# Bit Error Probability for 8PSK

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?

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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. 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}) +... P_{se} = 2\cdot Q(\frac{d_1\cdot \sqrt {2+\sqrt2}}{2 \sigma}) + 2 \cdot Q(\frac{\sqrt2\cdot d_1}{2 \sigma}) +... 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} • Don't forget to mark the answer as accepted if it answered everything you wanted. If it didn't answer everything you wanted, let me know and i'll try to make it more complete. Thanks! – gerrgheiser Mar 6 '16 at 14:36 • The symbol error probability calculation is incorrect (though the end result is nearly right because of the rapid decrease of Q(x)). The probability of a symbol error is the probability of that the received signal ((I,Q) output) is outside the octant. For example, as shown on pp.153-154 of this ancient lecture note of mine,$$Q\left(\sin\frac{\pi}{8}\sqrt{\frac{6E_b}{N_0}}\right)<P_{s}<2Q\left(\sin \frac{\pi}{8}\sqrt{\frac{6E_b}{N_0}}\right)$$The exact expression for$P_b\$ is also known. – Dilip Sarwate Mar 6 '16 at 15:19