With typical linear digital modulation schemes, we can, for a given bit rate, make a trade-off between modulation bandwidth and modulation order (effectively, how close the constellation points are).

For example, let's say I require a bit rate of 1 kbps. I can transmit with 4QAM, with two bits per symbol, yielding a symbol rate of 0.5 kbps and a bandpass signal bandwidth of around 0.5 kHz. Now suppose I instead used 8QAM, with three bits per symbol, allowing me to decrease the symbol rate to 0.33 kbps with a correspondingly smaller bandwidth. Of course, the trade-off I make is that the 8QAM constellation points are closer together, and I will have a higher bit error rate for a given SNR. Though we should note that since the 8QAM system uses a smaller bandwidth, it will also have a higher SNR, all things being equal. In theory we can keep increasing the modulation order to obtain an arbitrarily low bandwidth, provided we have enough Eb/N0 to support the modulation.

So my question is: can we make a similar trade-off with analog modulation schemes? The most obvious method that comes to mind is going from double-sideband AM to single-sideband AM, effectively halving the required bandwidth. But can we go further like we can with digital modulation?

For example, suppose I want to transmit an audio signal with 5 kHz bandwidth. Using SSB-SC modulation, I can transmit using only the 5 kHz. Is there any way to reduce the bandwidth requirement any further using some analog encoding scheme?

I can see problems arising when frequency modulation is used, since it is not a linear modulation scheme (and technically has infinite bandwidth anyways). But I don't see any reason why analog amplitude and phase modulation shouldn't both be fair game as they are for digital modulation.

  • \$\begingroup\$ QAM uses changes in the carrier's phase and amplitude to convey information. These are characteristics of an analog signal. Which types of modulation are you categorizing as "digital" and which "analog"? \$\endgroup\$
    – Sotto Voce
    Aug 7 at 5:49
  • \$\begingroup\$ @SottoVoce When people say QAM, they refer (or at least I do) specifically to encoding a digital bit stream into symbols which in turn correspond to phase and amplitude modulation of the carrier. So QAM specifically is a digital scheme. Though yes, analog modulation is merely the same thing except instead of symbols as data, we have a time- and amplitude-continuous waveform as the data. \$\endgroup\$ Aug 7 at 5:54
  • \$\begingroup\$ Though I'm now reading the Wikipedia page for QAM which notes that both digital and analog QAM exist... so I concede that the term can refer to both digital and analog. \$\endgroup\$ Aug 7 at 5:59
  • \$\begingroup\$ Maybe an analogue version of MP3 encoding? \$\endgroup\$
    – Andy aka
    Aug 7 at 8:52

3 Answers 3


AM and its reduced power and bandwidth version SSB were intended to transfer audio signals as is or at least as microphones captured them. The modulation does not at all worry what content the audio possibly contains for human beings. Everything which fits to the bandwidth limits is transferred. The transmission quality is as good as the noise and the quality of the devices allow.

Modulation schemes used in data transmission are thoroughly different. There's a limited set of discrete symbols. Only those symbols can be transferred. In QAM the symbols are presented with phase angle and amplitude. Even a narrow bandwidth can be enough if the noise is low enough and there's not too many symbols per a second. The sent symbol stream can be reconstructed well in the receiving end.

Audio signals in AM (or SSB) transmission channels do not have any symbol set limitation. A person can create neverherdbefore sounds and they are transferred if they fit to the bandwidth. Reducing bandwith causes a new limitation to the transferrable sounds. The only way I see it's possible to reduce the bandwidth without causing new limitations for the possible sounds is to record the sound and send it via AM with lower speed. Think it as playing a vinyl record by having a reduced rpm. In the receiving end the signal is re-recorded and played fast to restore all sounds.

In early military communication with AM (even the 1800's landline telephone was such system, carrier frequency only happened to be 0Hz) the messaging reliability was tried to get enhanced by allowing only sounds which can be written as text and preferably a certain, predefined set of words. The limitation was very useful. But that's a limited set of symbols and a jump to digital communication i.e. data transmission.

  • 2
    \$\begingroup\$ This seems to have the tone of disproving the idea of analog bandwidth tradeoff, but stops short of a proof, and that "The only way I see" seems to be pulling an awful lot of weight. Could you find some references to support this assertion? \$\endgroup\$ Aug 7 at 14:10
  • \$\begingroup\$ AM does not change the spectrum of the signal - it adds only a copy and possibly a carrier, if it's not SSB. The speed reduction is a scaling for time only and it's not in contradiction with the other aspects of the "keep the spectrum" idea. I cannot prove that nobody will ever find another transformation which makes possible to transfer EVERY possible voltage vs time function within a given bandwidth through a channel which has lower bandwidth. Some on statistics based compression is imaginable, but it's not valid for every possible signal. The already mentioned "analog MP3" is the same idea. \$\endgroup\$
    – Ormand
    Aug 7 at 15:52

I think I see the issue here:

The apparent paradox is not that more information can be transmitted at once with denser coding, but that less information can be received from the signal, when transmitted through a given code.

Consider the digital signal not as a bitstream, but an analog signal summed with quantization noise. When we choose a bit depth, or constellation or whatever, we are adding an LSB's noise onto the signal at the outset!

Note that this works perfectly well all the way down to 1 bit (e.g. PWM, sigma-delta modulation). Often, a simple analog filter (or digital equivalent) suffices to recover such a signal; quantization noise is either small relative to the signal bandwidth by way of a large symbol rate to signal bandwidth ratio (i.e. Fclk >> BW), or confining the quantization noise to high frequencies ("noise shaping").

If we grow the number of symbols, we reduce quantization noise, and can trade that for bandwidth; but we still have the same underlying signal, and we cannot sample that signal lower than Nyquist, regardless of method.

If we are required to convey a signal of given bandwidth and SNR, we would've failed manifestly with an overly-coarse digital method (unless it's clocked fast enough to filter down in turn, in which case the bandwidth spreading is considerable); or using large enough symbol sets, to basically mimic the analog signal as-is (which, if we consider a discrete-time sampled and voltage-quantized signal, is just PCM coding).

The key factor is that: when we transmit an analog signal, we transmit it at as high fidelity as possible (or practical), subject it to additive (channel) noise, and receive whatever is left. SNR is a property of transmitter power versus channel background level. When we transmit a digital signal, we transmit it with quantization noise already added to it, hopefully at a level just above the channel SNR so that we recover the bits with satisfactory error rate without squandering channel capacity. We are necessarily using more bandwidth in the process (Fs strictly greater than Fnyquist), and how much extra is a free variable (we can always sample faster, or clock the S-D converter faster, but never slower).

So you say, what if we take multiple successive samples, cram them together, and transmit one uber-symbol at a time? What if we take two samples? Or 16? Or 256?

Well, let's be careful here, because we assumed the analog system is subject to channel noise, and so too the digital system ought to be. If we take even a modest number of samples together, concatenate bits (or interleave or whatever) and cram that through a DAC, we now get an analog signal that is exponentially more precise than we started with, and, frankly, we'll very quickly run into fundamental limitations of the physical world itself if we try this with more than a few samples.

Consider that just three samples at 16 bits each, already vastly exceeds the accuracy of the best known physical constant (Rydberg constant is known to 2 parts per trillion, or ~39 bits). You're welcome to cram as many bits in theory as you like, but give me a call when you pull it off in practice!

So why do we use digital modulations with such flexibility? Because the channel itself varies over time, in a lot of applications. Cell phones and laptops move around in an RF field, getting different SNRs on different channels in their respective radio bands. Likewise, the data rates themselves are extremely variable -- and flexible, for the most part. We might grumble a bit when a page loads less than instantaneously, but it ultimately doesn't matter; but good luck holding any kind of conversation when your voice channel drops below a few kbps.

And, voice drops might manifest as a broad range of errors. This is an illustration of channel capacity and error rate. Modern codecs attempt to approximate the human vocal tract, so that slowly changing vowels, or some consonants, can be transmitted at impressively low rates (or, low compared to the complexity of turbulent fricatives and the like); but so too, when they glitch out, you get some strange stuttering, staring-off-into-the-distance mouth-gaping "Aaaaaaəəəee" sound, for example; or more traditional channels might suffer clipping or distortion (PCM bit errors, say), or popping or dropouts (packet loss, time division multiplex errors). So too, analog voice systems have exhibited similar faults; the previous analog telephone system used a sophisticated set of frequency and time division multiplexers to squeeze more voice channels onto trunk lines, and modulation errors, crosstalk and dropouts were among the possible defects.

So for bandwidth, I think it works much easier one way than the other. That is, you can always spread bandwidth, but you'll have a hard time ever reducing it. Notice this is mirrored by the digital case, you can always sample or modulate faster, or spread spectrum or code sequence or whatever, but you can't sample slower. Conversely, you might transmit symbols slower, but the bit rate is the same, given much higher channel SNR to support it.

Bandwidth spreading has the advantage that, it's more tolerant of poor SNR -- if the spectral noise floor is higher, but you're folding a larger swath of it into a tinier signal based on certain correlations in that spectrum, well, that works just like filtering the S-D bitstream does; doesn't matter if we do it in frequency or time domain, or any weird nonlinear mixture. Consider wideband FM for example: the bandwidth spreading is significant, but it's also renowned for its clarity of reception.

As for proof by example -- this seems the trickiest so I'll add it here.

We can indeed accomplish bandwidth compression in the analog domain.

Suppose we have a 16 ENOB channel, and wish to transmit an 8 ENOB signal of given bandwidth. If we can fold the signal "in half" somehow, we can utilize the excess SNR and reduce bandwidth (and, I guess let's say the channel only has BW/2 available for some reason).

First, filter the signal into upper (UFB) and lower frequency (LFB) bands, using a "brick wall" filter (and its complement). Of the LFB, quantize it: this chops it into 256 discrete levels (8 bits). This isn't "unfair" in any way: it's just a nonlinear transfer function, and it's not even a discrete-time function (but those are fair game, too!).

Of the upper band, treat it as a USB signal: mix it with Fc = BW/2 to shift it down to baseband. Attenuate it by 256x and sum with the quantized low band. Congrats, you now have an analog solution with half the bandwidth!

Notice some nonlinear function is required here, because if we have continuous signals \$x(t)\$ and \$y(t)\$, their sum \$x(t) + y(t)\$ is also a continuous function indistinguishable from them. We need to introduce some method of identifying one and separating it from the other. Since our channel has low noise, discretizing the voltage (making one a discontinuous function) seems an option.

Note that this is a toy example, as it makes no effort to address channel bandwidth (causing inter-symbol interference, i.e. the low-pass signal varies smoothly over time, not in satisfyingly hard steps), nor mixing (the UFB levels add with ISI and noise to fudge what are supposed to the LFB's discrete steps), nor reasonable noise statistics (it would be passable at best for evenly distributed noise -- not normal distribution!). But an example is sufficient for existence; it doesn't have to be a working prototype.

Any function that allows discriminating two signals (that need not be continuous), from one continuous starting function, will do here. I'm not real sure offhand what all kinds of things would fit here. But that seems one way.

Notice that QAM does not help us here:

Suppose we do the same band splitting trick, getting us two continuous signals of BW/2 each. If we SSB each one, we get BW/2 each, again. If we superimpose them, well, er, we just get plain old SSB again -- that is if both are USB modulated, and the UFB is placed at a center frequency exactly its offset above the LFB. We just made a lot of steps for no value, heh. Note we can't superimpose them with overlap, because there's no phase information to separate them with. And if there were, well..:

Suppose we DSB modulate each signal. Individually, we've doubled the bandwidth (we have LSB+USB of the LFB, and LSB+USB of the UFB). AM doesn't seem to have any advantage for us, and SSB is the most efficient such method for a single signal. Well, suppose these DSBs are placed at the same center frequency, with one carrier 90° phase shifted: now we can superimpose them, and the sidebands carry the sum and difference of the two signals; they are separable by orthogonality. We can demodulate them using the reverse process, and recover the original signal; and, notice we've used a total 2 * (BW/2) bandwidth in the process. Well, on one hand, it would sound weird to an AM radio -- without a coherent quadrature detector, the sidebands overlap, and it might not be intelligible. Could be interesting effects there, but, quirks aside, there isn't anything to be gained here, not from such simple transformations at least.

And, these arguments remain true no matter how we slice up the signal. Suppose we write the signal alternately to two tapes: while a given record head is active, its tape advances at normal speed, and freezes on the spot otherwise. Both tapes, after a short loop (there has to be a little slack in the system, of course), are read out, continuously, at half speed: we again have two signals of half bandwidth (give or take the discontinuity as they are switched). We've doppler-shifted and time division multiplexed them. But, we're stuck with two signals.


In digital transmission, we can freely choose how many different symbols (voltage levels) we use. Depending on SNR they may or may not be distinguishable at the receiving end. It is straightforward to map a pair of symbols in range (0..N) to a single symbol in range (0..N²)

But analog signals, by definition, use infinite number of different voltage levels. There are ways to map pair of real numbers to a single real number, but I couldn't find any that could be accomplished by analog electronics. Mathematically an infinite array of comparators could do that, but that is basically just an infinite-bit analog-to-digital converter, and of course not possible to implement in practice.

There is something that can be done in frequency domain: the vocoder divides analog signal to its frequency components. We could then connect the output to a set of oscillators with frequencies closer to each other than the original ones, to compress the frequency spectrum.


simulate this circuit – Schematic created using CircuitLab

But while this preserves the analog nature of the signal by allowing transmission of infinite number of different amplitudes, it only has a fixed set of different signal frequencies. Very similar to how e.g. MP3 compression divides sound signals to 576 frequency bands.


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