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I'm reading a paper [R1] where the authors propose a MAP estimator for the phase noise and frequency offset. However equation (17), which I reproduce below, represents a challenging step for me and I would appreciate any help on how this equality is obtained

\begin{align} \sum_{\bar{c}}p(\bar{c})\exp\left(\frac{2}{\sigma^2}\Re\left(\bar{r}^*\bar{c}e^{j\bar{\phi}}\right)\right)=\\ \prod_{k=1}^N\cosh\left(\frac{\lambda_k}{2}+\frac{2}{\sigma^2}\Re\left(r_k^*e^{j\phi_k}\right)\right) \end{align}

We can use the a priori information to calculate $p(\bar{c})$. For binary phase shift keying (BPSK), the a priori probability of each bit found by its log-likelihood ratio (LLR) in the previous iteration ($i-1$), can be expressed as

\begin{equation} p(c_n=1|\bar{z}^{(i-1)})=\frac{e^{\lambda_n}}{1+e^{\lambda_n}}, \end{equation}

\begin{equation} p(c_n=-1|\bar{z}^{(i-1)})=\frac{1}{1+e^{\lambda_n}}. \end{equation}

For BPSK, the LLR for each coded bit is given by $\lambda_n=\ln\frac{Pr(c_n=1|z_n)}{Pr(c_n=-1|z_n)}$ where $c_n$ is the $n$-th BPSK modulated symbol and $z_n$ is the corresponding equalizer output.

Note that $\bar{x}=[x_1,\ldots,x_N]$ and $\Re(\cdot)$ is the real part of a complex number.

[R1] Sabbaghian, M.; Falconer, D.; , "Joint Turbo Frequency Domain Equalization and Carrier Synchronization," Wireless Communications, IEEE Transactions on , vol.7, no.1, pp.204-212, Jan. 2008 doi: 10.1109/TWC.2008.060451 URL: http://ieeexplore.ieee.org/stamp/stamp.jsp?tp=&arnumber=4432262&isnumber=4432233

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  • $\begingroup$ Noticing that $cosh(\alpha)=\frac{1}{2}(e^{\alpha}+e^{-\alpha})$ and $p(c_n)=p(c_n=1)+p(c_n=-1)=\frac{1}{1+e^{\lambda_n}}(e^{\lambda_n/2}+e^{-\lambda_n/2})$ solves the problem. The normalization term $(2/(1+e^{\lambda_n}))^N$ is missing, though. $\endgroup$ – Pedro Pedrosa Aug 12 '17 at 22:50

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