Given a signal you are asked to perform baseband QPSK (not on a carrier).

Does this mean the following?

  1. In baseband QPSK the (possible) constellation points are $$\sqrt {E_s} e^{jn \pi / 2} \qquad E_s = A^2 T_s$$

and the frequency response has delta functions placed in the origin

  1. In QPSK modulation on carrier the (possible) constellation points are $$\sqrt {E_s} e^{jn \pi / 2} \qquad E_s = \frac {A^2 T_s} 2$$

and the frequency response has delta functions around the carrier frequency $\pm f_c$

Am I correct?

  • $\begingroup$ well, what use is it that we should confirm something that you can directly lift from your textbook? Power definition are always just conventions, so there's "wrong" or "right" here. There's no such thing as constellation points on a carrier. Constellation points are a baseband concept. $\endgroup$ Commented Jan 3, 2018 at 15:45
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    $\begingroup$ @MarcusMüller Since I am asking this question, it's taken as granted that I have looked for this definition in my academic textbook, so posting this question came after great consideration. This question is from an assignment and it is asked exactly as I have mentioned above; therefore I would like to get a more intuitive comment. :) $\endgroup$
    – bolzano
    Commented Jan 3, 2018 at 16:42
  • $\begingroup$ yes, but the textbook that introduces QPSK will probably have exactly this formula. Just posting "is this right?" bears no value in itself. Explain why you have a doubt. And, again, the 2. part makes no sense (and also, explicitly wasn't asked for). $\endgroup$ Commented Jan 3, 2018 at 16:43
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    $\begingroup$ I have updated the question (since it's assumed that setting the carrier frequency at 0 produces a baseband QPSK (as I have understood) and carrier frequency at $f_c \neq 0$ gives a passband QPSK modulation); therefore $E_s$ is correct) and got an accepted answer :) $\endgroup$
    – bolzano
    Commented Jan 3, 2018 at 17:01
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    $\begingroup$ I think the question is stated on point and it's already cleared by the OP that he could not find such information on his textbook/notes. Helping is never a bad thing. The accepted answer is also exactly what the OP seemed to be confused about. $\endgroup$
    – Rebellos
    Commented Jan 3, 2018 at 17:07

1 Answer 1


Yes, you are correct, though your formula puts the constellation points at 0, 90, 180, and 270 degrees, where usually the constellation points are represented as being at 45, 135, 225, and 315 degrees. Functionally it is equivalent though.

  • $\begingroup$ Thank you very much. You're also right about the constellation points in most convention (albeit my book puts them to 0, 90, 180 and 270 degrees) $\endgroup$
    – bolzano
    Commented Jan 3, 2018 at 17:02

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