Considered optimal system with "detector" and "decision"
The communication system given by the graph is to be investigated. The binary message $m ∈ \{m_0, m_1\}$ with equal occurrence probabilities
- $${\rm Pr} (m_0 ) = {\rm Pr} (m_1 ) = 0.5$$
is represented by the two signals
- $$s_0 = \sqrt{E_s} \hspace{0.05cm},\hspace{0.2cm}s_1 = -\sqrt{E_s}$$
where the assignments $m_0 ⇔ s_0$ and $m_1 ⇔ s_1$ are one-to-one.
The detector $($highlighted in green in the figure$)$ provides two decision values
- $$r_1 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} s + n_1\hspace{0.05cm},$$
- $$r_2 \hspace{-0.1cm} \ = \ \hspace{-0.1cm} n_1 + n_2\hspace{0.05cm},$$
from which the decision forms the estimated values $\mu ∈ \{m_0,\ m_1\}$ for the transmitted message $m$. The decision includes
- two weighting factors $K_1$ and $K_2$,
- a summation point, and
- a threshold decision with the threshold at $0$.
Three evaluations are considered in this exercises:
- Decision based on $r_1$ $(K_1 ≠ 0,\ K_2 = 0)$,
- decision based on $r_2$ $(K_1 = 0,\ K_2 ≠ 0)$,
- joint evaluation of $r_1$ und $r_2$ $(K_1 ≠ 0,\ K_2 ≠ 0)$.
Notes:
- The exercise belongs to the chapter "Structure of the Optimal Receiver" of this book.
- In particular, "the irrelevance theorem" is referred to here, but besides that also the "Optimal receiver for the AWGN channel".
- For more information on topics relevant to this exercise, see the following links:
- For the error probability of a system $r = s + n$ $($because of $N = 1$ here $s,\ n,\ r$ are scalars$)$ is valid:
- $$p_{\rm S} = {\rm Pr} ({\rm symbol\ \ error} ) = {\rm Q} \left ( \sqrt{{2 E_s}/{N_0}}\right ) \hspace{0.05cm},$$
- where a binary message signal $s ∈ \{s_0,\ s_1\}$ with $s_0 = \sqrt{E_s} \hspace{0.05cm},\hspace{0.2cm}s_1 = -\sqrt{E_s}$ is assumed.
- Let the two noise sources $n_1$ and $n_2$ be independent of each other and also independent of the transmitted signal $s ∈ \{s_0,\ s_1\}$.
- $n_1$ and $n_2$ can each be modeled by AWGN noise sources $($white, Gaussian distributed, mean-free, variance $\sigma^2 = N_0/2)$.
- For numerical calculations, use the values
- $$E_s = 8 \cdot 10^{-6}\,{\rm Ws}\hspace{0.05cm},\hspace{0.2cm}N_0 = 10^{-6}\,{\rm W/Hz} \hspace{0.05cm}.$$
- The "complementary Gaussian error function" gives the following results:
- $${\rm Q}(0) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.5\hspace{0.05cm},\hspace{1.35cm}{\rm Q}(2^{0.5}) = 0.786 \cdot 10^{-1}\hspace{0.05cm},\hspace{1.1cm}{\rm Q}(2) = 0.227 \cdot 10^{-1}\hspace{0.05cm},$$
- $${\rm Q}(2 \cdot 2^{0.5}) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} 0.234 \cdot 10^{-2}\hspace{0.05cm},\hspace{0.2cm}{\rm Q}(4) = 0.317 \cdot 10^{-4} \hspace{0.05cm},\hspace{0.2cm}{\rm Q}(4 \cdot 2^{0.5}) = 0.771 \cdot 10^{-8}\hspace{0.05cm}.$$
Questions
Solution
(1) The last alternative solution is correct:
- In general, the MAP receiver leads to a smaller error probability.
- However, if the occurrence probabilities ${\rm Pr}(m = m_0) = {\rm Pr}(m = m_1) = 0.5$ are equal, both receivers yield the same result.
(2) With $K_2 = 0$ and $K_1 = 1$ the result is
- $$r = r_1 = s + n_1\hspace{0.05cm}.$$
- With bipolar (antipodal) transmitted signal and AWGN noise, the error probability of the optimal receiver (whether implemented as a correlation or matched filter receiver) is equal to
- $$p_{\rm S} = {\rm Pr} ({\rm symbol\hspace{0.15cm} error} ) = {\rm Q} \left ( \sqrt{ E_s /{\sigma}}\right ) = {\rm Q} \left ( \sqrt{2 E_s /N_0}\right ) \hspace{0.05cm}.$$
- With $E_s = 8 \cdot 10^{\rm –6} \ \rm Ws$ and $N_0 = 10^{\rm –6} \ \rm W/Hz$, we further obtain:
- $$p_{\rm S} = {\rm Pr} ({\rm symbol\hspace{0.15cm}error} ) = {\rm Q} \left ( \sqrt{\frac{2 \cdot 8 \cdot 10^{-6}\,{\rm Ws}}{10^{-6}\,{\rm W/Hz} }}\right ) = {\rm Q} (4) \hspace{0.05cm}\hspace{0.15cm}\underline {= 0.00317 \%}\hspace{0.05cm}.$$
- This result is independent of $K_1$, since amplification or attenuation changes the useful power in the same way as the noise power.
(3) With $K_1 = 0$ and $K_2 = 1$, the decision variable is:
- $$r = r_2 = n_1 + n_2\hspace{0.05cm}.$$
- This contains no information about the useful signal, only noise, and it holds independently of $K_2$:
- $$p_{\rm S} = {\rm Pr} ({\rm symbol\hspace{0.15cm}error} ) = {\rm Q} (0) \hspace{0.15cm}\underline {= 50\%} \hspace{0.05cm}.$$
(4) Because of ${\rm Pr}(m = m_0) = {\rm Pr}(m = m_1)$, the decision rule of the optimal receiver (whether realized as MAP or as ML) is:
- $$\hat{m} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm arg} \max_i \hspace{0.1cm} [ p_{m \hspace{0.05cm}|\hspace{0.05cm}r_1, \hspace{0.05cm}r_2 } \hspace{0.05cm} (m_i \hspace{0.05cm}|\hspace{0.05cm}\rho_1, \hspace{0.05cm}\rho_2 ) ] = {\rm arg} \max_i \hspace{0.1cm} [ p_{r_1, \hspace{0.05cm}r_2 \hspace{0.05cm}|\hspace{0.05cm} m} \hspace{0.05cm} ( \rho_1, \hspace{0.05cm}\rho_2 \hspace{0.05cm}|\hspace{0.05cm} m_i ) \cdot {\rm Pr} (m = m_i)] = {\rm arg} \max_i \hspace{0.1cm} [ p_{r_1, \hspace{0.05cm}r_2 \hspace{0.05cm}|\hspace{0.05cm} s} \hspace{0.05cm} ( \rho_1, \hspace{0.05cm}\rho_2 \hspace{0.05cm}|\hspace{0.05cm} s_i ) ] \hspace{0.05cm}.$$
- This composite probability density can be rewritten as follows:
- $$\hat{m} ={\rm arg} \max_i \hspace{0.1cm} [ p_{r_1 \hspace{0.05cm}|\hspace{0.05cm} s} \hspace{0.05cm} ( \rho_1 \hspace{0.05cm}|\hspace{0.05cm} s_i ) \cdot p_{r_2 \hspace{0.05cm}|\hspace{0.05cm} r_1, \hspace{0.05cm}s} \hspace{0.05cm} ( \rho_2 \hspace{0.05cm}|\hspace{0.05cm} \rho_1, \hspace{0.05cm}s_i ) ] \hspace{0.05cm}.$$
- Now, since the second multiplicand also depends on the message ($s_i$), $r_2$ should definitely be included in the decision process. Thus, the correct answer is: YES.
(5) For AWGN noise with variance $\sigma^2$, the two composite densities introduced in (4) together with their product $P$ give:
- $$p_{r_1\hspace{0.05cm}|\hspace{0.05cm} s}(\rho_1\hspace{0.05cm}|\hspace{0.05cm}s_i) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{\sqrt{2\pi} \cdot \sigma}\cdot {\rm exp} \left [ - \frac{(\rho_1 - s_i)^2}{2 \sigma^2}\right ]\hspace{0.05cm},\hspace{1cm} p_{r_2\hspace{0.05cm}|\hspace{0.05cm}r_1,\hspace{0.05cm} s}(\rho_2\hspace{0.05cm}|\hspace{0.05cm}\rho_1, \hspace{0.05cm}s_i) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{\sqrt{2\pi} \cdot \sigma}\cdot {\rm exp} \left [ - \frac{(\rho_2 - (\rho_1 - s_i))^2}{2 \sigma^2}\right ]\hspace{0.05cm}$$
- $$ \Rightarrow \hspace{0.3cm} P \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \frac{1}{{2\pi} \cdot \sigma^2}\cdot {\rm exp} \left [ - \frac{1}{2 \sigma^2} \cdot \left \{ (\rho_1 - s_i)^2 + (\rho_2 - (\rho_1 - s_i))^2\right \}\right ]\hspace{0.05cm}.$$
- We are looking for the argument that maximizes this product $P$, which at the same time means that the expression in the curly brackets should take the smallest possible value:
- $$\mu = \hat{m} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}{\rm arg} \max_i \hspace{0.1cm} P = {\rm arg} \min _i \hspace{0.1cm} \left \{ (\rho_1 - s_i)^2 + (\rho_2 - (\rho_1 - s_i))^2\right \} $$
- $$\Rightarrow \hspace{0.3cm} \mu = \hat{m} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}{\rm arg} \min _i \hspace{0.1cm}\left \{ \rho_1^2 - 2\rho_1 s_i + s_i^2 + \rho_2^2 - 2\rho_1 \rho_2 + 2\rho_2 s_i+ \rho_1^2 - 2\rho_1 s_i + s_i^2\right \} \hspace{0.05cm}.$$
- Here $\mu$ denotes the estimated value of the message. In this minimization, all terms that do not depend on the message $s_i$ can now be omitted. Likewise, the terms $s_i^2$ are disregarded, since $s_0^2 = s_1^2$ holds. Thus, the much simpler decision rule is obtained:
- $$\mu = {\rm arg} \min _i \hspace{0.1cm}\left \{ - 4\rho_1 s_i + 2\rho_2 s_i \right \}={\rm arg} \min _i \hspace{0.1cm}\left \{ (\rho_2 - 2\rho_1) \cdot s_i \right \} \hspace{0.05cm}.$$
- So, correct is already the proposed solution 2. But after multiplication by $–1/2$, we also get the last mentioned decision rule:
- $$\mu = {\rm arg} \max_i \hspace{0.1cm}\left \{ (\rho_1 - \rho_2/2) \cdot s_i \right \} \hspace{0.05cm}.$$
- Thus, the solutions 2 and 3 are correct.
(6) Setting $K_1 = 1$ and $\underline {K_2 = \, -0.5}$, the optimal decision rule with realization $\rho = \rho_1 \, – \rho_2/2$ is:
- $$\mu = \left\{ \begin{array}{c} m_0 \\ m_1 \end{array} \right.\quad \begin{array}{*{1}c} {\rm f{or}} \hspace{0.15cm} \rho > 0 \hspace{0.05cm}, \\ {\rm f{or}} \hspace{0.15cm} \rho < 0 \hspace{0.05cm}.\\ \end{array}$$
- Since $\rho = 0$ only occurs with probability $0$, it does not matter in the sense of probability theory whether one assigns the message $\mu = m_0$ or $\mu = m_1$ to this event "$\rho = 0$".
(7) With $K_2 = \, -0.5$ one obtains for the input value of the decision:
- $$r = r_1 - r_2/2 = s + n_1 - (n_1 + n_2)/2 = s + n \hspace{0.3cm} \Rightarrow \hspace{0.3cm} n = \frac{n_1 - n_2}{2}\hspace{0.05cm}.$$
- The variance of this random variable is
- $$\sigma_n^2 = {1}/{4} \cdot \left [ \sigma^2 + \sigma^2 \right ] = {\sigma^2}/{2}= {N_0}/{4}\hspace{0.05cm}.$$
- From this, the error probability is analogous to subtask (2):
- $${\rm Pr} ({\rm symbol\hspace{0.15cm}error} ) = {\rm Q} \left ( \sqrt{\frac{E_s}{N_0/4}}\right ) = {\rm Q} \left ( 4 \cdot \sqrt{2}\right ) = {1}/{2} \cdot {\rm erfc}(4) \hspace{0.05cm}\hspace{0.15cm}\underline {= 0.771 \cdot 10^{-8}} \hspace{0.05cm}.$$
- Thus, by taking $r_2$ into account, the error probability can be lowered from $0.317 \cdot 10^{\rm –4}$ to the much smaller value of $0.771 \cdot 10^{-8}$, although the decision component $r_2$ contains only noise. However, this noise $r_2$ allows an estimate of the noise component $n_1$ of $r_1$.
- Halving the transmit energy from $8 \cdot 10^{\rm –6} \ \rm Ws$ to $4 \cdot 10^{\rm –6} \ \rm Ws$, we still get the error probability $0.317 \cdot 10^{\rm –4}$ here, as calculated in subtask (2). When evaluating $r_1$ alone, on the other hand, the error probability would be $0.234 \cdot 10^{\rm –2}$.
