickly switch between antennas without having to disconnect and connect coax connectors.
Because of their size difference, it's usual to have a multiband antenna that will work on 20m, 17m, 15m, 12m, and 10m (and maybe even 6m), and a second antenna for 40m, 80m (and maybe even 160m).
Where as a tuner for the upper band is optional if the antenna is well designed, the lower bands (specially 80m) are very wide compared to their frequencies so it's practically impossible to have an antenna that will work over the entire band. For that reason, a tuner for these bands is pretty much mandatory.
The Dummy Load is a 50Ω resistor that can dissipate all the power from the radio without converting any of it into a radio waves. It's useful for test purposes or to tune an amplifier.

All that being said, it's possible to go on HF with a transceiver that has a small integrated tuner in it, and a single antenna. It'll just be a matter of knowing which frequencies your system can transmit on, and stay within that range.

Transmitters

Review the Wave Modulation page. You should:

understand the difference between the baseband signal and the carrier, and how they combine to form the modulated signal.
understand the difference between AM and FM.

CW Transmitter

The master oscillator generates a stable sine wave signal at the rest frequency.
The driver / buffer is the first stage in the amplification process.
The key is basically an on/off switch used to let the signal through to the amplifier or block it.
The power amplifier is the last stage of amplification before the signal is released to the antenna.

FM Transmitter

The microphone picks up the sound waves carried by the air and converts them into a weak electrical signal that becomes the baseband.
The speech amplifier increases the strength of the baseband signal before it can modulate the carrier.
The modulator uses the baseband signal to change the carrier oscillator frequency in the next stage.
The oscillator generates a stable sine wave signal at the rest frequency when no modulation is applied. Its frequency changes linearly when fully modulated with no measurable change in amplitude. Note that the oscillator frequency can be much lower than the final transmitted frequency.
The frequency multiplier increases the frequency of the modulated signal by a factor of 2, 3, or 4. This is so that the oscillator frequency can be lower than the frequency transmitted at the antenna.
The amplifier increases the power of the signal to be transmitted by the antenna.

SSB Transmitter

The microphone picks up the sound waves carried by the air and converts them into a weak electrical signal that becomes the baseband.
The speech amplifier increases the strength of the baseband signal before it can modulate the carrier.
The RF oscillator generates a stable sine wave signal at the rest frequency. Note that the oscillator frequency can be much lower than the final transmitted frequency.
The outputs of the speech amplifier and the RF oscillator are mixed together in the balanced modulator. This produces a signal with two side bands, only one of which we want to keep.
The filter only lets one sideband through.
The VFO creates a sine wave frequency that can be varied.
The mixer uses the output from the filter and the VFO to generate a higher frequency signal.
The linear amplifier increases the power of the signal to be transmitted by the antenna.

Note that a Frequency Mixer multiplies two signals together, which usually results in a total of four (or more) frequencies. This is not to be confused with an Audio Mixer, which adds different signals together. See the optional Wave Modulation Math page for more details.

Receivers

FM Receiver

Before we look at individual components, there's an important step we should highlight: The received radio frequency (RF) is converted down to a predetermined intermediate frequency (IF) before being processed further. This solves a lot of problems that plague simpler receivers. Now let's look at the details.

The RF amplifier increases the strength of the weak radio signal received by the antenna.
The local oscillator creates pure sine wave that can be tuned by the listener.
The mixer takes in the RF signal and mixes it with the pure sine wave from the local oscillator. This produces two frequencies: one is the sum of both input frequencies, the other is the difference. Only one of these is the IF that we want to keep.
The filter removes unwanted frequency created by the mixer and lets the IF pass through to the next stage.
The IF amplifier increases the strength of the IF signal.
At this point, the signal's amplitude may vary due to noise. The limiter removes these amplitude modulations before the next stage.
The frequency discriminator extracts the baseband signal.
The audio amplifier increases the strength of the baseband signal before passing it to the speaker.

SSB/CW Receiver

The first part is the same as the FM Receiver above.
The Beat Frequency Oscillator: restores the carrier that was suppressed to create the SSB signal.
The Product Detector mixes the IF and the BF together to extract the audio frequency (AF = |IF - BF|).
For example, a receiver with a local oscillator at 3.995 MHz should have its IF tuned to 455 kHz to receive an incoming signal of 3.54 MHz.
The audio amplifier increases the strength of the baseband signal before passing it to the speaker.

Digital System

The important conceptual leap here that we want the computer to be able to transmit something. That something could be a Winlink message, APRS telemetry, a live JS8 chat, or any other kind of data.
That data needs to be modulated into a sound that can then be transmitted by the radio.
A modem (MOdulator/DEModulator) or a soundcard like the Signalink can do that. But some radios like the IC-7300 can connect directly to the computer using a USB cable.

Regulated Power Supply

The goal of a regulated power supply is to convert household voltage (120 VAC) into DC voltage for the radio (13.8 VDC).

The transformer converts 120 VAC to 12 VAC
The rectifier takes the negative values of the sign waves and flips them up.
The filter removes the small oscillations still present.
The regulator ensures that when there's a power draw, the voltage doesn't drop too much.

Questions

B-003-001-001 → B-003-008-006
B-003-010-004 → B-003-010-006

test

Optional Math of Waves

This is a brief survey of the math required to analyze waves at the first or second year university level. If you did well in grade 12 high school math, you'll probably be able to follow this and learn some new and really cool math.

Real numbers

If \$ (5)^2 = 25 \$ and \$ (-5)^2 = 25 \$, what number can you put in the box so that:

\$$ \Box ^2 = -25 \$$

It turns out that there is no real number such that when you multiply it by itself you get a negative number. But could we invent an imaginary one?

Complex Numbers

Let's create an imaginary number called \$ i \$ such that:

\$$ i = \sqrt{-1} \qquad \Rightarrow \qquad i^2 = -1 \$$

Even though \$ i \$ is nowhere on the real line (in math, we say that: \$ i \not\in \mathbb{R} \$), we can none-the-less perform interesting mathematical operations with it:

We can add it to a real number and create a complex number:

\$$(1 + i) \$$

We can multiply complex numbers together:

\begin{align*} (1+i)^2 &= (1+i)\cdot(1+i) \\ &= 1 + 2i + i^2 \\ &= 1 + 2i - 1 \\ &= 2i \end{align*}

We can find roots:

\begin{equation*} z^4 = 16 \Rightarrow z^2 = \left\{ \begin{array}{rl} 4 \Rightarrow z &= \pm 2 \\ -4 \Rightarrow z &= \pm 2i \end{array} \right. \end{equation*}

A Little Philosophy

If these weird numbers follow all of the algebra rules without inconsistencies, does it mean they exist as much as the real numbers? Aren't complex numbers a mere creation by mad mathematicians? How about mathematics itself: is it discovered or invented?⁴¹⁾

In a certain way, negative numbers are just as weird as complex numbers: after all, we know what 5 cars look like, but what does −5 cars mean? And yet, in certain context (like temperature), we have no problem using negative numbers. Could it be that there are contexts where complex numbers make sense?

The Complex Plane

In the same way that we can represent real numbers by a point on the real number line...⁴²⁾:

... we can also represent a complex number graphically on a complex plane by using the horizontal axis for the real part and the vertical axis for the imaginary part. For example, \$ (1 + i) \$ would be represented as a point 45° up the horizontal axis and \$ \sqrt{2} \$ away from the origin:

You can move the point around to look at other complex numbers on the plane.

Download polar.ggb

To convert between the Cartesian \$(a,b) \$ and the Polar \$ (r \angle \theta) \$ representations, only simple trigonometry and Pythagoras is needed.

\$$ (a, b) \rightarrow (r\angle \theta) \$$	\$$ (r\angle \theta) \rightarrow (a, b)\$$
\$$ r^2 = a^2 + b^2 \$$	\$$ a = r\cos\theta \$$
\$$ \tan \theta = \dfrac{b}{a} \$$	\$$ b = r\sin\theta \$$

Note that very often, we use radians instead of degrees for the angle. There are a total 360° or 2π radians in a circle. While most people are used to degrees, a radian is actually much easier to picture:

Imagine a circle.
Now imagine the length from the centre to the circle (along the “radius”).
Take that length and lay it down on the perimeter of the circle.
The angle that this length covers is 1 radian (because of the length of the radius on the circle).
That's why a circle has 2π radians (because the circumference is 2πr)

Roots

The complex plane has many useful applications, but one of them allows us to visualize roots of the form \$ z^n = w \$. For example, if we set \$ w = 9 \$ and \$ n = 2 \$ on the graph below, we'll see that the roots of \$z^2 = 9 \$ are \$ z = \pm 3 \$.

Download complexroots.ggb

Without using the graph above, what do you expect the solution(s) to \$ z^3 = 8 \$ will be? That is, what number(s), when multiplied by itself three times gives 8?

Now move \$ w = 8 \$ and \$n = 3 \$ to have a look at the solutions graphically, you might be surprised by what you find.

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So \$z = 2\$ was to be expected since \$ 2^3 = 8 \$ but it looks like there are two more solutions.

To find them, we first notice that the three solutions are spread out evenly around the circle, that is they are 120° apart. So in polar coordinates, the three solutions are \$z = 2\angle 0^\circ, \quad 2\angle 120^\circ,\quad 2\angle 240^\circ \$.

We can now convert them to Cartesian:

\$z = 2\angle 0^\circ = \big(2\cos(0^\circ), 2\sin(0^\circ)\big) = (2, 0) = 2\$
\$z = 2\angle 120^\circ = \big(2\cos(120^\circ), 2\sin(120^\circ)\big) = (-1, \sqrt{3}) = -1 + i\sqrt{3}\$
\$z = 2\angle 240^\circ = \big(2\cos(240^\circ), 2\sin(240^\circ)\big) = (-1, -\sqrt{3}) = -1 - i\sqrt{3}\$

Let's check that the second solution works: \begin{align*} (-1 + i\sqrt{3})^3 &= (-1 + i\sqrt{3})\cdot(-1 + i\sqrt{3})(-1 + i\sqrt{3}) \\ &= (-1 + i\sqrt{3})\cdot\big(1 -2i\sqrt{3} +(i^2)(3)\big) \\ &= (-1 + i\sqrt{3})\cdot\big(1 -2i\sqrt{3} +(-1)(3)\big) \\ &= (-1 + i\sqrt{3})\cdot(-2 -2i\sqrt{3}) \\ &= 2 + 2i\sqrt{3} - 2i\sqrt{3} -(2)i^2(3) \\ &= 2 -(2)(-1)(3) \\ &= 2 + 6 \\ &= 8 \end{align*}

The Euler Identity

The Euler identity exposes a deep relationship between trigonometric and exponential functions⁴³⁾:

\$$ e^{i \theta} = \cos \theta + i \sin \theta \$$

Let's use two different ways to verify that this mysterious identity is true.

The Derivatives

If we separate this identity into two functions and take their derivatives, we notice that: \begin{align*} && f(\theta) &= e^{i \theta} &\text{ & }&& g(\theta) &= \cos \theta + i \sin \theta \\ \Rightarrow && f'(\theta) &= i e^{i \theta} &\text{ & }&& g'(\theta) &= -\sin \theta + i \cos \theta \\ \Rightarrow && f'(\theta) &= i \cdot f(\theta) &\text{ & }&& g'(\theta) &= i \cdot g(\theta) \end{align*}

We know that there's only one functions \$ h(x) \$ that satisfies the differential equation \$ h'(x) = ah(x) \$, and it is \$ h(x) = A e^{ax} \$ What Euler discovered is that when \$ a = i \$ , there's a second function that also satisfies the same differential equation! These two functions must therefore be one and the same.

Taylor

Another method to verify the Euler identity is to use Taylor series:

\begin{align*} e^x &= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \cdots \\ \sin x &= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \\ \cos x &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \end{align*}

Replace \$ x \$ with \$ i \theta \$ in the Taylor series of \$ e^x \$
Replace \$ x \$ with \$ \theta \$ in the Taylor series for \$ \sin x \$ and \$ \cos x \$
Add the Taylor series of \$ \cos \theta \$ and \$ i \sin \theta \$ together and you'll get the Taylor series for \$ e^{i \theta} \$

Euler Identity and Polar-Cartesian Representations

In the previous section, we saw that a complex number \$z = a + ib \$ could be represented as a point \$(a, b)\$ on the complex plane, which could also be viewed in polar coordinates as (\$r\angle \theta) \$. We saw that to convert between the Cartesian \$(a,b) \$ and the Polar \$ (r \angle \theta) \$ representations, only simple trigonometry and Pythagoras is needed:

\$$ (a,b) \rightarrow (r\angle \theta) \$$	\$$ (r\angle \theta) \rightarrow (a,b)\$$
\$$ r^2 = a^2 + b^2 \$$	\$$ a = r\cos\theta \$$
\$$ \tan \theta = \dfrac{b}{a} \$$	\$$ b = r\sin\theta \$$

This means that:

\begin{align*} z &= a + ib \\ &= r\cos\theta + i r\sin\theta \\ &= r\big( \cos\theta + i \sin\theta \big) \\ &= r e^{i\theta} \end{align*}

This offers another interpretation of the Euler identity as the algebraic conversion between Cartesian and Polar coordinates:

	Cartesian	Polar
Graphical	\$$(a, b) \$$	\$$ (r\angle \theta) \$$
Algebraic	\$$z = a + ib\$$	\$$ z = re^{i\theta} \$$

This now allows us to simplify a lot of difficult mathematics. For example let's look at the root problem \$z^3 = 8 \$ again. Since the number 8 on the complex plane is the point \$(8,0)\$, in polar coordinates, it can be any of the following: \$8\angle 0, 8\angle 2\pi, 8\angle 4\pi, \cdots \$ This is because we can go around the circle as many times as we want and return to the same point. Since we expect three roots, let's use the first three polar representations of 8:

\$$ z^3 = 8 = \left\{ \begin{array}{c} 8e^{0 i} \\ 8e^{2\pi i} \\ 8e^{4\pi i}\\ \vdots \end{array} \right. \$$

\$$ \Rightarrow z = \left\{ \begin{array}{lcl} \left(8e^{0 i}\right)^{\frac{1}{3}} &=& 8^{\frac{1}{3}} e^{\frac{0}{3}} = 2\\ \left(8e^{2\pi i}\right)^{\frac{1}{3}} &=& 8^{\frac{1}{3}} e^{\frac{2\pi}{3}i} = 2 \left(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}\right) = 2 \left(-\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) = -1 + i\sqrt{3} \\ \left(8e^{4\pi i}\right)^{\frac{1}{3}} &=& 8^{\frac{1}{3}} e^{\frac{4\pi}{3}i} = 2 \left(\cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3}\right) = 2 \left(-\frac{1}{2} - i \frac{\sqrt{3}}{2}\right) = -1 - i\sqrt{3} \\ &\vdots& \end{array} \right. \$$

If we had used more than 3 numbers, the roots would have started repeating. If we had used less than 3, we would have missed some answers in the same way that there are two answers to \$z^2 = 9 \$ (namely \$z = \pm 3 \$).

Cartesian vs Polar

Which is the best representation: Cartesian, \$z = a + i b \$, or Polar, \$z = re^{i\theta}\$. As you might expect, it depends on what you're trying to do... For example, let's take:

\$$z_1 = 1 + i = \sqrt{2}e^\left(i\frac{\pi}{4}\right) \quad \text{and} \quad z_2 = -1 + i = \sqrt{2}e^\left(i\frac{3\pi}{4}\right) \$$

Imagine having to add, subtract, multiply, or divide these together. Or raise them to a power, or take a root of them. Which of the two representations do you think would be easiest to use for each operation?

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Operation	Cartesian	Polar
Addition	Easiest: \$$ z_1 + z_2 = (1 + i) + (-1 + i) = 2i \$$	I don't know how to do it. \$$ z_1 \pm z_2 = \sqrt{2}e^\left(i\frac{\pi}{4}\right) \pm \sqrt{2}e^\left(i\frac{3\pi}{4}\right) = ?? \$$
Subtraction	Easiest: \$$ z_1 - z_2 = (1 + i) - (-1 + i) = 2 \$$
Multiplication	Moderate: \$$ z_1 \cdot z_2 = (1 + i) \cdot (-1 + i) \$$ \$$ = (1)(-1) + (1)(i) + (i)(-1) + (i)(i) \$$ \$$ = -1 + i - i -1 = -2 \$$	Easiest: \$$ z_1 \cdot z_2 = \sqrt{2}e^\left(i\frac{\pi}{4}\right) \cdot \sqrt{2}e^\left(i\frac{3\pi}{4}\right) \$$ \$$ = \sqrt{2}\cdot \sqrt{2} e^\left(i\frac{\pi}{4} + i\frac{3\pi}{4}\right) \$$ \$$ = 2 e^{i\pi} = -2 \$$
Division	Tedious: \$$ \frac{z_1}{z_2} = \frac{(1 + i)}{(-1 + i)} = \frac{(1 + i)(-1 - i)}{(-1 + i)(-1 - i)} \$$ \$$ = \cdots = \frac{-2i}{2} = -i \$$	Easiest: \$$ \frac{z_1}{z_2} = \frac{\sqrt{2}e^\left(i\frac{\pi}{4}\right)}{\sqrt{2}e^\left(i\frac{3\pi}{4}\right)} = e^\left(i\frac{\pi}{4} - i\frac{3\pi}{4}\right) \$$ \$$ = e^\left(-i\frac{\pi}{2}\right) = -i \$$
Exponentiation	The bigger the exponent, the more tedious: \$$z_1^{100} = (1 + i)^{100} = \cdots \$$	Easy no matter how big the exponent: \$$z_1^{100} = \left(\sqrt{2}e^\left(i\frac{\pi}{4}\right)\right)^{100} \$$ \$$ = 2^{50}e^{25\pi i} = 2^{50}e^{12(2\pi i)} e^{\pi i} \$$ \$$ = 2^{50}(1)(-1) = -2^{50}\$$
Roots	I don't know if it's even possible: \$$ \sqrt[3]{z_2} = \sqrt[3]{-1 + i}\$$ \$$ (-1 + i)^\left(\frac{1}{3}\right) = ?? \$$	As easy as exponentiation: \$$ \sqrt[3]{z_2} = \left(\sqrt{2}e^\left(i\frac{3\pi}{4}\right) \right)^\left(\frac{1}{3} \right) \$$ \$$ = \sqrt[6]{2}e^\left(i\frac{\pi}{4}\right) = \frac{1 + i}{\sqrt[3]{2}} \$$

The lesson here is that since the polar representation uses exponents, and exponents turn multiplication into addition⁴⁴⁾, the polar representation is easiest for multiplication, division, exponentiation, and roots. It's essentially why the dB scale is so useful. But addition and subtraction is intrinsically easier in Cartesian coordinates.

Important Algebraic Results

Use the Euler identity to get the following two useful results:

\$$ \cos \theta = \dfrac{e^{i \theta} + e^{-i \theta}}{2} \qquad \text { & } \qquad \sin \theta = \dfrac{e^{i \theta} - e^{-i \theta}}{2i} \$$

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Modify the Euler identity to see what happens when the angle is \$ (-\theta) \$

\begin{align*} && e^{i \theta} &= \cos \theta + i \sin \theta \\ \Rightarrow && e^{i (-\theta)} &= \cos (-\theta) + i \sin (-\theta) \\ \Rightarrow && e^{-i \theta} &= \cos \theta - i \sin \theta \end{align*}

Add the original Euler identity to the new one for \$ (-\theta) \$ and simplify:

\begin{align*} && e^{i \theta} &= \cos \theta + i \sin \theta \\ + && \underline{e^{-i \theta}} &= \underline{\cos \theta - i \sin \theta} \\ \Rightarrow && e^{i \theta} + e^{-i \theta} &= 2\cos \theta \\ \Rightarrow && \cos \theta &= \frac{e^{i \theta} + e^{-i \theta}}{2} \\ \end{align*}

To get the second equation, take the original Euler identity and subtract the new one for \$ (-\theta) \$ from it and simplify:

\begin{align*} && e^{i \theta} &= \cos \theta + i \sin \theta \\ - && \underline{e^{-i \theta}} &= \underline{\cos \theta - i \sin \theta} \\ \Rightarrow && e^{i \theta} - e^{-i \theta} &= 2i\sin \theta \\ \Rightarrow && \sin \theta &= \frac{e^{i \theta} - e^{-i \theta}}{2i} \\ \end{align*}

Use the Euler identity to show that:

\$$ \cos (\theta + \phi) = \cos \theta \cos \phi + \sin \theta \sin \phi\\ \sin (\theta + \phi) = \cos \theta \sin \phi + \sin \theta \cos \phi \$$

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Multiply the Euler identities for \$\theta\$ and \$\phi\$ and simplify.

\begin{align*} && e^{i \theta} &= \cos \theta + i \sin \theta \\ \times && \underline{e^{i \phi}} &= \underline{\cos \phi + i \sin \phi} \\ \Rightarrow && e^{i \theta} e^{i \phi} &= \big(\cos \theta + i \sin \theta\big)\cdot \big(\cos \phi + i \sin \phi \big) \\ \Rightarrow && e^{i\theta + i\phi} &= (\cos \theta) (\cos \phi) + (\cos \theta) (i\sin \phi) + (i \sin \theta)(\cos \phi) + (i \sin \theta)(i \sin \phi)\\ \Rightarrow && e^{i (\theta + \phi)} &= \cos \theta \cos \phi + i\cos \theta \sin \phi + i \sin \theta \cos \phi + i^2 \sin \theta \sin \phi\\ \Rightarrow && e^{i (\theta + \phi)} &= \cos \theta \cos \phi + i (\cos \theta \sin \phi + \sin \theta \cos \phi) - \sin \theta \sin \phi\\ \Rightarrow && e^{i (\theta + \phi)} &= \color{green}{(\cos \theta \cos \phi - \sin \theta \sin \phi)} + i \color{blue}{(\cos \theta \sin \phi + \sin \theta \cos \phi)} \end{align*}

But the Euler identity for angle \$ (\theta + \phi) \$ is: \$$ e^{i (\theta + \phi)} = \color{green}{\cos(\theta + \phi)} + i\color{blue}{\sin(\theta + \phi)} \$$

So comparing the real and imaginary parts, we can conclude that: \$$ \color{green}{\cos(\theta + \phi) = \cos \theta \cos \phi - \sin \theta \sin \phi} \\ \color{blue}{\sin(\theta + \phi) = \cos \theta \sin \phi + \sin \theta \cos \phi} \$$

Use the two earlier results to show that:

\$$ \sin (\theta + \Delta \theta) + \sin (\theta - \Delta \theta) = 2 \cos \Delta \theta \sin \theta \$$

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Use the last result for \$ \sin(\theta + \phi) \$ but replace \$ \phi \$ with \$ \Delta \theta \$

\$$ \sin (\theta + \Delta \theta) = \cos \theta \sin \Delta \theta + \sin \theta \cos \Delta \theta \$$

Similarily, \begin{align*} && \sin (\theta - \Delta \theta) &= \cos \theta \sin(-\Delta \theta) + \sin \theta \cos (-\Delta \theta) \\ \Rightarrow && \sin (\theta - \Delta \theta) &= -\cos \theta \sin\Delta \theta + \sin \theta \cos \Delta \theta \\ \end{align*}

Adding both of these together, we get: \begin{align*} \sin (\theta + \Delta \theta) + \sin (\theta - \Delta \theta) &= \cos \theta \sin \Delta \theta + \sin \theta \cos \Delta \theta \\ & \quad - \cos \theta \sin\Delta \theta + \sin \theta \cos \Delta \theta \\ &= 2\sin \theta \cos \Delta \theta \\ &= 2\cos \Delta \theta \sin \theta \\ \end{align*}

This last result is the basis behind why modulating the amplitude of a carrier produces side bands.

Differential Equations

In the physics of wave, we often have to find solutions to the following type of differential equations:

\$$ a \ddot{x}(t) + b \dot{x}(t) + c x(t) = 0 \$$

A differential equation is an equation that relates a function to its derivatives in some ways and the question is: given some information about the system, what's the function (or family of functions) that satisfy the differential equation.

In physics we often use a dot above the function to indicate a derivative with respect to time, where as in math, we'll often use an apostrophe. Physicists don't like the apostrophe too much because they sometimes use it to denote a different coordinate system. So don't let the notation confuse you:

\$$ \dot{x}(t) = x'(t) = \frac{dx}{dt} \quad \text{and} \quad \ddot{x}(t) = x''(t) = \frac{d^2x}{dt^2} \$$

In our applications, the parameters \$ a, b, c \$ are all real and positive quantities. Even without having studied different equations in any depth, we can imagine that a possible solution to the above differential equation would be: \$ x(t)= e^{rt} \$ since the derivative of an exponential function is itself an exponential function, which is encouraging.

The next step is to try this ——test function'' in the differential equation and see if we can find the values of \$ r \$ that make it work. First we'll need derivatives of the test function:

\begin{align*} & x (t) = e^{rt} \\ \Rightarrow \qquad & \dot{x}(t) = r e^{rt} \\ \Rightarrow \qquad & \ddot{x}(t) = r^2 e^{rt} \end{align*}

When we put these into the differential equation, we get:

\begin{align*} & a \ddot{x}(t) + b \dot{x}(t) + c x(t) = 0 \\ \Rightarrow \qquad & a (r^2 e^{rt}) + b (r e^{rt}) + c (e^{rt}) = 0 \\ \Rightarrow \qquad & e^{rt} (a r^2 + b r + c ) = 0 \\ \Rightarrow \qquad & a r^2 + b r + c = 0 \\ \Rightarrow \qquad &r = - \dfrac{b}{2a} \pm \dfrac{\sqrt{b^2 - 4ac}}{2a} \end{align*}

So what does that result mean? Remember, what we're looking for is the function \$x(t)\$ that satisfies the differential equation \$ a \ddot{x}(t) + b \dot{x}(t) + c x(t) = 0 \$

What we've go so far says that our test function \$x(t) = e^{rt}\$ will satisfy the differential equation if \$r\$ is given by above equation. There is still a lot to unpack however. For example, since \$r\$ contains a square root, it could be real or complex depending on the values of \$a, b,\$ and \$c\$. And as we saw above, if \$r\$ is real, then \$x(t)\$ will be a real exponential function. But if \$r\$ is complex, then we can expect \$x(t)\$ to be some sort of sinusoidal function (recall the Euler Identity).

To simplify the notation, let's define \$\alpha\$ and \$\beta\$ as: \$$ \alpha = \dfrac{b}{2a} \qquad \text{and} \qquad \beta = \dfrac{\sqrt{|{b^2 - 4ac}|}}{2a} \$$

Notice how the absolute value under the square root ensures that \$\beta\$ is always real.

\$r\$ then:

\$$ r = \left\{ \begin{array}{ll} -\alpha \pm \beta & \text{if } b^2 - 4ac > 0,\\ -\alpha \pm i \beta & \text{if } b^2 - 4ac < 0, \end{array} \right. \$$

Let's examine both of these cases in more detail.

Case 1: Over Damped Oscillation

When \$ b^2 - 4ac > 0 \$ , \$ r \$ is real and the general solution is:

\begin{align*} x(t) &= A_1 e^{r_1 t} + A_2 e^{r_2 t} \\ &= A_1 e^{( -\alpha + \beta) t} + A_2 e^{( -\alpha - \beta) t} \\ &= A_1 e^{-\alpha t} e^{\beta t} + A_2 e^{-\alpha t} e^{-\beta t} \end{align*}

\$$ x(t) = e^{-\alpha t} ( A_1 e^{\beta t} + A_2 e^{-\beta t} ) \$$

It's normal to have two constants of integration since our differential equation has a second degree derivative in it. To find these constants, we'd need to know more about the system's initial conditions.

Case 2: Under Damped

When \$ b^2 - 4ac < 0 \$ , \$ r \$ is complex and we'll be using the Euler identity to simplify our solutions

\begin{align*} x(t) &= A_1 e^{r_1 t} + A_2 e^{r_2 t} \\ &= A_1 e^{( -\alpha + i \beta) t} + A_2 e^{( -\alpha - i \beta) t} \\ &= A_1 e^{-\alpha t} e^{i \beta t} + A_2 e^{-\alpha t} e^{-i \beta t} \\ &= e^{-\alpha t} ( A_1 e^{i \beta t} + A_2 e^{-i \beta t} ) \\ &= e^{-\alpha t} \Big( A_1 \big(cos( \beta t) + i \sin( \beta t) \big) + A_2 \big(cos( -\beta t) + i \sin( -\beta t) \big) \Big) \\ &= e^{-\alpha t} \Big( A_1 \big(cos(\beta t) + i \sin(\beta t) \big) + A_2 \big(\cos(\beta t) - i \sin(\beta t)\big)\Big) \\ &= e^{-\alpha t} \Big( (A_1 + A_2) \cos(\beta t) + i (A_1 - A_2) \sin(\beta t) \Big) \\ &= e^{-\alpha t} \Big( a_1 \cos(\beta t) + a_2 \sin(\beta t) \Big) \\ &= e^{-\alpha t} \Big( A \sin \phi \cos(\beta t) + A \cos \phi \sin(\beta t) \Big) \\ &= Ae^{-\alpha t} \Big(\sin \phi \cos(\beta t) + \cos \phi \sin(\beta t) \Big) \\ \end{align*}

In the last three lines, we've redefined the constants of integration a few times so that:

\begin{align*} a_1 &= A_1 + A_2 & , a_2 &= i(A_1 - A_2) \\ a_1 &= A \sin \phi & , a_2 &= A \cos \phi \end{align*}

And we finally use one of the trig identities we proved earlier to write the solution as: \$$ x(t) = A e^{-\alpha t} \sin(\beta t + \phi) \$$

Case 3: Critically Damped

When \$ b^2 - 4ac = 0 \$ , \$ r = -\frac{b}{2a} \$ is real and negative but our test solution is under determined. We'll instead propose a solution of the following type and test that it works: \begin{align*} && x(t) &= e^{rt}(A + Bt) \\ \Rightarrow && \dot x(t) &= re^{rt}(A + Bt) + Be^{rt} \\ && &= e^{rt}\big(r(A + Bt) + B)\big) \\ && &= e^{rt}(rA + B + Brt) \\ \Rightarrow &&\ddot x(t) &= re^{rt}(rA + B + Brt) + e^{rt}Br \\ && &= e^{rt}\big(r(rA + B + Brt) + Br\big) \\ && &= e^{rt}(Ar^2 + 2Br + Br^2t) \\ \end{align*}

Exemple

Nous avons donc deux types de solutions complètement différents qui dépendent de trois paramètres \$ a, b, c \$. Pour voir comment ces paramètres affectent le graphique, imaginons qu'une de nos conditions initiales est \$ \phi = \frac{\pi}{2} \$ . Ça veut dire que:

\$ \begin{align*} & a_1 = A \sin \pi/2 = A & & a_2 = A \cos \pi/2 = 0 \\ \Rightarrow \qquad & A_1 + A_2 = A & & A_1 - A_2 = 0 \\ \Rightarrow \qquad & A_1 = A/2 & & A_2 = A/2 \end{align*} \$

Dans ce cas particulier, nous avons donc:

\$ \begin{equation*} x(t) = \left\{ \begin{array}{rl} A e^{-\alpha t} \dfrac{e^{\beta t} + e^{-\beta t}}{2} & \text{si } b^2 - 4ac > 0 ,\\ A e^{-\alpha t} \cos(\beta t) & \text{si } b^2 - 4ac < 0 ,\\ \end{array} \right. \end{equation*} \$

Geogebra

⁴¹⁾

Whether math is discovered or invented is a famous philosophical problem. If you think it's invented: does that mean that 2 + 2 didn't equal 4 until someone invented that? If you think it's discovered, what about computer programs? At the root, all computing is basically just math.

⁴²⁾

Number line picture from Wikipedia: Number line

⁴³⁾

Note, to obtain the most beautiful equation in the world, set \$ \theta = \pi \$ in the Euler identity:
\$$ e^{i \pi} = \cos (\pi) + i \sin (\pi) = -1 + 0i = -1\$$ \$$ \Rightarrow e^{i \pi} = -1\$$ which is amazingly beautiful because it relates \$ e = 2.71828... \$, \$i = \sqrt{-1} \$, \$\pi = 3.14159... \$, and \$-1\$ in the most surprising and elegant way.

⁴⁴⁾

\$ x^a \cdot x^b = x^{a+b} \$

vhfops

Simplex

The most noticeable difference between a two-way radio and a cell phone is that with a two way radio, you can't talk and listen to the other person at the same time. The reason for this is that two-way radios typically share one frequency for transmitting and receiving. This mode of communication is called Simplex or Half-Duplex.⁴⁵⁾

Cellphones, on the other hand, use two frequencies at the same time: one for transmitting, and the other for receiving. This is called Full Duplex.

Full Duplex⁴⁶⁾		Simplex⁴⁷⁾

Full duplex is like a regular road where traffic has its own lane and can travel in both directions at the same time. Cells phones can do this.		Simplex (or Half-Duplex) is like only having one lane open and forcing traffic to alternate. Most two-way radios can only do this.

Since most of the voice communications we have with amateur radios are not full duplex, it's really important that operators wait until the other person is finished before beginning to speak otherwise they'll miss part of the conversation. When that happens, we say that the operators “doubled”, which means neither heard the other (although a third person may have heard both). We usually recommend waiting half a second or so before beginning to transmit. This also gives the chance to other stations to jump in if needed (with emergency traffic for example).

Range

A common question people often have is how far they can talk with a ham radio. This is a tricky question because the range depends on three main factors:

the antenna characteristics of both radios, their heights above the ground, and the line of sight between them.
the transmitting power of both stations.
the frequency band used (more on that later).

Depending on theses factors, two stations may barely be able to communicate to a kilometre, or have no problem making it all the way to the other side of the globe.

It's also a trick question because the range depends on the setup of both radio stations: it makes no sense to talk about the range of one radio.

Let's look at the first two factors now (we'll leave the third for later).

Antenna, Height, Line of Sight

The best way to improve your signal is to focus on the antenna system. A $2000 radio with a bad antenna system will do worse than a $100 radio with a great antenna system. The general principal is that an antenna should be the right type for the band used (much more on this later), as high as possible, and have a clear path to the other station.

For example:

Under “normal” conditions, two VHF hand held radios operating in simplex at 5 watts could probably communicate with one another to a distance of about 1 km. If the line of sight is obstructed by buildings or hills, this range would drop, but if one (or both) of the stations were at the top of a hill with a clear view of the valley below the range would increase.
A VHF base station with a good antenna up a tall tower might be able to communicate with a hand held radio 10 or 15 km away.
And two such base stations would probably be able communicate with one another to a distance of 50 to 100 km.

Again: it doesn't make sense to talk about the range of one radio; instead, the range depends on both stations. As soon as one station improves its antenna system and line of sight, both stations benefit.

Power

Personal radio services like CB or FRS require no license but have their power output regulated to a maximum of 4 Watts (for CB bands) and 2 Watts (for FRS bands). Ham radios, however, can be operated to hundreds of watts depending on the band, mode, and license type. This complicates the “range” question a bit.

For example, imagine that Alice and Bob can barely manage to hear each other at 5 W. What would happen if Alice cranks up her transmit (TX) power to 50 W? Would that help her hear Bob better?

Increasing your own power is like speaking louder: it helps others hear you better, but it doesn't help you hear them better. As such, one of the rules in ham radio is to use the minimum power required to make the contact. It's like adjusting the volume of your voice to the setting around you. You wouldn't shout to the person sitting in front of you in a quiet restaurant, but you might in a loud pub.

In general, it's best for both stations to try to match their TX power so that you're both “speaking” at an equal volume. If one person is transmitting at 50 W while the other is at 5 W, one person might not hear the other properly.

The rules⁴⁸⁾ for how much power can be used are a little complicated because they depend on where it is measured in the radio and what kind of transmission is used. We'll see what these mean in more detail later, but for now, here's the summary:

	Basic	Advanced
DC input power to the anode or collector circuit of the transmitter stage that supplies radio frequency energy to the antenna	250W	1000W
Peak envelope output power measured across an impedance-matched load for SSB emissions	560W	2250W
Carrier output power measured across an impedance-matched load for any other type of emission.	190W	750W

So while an advanced license allows the use of roughly four times more power (which only improves the received signal by one S unit), having a better antenna doesn't require any special licensing and helps both stations at once, which is why it should be the number one focus.

Personal Record

The furthest VA7FI has been able to communicate in simplex on VHF (146.520 MHz) was 300 km away. The other station (KG7EGT) was atop of Summit Lake Peak (47.03969,-121.83162) at an elevation of 1650m in Washington State and VA7FI was using his GP-9 antenna up a 20m tall tree with the base of the tree 100m above sea level with a direct view of the water. VA7FI could hear KG7EGT 57 and KG7EGT could hear VA7FI 59. This difference in signal strength was because VA7FI was using full power on his base station while KG7EGT was using a portable radio with less power.

Frequency Coordination

As we saw on the previous page, ham operators are allowed to use very specific bands of the electromagnetic spectrum, but each band is also further sub-divided for different usage. The following links from Radio Amateur of Canada (RAC) should be printed and used as reference:

And for British Columbia:

Plan "A" by Functional Allocation

Let's use the 2m band as an example. Even though, we are allowed to transmit between 144 and 148 MHz, only the following frequencies should be used for simplex, voice operation:

146.415, 146.430, 146.445, 146.460, 146.475, 146.490, 146.505, 146.520, 146.535, 146.550, 146.565, 146.580, 146.595, 147.420, 147.450, 147.480, 147.510, 147.540, 147.570

You'll notice that the first 13 channels are 0.015 MHz (or 15 kHz) apart and the last 6 channels are 30 kHz apart. That's to ensure that someone talking on one frequency doesn't interfere with someone talking on another. For example, if a group is using 146.415 MHz and another is using 146.430 MHz, they probably won't bother each other, (unless they are really close to each other), but if you tune your radio to 146.420 MHz or 146.425 MHz, you might hear noise from both groups. That's because communications on a frequency also have what we call bandwidth. Radio signals never only take just “one” frequency. Instead, they take up a certain amount of space on the electromagnetic spectrum.

The fourth column of the table on the previous page shows the maximum allowed bandwidth for each band. This bandwidth is regulated so that no one station takes more space than needed, which would cause interference on the other frequencies.

The Radio

Now that we understand a bit more about simplex communications, let's look at some buttons on an actual radio to see what each does.

This particular radio is the Icom IC-2730A. The first thing to notice about this radio is that it's actually two radios in one. You can listen to two different frequencies at the same time, with their own independent volumes and dials. So looking at the left side for example, we see...

VFO / MR

Two buttons labelled V/MHz and MR. They simply allow the main DIAL to tune the radio either by frequency or by channels saved by the operator. On other radios, these two buttons are sometimes a single button labelled VFO / MR.

Squelch

The SQL (squelch) dial is a circuit within the radio that keeps the speaker silenced (squelched) until the signal level exceeds a certain point set by the squelch control. Normally you set the squelch to just block out noise and allow signals to pass. The reason for this is that there is always some radio background noise on the air. Sometimes that noise is low, but sometimes it's high. Being able to silence it depending on its intensity ensures that we may be able to hear weak stations that are just above that noise level.

CTCSS

Another feature hidden in the MENU of the radio is called CTCSS (Continuous Tone-Coded Squelch System) or sub-audible tone (or sometimes just called “tone”). Imagine that you reside in a congested area where there is a lot of interference that opens the squelch unnecessarily. What you and the other operator can do is send a specific tone (which is just a pure note) on top of your voice when you transmit. That way, you can set your radio to only open the squelch if that note is heard. Typically, there are 50 tones available to choose from ranging from 67.0Hz to 254.1Hz. Have a listen at three of them here:

67.0 Hz	67 Hz
151.4 Hz	151.4 Hz
254.1 Hz	254.1 Hz

You probably noticed that you had to use headphones or good speakers to hear them (at least for the first two tones). That's because, although they are within our normal range of hearing (20 Hz ‒ 20 kHz), the lower frequencies require bigger speaker to play properly. Which is why, we tend not to actually hear these tones over the radio when we use them.

That being said, there is a trade off when picking a tone: the lower frequency tones are the less likely to be heard but they take longer to be recognized by the radio because they take more time to go through a full cycle. That means that it's even more important to wait a half a second before speaking after keying up the PPT (Push To Talk) button to give time to the other radio to recognize the tone and open the squelch for you.

Microphone

The microphone on this particular radio contains some new buttons, but some that are also on the radio itself like the VFO/MR, VOL and SQL

PTT

The PTT (push to talk) is the button used to engage the transmitter and talk on the air. This is the button you must wait half a second before talking after you press it.

DTMF

The DTMF (Dual-tone Multi-frequency) buttons are the same as on a regular phone. They allow radios to transmit numbers and four letters to activate various functions on remote radios. For example, it's possible to configure some radios to have their squelch on (to silence any incoming transmission) until a specific sequence of DTMF is heard. More often, they're used to control repeaters (which we'll see next).

From wikipedia:

“The DTMF system uses a set of eight audio frequencies transmitted in pairs to represent 16 signals, represented by the ten digits, the letters A to D, and the symbols # and *. As the signals are audible tones in the voice frequency range, they can be transmitted through electrical repeaters and amplifiers, and over radio and microwave links, thus eliminating the need for intermediate operators on long-distance circuits.

The *, #, A, B, C and D keys are still widely used worldwide by amateur radio operators and commercial two-way radio systems for equipment control, repeater control, remote-base operations and some telephone communications systems.”

They are created by playing two notes at the same time:

	1209 Hz	1336 Hz	1477 Hz	1633 Hz
697 Hz	1	2	3	A
770 Hz	4	5	6	B
852 Hz	7	8	9	C
941 Hz	*	0	#	D

Repeaters

⁴⁹⁾

To extend the range of VHF and UHF communications, repeater stations are often used. These stations are special radios located on mountain tops that have high quality antennas up tall towers. They work by listening on one frequency (the input) and automatically re-transmitting what they hear on another frequency (the output). The difference between the two frequencies is called the offset.

For example, to use VE7RXZ, individual radio stations need to transmit (TX) on 147.800 MHz (the repeater's input frequency) with a Tone of 100.0 Hz, but listen (RX) on 147.200 MHz (the repeater's output frequency).

In this case we'd say that the repeater frequency is:

147.200 MHz with a + offset of 600 kHz and a Tone of 100 Hz.

So two radio stations communicating through a repeater are not actually hearing each other's signals directly. Instead, they're hearing the repeater's signal, which is usually stronger and goes further.

At first glance, it might look like operating through a repeater would be a lot of work (with this constant flipping back and forth between two frequencies), but once radios are programmed with the proper RX frequency and offset, they will switch frequencies automatically when the Push-to-Talk (PTT) is button is pressed.

Repeater Frequency Coordination

In the same way that simplex frequencies are specified, the same is true for repeater frequencies. For example, here's a sample of http://wp.rac.ca/144-mhz-2m-page/:

With the following note:

“(10) Repeaters may include FM, ACSSB or digital modes of modulation. Consult with your local coordination body for frequency and modulation scheme allocations.”

What this means is that while there is a general agreement on which frequencies should be used for repeaters, the coordination of which repeater should use which specific frequency is managed at a more local level. In BC, the British Columbia Amateur Radio Coordination Council is responsible for coordinating repeater frequencies. Their website also includes pdf lists of repeaters.⁵⁰⁾ Other provincial Coordination Council can be found on the RAC website.

In general, though, radios sold in Canada come pre-programmed with the following repeater offset scheme:

VHF (offset = 600 kHz)⁵¹⁾	UHF (offset = 5 MHz)⁵²⁾
145.100 ‒ 145.500 MHz (-)	442.000 ‒ 445.000 MHz (+)
146.610 ‒ 146.970 MHz (-)
147.000 ‒ 147.390 MHz (+)

(-) means that the repeater's input is lower than its output and
(+) means that the repeater's input is higher than its output.

Repeater Use

To call someone, we always say their call sign first, then our own. To answer a call is more flexible, but the rules⁵³⁾ say that each station has to identify themselves in either English or French:

at the beginning of the conversation,
at least every 30 minutes during the conversation, and
at the end of the conversation.

Suppose that Graham, VE7ABC, wants to call Linda, VE7XYZ on a local VHF repeater.

Graham would first tune in to the repeater's output frequency and wait a few seconds to make sure the repeater is not in use.
When ready, Graham would then say: “VE7XYZ, VE7ABC”
If Linda is there, she would reply something like: “VE7ABC, this is VE7XYZ. Go ahead Graham.”

During their conversation, Linda and Graham should keep each transmission short and leave about half a second between them to allow others to break in. Some repeaters have a courtesy tone (or Roger beep) after each transmission to help operators slow down. Most also have a time-out timer to interrupt lengthy transmissions without pause. That way, if someone wants to join in, all they'd have to do is say their call sign between transmissions and wait to be called.

So when Linda and Graham are done, they could close with something like this:

“Ok Linda, talk to you later. VE7ABC clear on your final.”
“Very good Graham. 73. VE7XYZ clear.

The rules on station identification are very important. The only time a station is allowed to transmit without identification is to remote control a model craft. Even repeaters have to identify themselves (usually in morse code, but not always).

The procedure to send a general call varies a bit depending on whether we are using a repeater or calling in simplex:

In simplex, we use the code “CQ”. So if Graham wanted to talk with anyone, and Linda replied, the exchange would look like this:

“CQ CQ CQ, this is VE7ABC, VE7ABC, VE7ABC”
“VE7ABC, this is VE7XYZ.”

On repeaters, we usually simply state our callsign:

This is VE7ABC listening.

Of course, all call signs should be given in phonetic, especially when sending and replying to a CQ call since the two operators may not know each other.

Repeater Etiquette

Since repeaters extend the coverage of VHF/UHF transmissions, they also extend the number of operators monitoring or potentially needing to use a given frequency. As such, the following etiquette should be followed:

Keep transmissions short.
Give enough time between transmissions for others to break in.
Move to a simplex frequency whenever possible.

To know if it's possible for you to communicate with someone in simplex, you can listen on the repeater's input when the other station is transmitting. If you can hear their direct signal well, then you can reach them in simplex. When choosing a simplex frequency, remember to use one of the 13 dedicated frequencies otherwise you could fall on the input of another repeater, or between two pre-determined channels.

Questions

B-001-013-002 → B-001-013-010
B-001-017-001 → B-001-018-001
B-002-001-001 → B-002-001-011
B-002-003-001 → B-002-003-008
B-002-004-001 → B-002-004-002
B-002-004-010

⁴⁵⁾

Technically, simplex refers to “a communication channel that sends information in one direction only.” However, “the International Telecommunication Union definition is a communications channel that operates in one direction at a time, but that may be reversible; this is termed half duplex in other contexts.” Simplex_communication

⁴⁶⁾

Picture from: Duplex_(telecommunications)#Full-duplex

⁴⁷⁾

Picture from: Duplex_(telecommunications)#Half_duplex

⁴⁸⁾

See RBR-4 Sec. 10

⁴⁹⁾

Picture modified from Repeater

⁵⁰⁾

Another good place to find repeater frequencies is http://repeaterbook.com. They also have a phone app that's very convinient

⁵¹⁾

See http://wp.rac.ca/144-mhz-2m-page/

⁵²⁾

See http://wp.rac.ca/432-mhz-70-cm-page/

⁵³⁾

See RBR-3, p.1 and RBR-4, Sec 9

waveinteraction

Wave Interaction

When an electromagnetic wave (radio, light, etc) hits a surface, it can do one or a mix of six things⁵⁴⁾:

Let's start with refraction and reflection.

Principle of Least Time

Imagine you're on the beach when you suddenly notice a child in distress in the water. You're a good swimmer but you can run faster than you can swim. What do you do?

Option 1: You make a B-line for the child because the shortest distance between two points is a straight line. While it's true that this straight line is the shortest distance, it's not necessarily the fastest path. The problem here is that the water slows you down too much. It's better to cover more ground where you're faster and less where you're slower.

Option 2: You run until you're as close to the child as possible before jumping in the water to swim as little as possible. That path might be faster than the previous one, but it's not the fastest. Here's a thought experiment:

Imagine that you could swim as fast as you can run, then Option 1 would be the fastest path because there would be no difference between running or swimming so the shortest path would also be the quickest.
Now imagine that you could run only slightly faster than you can swim. Does that mean you should go all out and run all that distance to spend as little time in the water as possible? If so, how would running yet faster change the path?

Option 2 would be the path to take if you could instantly teleport on the beach (but not in the water). In this case, you'd want to teleport as close to the child as possible, then swim the rest of the way. This path is when you can run infinitely fast.

Option 3: For regular running speeds, the quickest path is to enter the water somewhere in between. It turns out that, people have a pretty good intuition of where that “somewhere” is. But using Calculus, it's possible to find exactly where to enter the water to get to the child as quickly as possible.

Refraction

In science classes, we learn that the speed of light is roughly 300,000,000 meters per second.⁵⁵⁾ But that's the speed of light in empty space. In air, glass, or water, light slows down. And since light has different speeds in different media, it means that, even for light, the quickest way to get from point A to point B is not necessarily a straight line.

If you shine a beam of light through a piece of glass, it will bend so as to get to the other side as quickly as possible.⁵⁶⁾

This principle is called Fermat's Principle of Least Time and in first year Calculus, students use this principle to derive Snell's Law of Refraction, taught in high school physics, which relates the angles of incidence and refraction to the refractive indices.

Qualitatively: If light enters a medium where it travels slower, it'll bend “inward” so as to spend less time in that medium (like the picture above).

But what if light goes into a medium where it can travel faster? Then this happens:

If this last one feels weird to you, imagine this: suppose you're a turtle who can swim faster than you can walk. It makes sense that you'd want to spend more time in the water and less on the beach:

To recap:

When going from a “quick” medium to a “slow” medium, light bends away from the surface to spend less time in the slow medium.
When going from a “slow” medium to a “quick” medium, light does the opposite and bends towards the surface.
Whatever it does, light always wants to spend less time in a “slow” medium and more time in a “fast” medium because that's the overall quickest way to get from A to B.

Total Internal Reflection

This second case (going from a “slow” medium to a “fast” medium) is really interesting because at some point, the light beam bends so much that it “exits” parallel to the surface, and then reflects like a mirror:⁵⁷⁾

This behaviour is a bit hard to explain without going into the math, but here's an animation that allows you to explore it:

You can move four points around to see how the refracted ray changes: “\$n_1\$”, “\$n_2\$”, “Laser”, and “Entry point”.
Note though that this particular animation only works if the laser is below the horizontal line.

Download snells2.ggb

\$n_1\$ and \$n_2\$ are the Refractive Indices of the media. They are defined as the ratio of the speed of light in vacuum to the speed of light in the media \$\left(n = \frac{c}{v}\right)\$. For example, if \$n = 2\$, then the speed of light is twice as slow in the medium as it is in vacuum. The bigger \$n\$ is, the slower the speed. \$n = 1\$ means that the speed is the same as the speed of light in a vacuum.

A few things to try:

Set \$n_1 = 1\$ and Set \$n_2 = 2\$ and move the Laser and the Entry Point around. These are the paths when you can run twice as fast as you can swim. Notice that if you set \$n_1 = 2\$ and \$n_2 = 4\$, or \$n_1 = 2.5\$ and \$n_2 = 5\$, it shouldn't matter. What really matters is the relative speeds between the two media.
Now move the laser in a straight line so that the angle \$\theta_1\$ doesn't change. The refracted ray shouldn't change either. So it doesn't matter how far the laser is from the surface. What matters is the angle at which the beam hits the surface.
Now move the laser back and forth in a semi circle around the Entry Point. Although the laser is the same distance away from the Entry Point, the angle of incidence changes so the refracted ray changes.
Now set \$n_1\$ = 1.5 and Set \$n_2\$ = 1 and play with the laser to change its angle of incidence (important). At what angle do you notice that the refracted ray goes parallel to the surface? This is called the critical angle. Passed that angle, the ray can't go through and gets reflected instead.

Example

Here's an underwater picture that VA7FI took in a lake with a waterproof camera. The camera is completely submerged under water looking up toward the surface. Above a certain angle, it's possible to see the beach, trees, and the sky. But below that angle, we see the reflection of his wetsuit.

Here's a sketch of the setup:

The other cool thing about that picture is that if you zoom in on the beach, you'll see the colours separate (as if through a prism). This indicates that the index of refraction, n, depends on the frequency. This will be important when we relate all of this back to radio waves.

Snell's Law (Optional)

Click to display ⇲

Click to hide ⇱

Snell's law gives the relationship between the angle of incidence and refraction depending on the refraction indices: \$$ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \$$

There are four interesting cases here:

If \$n_1 < n_2\$ (high speed to low speed), then the left hand side of the equation is in danger of being less than the right hand side. To maintain the equality, \$\theta_1 > \theta_2\$, which means that the path curves away from the surface.
If \$n_1 > n_2\$ (low speed to high speed), then the right hand side of the equation is in danger of being less than the left hand side. To maintain the equality, \$\theta_1 < \theta_2\$, which means that the path curves away from the surface.
If we keep increasing \$n_1\$ compared to \$n_2\$, then \$\theta_2\$ can increase to the point where it's going parallel to the surface (\$\theta_2 = 90^\circ\$), which means that: \$\frac{n_1}{n_2} \sin(\theta_1) = 1 \$. At this point, we call \$\theta_1\$ the critical angle.
If we keep increasing \$n_1\$ even further, then \$\frac{n_1}{n_2} \sin(\theta_1) >1 \$, which means that it's impossible for \$\theta_2\$ to keep up since \$\sin(\theta_2) \leq 1\$. This is when Total Internal Reflection occurs, which is what we use to “bounce” radio waves off the ionosphere (more on that next).

Scattering

Scattering occurs when an EM wave hits a bunch of “small particles” that in turn re-radiate the wave in all direction. Note that the “small particles” can be single atoms, molecules, dust, or pockets of gas with a different index of refraction. They can also be bigger objects like meteors or small planes! The size of the “particle” is always relative to the wavelength of the EM wave. To a 160m radio wave, a meteor is small, but to a laser beam (≈500nm), a dust particle is very big.

Here are a few examples⁵⁸⁾ in the visible light spectrum:

The first picture shows a laser beam shinning at the wall.

In the second picture, water is sprayed into the path of the laser beam.

The reason the beam is invisible in the first picture is that all the light from the laser travels toward the wall (and none toward the camera). But in the second picture, the water vapour scatters some of that light in random directions, allowing some of it to reach the camera. There's a subtle point here: light from a regular light bulb also does this. What I mean is this:

Look at the light bulb in the room you're in.
Now look at an object that the light bulb illuminates.
Now imagine a straight line between the light bulb and that object.

Just as with the laser in the first picture, you don't see any light along that line. If you did, the entire room would be glowing white from all the different light rays that the light bulb emits. In fact, you can see this on a foggy day when it's also sunny. Or on a foggy evening when you're driving with the high beam on.

Effect on Communications

We'll talk about the effects of scattering on communications in more detail later because we need to see a few more basics first. But for now, we'll just say that a radio signal received through scattering will generally be weak, and suffer from rapid flutter or hollow sounding distortion.

It'll be weak because only a small portion of the energy reaches you (think of how much weaker the scattered light from the laser beam is compared to the what reaches the wall directly). And it'll be distorted because your antenna will be receiving the signal from multiple directions (radio wave-paths) at once (think of how you can see the green laser scattered by the mist as an extended line, instead of a single point). As we'll next, when a signal splits and takes different path (of different lengths), they recombined with a sort of “echo” that can cause distortion.

Diffraction

Diffraction is the bending of waves around the corners of an obstacle or through an aperture. The diffracting object or aperture effectively becomes a secondary source of the propagating wave, which in turns can interact with the main wave or other diffracted waves.

All waves do this to an extent, but the phenomena is most pronounced when the the wavelength is of the same order as the size of the diffracting object.⁵⁹⁾

Effect on Communications

Since radio waves can bend around obstacles that are similar in size to the wavelength of the signal, lower frequencies can band over hills and travel beyond the horizon as ground waves because of diffraction (more on this later).⁶⁰⁾

And as we'll see in the next section, diffracted waves can also recombine with the original signal and create multi-path interference.

Interference

An important property of waves (radio, sound, water, quantum mechanical!, or otherwise) is that they can interfere with one another. Here's a Veritasium video showing how light going through two slits can interfere: In some places, the waves add up, in other places, they cancel out. Although not directly about radio waves, we saw in the intro that light and radio waves are in fact on the same electromagnetic spectrum.

Here's a computer animation from Wikipedia showing the same principle:

In terms of radio signals, every time you have more than one source (either because of reflection or because of another radio or antenna), you'll have regions where the signal fades and regions where it increases. Here's why...

Wave Reflection and Multipath

More commonly, radio waves often suffer from multipath interference caused by some sort of reflection, refraction, or diffraction (from mountains, the ground, buildings, the ionosphere, ...) This leads to fading (QSB) as either the transmitter, the receiver, or the reflective surface moves.

This next animation shows the direct wave going from the transmitter to the receiver, as well as a wave reflected by the horizontal axis.

The first thing to notice is that when a wave reflects off a surface, it suffers a half-wavelength phase shift. This means that if the receiver is right next to the “mirror”, the signal will cancel out.

Download multipath.ggb

If the receiver then moves away from the “mirror”, the reflected signal has to travel over a longer distance than the direct signal before reaching the receiver. This means that phase between the two waves will change, sometimes cancelling each other, sometimes reinforcing each other. When the path difference (Δ) between the reflected and direct waves is a whole number of the wave length, the two waves cancel each other because of the half-wavelength difference from the reflection. But when the difference is a multiple of a half wavelength, the two waves add up constructively and the resulting signal is stronger.

In this example, if the receiver moves straight up, the signals will interfere destructively every 5 wavelength-units or so. This means that on the 2m band, the signal will fade every 10 meters or so. This is why the signal strength of a mobile station sometimes goes up and down rapidly as the car moves, which we call picket fencing.

For more details, see the Fresnel Zone Wikipedia article.

Questions

B-007-004-003 → B-007-004-007
B-007-004-008 → B-007-004-011
B-007-008-002 → B-007-008-005

⁵⁴⁾

Picture from http://www.mrwaynesclass.com/lightOptics/reading/index02.html

⁵⁵⁾

It takes light roughly 8 minutes to travel from the Sun to the Earth

⁵⁶⁾

Picture modified from Wikipedia: Refractive Index

⁵⁷⁾

Picture from Wikipedia: Total Internal Reflection

⁵⁸⁾

The laser pictures were taken by Patrick, VA7FI with help from Justine. The picture of the forest is from: https://www.souvenirpixels.com/Photo-blog/i-cZgCHvZ

⁵⁹⁾

The two pictures of diffraction were adapted from http://www.saburchill.com/physics/chapters2/0008.html

⁶⁰⁾

Image of the radio tower and mountain is from https://kistodaynews.com/2017/09/12/scitech-magazine-waves/

wavemodulation

Properties of Waves

Here we dive a little more deeply into waves and look at three ways that a “pure” radio wave (called the carrier) can be modulated to encode a voice signal (called the baseband signal): AM, SSB, FM.

But first, let's look at the general characteristics of a wave.

Amplitude, Wavelength, Frequency, and Period

Here are two moving waves (press the play button on the bottom left corner of the picture). What's different about them? What's the same?

Download travelingwave.ggb

Imagine that the dots moving up and down create the waves that are travelling to the right (as we'll see later, this is kind of like how radio waves are created). Here are a few things to notice:

The Blue wave is twice as “tall” as the green wave.
Both waves are travelling to the right at the same speed.
The Blue dot is moving up and down three times as fast as the green dot.
The Blue wave is three times as compressed as the green wave.

To quantify these observations more precisely, let's look at a snapshot of both waves frozen in time.

the amplitude is the vertical height from the centre of the wave to its highest (or lowest) point. The blue wave has an amplitude of 2 and the green wave has an amplitude of 1.
the wavelength is the horizontal distance of one full cycle. The blue wave has a wavelength of 2m and the green wave has a wavelength of 6m.

Now imagine that the animation is in super slow motion and that the waves are actually travelling at the speed of light, which is roughly 300,000,000 metres per second: How many times does each dot go up and down in one second?

Another way of asking that question is: how many full cycles can you fit in 300,000,000 metres (since radio waves travel 300,000,000 metres each second).

Since the blue wave has a wavelength of 2m, it'll take 150,000,000 cycles to reach 300,000,000 metres. That means that the blue dot oscillates at 150,000,000 cycles per second, or 150,000,000 Hz, or 150 Mhz
Similarly, since the green wave has a wavelength of 6m, its frequency is 50 Mhz.

So a quick way to relate the frequency \$f\$ (in MHz) and the wavelength \$\lambda\$ (in metres) is:

\$$ \lambda = \frac{300}{f} \qquad \text{or} \qquad f = \frac{300}{\lambda}\$$

Note that the reason we're using just 300, instead of 300,000,000 is that we've cancelled six of the zeros so that the frequency is in MHz instead of in Hz.

Now, here's a related question: how long does it take for each wave to complete one cycle?

For the blue wave, we know that it oscillates 150,000,000 times / second, so only one of those time would take 150,000,000^th of a second, or \$\frac{1}{150,000,000}\$ s or 6.67×10^-9 s or 6.67 ns.⁶¹⁾
Similarly, the green wave oscillates at 50,000,000 cycles per second, so only one of those cycle would take \$\frac{1}{50,000,000}\$ s or 2×10^-8 s or 20 ns.

The time to complete one full cycle is called the period (T) and is the reciprocal of the frequency:

\$$T = \frac{1}{f} \qquad \text{or} \qquad f = \frac{1}{T}\$$

Wave Addition

When two waves overlap, they add up together at every point. Here, the blue and green waves are generated and add up together to form the red wave. You can move the blue and green waves and see the result. To convince yourself that the red wave is really the sum of the blue and green waves, look at points A, B, and C. You can move the blue or green waves by sliding their phase (φ and Φ) around. You'll see that point C is always the sum of A and B.

Download waveaddition.ggb

Where do the blue and green waves need to be so that...

the red wave is the biggest?
the red wave is cancelled out?⁶²⁾

If you press the play button on the bottom left corner, you'll see the blue wave travel to the right and the green wave travel to the left. The red wave, which is the sum of the forward and reflected waves, oscillates up and down but doesn't travel anywhere, which means it's not going into the antenna.

While the animation is running, slowly decrease the amplitude of the reflected wave (V_B) and you'll see that the red wave will start moving to the right. As you do that, notice how the SWR (Standing Wave Ratio) decreases toward 1:1. At this point, there is no reflected wave and all of the energy is going to the antenna (assuming no loss in the feedline).

Modulation

Modulation is the process of “encoding” a message (be it voice or digital) onto a radio wave.

Activity

If you have access to an HF radio:

Tune in to an AM signal and notice what happens as you slowly move off frequency:
1. How does the quality of the audio change?
2. How far can you go before you can't understand the audio anymore?
Tune in to an SSB signal and notice what happens as you slowly move off frequency:
1. How does the quality of the audio change?
2. How far can you go before you can't understand the audio anymore?

AM

AM stands for Amplitude Modulation. What this means is that the transmitted radio wave is obtained by changing the amplitude of a pure radio waves (the carrier) based on an audio signal (the baseband).

For example, let's transmit a single audio note of 10 kHz at a radio frequency of 200 kHz.⁶³⁾:

Before we modulate the carrier, we raise the audio signal above zero to get an envelope:

Finally, we multiply the envelope and the carrier, which gives us a wave that has the same frequency as the carrier, but an amplitude that varies like the voice signal:

So very roughly:

AM Radio Wave = (Audio Signal + 1) × Carrier Wave

Let's pause for a minute and highlight that here, we are multiplying two waves together (not simply adding them). Later on, we'll see that the electronic component that does that is called a mixer, not to be confused with a sound mixer, which does do addition.

The incredible thing about the resulting AM broadcast is that the transmitted radio signal can also be seen as the sum of three pure sine waves:

AM Radio Wave = LSB Wave + Carrier Wave + USB Wave

LSB means Lower Side Band
USB means Upper Side Band

This is absolutely not obvious but let's see why it's at least plausible. Imagine we start with the following three waves:

An LSB Wave oscillating at 190 kHz with an amplitude of 0.5
A Carrier Wave oscillating at 200 kHz with an amplitude of 1
A USB Wave oscillating at 210 kHz with an amplitude of 0.5

Now let's add them together. This is a bit of a mess, but let's look at specific places along the waves:

at point A, all three waves align so the sum is: 0.5 + 1 + 0.5 = 2
at point B, the two side bands are opposite and cancel each other and only the carrier remains: 0 + 1 + 0 = 1
at point C, the carrier is opposite the two side bands so the sum is: -0.5 + 1 - 0.5 = 0
at point D, the same as point B is happening
at point E, the same as point A is happening

The result is the same as final AM signal:

Note that the carrier has a frequency of 200 kHz just like the original carrier, but the two side bands are 10 kHz lower and higher with half of the amplitude. Notice also how the LSB Wave oscillates slower than the Carrier Wave, while the USB Wave oscillates faster.

Frequency Spectrum

An easier way to represent a radio signal is using a spectroscope, which shows the frequency spectrum of a wave. That is, instead of looking at the signal wave itself, the spectroscope shows the strength of each frequency that makes up the sum of the signal.

For example, the spectrum of our 10 kHz note transmitted over a 200 kHz carrier would look like this:

All this is saying is that the radio signal is composed of three pieces: a signal at 190 kHz with an amplitude of 0.5, a signal at 200 kHz with an amplitude of 1, and another at 210 kHz with an amplitude of 0.5.

There are three things to notice here:

The two side bands are 10 kHz on each side of the carrier (same as the baseband signal!). It is that distance away from the carrier that represents the audio signal we want to recover.
Most of the power is going into transmitting the carrier, which in itself doesn't carry any information, so that's a bit of a waste of energy.
More fundamentally: even though we say that the signal is transmitted at 200 kHz, in this example, it is really contained between 190 kHz and 210 kHz. That is, it has a bandwidth of 20 kHz (210 kHz - 190 kHz). This bandwidth is regulated and depends on the band used.

SSB

One way of saving power (and reduce bandwidth) is to only transmit one of the side bands. In this example, the radio would be tuned to 200 kHz, but...

...for LSB, the spectroscope would look like:

...for USB, the spectroscope would look like:

By itself, neither of these transmissions would carry the information we need (that the baseband signal was a 10 kHz note) since it's the difference between the sideband and the carrier that gives us that information. But if the receiver knows that this signal was generated by a transmitter at a frequency of 200 kHz, then the receiver can re-inject the missing carrier on its side.

This is why an AM signal is not too picky about being slightly off frequency (both the sidebands and the carrier are transmitted). But a SSB signal changes pitch if the receiver is not tuned precisely to the transmitter frequency.

In reality, the voice we transmit contains a whole group of “notes” typically between 300 Hz and 3000 Hz (music could range between 20 Hz and 20,000 Hz). A typical voice signal (baseband) could look something like this:⁶⁴⁾

So the AM signal would look like this:

Notice how:

the carrier is a single line in the centre because unlike the sidebands, it is a pure sine wave of only one frequency.
the two side bands are mirror images of each, which is why it's important that both the receiver be in the same mode as the transmitter.

And finally, each individual sideband would look like this:

The two main advantages of using SSB (LSB or USB) are that:

It takes less power to transmit the same information.
It takes half the bandwidth.

Here's a screenshot of VA7FI's IC-7300 scope showing both modes on the same screen:

The radio is tuned to 3.880 MHz (where no one is transmitting), but there are two neighbouring conversations going on:

one at 3.875 MHz using LSB,
and another at 3.885 MHz using AM.

The scope shows the recent history of the radio signal (called waterfall) where the present is at the top and the past at the bottom. Blue represent a weak signal strength and yellow or red represent a strong signal strength. Here are some things to notice:

	LSB (3.875 MHz)	AM (3.885 MHz)
Symmetry	The signal is on the left (low side) of where the carrier would be (at 3.875 MHz) and varies with speech.	The signal has a strong, constant carrier in the centre, and two symmetrical sides that vary with speech.
Bandwidth	About 2.7 kHz on the low side of 3.875 MHz	About 6 kHz (2.7 kHz on each side of 3.885 MHz, with two gaps near the carrier)
Pauses	During pauses, no radio signal is transmitted.	During pauses, the carrier is still transmitted.
Relationship	An AM signal can be understood in LSB mode because it contains the lower side band required. But an LSB signal can't be understood in AM mode because both sidebands and the carrier are needed to process the signal.

FM

FM stands for Frequency Modulation. What this means is that the transmitted radio wave is obtained by changing the frequency of the carrier based on the audio signal.

For example, let's again transmit a single audio note of 10 kHz at a radio frequency of 200 kHz using FM this time instead of AM:

This time, we don't simply multiply the baseband signal to the carrier (as in AM). Instead, we “compress” and “stretch” the carrier (ie, modulate its frequency) based on the baseband signal.

Here, the math is a bit more involved and requires at least 1^st year calculus to understand but in a nutshell, if the carrier is \$ c(t) = \cos(2 \pi f_c t) \$ and the baseband signal is \$s(t)\$, then the FM signal will be:

\$$ \cos\Big(2 \pi f_c t + 2 \pi k \int_0^t s(\tau) d\tau\Big) \$$

If this looks like Greek to you, don't worry; the math isn't important. The key concept to understand is that the highs and lows of the baseband signal are encoded in the horizontal compression (the frequency) of the radio wave: When the baseband is high, the radio signal is more compressed (its frequency is higher), and when the baseband is low, the radio signal is more stretched out (its frequency is lower).

Optional Details

For those interested in some of the mathematical details, see this optional page.

⁶¹⁾

“ns” means nanosecond. “Nano” means a billionth of ___

⁶²⁾

Fun fact: This is how noise cancelling headphones work. The headset has a microphone that picks up the noise, inverts the waves, and plays them back in the ear piece. The combination of the real life noise and the inverted noise being played in the speaker cancel out (somewhat).

⁶³⁾

Note that 200 kHz is lower than commercial AM broadcast and is not a ham radio frequency. I chose a ratio of 20:1 so we can see the effects on the graph

⁶⁴⁾

The next few images are from Single-sideband_modulation

wavemodulationmath

More Details: AM / FM

Here are a few more details about the AM, SSB, and FM modulation schemes introduced on the Wave Modulation page.

For both AM and FM examples, we'll let:

\$c(t) = \cos(2 \pi f_c t)\$ be the radio carrier with frequency \$f_c\$
\$s(t) = \cos(2 \pi f_s t)\$ be the baseband audio signal with frequency \$f_s\$

With the radio carrier frequency several times greater than the baseband audio signal.

AM

The resulting Amplitude Modulated radio wave is the product of the vertically shifted baseband signal and the radio carrier, which is also equal to the sum of the carrier and the two side bands:

\begin{align*} \Big(s(t)+1\Big) \times c(t) &= \Big(\cos(2 \pi f_s t) + 1\Big) \times \cos(2 \pi f_c t) \\ &= \cos(2 \pi f_s t)\cos(2 \pi f_c t) + \cos(2 \pi f_c t)\\ &= \underbrace{\frac{1}{2} \cos\Big(2 \pi (f_c + f_s) t\Big)}_\text{USB} + \underbrace{\frac{1}{2} \cos\Big(2 \pi (f_c - f_s) t\Big)}_\text{LSB} + \underbrace{\cos(2 \pi f_c t)}_\text{Carrier} \end{align*}

In line 1, I distributed the bracket, which, in line 2, gave us the carrier (last term) and a product (first term). To expend this product into the sum of the two side bands (line 3), I added these two trig identities together:

\begin{align*} \cos(A+B) =& \cos(A)\cos(B) - \sin(A)\sin(B) \\ \cos(A-B) =& \cos(A)\cos(B) + \sin(A)\sin(B) \end{align*}

Which gives:

\begin{align*} &\cos(A+B) + \cos(A-B) = 2 \cos(A)\cos(B) \\ \Rightarrow &\cos(A)\cos(B) = \frac{1}{2}\cos(A+B) + \frac{1}{2}\cos(A-B) \end{align*}

Use this animation to see what happens when you vary the individual frequencies. You can use the check boxes to show or hide different waves.

: animation in wrong place

Download am.ggb

Some things to try:

Set \$f_s\$ at 10 and \$f_c\$ at 200 and check only the transmitted signal. You can easily imagine what the envelope (the baseband signal) should be that produced that signal. But...
Decrease \$f_c\$ slowly. At some point (around 20 or 30) the baseband signal becomes unrecoverable. This illustrates the point that to transmit a high frequency baseband, a higher frequency carrier is needed (at least 3 to 4 times the frequency of the baseband signal. This is why with digital signals, the higher the transfer speed, the higher the carrier frequency needs to be.

Mixer

Later on, we'll see that a mixer is an electronic component that multiplies two waves together, resulting in four different frequencies: \$f_1, f_2, f_1+f_2, \text{ and } f_1 - f_2\$. Although it's not modulation, the math is very similar to the way AM is created, so this is a good place to have a look at it. So let's multiply two very general waves with the following properties:

Amplitudes \$A_1\$ and \$A_2\$
Frequencies \$f_1\$ and \$f_2\$
Phases (horizontal shift) \$\phi_1\$ and \$\phi_2\$
DC components (vertical shift) \$c_1\$ and \$c_2\$

\begin{align*} \Big( A_1 \cos(2 \pi f_1 t + \phi_1) + c_1 \Big) \times & \Big( A_2 \cos(2 \pi f_2 t + \phi_2) + c_2 \Big) \\ \\ = & A_1 A_2 \cos(2 \pi f_1 t + \phi_1) \cos(2 \pi f_2 t + \phi_2 ) \\ & + c_2 A_1 \cos(2 \pi f_1 t + \phi_1 ) + c_1 A_2 \cos(2 \pi f_2 t + \phi_2) + c_1 c_2 \\ \\ = & \frac{A_1 A_2}{2} \cos(2 \pi (f_1+f_2) t + (\phi_1+\phi_2)) + \frac{A_1 A_2}{2} \cos(2 \pi (f_1-f_2) t + (\phi_1-\phi_2)) \\ & + c_2 A_1 \cos(2 \pi f_1 t + \phi_1 ) + c_1 A_2 \cos(2 \pi f_2 t + \phi_2) + c_1 c_2 \end{align*}

The last line looks like a real mess, but all it says is that the result is:

A wave with amplitude \$\frac{A_1 A_2}{2}\$, frequency \$f_1 + f_2\$, and phase \$\phi_1 + \phi_2\$
A wave with amplitude \$\frac{A_1 A_2}{2}\$, frequency \$f_1 - f_2\$, and phase \$\phi_1 - \phi_2\$
A wave with amplitude \$c_2 A_1\$, frequency \$f_1\$, and phase \$\phi_1\$
A wave with amplitude \$c_1 A_2\$, frequency \$f_1\$, and phase \$\phi_2\$
A DC component of \$c_1 c_2\$

A mixer is useful to raise or lower the frequency of a signal. For example, if a signal at 13 Mhz is mixed with a local oscillator signal at 14 MHz, two new signals will be produced (in addition to the original two): one at 1 Mhz and the other at 27 Mhz. If we want the higher one, we can put the result through a high pass filter, which will discard the unwanted signals.

FM

Mathematically, FM is less intuitive and more complicated than AM to understand. The first step is to modulate the frequency by adding a scaled baseband function to it:

\$$2\pi f_c \quad \rightarrow \quad 2\pi f_c + 2\pi k s(t)\$$

Here, \$f_c\$ is the frequency of the carrier, which is a constant (this is important),
and \$k\$ is a scaling factor we can use to decide how much of a variation we allow the baseband signal to impart on the carrier frequency. When \$k = 0\$, there is no modulation, and the greater \$k\$ becomes, the bigger the effect is.

It might be tempting to simply substitute this sum in the wave like so:

\$$ \cos(2\pi f_c t) \quad \rightarrow \quad \cos\Big(\big(2\pi f_c + 2\pi k s(t)\big) t\Big) \$$

but that's not quite right because the frequency is derived from the change in angle.

To solve this properly, we need some calculus to deduce the angle from our new frequency:

\$$ \frac{d}{dt}\theta(t) = 2\pi f_c + 2\pi k s(t) \qquad \Rightarrow \qquad \theta(t) = 2\pi f_c t + 2\pi k \int_0^{t}s(\tau) d\tau \$$

The frequency modulated transmission is actually given by:

\$$ \cos\Big(2\pi f_c t + 2\pi k \int_0^{t}s(\tau) d\tau\Big) \$$

In our particular example, with \$s(t) = \cos(2 \pi f_s t)\$, the modulated radio signal becomes:

\begin{align*} \cos\Big(2 \pi f_c t + 2\pi k \int_0^{t}s(\tau)d\tau\Big) &= \cos\Big(2 \pi f_c t + 2\pi k \int_0^{t}\cos(2 \pi f_s \tau)d\tau\Big) \\ &= \cos\Big(2 \pi f_c t + k \sin(2 \pi f_s t)\Big) \end{align*}

For more details about FM, see: http://www.ece.umd.edu/~tretter/commlab/c6713slides/ch8.pdf

Use this animation to see what happens when you vary the individual frequencies. You can use the check boxes to show or hide different waves.

: animation in wrong place

Download fm.ggb

Some things to try:

Set \$f_s\$ at 10 and \$f_c\$ at 200 and check only the transmitted signal. Notice how when the baseband is high, the transmitted wave is “tight” (ie, its frequency is high), and vise-versa. But...
Decrease \$f_c\$ slowly. At some point (around 20 or 30) that pattern becomes unnoticeable. Again, this illustrates the point that to transmit a high frequency baseband, a higher frequency carrier is needed (at least 3 to 4 times the frequency of the baseband signal. This is why with digital signals, the higher the transfer speed, the higher the carrier frequency needs to be.
Increase and decrease k to see the effect it has on the transmitted wave. The greater k, the more bandwidth the resulting signal uses. This dictates the difference between “Narrow Band FM” and “Wide Band FM”.

See https://electronicspost.com/narrow-band-fm-wide-band-fm/

PM

Phase Modulation is not usually discussed in ham radio courses, but after understanding FM, we pretty much get PM for free... Recall that for the wave \$\cos(2\pi f + \phi)\$, \$f\$ is the frequency and \$\phi\$ is the phase shift. For a pure tone, both of these are constant.

With FM, we saw that modulating the frequency led to \$\cos\Big(2 \pi f_c t + 2\pi k \int_0^{t}s(\tau)d\tau\Big)\$.
With PM, it leads to \$\cos\Big(2 \pi f_c t + 2\pi k s(t)\Big)\$

Essentially, with PM, we simply let \$\phi\$ vary with the baseband \$s(t)\$. But the thing to notice is that PM looks a lot like FM. In fact, an FM signal modulated by \$s(t)\$ is the same as a PM signal modulated by \$\int_0^{t}s(\tau)d\tau\$. In other words, the receiver needs to know if the signal was modulated in FM or PM since both wave forms look similar.

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