Let the two axes (basis functions) be normalized respectively and denoted $I$ and $Q$ for simplicity. For further simplification, $E = 1$ can be set.
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.
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
The system parameter $R$ is to be determined in this exercise in such a way that the quotient
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
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 minimum distance $d_{\rm min}$, and
* the bit energy $E_{\rm B}$.
* 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.
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 [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation|"Carrier Frequency Systems with Coherent Demodulation"]].
Notes:
* Reference is made in particular to the sections [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#Quadrature_amplitude_modulation_.28M-QAM.29|"Quadrature amplitude modulation"]] and [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#M.E2.80.93level_amplitude_shift_keying_.28M.E2.80.93ASK.29|"Multilevel phase modulation"]].
* The exercise belongs to the chapter [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation|"Carrier Frequency Systems with Coherent Demodulation"]].
* Reference is made in particular to the sections [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#Quadrature_amplitude_modulation_.28M-QAM.29|"Quadrature amplitude modulation"]] and [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#M.E2.80.93level_amplitude_shift_keying_.28M.E2.80.93ASK.29|"Multi-level phase modulation"]].
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===Questions===
===Questions===
<quiz display=simple>
<quiz display=simple>
{Calculate the average energy $E_{\rm B}$ per bit depending on $R$, in particular for $R = 1$ and $R = \sqrt{2}$.
{Calculate the average energy $E_{\rm B}$ per bit depending on $R$, in particular for $R = 1$ and $R = \sqrt{2}$.
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.
(1) Because of $M = 8$ ⇒ $b = 3$, the mean signal energy per bit is $E_{\rm B} = E_{\rm S}/3$, where the mean 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:
Special cases of 8–QAM
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:
In the drawn example on the information section 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.
To calculate the minimum distance
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:
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):
For $R = R_{\rm max}= (\sqrt{3}+1)/\sqrt{2}$ exactly the same value results.
The (always desired) maximum of the power efficiency $\eta$ results, for example, for $R = R_{\rm max}$ – i.e. for the signal space constellation in the information section.
In this case all triangles of two neighboring blue points and the red point in between are equilateral.
Also for $R = R_{\rm min}$ there are equilateral triangles, but now each formed by two red and one blue point.
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.
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 $\eta = 0.5$, respectively (compared to $\eta = 0.634$).
