Mapping soft symbols in DQPSK
The paper actually deals with $\pi/4$-QPSK, not with plain DQPSK. In $\pi/4$-QPSK, the symbols are always rotated by a minimum of $\pm \pi/4$. Rappaport's Wireless Communications textbook has this table:

\begin{align*}
11 \rightarrow &amp; \, \pi/4 \\
01 \rightarrow &amp; \, 3\pi/4 \\
00 \rightarrow &amp; \, -3\pi/4 \\
10 \rightarrow &amp; \, -\pi/4
\end{align*}

Unfortunately, the paper you point to does not specify what mapping it's using; the only clue is equation (2), where we see that the rotations are $n\pi/4$ for $n=1,3,5,7$.

Let's assume that $\text{00} \rightarrow 7\pi/4$. Then, in your case you would have $s_{n-1} = \sqrt{2}/2 + i\sqrt{2}/2$, and $s_n = e^{i7\pi/4}s_{n-1} = 1$. Then, according to the paper,

\begin{align*}
b_1 = &amp; \text{Re}(s_n s_{n-1}^*) = \sqrt{2}/2 \\
b_2 = &amp; \text{Im}(s_n s_{n-1}^*) = \sqrt{2}/2
\end{align*}

I worked out $\text{00} \rightarrow 7\pi/4$ as the only rotation that produces results that make sense. However, it's not often that I see a paper with so many typos and so poorly typeset, and I wouldn't trust it without working out the details by myself first.