Offset Quadrature Phase Shift Keying (OQPSK) like QPSK transmits two bits per symbol. However, because of the offset on the quadrature component there are two transitions per symbol. Does this not double the bandwidth of each symbol, leading OQPSK to be more similar to BPSK in terms of spectral efficiency? This would make it 1 bit/s per Hz?


1 Answer 1


The spectral efficiency is the same as QPSK.

Consider the I and Q streams separately. Each has a certain bandwidth $B$. Shifting or delaying the stream does not change its bandwidth. So, shifting one of the streams relative to the other does not affect the overall bandwidth.

  • $\begingroup$ The I and Q signals are modulated separately and summed together (in this case, in quadrature with each other.) Summing two signals is never going to have a multiplicative effect on their bandwidth... the two signals just overlap each other. So the bandwidth has to be the same as the I or Q component by itself. Correct? $\endgroup$
    – Chris_F
    Commented Feb 9, 2023 at 2:57
  • $\begingroup$ @Chris_F That's correct. $\endgroup$
    – MBaz
    Commented Feb 9, 2023 at 14:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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