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I'm trying to learn about how SSB modulation works. My current understanding from reading and looking at spectrum plots from SDR software is that the sound is modulated with the carrier, but only half the carrier wave. (Alternatively, the sound's 0-point is offset either plus or minus a bit, and then modulated against the half-wave carrier.) Then, filters are used to remove the actual carrier, resulting in a much more efficient transmission at the expense of some bandwidth. Is this remotely accurate?

In any case, it seems clear that we're only using half of bandwidth. Does that mean we can use the other half, if we wanted to?

What I'm wondering is if it's possible to modulate a signal using both side bands but in a differential mode.

Differential Signaling

(Image from https://en.wikipedia.org/wiki/Differential_signaling.)

In professional audio applications, audio signal is sent down balanced cables twice, with one side inverted. When one side is then inverted on the receiving end and added to the other side, any noise that affected both sides in the same way is cancelled out.

If both side bands could be used simultaneously by the same transmitter, perhaps the lower side band could have the regular audio signal modulated on it, with the upper side band having the inverted audio signal modulated on it. Then the receiver could flip one side and add them together, eliminating a lot of noise.

Does this sort of modulation exist? Is it feasible? If not, what parts of SSB am I not understanding?

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    $\begingroup$ Unfortunately, there's no reason to assume that the noise added to one sideband has anything in common with the noise added to the other. $\endgroup$ – Dave Tweed N3AOA May 8 '16 at 22:15
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My current understanding from reading and looking at spectrum plots from SDR software is that the sound is modulated with the carrier, but only half the carrier wave.

That's — not really accurate. A SSB signal uses half the bandwidth of a double-sideband (whether AM or DSB-SC) signal, but there's no sense in which it uses "half the carrier" in the modulation process.

Since you're starting from a SDR perspective, let's talk DSP and complex (IQ) signals. Any real signal, such as that obtained from your microphone, has, from the complex-signal perspective, a spectrum which is symmetric about zero Hz. All methods of generating a SSB signal are essentially methods of removing one half of that symmetric signal. Some of them are more like “filter it out” (complex bandpass filter) and some of them are more like “synthesize a signal which cancels out only the unwanted half” (phasing), but they all are taking what is “really” a symmetric signal and making an asymmetric signal out of it.

(Side note: In DSP, there is no reason to think about “carrier waves” as an essential ingredient. Because we can work at complex baseband, we can work with (de)modulation completely separately from frequency-shifting, whereas electronic circuits largely have to use nonzero “intermediate frequencies”. On the other hand, the “carrier frequency” is meaningful — it's a convenient reference, it's what we shift down from or up to vs. a baseband signal.)

In any case, it seems clear that we're only using half of bandwidth. Does that mean we can use the other half, if we wanted to?

Yes, but:

  • If you use the other half for an additional independent signal, then you're just transmitting USB and LSB with the same carrier frequency.

  • If you use the other half for the same signal, then you're transmitting DSB-SC (double sideband, suppressed carrier) which is exactly the same as AM without a transmitted carrier wave. DSB-SC has most of the same properties as SSB, but of course does use the double bandwidth.

  • If you use the other half for an inverted (180° phase shifted) signal:

What I'm wondering is if it's possible to modulate a signal using both side bands but in a differential mode.

Not usefully. As you mentioned, differential signaling is a technique to use against “noise that affected both sides in the same way”, but in the case where it's useful, you have two signals at the same frequency traveling on different wires. Here, you have only one “wire” (empty space), and the two signals you propose are at different frequencies. And that's the problem: in most cases, the RF noise affecting the upper sideband will not be the same as the noise in the lower sideband.

(And because the receiver must receive twice the bandwidth of SSB, it receives additional noise as well, so it has a worse signal-to-noise ratio than doubling the power of a regular SSB signal.)

This mode would help with some types of noise sources — for example, broadband impulses like those a noise blanker is used against, but that type of noise isn't common enough to motivate modes for its sake.

But if you and a friend have open-source SDR transceivers, maybe it's worth experimenting with! Just remember that you're using twice the bandwidth, and don't use it on a crowded band.

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  • $\begingroup$ This seems like a long-winded way of saying the same thing I did in my comment ... but +1 for making the effort. One minor nit: transmitting DSB-SC is not exactly equivalent to transmitting SSB at twice the power, because you're also admitting twice as much noise bandwidth into the receiver. $\endgroup$ – Dave Tweed N3AOA May 9 '16 at 2:08
  • $\begingroup$ @DaveTweed Fixed the mistake. I admit to being inspired by your comment, but I felt it needed some explication. And if you thought what you wrote is a sufficient answer, why'd you put it in the comment box instead of the answer box? :) $\endgroup$ – Kevin Reid AG6YO May 9 '16 at 3:36
  • $\begingroup$ I was hoping to elicit more details from the OP about his proposed scheme. But your point about the DSB-SSB tradeoff is still not quite right: You double the signal power, but the noise power only goes up with the square root of the bandwidth -- by 1.5 dB. (Remember, the signal power is correlated, but the noise power is uncorrelated.) So, SSB at double the power has a 1.5 dB SNR advantage over DSB. $\endgroup$ – Dave Tweed N3AOA May 9 '16 at 4:40
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    $\begingroup$ sigh I knew that. Sort of. Fixed. $\endgroup$ – Kevin Reid AG6YO May 9 '16 at 4:49

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