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Both BPSK and 2-VSB transmit one bit of information per smybol and both modulate the real part of the signal. The difference is, as far as I understand, that BPSK does not modulate the imaginary part of the complex signal at all, while 2-VSB uses hilbert transformation to suppress the lower sideband of the transmitted signal.

Are there any advantages of doing this or is it just adding complexity without any advantages?

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Suppressing the lower sideband decreases the bandwidth by a factor of $2$ compared to BPSK. Since we're not talking about single-sideband (SSB) modulation, but vestigial sideband (VSB) modulation, we don't gain a factor of $2$, but a factor that is slightly smaller than two, because a part of one of the sidebands is retained.

Note that both BPSK and 2-VSB carry the information only in the (bi-polar) amplitude of the signal, i.e., only in the in-phase component, not in the quadrature component. The complex baseband signal for 2-VSB only occurs because the larger part of one of the sidebands is filtered out. This, however, is not necessarily done by complex baseband processing (using the Hilbert transform), but it can be done by applying a band pass filter after modulating the baseband signal. The latter option only uses real-valued processing.

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  • $\begingroup$ Suppressing the lower sideband decreases the bandwidth by a factor of 2 compared to BPSK is this equivalent to 4-qam at half the symbol rate? $\endgroup$
    – taffer
    Jun 29, 2016 at 14:30
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    $\begingroup$ Binary Phase Shift Keying doesn't carry the information in the phase? Ahem. Sure, it's the sign which is pretty simple, but it's still the phase rather than the magnitude. I suppose you could say it's the amplitude, but ... your statement seems a little contradictory. ;-) $\endgroup$
    – Peter K. ♦
    Jun 29, 2016 at 15:11
  • $\begingroup$ @PeterK.: I agree that my statement is a bit confusing. I meant (and wrote) that the information is carried in the amplitude (which can have both signs), and of course not in the magnitude. Anyway, thanks for pointing that out, I'll edit. $\endgroup$
    – Matt L.
    Jun 29, 2016 at 16:02
  • $\begingroup$ @taffer: If you only consider bandwidth and bit rate, then you're right. But note that BPSK and 4-QAM are equivalent in the sense that they achieve very similar bit error rates at a given SNR. This is due to the fact that you need more transmit power for 4-QAM than for BPSK if you want the same minimum distance between symbols. $\endgroup$
    – Matt L.
    Jun 29, 2016 at 16:05
  • $\begingroup$ In fact, BPSK can use the imaginary part. For example, LTE uses a constellation rotated by $\pi/4$. Because as Peter said, the information is carried in the phase. $\endgroup$
    – vaz
    Jun 29, 2016 at 21:30

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