Aufgaben:Exercise 4.15: Optimal Signal Space Allocation

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Considered  "8–QAM"

A signal space constellation with  $M = 8$  signal space points is considered here:

• Four points lie on a circle with radius  $r = 1$.

• Four further points lie offset by  $45^\circ$  on a second circle with radius  $R$,  where the following shall hold:

$$R_{\rm min} \le R \le R_{\rm max}\hspace{0.05cm},\hspace{0.2cm} R_{\rm min}= \frac{ \sqrt{3}-1}{ \sqrt{2}} \approx 0.518\hspace{0.05cm},\hspace{0.2cm}R_{\rm max}= \frac{ \sqrt{3}+1}{ \sqrt{2}} \approx 1.932\hspace{0.05cm}.$$

Let the two axes  ("basis functions")  be normalized respectively and denoted  $I$  and  $Q$  for simplicity.  For further simplification,  $E = 1$  can be set.

In the question section,  we speak of  "blue"  and  "red"  points.  According to the diagram,  the blue points lie on the circle with radius  $r = 1$,  the red points on the circle with radius  $R$.  The case  $R = R_{\rm max}$ is drawn.

The system parameter  $R$  is to be determined in this exercise in such a way that the quotient

$$\eta = \frac{ (d_{\rm min}/2)^2}{ E_{\rm B}} $$

becomes maximum.  $\eta$  is a measure for the quality of a modulation alphabet at given transmission energy per bit  ("power efficiency").  It is calculated from

• the minimum distance  $d_{\rm min}$,  and

• the average bit energy  $E_{\rm B}$.

It must be ensured that  $d_{\rm min}^2$  and  $E_{\rm B}$  are normalized in the same way,  but this is already implicit in the exercise.

Notes:

• The exercise belongs to the chapter  "Carrier Frequency Systems with Coherent Demodulation".

• Reference is made in particular to the sections  "Quadrature amplitude modulation"  and  "Multi-level phase modulation".

Questions

1 Calculate the average energy  $E_{\rm B}$  per bit depending on  $R$,  in particular for  $R = 1$  and  $R = \sqrt{2}$.

$R = 1 \text{:} \hspace{0.55cm} E_{\rm B}\ = \ $

$R = \sqrt{2} \text{:} \hspace{0.2cm} E_{\rm B}\ = \ $

2 Which statements are true for the minimum distance between two signal space points?

	For  $R < R_{\rm min}$,  the minimum distance occurs between two red points.
	For  $R > R_{\rm max}$,  the minimum distance occurs between two blue points.
	For  $R_{\rm min} ≤ R ≤ R_{\rm max}$,  the minimum distance occurs between  "red"  and  "blue".

3 Calculate the minimum distance depending on  $R$,  in particular for

$R = 1 \text{:} \hspace{0.55cm} d_{\rm min}\ = \ $

$R = \sqrt{2} \text{:} \hspace{0.2cm} d_{\rm min}\ = \ $

4 Give the power efficiency  $\eta$  in general terms.  What  $\eta$  results for  $R = 1$?

$\eta\ = \ $

5 What power efficiency values result for  $R = R_{\rm min}$  and  $R = R_{\rm max}$?  Interpretation.

$R = R_{\rm min} \text{:} \hspace{0.35cm} \eta\ = \ $

$R = R_{\rm max} \text{:} \hspace{0.2cm} \eta\ = \ $

Solution

Solution

Special cases of  "8–QAM"

(1)  Because of  $M = 8$   ⇒   $b = 3$,  the average signal energy per bit is  $E_{\rm B} = E_{\rm S}/3$,  where the average signal energy per symbol  $(E_{\rm S})$  is to be calculated as the mean square distance of the signal space points from the origin.  With  $r = 1$  one obtains:

$$E_{\rm S} = {1}/{8 } \cdot ( 4 \cdot r^2 + 4 \cdot R^2) = ({1 + R^2})/{2 }$$

$$\Rightarrow \hspace{0.3cm} E_{\rm B} = {E_{\rm S}}/{3} = ({1 + R^2})/{6}\hspace{0.05cm}.$$

In particular:

• For  $R = 1$,  there is an  "8–PSK"   ⇒   $E_{\rm S} = 1$  and  $E_{\rm B} \ \underline {= 0.333}$  (see left graph).

• The right graph is valid for  $R = \sqrt{2}$.  In this case,  $E_{\rm B} \ \underline {= 0.5}$.

Note that these energies actually still have to be multiplied by the normalization energy $E$.
(2)  All statements are true:

To calculate minimum distance

• In the drawn example on the front page with  $R = R_{\rm max}$,  the distance between two neighboring blue points is exactly the same as the distance between a red (outer) and a blue (inner) point.

• For $R > R_{\rm max}$,  the distance between two blue points is the smallest.

• For $R < R_{\rm min}$,  the minimum distance occurs between two red points.

(3)  The graphic illustrates the geometric calculation.  With  "Pythagoras"  one obtains:

$$d_{\rm min}^2 =(R/\sqrt{2})^2 + (R/\sqrt{2}-1)^2 = 1 - \sqrt{2} \cdot R + R^2$$

$$ \Rightarrow \hspace{0.3cm}d_{\rm min} = \sqrt{ 1 - \sqrt{2} \cdot R + R^2}\hspace{0.05cm}.$$

• In particular,  for $R = 1$  ("8–PSK"):

$$d_{\rm min} = \sqrt{ 2 - \sqrt{2} } \hspace{0.1cm} \underline{= 0.765} \hspace{0.1cm} (= 2 \cdot \sin (22.5^{\circ}) )\hspace{0.05cm}.$$

• In contrast,  for  $\underline {R = \sqrt{2}}$  corresponding to the right graph for subtask  (1),  the minimum distance is  $d_{\rm min} \ \underline {= 1}$.

(4)  Using the results of  (1)  and  (3),  we obtain in general or  for $R = 1$ ("8–PSK"):

$$\eta = \frac{ d_{\rm min}^2}{ 4 \cdot E_{\rm B}} = \frac{ 1 - \sqrt{2} \cdot R + R^2}{ 4 \cdot (1 + R^2)/6}= \frac{ 3/2 \cdot(1 - \sqrt{2} \cdot R + R^2)}{ 1 + R^2}\hspace{0.3cm}\Rightarrow \hspace{0.3cm} R = 1: \hspace{0.2cm}\eta =\frac{ 3/2 \cdot(2 - \sqrt{2}) }{ 2} = 3/4 \cdot(2 - \sqrt{2})\hspace{0.1cm} \underline{\approx 0.439}\hspace{0.05cm}.$$

(5)  For  $R = R_{\rm min} = (\sqrt{3}-1)/\sqrt{2}$,  the following value is obtained:

$$\eta = \frac{ 3/2 \cdot(1 - \sqrt{2} \cdot R + R^2)}{ 1 + R^2} = 3/2 \cdot \left [ 1 - \frac{ \sqrt{2} \cdot R }{ 1 + R^2}\right ]\hspace{0.05cm},$$

$$\sqrt{2} \cdot R = \sqrt{3}- 1\hspace{0.05cm},\hspace{0.2cm} 1 + R^2 = 3 - \sqrt{3} \hspace{0.3cm}\Rightarrow \hspace{0.3cm}\eta = 3/2 \cdot \left [ 1 - \frac{ \sqrt{3}- 1 }{ 3 - \sqrt{3}}\right ]\hspace{0.1cm} \underline{\approx 0.634}\hspace{0.05cm}.$$

• For  $R = R_{\rm max}= (\sqrt{3}+1)/\sqrt{2}$  exactly the same value results.

1. The  (always desired)  maximum of the power efficiency  $\eta$  results e.g. for  $R = R_{\rm max}$ – i.e. for the signal space constellation in the information section.
2. In this case all triangles of two neighboring blue points and the red point in between are equilateral.
3. Also for  $R = R_{\rm min}$  there are equilateral triangles,  but now each formed by two red and one blue point.
4. In this case the edge length  $d_{\rm min}$  is clearly smaller,  but at the same time a smaller  $E_{\rm B}$  results,  so that the power efficiency  $\eta$  has the same value.
5. The previously considered special cases  $R = 1$  ("8–PSK",  left graph in the first subtask)  and  $R = \sqrt{2}$  (right graph)  have a noticeably smaller  $\eta$  with  $\eta = 0.439$  and  $\eta = 0.5$,  resp.  $($compared to  $\eta = 0.634)$.