QPSK Data Rate vs Bandwidth at Passband: Why can it only transmit 1 bit per Hz and not 2 bits per Hz?

Question

1) This is a follow up question to the most up-voted answer to this question.

2) The accepted answer is that "QPSK transmits 1 bit per Hz at passband".

3) The theory found in books seems to agree with this answer.

4) But this defies my logic and my experience.

5) My logic and experience tells me that QPSK transmits 2 bits per Hz at passband.

6) From experience, I know that in order to acquire 1 MHz of complex bandwidth (at baseband), I need to set the IQ rate of my receiver to 1 Mega samples per second. (i.e. IQ rate = Complex Bandwidth, and keep in mind that the IQ rate usually needs to be slightly higher to account for the roll-off of an anti-aliasing filter. But let's assume that there is no filter for simplicity.)

7) At baseband, this is the equivalent of 500 KHz of bandwidth if you don't consider the negative frequencies.

8) And at passband, the total bandwidth will be 1 MHz because the negative frequencies "appear" as the frequencies on the "lower" side of the carrier frequency and they are now taken into consideration on the amount of bandwidth, effectively "doubling" the bandwidth.

9) With an IQ rate of 1 Mega samples per second, I can send 2 bits on each sample (because it is QPSK).

10) So the effective data rate is 2 Mbps for a bandwidth of 1 MHz at passband (in my experience).

11) Refer to image below, describing relationship between data rate and bandwidth in my experience.

12) Other people seem to have obtained similar results to mine:

A) Refer to the last video on these student experiments: They show a transmission of 2 Mbps with a 1 MHz QPSK signal.

B) In this video a 4 KHz data rate signal results in a 2 KHz QPSK 3dB spectrum.

Answer 1

Let us assume you have a bandwidth $W = 1\,\text{Hz}$ in passband. This means that you have a bandwidth $B = W/2 = 0.5\,\text{Hz}$ in baseband.

In baseband, using sinc pulses, we can transmit at a symbol rate $R_p = 2B = 1\,\text{Bd}$, where $\text{Bd}$ (baud) is the SI unit for "symbols (or pulses) per second".

Using BPSK modulation, we can transmit one bit per symbol, for a bit rate $R_b = 1\,\text{b/s}$.

Now, QPSK is esentially two BPSK signals in parallel. In other words, we have two BPSK baseband signals $s_I(t)$ and $s_Q(t)$, each having a bandwidth of 0.5 herz and transmitting one bit per second.

By using quadrature, we transmit both $s_I(t)$ and $s_Q(t)$ at the same time in passband, using the signal $$ s(t) = s_I(t)\cos(2\pi f_c t) - s_Q(t)\sin(2\pi f_c t). $$

The QPSK signal $s(t)$ has a bandwdith of 1 herz and transmits 2 bits per second (one over the in-phase component and one over the quadrature component).

However, note that the relationship $R_p = 2B$ is valid only for sinc pulses, which are almost never used in practice. If you use other pulses, then the answer will be different. Many introductory textbooks are not explicit about this.

Let us say that you wish to use a pulse $p(t)$. Furthermore, $p(t)$ is a Nyquist pulse. Then, the spectrum of the pulse determines the pulse rate and subsequently the bit rate.

As an example, let us repeat the exercise above for pulses shaped like a half-sine wave. In this case, the pulse rate is $R_p = B/2.5$. Then, if $B=0.5\,\text{Hz}$, $R_p = 0.2\,\text{Bd}$ and, using QPSK, you will only be able to transmit at a bit rate $R_b = 0.4\,\text{b/s}$ per herz in passband.

Note that the $2.5$ figure I used above is my (conservative) personal measurement and reflects what I'm comfortable with. Many introductory textbooks are extremely optimistic about the actual bandwidth of different pulses. In particular, it is common for textbooks to claim that $R_p = B$ for rectangular pulses, which I believe borders on the dishonest (Stallings and Tomasi do this, among others). This is what is behind the claim that "QPSK transmits 1 b/s/Hz" that you (very sensibly) suspect is false.

Note also that the answer you linked to is correct, for the specific pulse shape that is assumed. My answer is more general and allows you to calculate the bit rate for any Nyquist pulse you choose to use.