How does single-sideband (SSB) work theoretically? If a theoretical SSB transceiver is a "black box" and only its inputs and outputs can be analyzed, not the way it works internally, what is happening?
6 Answers
It's quite simple, really. If we're describing a black box, we will have to describe the blackbox in terms of the things going in and out. So, we start by giving them names:
| Signal | Description |
|---|---|
| $m(t)$ | Message signal in time domain – the audio to be transmitted |
| $M(f)$ | Message signal in frequency domain – spectrum of the audio |
| $s(t)$ | Transmitted (passband) RF signal in time domain |
| $S(f)$ | Transmitted (passband) RF signal in frequency domain |
Note that $s$ and $S$ are the same signal – just that the first describes the signal as how it is over time, and the other how it is over frequency. Both representations contain the exact same information – you can convert between them back and forth as you want, using the Fourier transform. The same is true for $m$ and $M$.
What every mixer does is simply shift a signal in spectrum. What that means is that it takes some signal (like our message signal $m(t)$ / $M(f)$, for example) and moves it to a different frequency (same example: it simply makes a different signal $Q(f) := M(f+f_{\text{mixer}})$). That's it. If we do that to our message signal (audio) with an RF frequency, we end up with AM with a "suppressed carrier" (there's no carrier, but it's the name hams tend to give the mode). But that's a double-sideband AM:
Because the message signal $m(t)$ is a real-valued signal (i.e., the pressure at the microphone, and consequently the voltage in your mic amp, are real numbers at any point in time), the spectrum is (hermitian) symmetrical to the $f=0$ line – that's a direct consequence of how the Fourier transform defines what the spectrum is. By shifting it up to some $f_{\text{mixer}}$, we shifted both the positive frequencies to $f_{\text{mixer}} + \text{something}$ and the negative frequencies to $f_{\text{mixer}} - \text{something}$.
So, all we need to do is get rid of everything below $f_{\text{mixer}}$ to get the upper sideband or everything above $f_{\text{mixer}}$ to get the lower sideband modulation.
There's multiple ways to represent that in a blackbox model: we can say that after mixing, a low-pass filter with cutoff at $f_{\text{mixer}}$ is convolved with the time-domain signal for USB (high-pass for LSB); we could say the spectrum is multiplied with a mask (which is the same filtering operation).
The way "modern" (read: after 1940) communications technology would write that is probably that instead of "killing" half of the transmit RF signal after mixing, you just "kill" half of the message signal: A complex filter can filter out the negative (for USB) or positive (for LSB) half of the spectrum, before mixing.
So, for me, I'd write the blackbox model of SSB:
$$s(t) = \cos (2\pi f_{\text{mixer}}) \cdot \left(h_{\text{complex half-band filter}}*m(t)\right),$$ where $*$ is the convolution operation and $h$ is the above-mentioned "kills all negative (or positive) frequencies" filter's impulse response.
Technologically, that's how I'd actually implement the SSB modulator, myself. Because: Having a sharp filter at RF is harder (both if you want to build that as analog filter, and if you want to implement that as digital filter in a microcontroller, DSP, FPGA or ASIC) than at low frequencies. (Generally, "hardness" of a filter is pretty well-described by how quickly it goes from passband to stopband, relative center frequency it works at.)
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Isn't this mostly explaining how transceivers work, when the question is explicitly not about actual transceivers ? – Someone Jan 01 '23 at 23:21
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No, the mixer part is quite universal, but this specifically answers the Blackbox model of SSB and little else. General part was just necessary for explaining the definitions. – Marcus Müller Jan 02 '23 at 07:57
It's extremely simple. Simpler than AM.
If you take a X Hz audio signal and feed it through an SSB transceiver set to upper sideband (USB) and 14.3 MHz, you get a (14300000 + X) Hz RF signal.
You can modulate the X input, and the output of the black box will match the modulation of X, but be offset by 14300 kHz in frequency. If your audio is the superposition of lots of frequencies, the resulting RF will also be a similar superposition. e.g. X Hz audio + Y Hz audio gets converted (14300000 + Y) Hz superimposed on top of (14300000 + X) Hz RF. And so on for much fuller audio spectrums becoming richer RF spectrum, but with similar bandwidths (usually limited to 2.7 kHz in amateur HF bands).
Your blackbox can use any of complex modulation, FFT hacking, subtraction filtering of AM, a mass of symbols on your math class chalkboard, or some form of black magic, to achieve this result.
SSB only seems more complicated than AM because ancient radio technologies did not have enough GFLOPS of DSP processing resources to do it the simple way, shift a spectrum uphill.
Added: for lower sideband (LSB), X Hz audio gets converted by the black box into (transceiver_frequency_dial_setting - X) Hz RF. e.g. mirrored/upside-down stuff slightly below the transceiver's dial frequency.
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well as you say, it's hard to assign a "hardness" to a black box, but your description is indeed missing the aspect that the original audio signal is DSB, not only the positive frequencies. – Marcus Müller Jan 02 '23 at 21:57
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A 1 kHz audio signal is a 1 kHz audio signal, not a -1 kHz audio signal. The frequency of an audio spectrum is described differently than the frequency of audio (what you set on the dial of your tone generator box), and can include stuff that doesn't exist in the real world. – hotpaw2 Jan 02 '23 at 22:00
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1the $e^{+jx}$ in $\cos(x) = \frac12(e^{jx} + e^{-jx})$ doesn't go away just because you decide to look at only the positive frequencies of an audio cosine; looking only at the positive frequencies only works because the signal is real and thus spectrally symmetric! So, yes, your 1 kHz tone ($x=2\pi 1000\text{ Hz} t$) is also a -1 kHz tone. – Marcus Müller Jan 02 '23 at 22:03
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No one tunes their guitar to e^-jx, even if it shows up that way on some chalkboard. Two different domains. – hotpaw2 Jan 02 '23 at 22:07
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you're right, people tune their guitar to actually correlate maximally with a tone, which again has positive and negative frequency components. Yes, it's a 1 kHz tone, but that doesn't make it any less of a -1 kHz tone. And: if the math done by any cheap digital guitar tuner didn't account for both the real and imaginary part (and thus, negative and positive frequencies separately), that tuning would end up being phase-dependent (and would change depending on when you plucked the string relative to the "reference oscillator"); – Marcus Müller Jan 02 '23 at 22:10
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I note your own answer above does not say the the radio also transmits an equal amount of RF at -14.3 MHz. – hotpaw2 Jan 02 '23 at 22:12
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note that multiplying with a cosine is shifting one copy up to the positive cosine frequency, and one down to the negative frequency – Marcus Müller Jan 02 '23 at 22:15
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Only for signals that are asymmetric with reference to whatever you consider to be T0. – hotpaw2 Jan 02 '23 at 22:17
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I don't understand that sentence – any real signal is always symmetrical to f=0 in frequency domain. There's nothing there "relative" to any T0. – Marcus Müller Jan 02 '23 at 22:18
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If your signal is both real in time and symmetrical to t=0, then it's not only symmetrical in spectrum, but also purely real in spectrum. But that's a very specific case that we do not need to care about here :) – Marcus Müller Jan 02 '23 at 22:20
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Actually, professional piano tuners do tune by listening for beats, the phase dependance against the reference oscillator. And using fairly expensive, not cheap, tuning forks and strobe devices. – hotpaw2 Jan 02 '23 at 22:22
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heh, that's a good illustration of positive and negative frequencies that you mentioned there with the strobe ;) remember vinyl disk players with their arrays of mirrors at the edge of the turntable? Say, there's 100 mirrors equally spaced on that edge, and say the nominal rotational rate of the LP is 30 RPM = 0.5 Hz. If you strobe a shining at a small spot at the turntable edge exactly at 50 Hz, you would always hit the mirrors in the exact same spot, right? Now, say the rotation is a little slow, 0.4995 Hz, then the spot the light shines at on each mirror wanders a bit in one direction, – Marcus Müller Jan 02 '23 at 22:28
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I think math professors sometime are confused that because a certain formulation is far more elegant, and very importantly uses far less chalk on the chalkboard, when using complex exponentials instead of a mess of sines and cosines, that only their form of description is correct. – hotpaw2 Jan 02 '23 at 22:32
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and if it's rotating a bit too fast, say at 0.5005 Hz, it also wanders a bit, but in the other direction. Now, we both know exactly that the mirror positions start repeating every 20 s! But the cyclic wandering goes in opposite directions. So, the person adjusting the turntable speed can directly infer whether they need to speed up or slow down the motor, simply from the wandering direction – the frequency difference is positive or negative. We can assign this notion to the frequency of the light phenomenon itself. That's the reason strobe instrumentation are great for tuning instruments: – Marcus Müller Jan 02 '23 at 22:32
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1you see positive and negative frequencies – they're not some poor professor's imagination, they're right there, in front of your eyes, flashing as the little dots on the edge of your turntable. I think non-mathy people just don't like the idea of negative frequencies, because they consider that to be unelegant, but actually it's a very easy and intuitive thing to look at in real life, more than on a blackboard ;) – Marcus Müller Jan 02 '23 at 22:34
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(by the way, I don't consider you to be un-math-y; I think your arguments here show quite the opposite! It's just that the notion of "no, all real-world frequencies are positive" make the real world harder to grasp for me, because I tried working with that definition; I'm quite fine with saying "whenever I observe a real signal, there's negative frequencies", though, once I stopped caring about whether "negative frequencies" felt bad.) – Marcus Müller Jan 02 '23 at 22:36
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You seem to be confusing frequency with frequency differences. Two people with different heights do not have negative heights on their medical records. Even when comparing them back to back. – hotpaw2 Jan 02 '23 at 22:36
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A difference between lengths is still a length, and can have positive or negative sign. That's really the point here. Defining a length could only be positive suddenly says "when I subtract a length in meters, I need to introduce a sign check and if its negative, I need to make it positive"; that's more complicated than saying the "height difference between the heighest summits of the Alps and summits of Himalaja is -6km"; and especially in electronics, you see this a lot – depending on which direction I measure the voltage of a battery, I get +1.5V or -1.5V. – Marcus Müller Jan 02 '23 at 22:37
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1I sell strobe tuner software. The new apps use complex FFTs. The really old versions just used strictly real sine and cosine math. The new ones use far less code, and the code is far more readable. BUT, they worked equally well. – hotpaw2 Jan 02 '23 at 22:40
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isn't the FFT a beautiful illustration of how your instrument strings produce both positive and negative frequencies? Sure, you ignore (or even avoid computing) the negative frequencies in any tuner application, because you know they're "the same" as the positive ones, but your analysis method gives you exactly that: both :) – Marcus Müller Jan 02 '23 at 22:42
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That's one possible description (likely preferred by anybody who's studied complex algebra), but not the only description (as typically used by "normal" humans). The assumption is that the OP is part of the majority of humans. – hotpaw2 Jan 02 '23 at 22:46
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and the thing also matches reality: if you take a microphone, add a preamp, add a DC offset and then multiply that with a cosine tone, you get DSB, not SSB, exactly as FFT of your audio signal foretold! – Marcus Müller Jan 02 '23 at 22:47
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1The OP likely wanted a description that did not involve implementation details, such as what type of multiplication might be involved. – hotpaw2 Jan 02 '23 at 22:49
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And that even works within the audible range, right? If you make a 200 Hz tone, and multiply it with a 700 Hz tone, you get an FFT that has components at -900 Hz, -500 Hz, + 500 Hz and 900 Hz. All is consistent if, and only if, your original 200 Hz tone had a negative frequency component at -200 Hz. (if the -200 Hz wasn't in there, where would the 500 Hz come from?) – Marcus Müller Jan 02 '23 at 22:50
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The OP likely wanted a description… you're right! But the black box model needs to deal with the input as it is – and the input has negative frequency components and will lead to DSB when just directly being upconverted. That's where my original comment under your question came from, exactly that blackbox aspect :) – Marcus Müller Jan 02 '23 at 22:51
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Might just be a mix or real cosine waves and real sine waves, no imaginary signal generator boxes required. – hotpaw2 Jan 02 '23 at 22:54
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I'm not saying you need to calculate this with complex sinusoids; I'm not even claiming it's more elegant. It's only that "shifting something in frequency" works perfectly well in the real world, and makes the fact visible that the original audio signal had negative frequencies. – Marcus Müller Jan 02 '23 at 22:58
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you certainly could :) though I think we're about through, in a positive way! – Marcus Müller Jan 02 '23 at 23:37
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You posted a mathematical identity, one side used negative numbers, one side did not. How can only one side of an identity be the correct description? – hotpaw2 Jan 03 '23 at 00:50
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1Comments are not for extended discussion; this conversation has been moved to chat. – David Hoelzer Jan 03 '23 at 14:29
I won't read through all the lengthy answers which seem to attempt to explain SSB well outside of your black-box question. So if SSB were a black-box problem, the answer is, an audio signal drives a radio frequency signal off of its "base" or "center" frequency, by whatever the audio frequency signals frequency is, in the direction determined by whether the black-box is outputting upper sideband, or lower sideband.
Disregarding the electronics to make this happen, as an example, if you have a 1 MHz (1,000,000 Hz) radio signal and you impose a 1kHz (1,000 Hz) audio tone on that radio frequency, and the black-box is outputting on upper sideband, your output of that black-box will be 1.001 MHz (1,001,000 Hz). By the same token, if your black-box were set to lower sideband, the signal exiting the black-box would be 0.999,000 MHz, or 999,000 Hz. Upper sideband drive the "base" or "center" frequency up, by whatever the audio frequency is, and lower sideband drive the "base" or "center" frequency down, by whatever the audio frequency is. It is also the reason that sideband is so sensitive to precise tuning, as the audio frequencies coming out of a receiver's speaker, are the actual radio frequency (minus) the "base/center" frequency of the radio signal. So if you are tuned to 1,000,500 Hz, a 1,000 Hz upper sideband signal being transmitted on a base/center frequency of 1,000,000 Hz will actually sound like a 500 Hz tone, because your receiver is tuned to the wrong base/center frequency.
Furthermore, this is why if someone is communicating on upper sideband where the "base" or "center" frequency is 1 MHz, they should not interfere with someone else on lower sideband using 1 MHz as their "base" or "center" frequency; also, it is why many describe single sideband communications as being able to "double" the number of "channels" that can be used in a given band-plan, over AM communications, since AM by its nature produces both upper and lower sidebands as a result of its modulation technique, and takes up twice the bandwidth.
Final note on the black-box. Audio frequency is only one component that is needed for intelligible analog audio, the other aspect that is needed in analog communications is amplitude (power level), or changes in "loudness"; so the amplitude of your output sideband signal will not only have a frequency, but it will have an amplitude as well, and that amplitude (power level) is directly proportional to the "loudness" of the audio frequency on the input of the black-box; i.e., the stronger/louder the input audio, the higher the voltage (power level) output of the sideband signal of the black-box.
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Single-sideband consists of two separate but similar modes: upper sideband (USB; not to be confused with the Universal Serial Bus computer protocol) and lower-sideband (LSB). I will describe USB first.
What is USB?
A theoretical USB transmitter is merely an upconverter, and a theoretical receiver is a downconverter.
A USB transmitter takes the baseband (a.k.a. audio [frequency] or AF) signal and shifts its frequency up to the desired output frequency, and transmits it on the radio-frequency (RF) output to the antenna. If the baseband signal is 3 kHz wide, a typical bandwidth for amateur SSB, it ranges from 0 to 0.003 MHz (3 kHz = 0.003 MHz). If the radio is set to 14.300 MHz, the RF signal will range from 14.300 to 14.303 MHz.
A receiver does basically the same thing, but in reverse. If you set the dial to 14.300 MHz, it will take the range of frequencies from 14.300 to 14.303 MHz from the RF input from the antenna and downshift them to 0 to 0.003 MHz, creating the baseband audio signal which you hear. When you listen to a USB signal, you are actually listening to the radio waves, just shifted down so you can hear them.
The reason that I specified "theoretical" receiver and transmitter is that real ones do a lot more than shift frequencies, such as amplify the signal and do processing to improve intelligibility (such as speech compression on the transmitter and noise reduction on the receiver).
What about LSB?
Lower sideband (LSB) is similar, except that the signal is reversed. I will only describe the transmitter; as with USB, the receiver is basically the same, except in reverse.
As with the USB example, I will be using a 3 kHz wide signal with the radio set to 14.300 MHz. (In reality LSB is not normally used on 20 meters, but it simplifies the explanation to use the same frequency. There is no technical reason why LSB cannot be used on 20 meters with modern transceivers, so the explanation is not affected.)
We have the same baseband signal ranging from 0 to 3 KHz, or 0 to 0.003 MHz. As with USB, the frequency of 0 Hz in the baseband signal corresponds to the radio frequency to which the transmitter is tuned, 14.300 MHz in this case. However, the frequency of 0.003 MHz in the baseband signal does not correspond to 14.300+0.003=14.303 MHz, but to 14.300-0.003=14.297 MHz. Notice that this is below the frequency corresponding to 0 Hz in the baseband signal. Thus, LSB inverts the frequencies of the baseband signal in addition to upshifting it. If you look at a spectrogram of the baseband and RF signals, the RF signal from 14.297 to 14.300 MHz will be a mirror image of the baseband signal from 0 to 0.003 MHz.
Why can't I transmit directly on the band edges?
Consider an Amateur Extra-class licensee operating on 20 meters. The phone segment of 20 meters for Extra licensees is from 14.150 MHz to 14.350 MHz. If I transmit on 14.350 MHz upper-sideband, however, my 0.003 MHz-wide signal will actually extend from 14.350 to 14.353 MHz, putting it out of band. If I am transmitting a 3 KHz-wide signal on upper-sideband, I should set my dial frequency no higher than 14.347. In reality, I should leave a "buffer" of several kilohertz to allow for mistuning of the radio and other issues that may cause transmissions on unexpected frequencies.
The issue is reversed with LSB. If I am transmitting on 14.150 MHz LSB, my signal will extent from 14.150 to 14.147 MHz, putting it in the digital/CW segment rather than the phone segment. I must set the dial to no lower than 14.153 MHz; again, I should leave a larger buffer.
Which mode should I use?
Normally, LSB is used below 9 MHz and USB is used on higher frequencies. This is because it was historically simpler to build USB transceivers above 9 MHz and USB on lower frequencies. This is no longer relevant with modern transceiver design; most transceivers will work just fine on the "wrong" sideband, although you will, of course, not be able to talk to anyone unless they are also using the wrong sideband. To avoid this issue, it is still conventional to use LSB on 40 meters and lower bands and USB on 20 meters and higher.
The exception is for digital modes. For sound-card digital modes such as RTTY, PSK31, and FT8, the radio is normally set to USB on all bands.
Is this what's actually happening in my transmitter?
I'm not familiar enough with how modern transmitters work to answer this definitively, but I do not believe that actual transceivers, except possibly SDRs, actually work this way. The design with which I am familiar creates a double-sideband full-carrier amplitude-modulated signal (a.k.a. normal AM) and filters out the carrier and unwanted sideband. However, I am not sure that this is the design used in modern transceivers.
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3The issue with viewing USB modulation as pure upconversion is that when you're upconverting from 0, you can't disentangle the "positive" from the "negative" frequencies, so you end up with both sidebands unless you do something else interesting (and a brickwall highpass filter at the suppressed carrier frequency isn't a practically realizable option). – hobbs - KC2G Dec 31 '22 at 23:59
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1@hobbs-KC2G isn't that an "implementation detail" of the transmitter, not a part of the modulation itself? i.e. if the transmitter is a black box and you're only analyzing its inputs and outputs, the negative frequencies don't matter, right? – kj7rrv Jan 01 '23 at 03:49
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4If AM is a modulation, how is SSB (half of AM) not a modulation? This answer defines SSB without defining modulation, and without explaining how this doesn't fit that definition. – user10489 Jan 01 '23 at 09:22
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1@user10489 you're right; I'll change the question to "How does single-sideband work?" – kj7rrv Jan 01 '23 at 17:56
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1Hey coming to chime in here: neither USB not LSB is pure up conversion, as explicitly claimed! Unless the baseband that you up convert is already only positive or only negative frequency contents. So, you should really change that statement :) – Marcus Müller Jan 02 '23 at 08:00
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@MarcusMüller isn't a typical baseband signal going to have only positive frequencies? – kj7rrv Jan 02 '23 at 18:50
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@MarcusMüller so it could be understood as shifting the positive frequencies up and the negative frequencies down? I don't really understand this either; the OP's explanation seems accurate to me based on my understanding. – Someone Jan 02 '23 at 20:15
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I thought that negative frequencies didn't even exist but were just a useful fiction in digital signal processing. When i tune my radio to 14.074 MHz, am I actually transmitting on 14.074 and -14.074? – Someone Jan 02 '23 at 20:21
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2@Someone but you cannot just "shift one half down and the other up"; that's two separate "filtering away a half + shifting it" operations. OP's description is plain wrong – the baseband is not just positive; yes, when you transmit at 14.074 MHz, you're actually transmitting at both positive and negative frequencies. When you do a Fourier transform of the signal coming out of your microphone, you'll see frequencies between - 20 kHz and + 20 kHz, which happen to be symmetrical to 0 Hz, because both the RF signal is inherently real-valued, and the audio signal is inherently real-valued and, – Marcus Müller Jan 02 '23 at 20:48
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thus, symmetrical in spectrum. This is really signals & systems basics – only complex-valued signals can be asymmetrical in spectrum. – Marcus Müller Jan 02 '23 at 20:48
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"negative frequencies are just useful fiction": um, no; at least not how you usually define "frequencies" and "spectrum": Let's look at the function $\cos (2 \pi f t)$ with some positive $f$: Intuitively, we know how many periods per time unit that has, right. If we use Euler's Formula, we know that this is $$\cos (2 \pi f t)=\frac12\left(e^{+j2\pi f t}+ e^{-j2\pi f t}\right);$$ and we know that the term with the positive exponent describes a circle for every full increase of $t$ by one counterclockwise, whereas – Marcus Müller Jan 02 '23 at 20:59
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the term with the negative exponential term is performing a circle in the same time – but clockwise. Hence, both have a frequency of $|·| = 1$, but one with a positive frequency, and the other with a negative frequency. No DSP involved at all, just the basic definition of trigonometric functions according to some 18th century Swiss guy! Now here's the funky thing about the Fourier transform: It takes a function, and decomposes it into $e^{j 2 \pi f_? t}$ terms, and gives back a new function that tells you for every (negative or positive, separately!) $f_?$, how much amplitude and phase – Marcus Müller Jan 02 '23 at 21:03
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the $e^{j2\pi f_?t}$ has. That's especially easy for the $\cos$ine above – we can see that we have phase 0 (for $t=0$, the exponent is $j\cdot 0$), and amplitude $\frac 12$ (the factor before); so the Fourier transform of the cosine is just two peaks at $+f$ and $-f$, and zero everything elsewhere. For the sine, it's the same positions, but the amplitude happens to be $\frac1{2j}$ and $-\frac1{2j}$. That's it – no DSP at all! And since Fourier promises us that we can decompose any "nicely behaving" real signal into a (even) time-symmetric, i.e. cosine-coefficients, and an odd-symmetric, i.e., – Marcus Müller Jan 02 '23 at 21:06
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sine-coefficients, part, we see that you cannot have anything on the positive frequencies if you've got nothing on the negative frequencies, and vice versa, as long as your signal is real-valued. – Marcus Müller Jan 02 '23 at 21:09
USB Upper sideband tx and RX is AM with the lower side band auttuated.
Background
When a fm or am signal is plotted the carrier frequency is a large peak in the middle this falls down and platues and it forms two "wings"
These wings are the result of
Upper = audio + carrier
Lower = audio - carrier.
These sidebands carry the goldmine of information. People figured out In single side band transmition not only you filter out the unwanted band but also subdue the carrier as well.
USB = lower band missing
LSB = upper side band missing.
Two mixers are involved with this your IF mixer and your tunning mixer. Your tunning VFO, mixer and filter are the only thing needed to demodulate a ssb signal.
This is why if you are to high the operator sounds like donald duck but if you are low, it sounds like darth vader.
To answer your question ssb is the result of modulation (am maybe) with a the unwanted sideband severly dimished and the carrier taken down with the filtering in place.
In elder radios like drakes and swans it was acomplished by crystal filters now adays we use DSP to carryout what a swan can do.
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This is one of the cases where an acronym has two completely different meanings. In the amateur realm USB is (upper side band) and LSB is (lower side band). To computer people USB is (universal serial bus).
Note: When single-sideband transmission is used in amateur radio voice communications, it is common practice that for frequencies below 10 MHz, lower sideband (LSB) is used. Conversely for frequencies of 10 MHz and above, upper sideband (USB) is used. The advantage of using the sidebands was that more peak power could be put into the transmission envelope getting better reception at longer differences. This was even implemented in the CB transceivers such as the Browning Eagle (that should take you way back). Note the Browning was in two boxes, i.e., separate transmitter and receiver. It had a distinctive ping when turning on the transmitter if the antenna relay timing was a bit off. This ping was a pseudo trademark of the Browning. They were available from about 1964 to about 1978 (dates just an old memory not fact)
Amateurs were limited to about 1 kW depending on band licence etc while CB was limited to 5W (cough cough). These power limitations helped make SSB (single side band) popular.