Question

Consider the signal space representation of two 4-ary modulation schemesas shown below:



For modulation scheme of constellation-1, the average symbol energy is $E_1$ and the average symbol error is $p_1$. For modulation scheme of constellation-2, the average symbol energy is $E_2$ and the average symbol error is $p_2$. If the message symbols are ocurring with equal probability, then select the correct one of the following relations.

• A. $E_1<E_2$ and $p_1=p_2$
• B. $E_1<E_2$ and $p_1<p_2$
• C. $E_1>E_2$ and $p_1>p_2$
• D. $E_1>E_2$ and $p_1=p_2$

Answer 1

The correct option is D $E_1>E_2$ and $p_1=p_2$

$E_1=\frac{1}{4}\left[\right(0{)}^2+(2A{)}^2+(\sqrt{2}A{)}^2+\left(\sqrt{2}A{)}^2\right]=2A^2$
$E_2=\frac{1}{4}\left[\right(A{)}^2+(A{)}^2+(A{)}^2+\left(A{)}^2\right]=A^2$
So, $E_1>E_2$
The distance between adjacent symbols is same in both the constellations. So, both the modulation schemes provide same average symbol error under similar circumstances.
Hence, $p_1=p_2$