## 6 Bit Error Rate in Single-Bit Sampling Transceivers
\begin{cases}\int_{(-1)^{\infty}\cdot U_{\text{H}}/2}^{ \infty}p(u_{\text{c,in}}(t_{s}=i\cdot T_{i}))&\text{for $x_{i}=1$}\\ \int_{-\infty}^{-1}\cdot U_{\text{H}}/2}p(u_{\text{c,in}}(t_{s}=i\cdot T_{i}) )&\text{for $x_{i}=0$.}\end{cases}  \tag{35}\]  
Figure 13: A sinusoidal RF input signal \(u_{\text{c,in}}(t)\) is sampled at eight points per period. As the voltage is subject to phase and amplitude noise, the acquired voltage shows a certain probability to deviate from the nominal value, expressed by a probability density function \(p(u_{\text{c,in}})\) being indicated by the Gaussian curves at any sampling point. Due to the hysteresis thresholds \(\pm U_{\text{H}}/2\) of the comparator, illustrated by the two dashed lines, the probability of a value \(x_{i}\) being sampled as ‘0’ or as ‘1’ is expressed by a conditional probability \(P(x_{i}|x_{i-1})\) is taking the previous value \(x_{i-1}\) into account. The graph is plotted for \(\hat{u}_{\text{c,in}}=20\) mV, \(\sigma_{\text{P}}=2\pi\times 0.03\), \(\sigma_{\text{A}}=2\) mV and \(U_{\text{H}}=10\) mV. The given approach is a simplification, as we only consider the probability density function at the sampling points, but not in the time slots in-between the sampling points. This procedure can be refined if fewer sampling points are taken into account. Knowing the conditional probabilities of every sampling point allows for determining the probability of any possible output sequence \(X_{n}=\{x_{1},x_{2},\ldots,x_{N}\}\) of length \(N\). All possible sequences can graphically be represented by the tree diagram of Figure 14: Every logical value at any sampling point can be followed by a '0' or a '1' with a certain probability, resulting in \(n=2^{(N+1)}\) overall branches, as the beginning of the sequence also has two potential starting points. The probability of a specific output sequence \(X_{n}=\{x_{1},x_{2},\ldots,x_{N}\}\) can then be calculated by multiplying the conditional probabilities along its path, and by adding the results of the path starting with an initial '0' and that starting with a '1':  
\[P_{\text{seq}}(\{x_{1},x_{2},\ldots,x_{N}\}) =P(\{0,x_{1},x_{2},\ldots,x_{N}\})+P(\{1,x_{1},x_{2},\ldots,x_{N}\})\] \[=P_{0}(x_{0})\cdot\prod_{i=1}^{N}P_{i}(x_{i}|x_{i-1})\bigg{|}_{x_ {0}=0}+P_{0}(x_{0})\cdot\prod_{i=1}^{N}P_{i}(x_{i}|x_{i-1})\bigg{|}_{x_{0}=1} \tag{36}\]  
Here, we assume that the transition from the previous symbol is instantaneous and that '0' and '1' are equally likely, which results in \(P_{0}(x_{0})=0.5\). Knowing the probability of all sequences allows for selecting the carrier samples \(C_{\text{seq}}=\{c_{1},c_{2},\ldots,c_{N}\}\) to be synchronized on, the cyclic permutation of the sampling sequence \(\{1,1,1,0,0,0,0\}\) having the largest value of \(P_{\text{seq}}\). Recalling the XNOR operation from (27) and the summation of eight samples from (29) together with the carrier sequence \(C_{\text{seq}}\), every input sequence \(X_{n}\) can be mapped to a certain demodulator output value \(s_{0}\in\{0,\ldots,8\}\):  
\[s_{0}(\{x_{1},x_{2},\ldots,x_{N}\})=\sum_{i=1}^{8}s_{\text{m},i}=\sum_{i=1}^{8 }\
eg(x_{i}\oplus c_{i}). \tag{37}\]  
Further defining a set \(\mathbf{X_{s}}\) that includes all sequences \(X_{n}\) resulting in an output sum of \(s_{0}\) with  
\[\mathbf{X_{s}}=\{X_{n}\mid s_{0}(X_{n})=s\}, \tag{38}\]  
we can express the probability of a certain output sum \(s_{0}\) by adding up the sequence probabilities \(P_{\text{seq}}\) that lead to the given output value:  
Figure 14: The possible sequences of samples \(X_{n}\) acquired for one period can be represented as a binary tree. The transition probabilities are given by the conditional probabilities \(P(x_{i}|x_{i-1})\). \[P_{\text{out}}\left(s\right)=\sum_{X_{n}\in X_{\mathbf{s}}}P_{\text{seq}}(X_{n}) \tag{39}\]  
With the symbol mapper defined in Section 5.3.2, an output of \(s_{o}>4\) is mapped to a symbol value '1', and \(s_{o}<4\) to '0', while \(s_{\text{o}}=4\) does not evoke a change in the symbol value, statistically leading to a bit error in half of the cases.