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Antipodal Signalling has a bit error probability of $P_b = \frac{1}{2} \mathrm{erfc}\big(\sqrt{\frac{E_s}{N_0}}\big)$ and on-off-signalling has a bit error probability of $\frac{1}{2} \mathrm{erfc}\big(\sqrt{\frac{E_s}{2 N_0}}\big)$. Why does this correspond to a higher SNR of 3 dB which we need for on-off-signalling to achieve the same bit error probability as with antipodal signalling?

I know that the factor 2 corresponds to 3 dB. But the complementary error function does not decrease by a factor of 2 but by a factor of $\sqrt{2}$

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Assuming your bit error probability formulas are correct, for antipodal signalling

$$P_{b, aS} = \frac{1}{2} \mathrm{erfc}\left(\sqrt{\frac{E_{s,aS}}{N_0}}\right)$$

and for on-off signalling

$$P_{b, oS} = \frac{1}{2} \mathrm{erfc}\left(\sqrt{\frac{E_{s,oS}}{2N_0}}\right)$$

If these two error probability are identical, i.e. $P_{b, aS} = P_{b, oS}$,

$$\frac{1}{2} \mathrm{erfc}\left(\sqrt{\frac{E_{s,aS}}{N_0}}\right) = \frac{1}{2} \mathrm{erfc}\left(\sqrt{\frac{E_{s,oS}}{2N_0}}\right) \tag{1}$$

Because both $\mathrm{erfc(.)}$ and $\sqrt{.}$ are bijective functions, $(1)$ implies $\frac{E_{s,oS}}{N_0} = 2\frac{E_{s,aS}}{N_0}$ in linear scale and 3dB difference in log scale.

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  • $\begingroup$ great, thx very much! $\endgroup$ – user32151 Nov 22 '17 at 15:54
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Looking at the error probability formula in terms of the average energy only makes one miss important aspects of the comparison. In antipodal signaling, the signal output (at the sampling time) of the matched filter is $\pm A$ and is achieved by transmitting energy $E_b$ during each signaling interval. The (noisy) matched filter output is compared to a threshold of $0$ to decide whether the noise-free output was supposed to be $+A$ or $-A$. If the (zero-mean Gaussian) noise has standard deviation $\sigma$, the error probability is $Q\left(\frac{A}{\sigma}\right)$ where $Q(\cdot)$ is the complementary standard Gaussian cumulative distribution function.

In on-off signaling, the signal outputs are $2A$ and $0$ and are compared to a threshold of $A$ to determine if the noise-free output should have been $2A$ or $0$. This gives the same error probability as antipodal signaling. But, it is necessary to transmit four times as much energy ($4E_b$) in the ON state to get the signal output level to be double that of the level with antipodal signaling. Assuming ZEROes and ONEs to be equally likely to be transmitted, the average energy transmitted per bit in on-off signaling is $2E_b$ as compared to $E_b$ in antipodal signaling, which is the doubling referred to in your question. But the peak energy is four times as large which means that more powerful (and thus, more likely, larger) transmitters are needed; something to keep in mind in applications where power consumption and weight of the transmitter electronics is an issue.

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