Determining PSD of 1uaternary (4-ary) line code

[Evo8]

Homework Statement

Ok So this is an example from my text. I am having trouble following a portion of it and was curious if anyone could shed some light on it. I've typed most of it and posted a screen shot of the second portion since there is a table involved.

Determine the PSD of the quaternary (4-ary) baseband signaling. The 4-ary line code has four distinct symbols corresponding to the four different combinations of two message bits. One such mapping is:

$$a_k= \begin{cases}-3 \ \ message \ bits \ 00\\
-1 \ \ message \ bits \ 01\\
+1 \ \ message \ bits \ 10\\
+3 \ \ message \ bits \ 11 \end{cases}$$

Therefore, all four values of ##a_k## are equally likely, each with a chance of 1 in 4. Recall that
$$R_0=lim_{n→∞} \frac{1}{N} \sum_{k} a_k^2$$

Within the summation, 1/4 of the ##a_k \ will \ be \ ±1, \ and \ ±3## thus,

$$R_0=lim_{N→∞} \frac{1}{N} [\frac{N}{4}(-3)^2+\frac{N}{4}(-1)^2++\frac{N}{4}(1)^2++\frac{N}{4}(3)^2]=5$$

Up to this poing I understand.

Here is the second portion:

Homework Equations

The Attempt at a Solution

I understand how to calculate ##R_0## from this example. I get lost when the text calculated R_n. I see how they created the table of all possible values for ##a_k*a_k+n## but when the ##R_n## equation is put together i get confused.

In the paragraph below it they attempt to explain. Why are ±1 and ±9 equally likely (1 in 8) and ±1 are equally likely (1 in 4). I would have expected them to be the other way around.

Can anyone shed some light or offer a better explanation?

It would be much appreciated!

 

[I like Serena]

The table contains 16 entries. Each is equally likely.
The value 1 appears twice in the table, so the corresponding probability is ##\frac{2}{16}=\frac{1}{8}##.
However, the value 3 appears four times in the table, so its probability is ##\frac{4}{16}=\frac{1}{4}##.

 

[Evo8]

Simple! Man I really should have seen that.

Anyway thanks again for your help ILS!