What is the difference between a $\pi/4$-QPSK modulation and a QPSK differential modulation (DQPSK)?

Are these modulations identical?

  • 1
    $\begingroup$ They're not identical at all. The differential modulator has memory, the other one doesn't. $\endgroup$
    – MBaz
    May 13, 2018 at 16:47

1 Answer 1


No, the modulations are not identical; they are quite different.

Differential QPSK uses one of 4 different signals $$\sqrt{2}\cos\left(2\pi f_c t + \frac{\pi}{4}\right), ~ \sqrt{2}\cos\left(2\pi f_c t + \frac{3\pi}{4}\right), ~ \sqrt{2}\cos\left(2\pi f_c t + \frac{5\pi}{4}\right), ~\sqrt{2}\cos\left(2\pi f_c t + \frac{7\pi}{4}\right)$$

with phase angles $\displaystyle \frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},$ and $\displaystyle \frac{7\pi}{4}$ to transmit two information bits, but the information is carried in the change of phase from the previous signaling interval, and not in the absolute value of the phase. That is, the information bits are $(0,0), (1,0), (1,1), (0,1)$ according as the phase changed by $\displaystyle0, \frac{\pi}{2}, \pi,$ or $\displaystyle\frac{3\pi}{2}$ in comparison to the previous signaling interval. Note that it is possible to have no phase change or a phase change of $\pi$ (inversion of polarity of the signal) both of which can undesirable for different reasons.

In contrast, $\displaystyle \frac{\pi}{4}$-QPSK uses the four signals shown above and the same signals shifted in phase by $\displaystyle \frac{\pi}{4}$, namely, $$\sqrt{2}\cos(2\pi f_c t), ~ -\sqrt{2}\sin(2\pi f_c t), ~ -\sqrt{2}\cos(2\pi f_c t), ~\sqrt{2}\sin(2\pi f_c t)$$ but it uses them in alternation: in one signaling interval, the signal constellation is $$\sqrt{2}\cos\left(2\pi f_c t + \frac{\pi}{4}\right), ~ \sqrt{2}\cos\left(2\pi f_c t + \frac{3\pi}{4}\right), ~ \sqrt{2}\cos\left(2\pi f_c t + \frac{5\pi}{4}\right), ~\sqrt{2}\cos\left(2\pi f_c t + \frac{7\pi}{4}\right)$$ and in the next signaling interval, the constellation is $$\sqrt{2}\cos(2\pi f_c t), ~ -\sqrt{2}\sin(2\pi f_c t), ~ -\sqrt{2}\cos(2\pi f_c t), ~\sqrt{2}\sin(2\pi f_c t),$$ followed by $$\sqrt{2}\cos\left(2\pi f_c t + \frac{\pi}{4}\right), ~ \sqrt{2}\cos\left(2\pi f_c t + \frac{3\pi}{4}\right), ~ \sqrt{2}\cos\left(2\pi f_c t + \frac{5\pi}{4}\right), ~\sqrt{2}\cos\left(2\pi f_c t + \frac{7\pi}{4}\right)$$ again, and so on, switching back and forth between the two choices. Effectively, the overall constellation is that of $8$-PSK but only four points can be used during any given signaling interval. Note that it is guaranteed that the phase will change at the transition from one signaling interval to the next, and also that the phase will not change by $\pi$ and so the undesirable phase transitions mentioned in differential QPSK (they also exist in plain vanilla QPSK) are avoided. Of course, the $\displaystyle \frac{\pi}{4}$-QPSK receiver (and transmitter) is more complicated than a QPSK receiver or transmitter. Whether it is a case of "What you gain on the swings, you lose on the roundabouts" is something that needs to be considered carefully.

For the signals mentioned above, see also this answer of mine.


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