# Why the error probability is given by this expression?

I have seen this equation:

$$\Pr({\epsilon}) = Q({\sqrt{2E_b/N_0}}) \,.$$

where these definitions hold:

$$N_0/2 - the \ spectral \ density \ of \ white \ noise$$ $$E_b = PT_b - the \ energy \ per \ transmitted \ bit$$ $$P - Power$$ $$T_b - the \ time\ the \ bit\ transmitted \ in$$

Q - is (this) function, how do I derive this equation?

• Your error probability expression is incorrect; the square-root sign needs to extend over $N_0$. You can find a derivation of the correct expression in a thread titled Antipodal BER - help" on groups.google.com and also here. – Dilip Sarwate Feb 4 '12 at 14:03
• "corrected,thanks" Good. Did you find what you needed in the material I referred you to? For a lot more detailed analysis, you may want to read the notes on matched filters found here – Dilip Sarwate Feb 4 '12 at 22:59
• Great material; thanks for making it available. I would also recommend Lecture 3 Appendix A for what I found to be the most important "missing link" in communication theory: how the continuous-time "waveform channel" maps to the discrete-time "vector channel." That is, how does the waveform observed by a receiver over a symbol period map to a discrete decision problem? Although a bit tough to read, my favorite treatment is the classic Wozencraft and Jacobs. – Jason R Feb 5 '12 at 1:53
• The URL given in my previous comment has been changed by the powers that be. The materials referred to are available at this corrected location – Dilip Sarwate Jan 25 '13 at 17:37

## 1 Answer

A whole chapter of a book can be written to answer the "Why" in this question but the short answer is that if the signals used are an antipodal signal set (meaning bits $0$ and $1$ are transmitted using signals $s_0(t)$ and $s_1(t)$ respectively where $s_1(t) = -s_0(t)$ and the signals are of duration $T$ and energy $E$, then with matched filtering, the receiver makes a decision that $0$ was transmitted or a $1$ was transmitted according as the sample value from the matched filter is positive or negative. The sample value is a random variable $X$ whose distribution is $\mathcal N\left(\pm \sqrt{E}, N_0/2\right)$ depending on whether a $0$ or a $1$ was transmitted. If $X$ has mean $-\sqrt{E}$ when a $1$ is transmitted, then the probability of error is \begin{align*} P_{e,1} &= P\{X > 0 \mid 1~\text{transmitted}\}\\ &= P\left\{\mathcal N\left(-\sqrt{E}, N_0/2\right) > 0\right\}\\ &= Q\left(\frac{0 - \left(-\sqrt{E}\right)}{\sqrt{N_0/2}}\right)\\ &= Q\left(\sqrt{\frac{2E}{N_0}}\right). \end{align*} Similarly, $X$ has mean $\sqrt{E}$ when a $0$ is transmitted, and so the probability of error is \begin{align*} P_{e,0} &= P\{X < 0 \mid 0~\text{transmitted}\}\\ &= P\left\{\mathcal N\left(\sqrt{E}, N_0/2\right) < 0\right\}\\ &= \Phi\left(\frac{0 -\sqrt{E}}{\sqrt{N_0/2}}\right)\\ &= \Phi\left(-\sqrt{\frac{2E}{N_0}}\right)\\ &= Q\left(\sqrt{\frac{2E}{N_0}}\right). \end{align*} Finally, using the law of total probability, the probability of error is $$P_e = P_{e,0}P\{0~\text{transmitted}\} + P_{e,1}P\{1~\text{transmitted}\} = Q\left(\sqrt{\frac{2E}{N_0}}\right)$$ regardless of the probabilities of transmitting $0$'s and $1$'s. For more details such as why the matched filters outputs are as stated above, see, for example, this lecture note and this one of mine.