I am reading Digital Communication Systems by Simon Haykin and I am stuck at one point.

Consider a two-dimensional signal space that has a message constellation of four points, given by $s_1,s_2,s_3,s_4$. We need to calculate pairwise error probability in the presence of additive white Gaussian noise ($w$). The signal vectors are arranged so as to be aligned with the basis vectors, as shown in the figure


The observed vector $x$ is given as $$x=s_1+w$$ where $s_1$ is the transmitted message. We can calculate the probability of error using $\int_{Z_j} f_x(x|s_i)dx$ where $j\neq i$ and $Z$ is the observation space which is divided into various small $Z_i$'s for detection purposes. Assuming variance of noise to be $N_0/2$, we can write the error as: $$\int_{Z_j}\frac{1}{\sqrt{\pi N_0}}\exp(-(x-s_1)^2/N_0) dx$$ As shown in the figure


we take the bisector and define the distance between $s_1$ and $s_4$ as $d_{14}$, which gives the pairwise probability expression as: $$\int_{d_{14}/2}^{\infty}\frac{1}{\sqrt{\pi N_0}}\exp(-v^2/N_0) dv$$ where $d_{14}$ is the distance between constellation points $s_1$ and $s_4$ However, my confusion is what if we exceed the limit $d_{14}/2$ in another location as shown in the figure


As Euclidean distance is always positive, does the expression still hold? Please note I know that detection is going to work as the distance between $x$ and $s_1$ is smaller than between $x$ and $s_4$, but I am more concerned about this integral expression, how does it work here? $$\int_{d_{14}/2}^{\infty}\frac{1}{\sqrt{\pi N_0}}\exp(-v^2/N_0) dv$$ because the distance is greater than $d_{14}/2$.


1 Answer 1


If you read the book carefully, you will notice and understand that the probability given by the equation

$$\int_{d_{14}/2}^{\infty}\frac{1}{\sqrt{\pi N_0}}\exp(-v^2/N_0) dv$$

is the error probability for the event that the received signal x is closer to $s_k$ than $s_i$, when $s_i$ is sent. In your figure, it is the error probability that $x$ is closer to $s_4$ than $s_1$ when $s_1$ is sent, or $x$ is closer to $s_1$ than $s_4$ when $s_4$ is sent.

The situation which makes you confused is a different situation, in which $x$ is closer to $s_1$ than $s_4$ when $s_1$ is sent. This situation does not create an error decision, and makes no contribution to the error probability.

  • $\begingroup$ The other thing not being taken into account is the decision boundaries. For the four points show and zero mean, symmetric noise, I believe the decision boundaries will be the lines $y=x$ and $y=-x$. The given integral formula for the error is only applicable if the measurement point $x$ is on the line between $s_1$ and $s_4$. If it's not, $x$ may be closer to $s_2$ or $s_3$. $\endgroup$
    – Peter K.
    Jun 23, 2023 at 2:05
  • $\begingroup$ @Peter K. Please note we are talking about pairwise error probabilities here and therefore, the decision boundary is just between two signal points and not more than that. $\endgroup$
    – Userhanu
    Jun 23, 2023 at 4:02
  • $\begingroup$ @Userhanu OK. Seems like a toy problem, though, that could have been done with just two signal points $s_1$ and $s_2$. $\endgroup$
    – Peter K.
    Jun 23, 2023 at 18:17

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