I am currently reading Pasupathy, Subbarayan. "Minimum shift keying: A spectrally efficient modulation." IEEE Communications Magazine 17.4 (1979): 14-22.

The paper explains MSK as a form of OQPSK with a half sinusoid instead of rectangular weighting with a figure below:

enter image description here

But in my opinion, this figures seems incorrect. Because figure 5(a) shows a bit stream +1, -1, -1 and +1 inside a waveform, but the actual waveform is drawn as if its corresponding bit stream is +1, +1, -1, -1.

Am I missing something?


Maybe I learned that the original problem is solved by taking a phase into account.

Another problem is, a waveform in the IEEE 802.15.4 standard differs from the original one.

enter image description here

It seems that a direction (upward/downward) of a waveform indicates a bit.

Are both acceptable? Is 802.15.4's O-QPSK a variant of the original one?

  • \$\begingroup\$ Maybe I got it. a direction of a waveform (upward/downward) does not indicate a bit, but a phase of waveform does. \$\endgroup\$
    – Jeon
    Oct 20, 2017 at 8:22
  • \$\begingroup\$ Now the question is the original O-QPSK vs 802.15.4's O-QPSK \$\endgroup\$
    – Jeon
    Oct 20, 2017 at 9:16

1 Answer 1


i think, for OQPSK with the convention that the even-indexed data samples go into \$i(t)\$ and the odd-indexed data samples go into \$q(t)\$ , the IQ waveforms, in terms of the data is :

$$\begin{align} i(t) &= \sum\limits_{m=-\infty}^{\infty} x[2m] \, p(t-2m T_c) \\ \\ q(t) &= \sum\limits_{m=-\infty}^{\infty} x[2m+1] \, p(t-(2m+1) T_c) \\ \end{align}$$

in the waveforms shown in the OP, the definition of the pulse shape \$p(t)\$ is:

$$ p(t) = \begin{cases} 0 \qquad & t < 0 \\ \sin\left(\tfrac{\pi}{2 T_c} t \right) \qquad & 0 \le t < 2T_c \\ 0 \qquad & 2T_c \le t \\ \end{cases} $$

and the bipolar binary data is:

$$ x[n] = 2 a[n] - 1 $$

and \$a[n] \in \{0,1\}\$ are the raw bit data.

The nice thing about Offset-QPSK is that it naturally separates the even and odd samples because the \$i(t)\$ depends only on the even samples and switches at times of even multiples of \$T_c\$. \$q(t)\$ depends only on the odd samples and switches at times of odd multiples of \$T_c\$.

A smarter pulse shape to simply constrain the IQ signal to a bandwidth of \$\frac{1}{T_c}\$ is:

$$ p(t) = \operatorname{sinc}\left(\tfrac{t - T_c}{2 T_c} \right) w\left(\tfrac{t-T_c}{N T_c}\right) $$


$$ \operatorname{sinc}(u) \triangleq \begin{cases} 1 \qquad & u = 0 \\ \frac{\sin(\pi u)}{\pi u} \qquad & 0 < |u| \\ \end{cases} $$.

I didn't bother to define the window function here, but suggest a Kaiser of width \$2N\$ samples.


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