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Consider a binary-input additive white Gaussian noise channel.

Let $\mathbf{x}_0 = (\sqrt{Es},\sqrt{E_s},⋯,\sqrt{E_s})$ and $\mathbf{x}_1 = (-\sqrt{E_s},-\sqrt{E_s},⋯,-\sqrt{E_s})$, be two codewords of length $d$. Suppose $\mathbf{x}_0$ is transmitted and the received vector is given by $\mathbf{y} = \mathbf{x}_0 + \mathbf{n}$, where the noise vector $\mathbf{n}$ is an i.i.d. Gaussian random vector with mean zero and variance $\frac{N_0}{2}$ . Suppose that maximum-likelihood decoding is employed. The pairwise error probability $P_d$ is the probability that the decoder chooses the incorrect codeword $\mathbf{x}_1$ as the decision instead of the originally transmitted $\mathbf{x}_0$. Show that $P_d=Q(\sqrt{\frac{2dE_s}{N_0}})$, where $Q(x) = (\frac{1}{\sqrt{2 \pi}})\int ^{\infty}_{x}e^{-\frac{t^2}{2}}dt$

I know how to calculate the error probability for bpsk in the other type,but not codewords type,does anyone know how to calculate the probability for codeword type?

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The probability of error of the ML detector is equal to the probability that the received vector $\mathbf{y}$ is closer to $\mathbf{x}_1$ than to $\mathbf{x}_0$, which is equal to the probability that the noise component in the direction of $\mathbf{x}_0-\mathbf{x}_1$ is greater than half of the Euclidean distance between $\mathbf{x}_0$ and $\mathbf{x}_1$. The Euclidean distance between $\mathbf{x}_0$ and $\mathbf{x}_1$ is

$$D=||\mathbf{x}_0-\mathbf{x}_1||=\sqrt{\sum_{i=1}^d(x_{0,i}-x_{1,i})^2}=\sqrt{\sum_{i=1}^d4E_s}=2\sqrt{dE_s}\tag{1}$$

The noise variance in the direction of $\mathbf{x}_0-\mathbf{x}_1$ equals $N_0/2$ (as in any other direction). The probability that a zero mean Gaussian noise variable with variance $N_0/2$ assumes a value greater than $D/2$ is given by

$$P_E=Q\left(\frac{D/2}{\sqrt{N_0/2}}\right)=Q\left(\frac{\sqrt{dE_s}}{\sqrt{N_0/2}}\right)=Q\left(\sqrt{\frac{2dE_s}{N_0}}\right)\tag{2}$$

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  • $\begingroup$ Is it $\sqrt{\sum\limits_{i=1}^{d}(x_{0,i}-x_{1,i})^2}=\sqrt{\sum\limits_{i=1}^{d}4E_s}$? $\endgroup$ – electronic component Aug 19 '18 at 10:24
  • $\begingroup$ @electroniccomponent: Yes, that was a typo, corrected. $\endgroup$ – Matt L. Aug 19 '18 at 10:27

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