Question

Consider the two octal signal point constellation shown in the figure below.

The nearest neighbour signal points in the signal constellation are separated in distance by $A$ units. Then

• A. The value of $b=\frac{1+\sqrt{3}}{2}A$
• B. The value of $\frac{r}{b}=\frac{\sqrt{2-\sqrt{2}}}{2(1+\sqrt{3})}$
• C. The value of $r=\frac{A}{\sqrt{2-\sqrt{2}}}$
• D. The value of $a=\frac{A}{\sqrt{2}}$

Answer 1

Answer: (A), (C), (D)

The value of $a=\frac{A}{\sqrt{2}}$

(i) To calculate value of $a,$ we have,

by using Pythagoras theorem, we get,
$a^2+a^2=A^{2}a=\frac{A}{\sqrt{2}}$
For calculating ${}^{\prime }{b}^{\prime }$ we can draw the constellation diagram as

the angle $x$ is ${90}^{\circ }$ and angle $y$ is equal to ${60}^{\circ }.$ thus,
$2z+x+y={360}^{\circ }z={105}^{\circ }$
by using cosine formula, we get
$b^2=a^2+A^2-2aA\cos {105}^{\circ }b=\frac{1+\sqrt{3}}{2}A$
For figure(3)

$\therefore A^2=r^2+r^2-2r^2\cos {45}^{\circ }r=\frac{A}{\sqrt{2-\sqrt{2}}}$