Bit Error Probability for 8PSK

Question

I'm trying to understand how to derive the BEP for 8PSK by geometrically looking at the decision regions.

Say I have this 8PSK signal set and the corresponding decision boundaries.

I assume that '000' is sent, and I want to calculate the BEP. So I have

$$P_b(e) = \frac{1}{3}(1 \times R_1 + 1 \times R_2 + 2 \times R_3 + 1 \times R_4 + 2 \times R_5 + 2 \times R_6 + 3 \times R_7)$$

Since $R_1 = R_4$, $R_3 = R_5$ and $R_2 = R_7$ by symmetry, the above can be simplified to

$$P_b(e) = \frac{2}{3}(R_1 + 2R_2 + 2R_3 + R_6)$$

Then we have

$$R_1 + R_2 + R_3 + R_6 = P\Big(n > d_1\sin\frac{\pi}{8}\Big)\tag{1}$$

and

$$R_2 + R_3 = P\Big(n_0 > d_1\sin\frac{3\pi}{8}\Big) \times P\Big(n_1 < d_1\sin\frac{\pi}{8}\Big)\tag{2}$$

I'm having a hard time visualizing and understanding how $(1)$ and $(2)$ are obtained.

I understand that $n$ is the noise vector, and $n_0$ and $n_1$ are the horizontal and vertical components of the noise vector, and we are trying to find out how far $n$ or $n_0$ and $n_1$ will go to "push" the signal into the wrong decision region.

I've seen the following from this website, and it seems to explain $(1)$, that is, the noise vector $n$ "pushes" the signal past the decision line (dl) into the wrong decision regions $R_1 + R_2 + R_3 + R_6$.

But what about $(2)$? I don't know why we have $\sin\frac{3\pi}{8}$ and $\sin\frac\pi8$.

Also, is the noise vector always perpendicular to the decision boundaries?

Answer 1

The way I learned this is there's two ways to find the probability of error. You can find the exact or as estimated probability. Exact is typically calculated for signal space where the points are in a rectangular pattern, because you're decision regions are nice and easy. You would see this rectangular pattern in a 4-QAM, 16-QAM, etc.

For estimated probability, we use the Union Bound. This states that the symbol probability of error = $P_{se|1} = \sum P_{se}($between point 1 and each other individual point$)$. Out total symbol probability of error

$P_{se} = $$\frac{\sum_{i=1}^M P_{es|i} }{M} $ where M is the number of points on our signal space ($M= 8$ in this case)

This means we're take a single point, and then with each other individual point we're going to create a new decision boundaries exactly half way between to the two and find the probability of error for just those two points. We do that for all the points, add them up, and now we have an estimated probability of error for the symbol. So let's say our noise is Gaussian, the variance of our noise $\sigma^2 = \frac{N_0}{2}$. Let's say D is the distance between two points. The probability of error between those two points $= Q(\frac{D}{2\sigma}) = Q(\frac{\sqrt2 \cdot D}{\sqrt N_0})$ So, let's look at point 000 in your examples. We can use cosine theory to find the distance between two points.

$D^2 = D_1^2 + D_2^2 -2\cdot D_1 \cdot D_2 \cdot cos(\theta)$ where $\theta$ is the angle between two points. Note $D_1$ and $D_2$ are the distances from the origin of our points, and in this case $D_1 = D_2$ because all points are the same distance from the origin.

\begin{align} P_{se} &= Q(\frac{d_1\cdot \sqrt {2+\sqrt2}}{2 \sigma}) + Q(\frac{d_1\cdot \sqrt {2+\sqrt2}}{2 \sigma}) +Q(\frac{\sqrt2\cdot d_1}{2 \sigma}) + Q(\frac{\sqrt2\cdot d_1}{2 \sigma}) +\ldots\\ P_{se} &= 2\cdot Q(\frac{d_1\cdot \sqrt {2+\sqrt2}}{2 \sigma}) + 2 \cdot Q(\frac{\sqrt2\cdot d_1}{2 \sigma})+\ldots \end{align}

We could go on, but since the $Q$ function falls off so quickly, we really only care about the point(s) closes to what we're comparing too. All the other terms are insignificant cared to that. So we end up with $P_se = 2\cdot Q(\frac{d_1\cdot \sqrt {2+\sqrt2}}{2 \sigma})$.

That's the symbol probability of error if we look at the point 000. We can calculate this for the other points, but it's just going to be the exact same due to symmetry. Since the Probability of error is the same for each symbol, we know the probability of error for all symbols is what we found above.

This is for symbol probability of error. To get to bit probability of error we have to look at our bit mapping. Since each point will only have two points that are closest to it, we can implement Gray coding. That means we can divide our symbol probability by $log_2(M)$ where $M =$ the number of different points we have on our signal space, which is 8 in this case. Therefore our $P_{be}$ is

$P_{be} = \frac{P_{es}}{log_2(8)} = \frac{2\cdot Q(\frac{d_1\cdot sqrt(2+sqrt(2))}{(2 \sigma)})}{3}$

Answer 2

I will borrow the 8PSK constellation diagram and bit-labeling used by the OP and also the names $R_1$ through $R_8$ of the 8 decision regions, but not the meanings (probabilities) used by the OP in his explanations of $R_1$ through $R_8$. That is, if the observation $(X,Y)$ lies in $R_i$, it is demodulated/decoded into the three bits that label the constellation point in the region $R_i$.

The unstated (but standard) assumptions are that we are looking at 8PSK signaling on an AWGN channel with two-sided power spectral density $\frac{\mathscr N_0}{2}$ and thus the 8 constellation points lie on a circle of radius $\sqrt{\mathscr E_s} = \sqrt{3\mathscr E_b}$ centered at the origin. Here $\mathscr E_s$ is the received energy per symbol and $\mathscr E_b$ the received energy per bit. Conditioned on which symbol has been transmitted, the receiver observation is $(X,Y)$ where $X$ and $Y$ are independent Gaussian random variables with variance $\frac{\mathscr N_0}{2}$ and mean point the constellation point corresponding to the given transmitted signal. Put another way, the joint distribution of $X$ and $Y$ has circular symmetry about the mean point $(\mu_X, \mu_Y)$, and it is a standard result in signal space theory that the probability that $(X,Y)$ lies on the other side of a straight line is $Q\left(\dfrac{d}{\sqrt{\mathscr N_0/2}}\right)$ where $d$ is the distance of the line from the mean point.

There are three different bit error events whose probabilities that we need to calculate: for $i=1,2,3$, $E_i$ is the event that the $i$-th bit is in error. These events are not independent, nor do they have identical probabilities.

Analysis for $E_1$: Suppose that $000$ is transmitted. Then, looking at the borrowed diagram above, we see that event $E_1$ occurs if $(X,Y) \in R_4\cup R_5\cup R_6 \cup R_7$ which compound region has boundary the straight line of slope $-\frac{\pi}{8}$ through the origin. This straight line is at distance $\sin\left(\frac{\pi}{8}\sqrt{\mathscr E_s}\right)$ from the mean point $\left(\sqrt{\mathscr E_s},0\right)$. Hence, $$P(E_1\mid 000) = Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right).\tag{1}$$ Using symmetry, we get that $$P(E_1\mid 010) = Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)\tag{2}$$ also, while $$P(E_1\mid 001) = P(E_1\mid 011) = Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)\tag{3}$$ because the constellation points $001$ and $011$ are at distance $\sin\left(\frac{3\pi}{8}\sqrt{\mathscr E_s}\right)$ from the boundary of $R_4\cup R_5 \cup R_6 \cup R_7$. Note that $$Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right) < Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)$$ which makes sense; the boundary is farther away from these points.

A similar analysis can be carried out for the other $4$ constellation points to arrive at the conclusion that $$P(E_1\mid 110) = P(E_1\mid 100) = Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)\tag{4}$$ $$P(E_1\mid 101) = P(E_1\mid 111) = Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right).\tag{5}$$ Under the usual assumption that all $8$ constellation points have equal probability $\frac 18$, we conclude that $$P(E_1) = \frac{1}{2}\left[Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right) + Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)\right].\tag{6}$$

Analysis for $E_2$: Suppose that $000$ is transmitted. Then, looking at the borrowed diagram above, we see that event $E_2$ occurs if $(X,Y) \in R_2\cup R_3\cup R_6 \cup R_7$ which compound region has boundary which is the straight-line of slope $\frac{3\pi}{8}$ through the origin. Going through a similar analysis mutatis mutandis, we get that $$P(E_2\mid 000) = P(E_2\mid 100) = Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right), \tag{7}$$ and $$P(E_2\mid 001) = P(E_2\mid 101) = Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right).\tag{8}$$ The conditional probability of $E_2$ given the other points can be calculated in a similar fashion and thus we arrive at $$P(E_2) = P(E_1) = \frac{1}{2}\left[Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right) + Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)\right].\tag{9}$$

Analysis for $E_3$: This one is more complicated.

Suppose that $000$ is transmitted. Then, looking at the borrowed diagram above, we see that event $E_3$ occurs if $(X,Y) \in R_1\cup R_3\cup R_5 \cup R_7$, which is two quadrants that touch at the origin. But note that the region of interest can be expressed as the Exclusive-OR union of the two half-planes whose boundaries are the lines through the origin of slopes $\frac{\pi}{8}$ and $-\frac{3\pi}{8}$. In symbols, $$R_1\cup R_3\cup R_5 \cup R_7 = (R_1\cup R_3\cup R_2 \cup R_6) \oplus (R_5 \cup R_7 \cup R_2 \cup R_6).$$ Now, \begin{align} P(A\oplus B) &= P(A) + P(B) -2P(A\cap B)\\ &= P(A) + P(B) - 2P(A)P(B) & \scriptstyle{\text{for independent $A$ and $B$}} \end{align} and since $000$ is at distance $\sin\left(\frac{\pi}{8}\sqrt{\mathscr E_s}\right)$ from one boundary and at distance $\sin\left(\frac{3\pi}{8}\sqrt{\mathscr E_s}\right)$ from the other boundary, we get that \begin{align} P(E_3 \mid 000) &= Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right) + Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)\\ &\;\;- 2 Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right) Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right).\tag{10} \end{align} Those who have followed the analysis thus far should have no trouble understanding why $$P(E_3 \mid 100) = P(E_3 \mid 000) = P(E_3\mid 010) = P(E_3 \mid 110)$$ and indeed why \begin{align}P(E_3) &= Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right) + Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)\\ &\;\;- 2 Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right) Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right).\tag{11} \end{align} Note that $P(E_3)$ is larger than $P(E_1) = P(E_2)$. The latter probability is the arithmetic average of $Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)$ and $Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)$ while the former is slightly smaller than the sum of $Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)$ and $Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)$. Thus, we have the approximation that is worth keeping in mind: $$P(E_3)\approx 2P(E_1) = 2P(E_2).$$

Those interested in finding a single formula for the BER of a Gray-coded 8PSK system operating on an AWGN channel can take the average of the three error probabilities found above to arrive at \begin{align} \overline{P(E)} &= \left.\left.\frac 13\right[P(E_1) + P(E_2) + P(E_3)\right]\\ &= \left.\left.\frac 23\right[Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right) + Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)\right.\\ &\qquad\quad - Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)\cdot Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{2\mathscr E_s}{\mathscr N_0}}\right)\bigg]\tag{12}\\ &= \left.\left.\frac 23\right[Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{6\mathscr E_b}{\mathscr N_0}}\right) + Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{6\mathscr E_b}{\mathscr N_0}}\right)\right.\\ &\qquad\quad - Q\left(\sin\left(\frac{\pi}{8}\right)\sqrt{\frac{6\mathscr E_b}{\mathscr N_0}}\right)\cdot Q\left(\sin\left(\frac{3\pi}{8}\right)\sqrt{\frac{6\mathscr E_b}{\mathscr N_0}}\right)\bigg]\tag{13}.\\ \end{align} In the form $(13)$ above, the average error probability formula can be found (without the above details) at the bottom of page 340 of M.B. Pursley's Introduction to Digital Communications, Pearson Prentice-Hall, 2005, with a citation of "Computation of the Bit Error Rate of Coherent M-ary PSK with Gray Code Bit Mapping" by P.Lee, IEEE Transactions on Communications, May 1986 (which is behind IEEE's paywall for many people) as the source of the result. Note that since the last term in $(12)$ and $(13)$ is considerably smaller than the first two terms, $$\overline{P(E)} \approx \frac 43 P(E_1)\approx \frac 23 P(E_3),$$ but I think considering the three bit error probabilities separately is more informative than looking at just the average BER. Bear in mind that the three error events are not independent events.

So, there you have it, folks, the exact expressions for the BER(s) sustained by the three bits transmitted in a 8PSK scheme with Gray coding operating on an AWGN channel right here on dsp.SE without the need to look elsewhere. I know it won't satisfy @Loran but I hope the rest of you find it useful.