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

1. Apr 10, 2013

### Evo8

1. The problem statement, all variables and given/known data
Ok So this is an example from my text. Im having trouble following a portion of it and was curious if anyone could shed some light on it. Ive 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:

2. Relevant equations

3. 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!

#### Attached Files:

• ###### example 7.4.jpg
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2. Apr 11, 2013

### 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 $\frac{4}{16}=\frac{1}{4}$.

3. Apr 11, 2013

### Evo8

Simple! Man I really should have seen that.

Anyway thanks again for your help ILS!