2
$\begingroup$

While reading Bernard Sklar's Digital Communications book I've got confused with ISI Nyquist criterion (not Nyquist rate criterion). There is stated (Sklar, 3.3 section) for symbol rate $R_s$ minimum bandwidth $W = R_s/2$ and for practical cases we have equation:

$W = 1/2 \cdot (R_s + r \cdot R_s)$,

where $r = 0...1$. So theoretically we have 2 symbols per Hz. For BPSK it is 2 bits/s per Hz. For QAM-64 it is 12 bits/s per Hz. So this statement I can't understand, because through my practice I used to think if we use BPSK and we want to achive e.g. 6 Mbps we need to allocate about 6 MHz of bandwidth (actually more because of rounding and synchronization/code redundancy). Where is my mistake in understanding of Nyquist theory?

Upd. Here is the picture from the book, where Nyquist channel is pictured enter image description here

It seems the bandwidth of such a channel is calculated from $0$ to $1/2T$ and negative frequencies are neglected. Why in the baseband case we need to determine bandwidth in such a way? This can be an answer for the question above.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
3
$\begingroup$

I don't have Sklar's text but I guess he is referring to the baseband case. For bandpass signals the required bandwidth is doubled. So in the bandpass case you get a maximum spectral efficiency of 1 bit/s/Hz for BPSK, and $\log M$ bits/s/Hz for $M$-ary QAM.

Just as a side note, spectral efficiency is measured in bits/s/Hz (and not in bits/Hz as you seem to suggest).

Improve this answer
$\endgroup$
2
  • $\begingroup$ Matt, thanks, I've substituted bits/Hz for bits/s/Hz, my mistake. And also I add a picture for explanation of Nyquist bandwidth in baseband case. This can be an answer, but I can't understand why it is different bandwidth value for baseband and bandpass cases. $\endgroup$
    – Serj
    Apr 26, 2015 at 18:30
  • 1
    $\begingroup$ @Serj: Bandwidth is measured for positive frequencies. So imagine you modulate your baseband signal of bandwidth B (i.e. it extends from -B to +B) up to some carrier frequency. Then you get a bandpass signal of bandwidth 2B, because it extends from (carrier - B) to (carrier + B). $\endgroup$
    – Matt L.
    Apr 26, 2015 at 19:27
1
$\begingroup$

I agree that this topic is somewhat confusing, it also confused me at times, like this morning when i looked around for an answer, and resorted to answer to it for myself.

I believe the root cause of this halving of the spectral efficiency when going from baseband to passband is that when we modulate the signal onto a real carrier, the negative frequency spectrum is unavoidable, thereby halving the efficiency.

Whether we consider baseband signals or bandpass signals, spectral efficiency is always only concerned with positive frequencies.

So, whereas in baseband you have a spectrum from [-B,B], in band-pass you have two spectra in the intervals [-fc-B, -fc+B] and [fc-B, fc+B]. In the case of SSB (Single Side Band) modulation the baseband bandwidth will be [0,B] and the bandpass ones [-fc-B,fc] and [fc,fc+B]. Assuming your shaped pulses are sent at a symbol rate 1/T, for an M-ary modulation we get

$$ \rho_{Nyq,BB} = \frac{R}{B} = \frac{\log_2(M)/T}{1/2T} = 2\log_2(M) [b/s/Hz], or [b/2D] $$

$$ \rho_{Nyq,BP} = \frac{R}{2B} = \frac{\log_2(M)/T}{1/T} = \log_2(M) [b/s/Hz], or [b/2D] $$

On the receiving side a coherent demodulator can reconstruct the complex envelope by analytic extension, thereby removing the negative frequencies, but the resulting signal becomes 2-dimensional.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.