In real signal modulation, e.g. ofdm based on DCT transform, the modulation order $m$ should be $\sqrt{M}$ where $M$ is the modulation order of the complex signal. for example, if we are using QPSK DFT-OFDM, it's equivalent into BPSK in real signal$[1]$

My question, what's about if the complex modulation signal is BPSK or 8-PSK, what is the equivalent modulation order of real signal?

[1] F. Xiong, “M-ary amplitude shift keying OFDM system,”IEEE Trans.Commun., vol. 51, no. 10, pp. 1638–1642, 2003.

  • In the case of BPSK, the quadrature signal is zero. Then, the order of the real part of the signal is 2, and the order of the imaginary part is zero.

  • I'm not sure what you mean by "8QPSK", but normally there are two cases with $M=8$:

    • A rectangular constellation. In this case, the real part may have order 2, and the quadrature part order 4; or vice versa.
    • 8-PSK, in which constellation points are placed on a circle, or two rotated QPSK constellations with different amplitudes. In these cases there is no way to define an integer order for each of the real and imaginary signals, because they are not independent. One possible approach is to say that each component transmits $\log_2(8)/2=1.5$ bits, and has order $2^{1.5} = 2.83$.
  • $\begingroup$ The meaning of 8QPSK is 8-PSK. sorry for that error. I will edit it. But in real signal, we don't have imaginary part, I mean if complex modulation order $M=8$, what's the equivalent modulation order when using real signal modulation? $\endgroup$ Jun 26 at 1:47
  • $\begingroup$ @New_student I'm not sure I understand your question, but here are some ideas. (A) For any complex modulation with order $M=2^n$, you can define a purely real $M$-PAM modulation wih the same order $M$. (B) Some complex modulations, like 16-QAM, are made up of two real modulations, each with order $sqrt{M}$; one is transmitted in-phase and the other in quadrature. For example, 16-QAM is made up of two 4-PAM real signals. (C) Some complex modulations, like 8-PSK, cannot be neatly expressed as two real PAM signals, because their in-phase and quadrature signals are not independent. $\endgroup$
    – MBaz
    Jun 26 at 15:19
  • $\begingroup$ OK, got it. thank you so much $\endgroup$ Jun 27 at 14:28
  • $\begingroup$ You're welcome :) $\endgroup$
    – MBaz
    Jun 27 at 15:26
  • 1
    $\begingroup$ @New_student I don't use pammod, so I'm not sure how it works. However, keep in mind that the normalization factor only affects the signal energy. Figure out the energy you want the signal to have, and then normalize accordingly. $\endgroup$
    – MBaz
    Jun 28 at 13:32

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