Return to Question

The signal constellation for a communication system with 16 equiprobable symbols is shown as below. The channel is AWGN with noise power spectral density of $N_ 0/2.$Using the union bound, find a bound in terms of $A$ and $N_0$ on the error probability for this channel.

Does the union bound means error probability?the solution said $$P_e \le 15Q\left(\sqrt{\frac{d^2_{min}}{2N_0}}\right)=15Q\left(\sqrt{\frac{2A^2}{N_0}}\right) $$

but why is

\begin{align} &2\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+3\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+3\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+2\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)\\ &+3\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+4\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+4\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+3\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)\\ &+3\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+4\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+4\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+3\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)\\ &+2\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+3\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+3\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+2\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)\\ &\neq\frac{38}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)\quad{?} \end{align}

The signal constellation for a communication system with 16 equiprobable symbols is shown as below. The channel is AWGN with noise power spectral density of $N_ 0/2.$Using the union bound, find a bound in terms of $A$ and $N_0$ on the error probability for this channel.

Does the union bound means error probability?the solution said $$P_e \le 15Q\left(\sqrt{\frac{d^2_{min}}{2N_0}}\right)=15Q\left(\sqrt{\frac{2A^2}{N_0}}\right) $$

but why is

\begin{align} &2\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+3\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+3\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+2\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)\\ &+3\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+4\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+4\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+3\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)\\ &+3\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+4\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+4\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+3\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)\\ &+2\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+3\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+3\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+2\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)\\ &=\frac{48}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)\quad{?} \end{align}

The signal constellation for a communication system with 16 equiprobable symbols is shown as below. The channel is AWGN with noise power spectral density of $N_ 0/2.$Using the union bound, find a bound in terms of $A$ and $N_0$ on the error probability for this channel.

Does the union bound means error probability?the solution said

$P_e \le 15Q(\sqrt{d^2_{min}/2N_0})=15Q(\sqrt{2A^2/N_0})$,

but why is not

$2 \times \frac{1}{16}Q(\frac{A}{\sqrt{\frac{N_0}{2}}})+3 \times \frac{1}{16}Q(\frac{A}{\sqrt{\frac{N_0}{2}}})+3 \times \frac{1}{16}Q(\frac{A}{\sqrt{\frac{N_0}{2}}})+2 \times \frac{1}{16}Q(\frac{A}{\sqrt{\frac{N_0}{2}}})$

$+3 \times \frac{1}{16}Q(\frac{A}{\sqrt{\frac{N_0}{2}}})+4 \times \frac{1}{16}Q(\frac{A}{\sqrt{\frac{N_0}{2}}})+4 \times \frac{1}{16}Q(\frac{A}{\sqrt{\frac{N_0}{2}}})+3 \times \frac{1}{16}Q(\frac{A}{\sqrt{\frac{N_0}{2}}})$

$+3 \times \frac{1}{16}Q(\frac{A}{\sqrt{\frac{N_0}{2}}})+4 \times \frac{1}{16}Q(\frac{A}{\sqrt{\frac{N_0}{2}}})+4 \times \frac{1}{16}Q(\frac{A}{\sqrt{\frac{N_0}{2}}})+3 \times \frac{1}{16}Q(\frac{A}{\sqrt{\frac{N_0}{2}}})$

$+2 \times \frac{1}{16}Q(\frac{A}{\sqrt{\frac{N_0}{2}}})+3 \times \frac{1}{16}Q(\frac{A}{\sqrt{\frac{N_0}{2}}})+3 \times \frac{1}{16}Q(\frac{A}{\sqrt{\frac{N_0}{2}}})+2 \times \frac{1}{16}Q(\frac{A}{\sqrt{\frac{N_0}{2}}})$

=$\frac{38}{16}Q(\frac{A}{\sqrt{\frac{N_0}{2}}})$ ?

The signal constellation for a communication system with 16 equiprobable symbols is shown as below. The channel is AWGN with noise power spectral density of $N_ 0/2.$Using the union bound, find a bound in terms of $A$ and $N_0$ on the error probability for this channel.

Does the union bound means error probability?the solution said $$P_e \le 15Q\left(\sqrt{\frac{d^2_{min}}{2N_0}}\right)=15Q\left(\sqrt{\frac{2A^2}{N_0}}\right) $$

but why is

\begin{align} &2\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+3\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+3\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+2\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)\\ &+3\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+4\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+4\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+3\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)\\ &+3\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+4\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+4\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+3\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)\\ &+2\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+3\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+3\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)+2\cdot\frac{1}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)\\ &\neq\frac{38}{16}Q\left(\frac{A}{\sqrt{\frac{N_0}{2}}}\right)\quad{?} \end{align}