When you listen to an LSB (Lower Side Band) signal with your radio set to USB (Upper Side Band) or vice versa, the speech you hear is malformed. What causes this behaviour?

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    \$\begingroup\$ Whistle an ascending scale (as a thought experiment, do not do it on the air) and you would hear it received as a descending scale. \$\endgroup\$ Jun 24, 2013 at 20:28
  • \$\begingroup\$ @ChrisStratton that's a great way of explaining it! Now you've visualized the theory jippie gave me :) thanks! \$\endgroup\$
    – user17592
    Jun 24, 2013 at 20:29
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    \$\begingroup\$ Actually, I'm somewhat wrong - it would also be horribly out of tune, as a scale is an exponential sequence, while the frequency inversion is a linear effect. \$\endgroup\$ Jun 24, 2013 at 20:31
  • \$\begingroup\$ @ChrisStratton I get that, the frequency is inversed, not the pitch \$\endgroup\$
    – user17592
    Jun 24, 2013 at 20:31
  • \$\begingroup\$ Does it do crazy sound things with DSBSC (rather than plain ordinary 2 quadrant AM) I wonder? If the "retrieved" or "supposed" carrier is offset? \$\endgroup\$
    – Andy aka
    Jun 24, 2013 at 21:43

3 Answers 3


The spectrum is reversed. Nice images at Wikipedia

  • LSB is short for Lower Side Band and
  • USB is short for Upper Side Band.

  • Normal AM transmissions are Dual Side Band,

  • but with filtering you can create Single Side Band (either LSB or USB).

Basically with AM modulation the audio spectrum is 'shifted' right with an equal amount as the carrier wave to the USB and then mirrored from USB to LSB, with the carrier as mirror.

Or you can reason that negative frequencies the give an identical signal as positive frequencies and hence when you start to shift it up the spectrum by amplitude modulation you get two lobes, USB and LSB.


I thought it would be nice to see how the signal looks in the time domain (mathematically, I mean). What you get when you treat a USB signal as an LSB signal and vice-versa is that in the baseband the positive frequencies appear as negative frequencies and the other way around. The (complex) signal corresponding to the positive frequencies (with no negative frequencies) is

$$x_+(t)=\frac{1}{2}\left[ x(t)+j\hat{x}(t)\right]$$

where \$x(t)\$ is the original baseband signal, and \$\hat{x}(t)\$ is its Hilbert transform

$$\hat{x}(t)=x(t)*\frac{1}{\pi t}=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{x(\tau)}{t-\tau}\;d\tau$$

The signal \$x_+(t)\$ is basically the USB shifted down to baseband. The signal corresponding to the negative frequencies (with no positive frequencies) is

$$x_-(t)=\frac{1}{2}\left[ x(t)-j\hat{x}(t)\right]=x_+^*(t)$$

where \$^*\$ denotes complex conjugation.

Let \$B\$ be the bandwidth of the baseband signal \$x(t)\$. Then the resulting signal with exchanged positive and negative frequency components can be written as

$$\tilde{x}(t)=x_+(t)e^{-j2\pi Bt}+x_-(t)e^{j2\pi Bt}=2\text{Re}\{x_+(t)e^{-j2\pi Bt}\}=\\ =x(t)\cos(2\pi Bt)+\hat{x}(t)\sin(2\pi Bt)$$

So the signal you hear is the original signal modulated by a cosine with frequency \$B\$ plus (and that's probably the most annoying part) its Hilbert transform modulated by a sine carrier with frequency \$B\$.


The simple explanation is an AM signal transmits the carrier and modulation, and the modulation is both above and below the carrier frequency. During sideband transmission, the radio only transmits the upper or lower half of the modulation as they are exact mirror images, and the receiver adds in the missing parts (the carrier and the other half) to recreate the audio.

So if you have your reciever set incorrectly, it is attempting to recreate the missing portion at a higher or lower frequency than when the sideband transmission was made.

Source: My layman's understanding gained from studying the ARRL Technician and General Class Licence manuals.

  • \$\begingroup\$ There isn't actually a missing portion. \$\endgroup\$ Jun 27, 2013 at 22:36

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