7

I have come across modulation techniques that have $\pi/2$ phase shifts from the regular constellations.

For example $\pi/2 $ BPSK vs constellation the regular BPSK.

Other examples include $\pi/2 $ QPSK vs. QPSK

I am wondering what are the benefits of phase shifts like this?

Thanks looking forward for your answers

Tyrone
  • 781
  • 1
  • 11
  • 23

1 Answers1

11

The signal always changes phase at bit transition times. This makes it easier to achieve bit synchronism at the receiver. In plain BPSK, long runs of 0 or 1, with no phase transitions occurring at the boundary between two identical bits, the phase-lock loop that is generating the signal telling the receiver where the bit transitions are, can lose lock; or the initial acquisition is more difficult. Also, in satellite channels with nonlinear amplifiers such as traveling-wave-tube (TWT) amplifiers, smaller phase transitions are better. In particular, if the RF carrier phase makes a transition of $\pi$ radians (meaning the carrier envelope makes a transition through the origin), the TWT amplifier output contains all the sidebands that were carefully suppressed in the uplink. So, filtering in the satellite is required to suppress these sidebands. Hence, $\pi/2$-BPSK and $\pi/4$-QPSK are preferred over their plain cousins. Note 4, not 2, with QPSK: $\pi/2$-QPSK does not avoid the transitions through $0$, and adds complexity without necessarily adding benefits.

Note: in plain BPSK, the signals are $\pm\cos(\omega_0 t) = \{\cos(\omega_0 t), \cos(\omega_0t + \pi)\}$, and so, at bit transitions, the phase either stays the same, or changes by $\pi$. in $\pi/2$-BPSK, the signal set is $\pm\cos(\omega_0 t)$ for one bit and $\pm\sin(\omega_0 t)$ for the next. Thus, regardless of whether we have two successive identical bits or two different successive bit, the signal phase changes by $\pm \pi/2$ since the signal changes from a cosine to a sine. Figure out for yourself why $\pi/2$-QPSK has transitions through $0$ while $\pi/4$-QPSK does not; then think of offset QPSK a little.

Dilip Sarwate
  • 20,349
  • 4
  • 48
  • 94
  • thanks alot. But two questions what do you mean by it is easier to achieve bit sync? Do you mean at reciever? Also what do you mean by smaller phase transitions are better? In my understanding whether $ \pi/2$ BPSK or BPSK the phase transition is the same betren symbols? – Tyrone May 23 '15 at 17:44
  • thanks, so for $\pi/2$ BPSK, if I transmit a zero bit, do I send $+\cos(\omega_0t)$ or $-\cos(\omega_0t)$? What if I transmit a one bit, do I send $+\sin(\omega_0t)$ I am not sure why you have four options for two bits? @dilipsarwate – Tyrone May 23 '15 at 22:11
  • Please I would like to have your opinion on my comment, I am not sure why we have four signals for two bits with $\pi/2$ BPSK @DilipSarwate – Tyrone May 24 '15 at 03:12
  • 1
    There are two kinds of people: those who transmit a bit $b\in{0,1}$ by transmitting $(-1)^b\cos(\omega_0t)$ and those who transmit it via $(2b-1)\cos(\omega_0t)$. Choose which group you belong to. So, in the first transmission interval, you send $b_0$ and you need to have on hand two signals $\pm\cos(\omega_0t)$ even though you will use only one of them to transmit your given value of $b_0$. For the next bit $b_1$, advance the phase by $\pi/2$. Note that $\pm\cos(\omega_0t+\pi/2)=\mp\sin(\omega_0t)$. Again, two signals, though you will use only one of them to transmit $b_1$.. – Dilip Sarwate May 24 '15 at 04:26
  • 1
    Most underrated comment: "There are two kinds of people: those who transmit a bit $b \in {0, 1}$ by transmitting $(-1)^b\cos(\omega_0 t)$ and those who transmit it via $(2b - 1)\cos(\omega_0 t)$. Choose which group you belong to." – Gilles Jan 15 '21 at 22:14
  • 1
    @Gilles I try to add a bit of levity to otherwise boring responses. Glad that you found it amusing. – Dilip Sarwate Jan 15 '21 at 23:55
  • This link helped me visualize the constellation https://www.gaussianwaves.com/2022/02/%CF%80-2-bpsk-pi-2-bpsk-5g-nr-phy-modulation/ – VMMF Nov 08 '22 at 18:47