signal analysis - Probability error for antipodal systems and orthogonal systems

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

Answer

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):

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 $\sqrt{2}$.