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I know that the error probability for antipodal systems is Q (√((2Fs)/No)) and for orthogonal systems it is Q (√(Fs/No)), however, can anybody give their derivation? Because I am finding different ones on the internet I am getting confused.

I found pages 74-75 useful but I think their derivations are quite short here: http://www.sps.ele.tue.nl/members/F.M.J.Willems/TEACHING_files/5JK00/signalenergyorthogonalsignals.pdf

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    $\begingroup$ Do you know what you mean by $F_s$? $\endgroup$ – Matt L. Apr 9 '14 at 20:31
  • $\begingroup$ usually it's the sampling frequency however in this case it can be Eb but I also found it Fs online $\endgroup$ – user1930901 Apr 9 '14 at 20:42
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I believe that Dilip Sarwate's answer has all the necessary information, but I would like to add a few things that might make it a bit easier for the less initiated to get a better understanding of the formula for the error probability. Furthermore, I want to point out one reason why you can find in different texts slightly different versions of this formula.

First, I think it is important to see that the term in the numerator of the argument of the $Q(\cdot)$ function is simply the distance between the two signals $s_0(t)$ and $s_1(t)$:

$$\sqrt{E_0 + E_1 - 2 \langle s_0, s_1 \rangle}=\sqrt{||s_0||^2+||s_1||^2-2\langle s_0, s_1 \rangle}=\sqrt{||s_0-s_1||^2}=||s_0-s_1||$$

For $||s_0||^2=||s_1||^2=E$, this distance is $2\sqrt{E}$ for antipodal signaling, and $\sqrt{2E}$ for orthogonal signaling, as shown in the figure below (from Digital Communication by E.A. Lee and D.G. Messerschmitt):

signal distances for antipodal and orthogonal signaling

This picture very cleary illustrates the reason why orthogonal signaling requires 3dB more SNR for the same probability of error as antipodal signaling.

Noting that the argument of the $Q(\cdot)$ function is actually half the distance between the signals divided by the square root of the noise variance (which, admittedly, is not rigorously shown in this answer), the value of the $Q(\cdot)$ function is simply the probability that the noise is greater than half the distance between the two signals. Only then will the received signal be closer to the wrong signal, and, consequently, the decision will be wrong.

As for the formula for the error probabiltiy, a source of confusion is the definition of the noise power spectral density $N_0$. There is the one-sided spectral density $N_{0,1}$ and the two-sided density $N_{0,2}$, and they are related by $N_{0,1}=2N_{0,2}$. Unfortunately, everybody just denotes them by $N_0$, so you always have to check which one is meant. For antipodal signaling, the formula for the error probability is

$$P_e=Q\left(\sqrt{\frac{2E}{N_{0,1}}}\right)=Q\left(\sqrt{\frac{E}{N_{0,2}}}\right)$$

And for orthogonal signaling we have

$$P_e=Q\left(\sqrt{\frac{E}{N_{0,1}}}\right)=Q\left(\sqrt{\frac{E}{2N_{0,2}}}\right)$$

The formulas in Dilip Sarwate's answer use $N_0$ as a one-sided power spectral density, or - equivalently (as stated in his answer) - $\frac{N_0}{2}$ as a two-sided density. In any case, the difference between antipodal and orthogonal signaling is always a factor of $\sqrt{2}$.

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  • $\begingroup$ I don't think that the difference between $N_{0,1}=N_0$ and $N_{0,2}=\frac{N_0}{2}$ that you discuss is something people use in practice. The key number is $N_0$, measured in watts per Hertz, and it means that an ideal brickwall filter of bandwidth $B$ Hz has noise power $N_0B$ at its output. Ditto for a practical filter with _noise bandwidth $B$ Hz. (This works for low-pass as well as bandpass filters; $N_0B$ always). Variations of the $P_e$ formula sometimes are stated in terms of signal power, bit duration, or signal bandwidth, and in these cases, it is always $N_0B$ that shows up. $\endgroup$ – Dilip Sarwate Apr 10 '14 at 14:46
  • $\begingroup$ @DilipSarwate My point is that different textbooks use the symbol $N_0$ to denote different things. In some, $N_0$ is the two-sided noise power spectral density, in others it is the one-sided noise psd. One annoying consequence is that the formulas for the error probability for antipodal and orthogonal signaling may look exactly the same, if in one case $N_0$ is used as one-sided psd, and in the other as two-sided psd. This might be the difference that the OP refers to. $\endgroup$ – Matt L. Apr 11 '14 at 7:05
  • $\begingroup$ Could you provide citations for textbooks that use $N_0$ as the value of the two-sided noise psd? I own most of the textbooks in communications theory published over the past thirty years (and have taught out of several of them too) and I don't recall $N_0$ being the value of the two-sided psd of white noise in any of them. So I am curious to see which communications theory texts I haven't read at all. $\endgroup$ – Dilip Sarwate Apr 11 '14 at 13:10
  • $\begingroup$ Sure! (I wouldn't make up something like this, would I?) Take a look at "Digital Communication" by Lee and Messerschmitt (I'm referring to the second printing of the first edition, 1990). They use $S_X(j\omega)=N_0$ for the power spectrum of a white random process $X(t)$, so the noise power in a frequency band of width $B$ Hz is $2BN_0$. $\endgroup$ – Matt L. Apr 11 '14 at 13:59
  • $\begingroup$ Weird! Thanks for pointing that out. I have 28 books on communication theory (yes, I counted them) in my bookcase, and Lee and Messerschmidt is the only one that uses $N_0$ as the value of the two-sided psd. It just adds to the many reasons why I never ever used it as a textbook in my classes. $\endgroup$ – Dilip Sarwate Apr 11 '14 at 19:00
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For equally likely binary signals $s_0(t)$ and $s_1(t)$ of energies $E_0$ and $E_1$ respectively in an AWGN channel with two-sided power spectral density $\frac{N_0}{2}$, the error probability is given by $$P_e = Q\left(\sqrt{\frac{E_0 + E_1 - 2 \langle s_0, s_1 \rangle}{2N_0}}\right) \tag{1}$$ where $\displaystyle \langle s_0, s_1 \rangle = \int_{-\infty}^\infty s_0(t)s_1(t)\,\mathrm dt$ is the inner product of the two signals.

For antipodal signals, $E_0 = E_1 = E$ and $\langle s_0, s_1 \rangle = -E$ giving everyone's favorite answer $Q\left(\sqrt{2E/N_0}\right)$ while for equal energy orthogonal signals, the inner product is $0$ giving everybody's second favorite answer $Q\left(\sqrt{E/N_0}\right)$. For some details of the proof of $(1)$, see this answer that I wrote some time ago. For even more detail, see this Lecture Note of mine.

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