Question

Consider the two 8-point QAM signal constellations shown in the figure below:




In both the systems, the signal points are equally probable. The minimum distance between adjacent points in both the systems is same if the average symbol energy is transmitted in system-1 is $E_1$ and that of in system -2 is $E_2$ then select the correct one of the following statements.

• A. $E_2$ is larger than $E_1$ by 1.33dB
• B. $E_2$ is larger than $E_1$ by 1.25dB
• C. $E_1$ is larger than $E_2$ by 1.25dB
• D. $E_1$ is larger than $E_2$ by 1.33dB

Answer 1

The correct option is C $E_1$ is larger than $E_2$ by 1.25dB

For system-1:
$r_1=2a$
$r_2=\sqrt{(2a{)}^2+(2a{)}^2}=2\sqrt{2}a$
In this system, 4 points are at a radial distance of "2a"
and 4 points are at a radial distance "
$2\sqrt{2}a$", from the origin.
So,
$E_1=\frac{4(2a{)}^2+4(2\sqrt{2}a{)}^2}{8}=6a^2$


For system-2
$r_1=a$
$(2a{)}^2=r_2^2+a^2=\sqrt{3}a$
$r_3=\sqrt{r_2^2+\left(2a^2\right)}=\sqrt{7}a$
In this system, 2 points are at a radial distance of "a"
and 2 points are at a radial distance of "
$\sqrt{3}a$" and 4 points are at a radial distance of $\sqrt{7}a$ , from the origin.
So,
$E_2=\frac{2(a{)}^2+2(\sqrt{3}a{)}^2+4(\sqrt{7}a{)}^2}{8}=4.5a^2$
$\frac{E_1}{E_2}=\frac{6a^2}{4.5a^2}=\frac{4}{3}=1.33$
$E_1\left(dB\right)-E_2\left(dB\right)=10\left(log{)}_{10}\right(\frac{4}{3})\approx 1.25dB$