A USB signal is basically a baseband signal shifted up in frequency. However, an LSB signal is the mirror image of a baseband signal then shifted up in frequency. How is this 'mirroring' modulation done in a transmitter and its demodulation done in a receiver? Also why is LSB preferred below 9 MHz and USB preferred above 9 MHz.

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    – rclocher3
    Apr 19, 2020 at 17:16

3 Answers 3


USB is basically AM with the carrier suppressed and the lower sideband filtered out. LSB is basically AM with the carrier suppressed and the upper side and filtered out.

To demodulate either, you need to reinsert a carrier, and the signal becomes intelligible. To demodulate USB you insert the carrier below the signal, and for LSB you insert the carrier above.

As to why LSB is used below 10MHz and USB above, it’s basically a convention. Nothing more than that. There are other questions on this site where this is addressed. Here is one: Why do we use LSB below 10 MHz and USB above 10 MHz when operating SSB HF?


The mirroring is achieved through negative frequencies.

This seems like a bit of magic, but the math checks out. We can represent an unmodulated carrier with angular frequency $\omega$ as:

$$ s(t) = \cos(\omega t) $$

If $\omega$ is negative, the result is the same as if it's positive due to the basic trigonometric identity:

$$ \cos (x) = \cos (-x) $$

Weird but true: negative frequencies work.

So to demodulate LSB you just select the LO frequency such that the carrier frequency ends up at zero and the lower sideband ends up at the negative frequencies just below zero. Then as a consequence of the trig identity above, the spectrum gets "flipped".

Of course you also have to design everything so other mixing products don't interfere, but that's all the same things you would to do to design an upper sideband demodulator.

To drive the point home, consider how a frequency mixer works. Usually it's described like this:

$$ f_\text{out} = \begin{cases} f_1 \pm f_2 & (f_1 > f_2)\\ f_2 \pm f_1 & (f_2 > f_1) \end{cases} $$

If we say $f_1$ is 9999 kHz and $f_2$ is 10,000 kHz, we choose the second case and we end up with mixing products at 1 kHz and 19,999 kHz.

But wait, how does the frequency mixer "know" which case to choose?

Say we instead pick the first case, we instead end up with mixing products at -1 kHz and 19,999 kHz. If you accept that negative frequencies work just like positive frequencies, it means that don't the two cases because it doesn't matter which one you pick.

Modulation is the same, in reverse. If you want to transmit LSB at 10 MHz, you can think of it as mixing the baseband signal to -10 MHz.

  • $\begingroup$ Phil: "Then as a consequence of the trig identity above, the spectrum gets "flipped"." So then demodulation for LSB is the same as demodulation for USB? $\endgroup$
    – KevinHJ
    Dec 17, 2020 at 15:43
  • $\begingroup$ @Allasso Pretty much, yeah. The only difference is whether you are filtering the frequencies above or below the carrier. $\endgroup$ Dec 17, 2020 at 15:48

A USB signal is basically a baseband signal shifted up in frequency.

This is not accurate. @ScottEarle points out that SSB can be produced by filtering out the carrier and undesired sideband from an AM signal, but that AM signal was generated at an intermediate frequency (IF), not at baseband; the baseband signal was used to modulate an IF carrier to produce the AM which was filtered to produce SSB at the IF before being heterodyned to the final signal frequency.

In addition to the "filter method" described earlier, SSB can be generated by the "phasing method." As such, it is a special case of quadrature amplitude modulation (QAM). This implementation - adding or substracting quadrature versions of a signal to produce upper or lower sideband - may more clearly illustrate the distinction between the two.

  • $\begingroup$ An AM signal at baseband can be filtered by a complex kernel (either IIR or FIR/FFT) to create a baseband SSB USB or LSB signal (in IQ form). No IF needed. $\endgroup$
    – hotpaw2
    Apr 20, 2020 at 5:40
  • $\begingroup$ The equation for amplitude modulation requires a carrier wave with a non-zero angular frequency that is modulated by the baseband input signal. Are you referring to the fact that, as @PhilFrost-W8II pointed out, the baseband signal naturally comprises negative frequency components that mirror the positive frequency components? $\endgroup$
    – Brian K1LI
    Apr 20, 2020 at 7:56
  • $\begingroup$ The spectrum (FFT) of a strictly real signal is conjugate symmetric. Thus the carrier can be at DC (0 Hz) in a complex (or IQ) signal. $\endgroup$
    – hotpaw2
    Apr 20, 2020 at 14:55
  • $\begingroup$ Isn't that just the baseband signal that would be used to modulate a carrier? $\endgroup$
    – Brian K1LI
    Apr 20, 2020 at 20:49
  • $\begingroup$ Yes, and a baseband IQ signal can be AM, or (just by filtering) USB, or LSB. No IF needed. Unless you consider 0 kHz to be an IF. $\endgroup$
    – hotpaw2
    Apr 20, 2020 at 20:57

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