Difference between revisions of "Aufgaben:Exercise 4.10Z: Signal Space Constellation of the 16-QAM"
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− | [[File:P_ID1719__Mod_Z_4_9.png|right|frame| | + | [[File:P_ID1719__Mod_Z_4_9.png|right|frame|Signal space constellation]] |
− | + | We now consider the 16-QAM method according to the block diagram given in the theory section. The diagram shows the possible complex amplitude coefficients $a = a_{\rm I} + {\rm j} · a_{\rm Q}$. | |
− | + | As in [[Aufgaben:Exercise_4.10:_Signal_Waveforms_of_the_16-QAM|Exercise 4.10]], the following should be assumed: | |
− | * | + | * The possible amplitude coefficients $a_{\rm I}$ and $a_{\rm Q}$ of the two component signals are $ ±1$ and $±1/3$, respectively. |
− | * | + | * The basic transmission pulse $g_s(t)$ is rectangular with amplitude $g_0 = 1\ \rm V$ and duration $T = 1 \ \rm µ s$. |
− | * | + | * The source signal $q(t)$ before the serial-to-parallel converter is binary and redundancy-free. |
− | |||
− | |||
− | |||
− | |||
− | |||
− | === | + | |
+ | Hints: | ||
+ | *This exercise belongs to the chapter [[Modulation_Methods/Quadrature_Amplitude_Modulation|"Quadrature Amplitude Modulation"]]. | ||
+ | *The page [[Modulation_Methods/Quadrature_Amplitude_Modulation#Quadratic_QAM_signal_space_constellations|"Quadratic QAM signal space constellations"]] is helpful for completing this exercise. | ||
+ | *The signals belonging to the colored points are shown in the same colors as in [[Aufgaben:Exercise_4.10:_Signal_Waveforms_of_the_16-QAM|Exercise 4.10]]. | ||
+ | |||
+ | |||
+ | |||
+ | ===Questions=== | ||
<quiz display=simple> | <quiz display=simple> | ||
− | { | + | {What is the bit rate $R_{\rm B}$ of the binary source signal $q(t)$? |
|type="{}"} | |type="{}"} | ||
$R_{\rm B}\ = \ $ { 4 3% } $\ \rm Mbit/s$ | $R_{\rm B}\ = \ $ { 4 3% } $\ \rm Mbit/s$ | ||
− | { | + | {Give the magnitude and the phase $($between $±180^\circ)$ for the red symbol ⇒ $a = 1 +{\rm j}$. |
|type="{}"} | |type="{}"} | ||
$|a| \ = \ $ { 1.414 3% } | $|a| \ = \ $ { 1.414 3% } | ||
− | ${\rm arc} \ a \ = \ $ { 45 3% } $\ \rm | + | ${\rm arc} \ a \ = \ $ { 45 3% } $\ \rm degrees$ |
− | { | + | {Give the magnitude and the phase for the blue symbol ⇒ $a = 1/3 +{\rm j}/3$. |
|type="{}"} | |type="{}"} | ||
$|a| \ = \ $ { 0.471 3% } | $|a| \ = \ $ { 0.471 3% } | ||
− | ${\rm arc} \ a \ = \ $ { 45 3% } $\ \rm | + | ${\rm arc} \ a \ = \ $ { 45 3% } $\ \rm degrees$ |
− | { | + | {Give the magnitude and the phase for the green symbol ⇒ $a = -1 +{\rm j}/3$. |
|type="{}"} | |type="{}"} | ||
$|a| \ = \ $ { 1.054 3% } | $|a| \ = \ $ { 1.054 3% } | ||
− | ${\rm arc} \ a \ = \ $ { | + | ${\rm arc} \ a \ = \ $ { 161.57 } $\ \rm degrees$ |
− | { | + | {Give the magnitude and the phase for the purple symbol ⇒ $a = -1 -{\rm j}/3$. |
|type="{}"} | |type="{}"} | ||
$|a| \ = \ $ { 1.054 3% } | $|a| \ = \ $ { 1.054 3% } | ||
− | ${\rm arc} \ a \ = \ ${ -166.57--156.57 } $\ \rm | + | ${\rm arc} \ a \ = \ ${ -166.57--156.57 } $\ \rm degrees$ |
− | { | + | {How many different magnitudes ⇒ $N_{|a|}$ and phase positions ⇒ $N_{arc}$ are possible? |
|type="{}"} | |type="{}"} | ||
$N_{|a|}\ = \ $ { 3 } | $N_{|a|}\ = \ $ { 3 } | ||
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</quiz> | </quiz> | ||
− | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''1 | + | '''(1)''' Each $\log_2 \ 16 = 4$ bits of the source signal are represented by one symbol, two bits by the four-level coefficient $a_{\rm I}$ and two more by $a_{\rm Q}$. |
+ | *Thus, the bit duration is $T_{\rm B} = T/4 = 0.25 \ \rm µ s$. | ||
+ | *And the bit rate is $R_{\rm B} = 1/T_{\rm B}\hspace{0.15cm}\underline { = 4 \ \rm Mbit/s}$. | ||
+ | |||
+ | |||
+ | |||
+ | '''(2)''' From geometry, for $a = 1 + {\rm j}$ it follows: | ||
+ | :$$a| = \sqrt{1^2 + 1^2}= \sqrt{2}\hspace{0.15cm}\underline { =1.414}\hspace{0.05cm}, \hspace{0.5cm} {\rm arc}\hspace{0.15cm} a = \arctan \left ({1}/{1} \right ) \hspace{0.15cm}\underline {= 45^{\circ}}\hspace{0.05cm}.$$ | ||
+ | |||
+ | |||
+ | |||
+ | '''(3)''' The angle is obtained as in subtask '''(2)''', the magnitude is smaller by a factor of $3$: | ||
+ | :$$|a| = \sqrt{(1/3)^2 + (1/3)^2}= \sqrt{2}\hspace{0.15cm}\underline { =0.471}\hspace{0.05cm}, \hspace{0.5cm} {\rm arc}\hspace{0.15cm} a \hspace{0.15cm}\underline {= 45^{\circ}}\hspace{0.05cm}.$$ | ||
+ | |||
+ | |||
+ | |||
+ | '''(4)''' For the complex amplitude coefficient $a = -1 + {\rm j}/3$, geometry gives us: | ||
+ | :$$|a| = \sqrt{1^2 + (1/3)^2}\hspace{0.15cm}\underline {= 1.054}\hspace{0.05cm},\hspace{0.5cm} | ||
+ | {\rm arc}\hspace{0.15cm} a = 180^{\circ} - \arctan \left ( {1}/{3} \right ) = 180^{\circ} - 18.43^{\circ} \hspace{0.15cm}\underline {= 161.57^{\circ}}\hspace{0.05cm}.$$ | ||
+ | |||
+ | |||
+ | |||
+ | '''(5)''' The purple symbol $a = -1 - {\rm j}/3$ has the same magnitude as the green symbol according to subtask '''(4)''', while the phase angle changes sign: | ||
+ | :$$|a| \hspace{0.15cm}\underline {= 1.054}\hspace{0.05cm},\hspace{0.5cm} | ||
+ | {\rm arc}\hspace{0.15cm} a \hspace{0.15cm}\underline {= -161.57^{\circ}}\hspace{0.05cm}.$$ | ||
− | |||
− | |||
− | |||
− | ''' | + | '''(6)''' For the magnitude: $N_{|a|}\hspace{0.15cm}\underline { = 3}$ different results are possible: $1.414$, $1.054$ and $0.471$. |
− | |||
− | |||
− | |||
− | + | *In contrast, there are $N_{\rm arc}\hspace{0.15cm}\underline { = 12}$ possible phase positions, namely: | |
− | $$ \pm \arctan (1/3) = \pm 18.43^{\circ}, \hspace{0.2cm}\pm \arctan (1) = \pm 45^{\circ}, \hspace{0.2cm}\pm \arctan (3) = \pm 71.57^{\circ}\hspace{0.05cm},$$ | + | :$$ \pm \arctan (1/3) = \pm 18.43^{\circ}, \hspace{0.2cm}\pm \arctan (1) = \pm 45^{\circ}, \hspace{0.2cm}\pm \arctan (3) = \pm 71.57^{\circ}\hspace{0.05cm},$$ |
− | $$\pm (180^{\circ}-71.57^{\circ}) = \pm 108.43^{\circ}, \hspace{0.2cm}\pm (180^{\circ}-45^{\circ}) = \pm 135^{\circ}, \hspace{0.2cm}\pm 161.57^{\circ} \hspace{0.05cm}.$$ | + | :$$\pm (180^{\circ}-71.57^{\circ}) = \pm 108.43^{\circ}, \hspace{0.2cm}\pm (180^{\circ}-45^{\circ}) = \pm 135^{\circ}, \hspace{0.2cm}\pm 161.57^{\circ} \hspace{0.05cm}.$$ |
{{ML-Fuß}} | {{ML-Fuß}} | ||
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− | [[Category: | + | [[Category:Modulation Methods: Exercises|^4.3 Quadrature Amplitude Modulation^]] |
Latest revision as of 17:10, 16 April 2022
We now consider the 16-QAM method according to the block diagram given in the theory section. The diagram shows the possible complex amplitude coefficients $a = a_{\rm I} + {\rm j} · a_{\rm Q}$.
As in Exercise 4.10, the following should be assumed:
- The possible amplitude coefficients $a_{\rm I}$ and $a_{\rm Q}$ of the two component signals are $ ±1$ and $±1/3$, respectively.
- The basic transmission pulse $g_s(t)$ is rectangular with amplitude $g_0 = 1\ \rm V$ and duration $T = 1 \ \rm µ s$.
- The source signal $q(t)$ before the serial-to-parallel converter is binary and redundancy-free.
Hints:
- This exercise belongs to the chapter "Quadrature Amplitude Modulation".
- The page "Quadratic QAM signal space constellations" is helpful for completing this exercise.
- The signals belonging to the colored points are shown in the same colors as in Exercise 4.10.
Questions
Solution
- Thus, the bit duration is $T_{\rm B} = T/4 = 0.25 \ \rm µ s$.
- And the bit rate is $R_{\rm B} = 1/T_{\rm B}\hspace{0.15cm}\underline { = 4 \ \rm Mbit/s}$.
(2) From geometry, for $a = 1 + {\rm j}$ it follows:
- $$a| = \sqrt{1^2 + 1^2}= \sqrt{2}\hspace{0.15cm}\underline { =1.414}\hspace{0.05cm}, \hspace{0.5cm} {\rm arc}\hspace{0.15cm} a = \arctan \left ({1}/{1} \right ) \hspace{0.15cm}\underline {= 45^{\circ}}\hspace{0.05cm}.$$
(3) The angle is obtained as in subtask (2), the magnitude is smaller by a factor of $3$:
- $$|a| = \sqrt{(1/3)^2 + (1/3)^2}= \sqrt{2}\hspace{0.15cm}\underline { =0.471}\hspace{0.05cm}, \hspace{0.5cm} {\rm arc}\hspace{0.15cm} a \hspace{0.15cm}\underline {= 45^{\circ}}\hspace{0.05cm}.$$
(4) For the complex amplitude coefficient $a = -1 + {\rm j}/3$, geometry gives us:
- $$|a| = \sqrt{1^2 + (1/3)^2}\hspace{0.15cm}\underline {= 1.054}\hspace{0.05cm},\hspace{0.5cm} {\rm arc}\hspace{0.15cm} a = 180^{\circ} - \arctan \left ( {1}/{3} \right ) = 180^{\circ} - 18.43^{\circ} \hspace{0.15cm}\underline {= 161.57^{\circ}}\hspace{0.05cm}.$$
(5) The purple symbol $a = -1 - {\rm j}/3$ has the same magnitude as the green symbol according to subtask (4), while the phase angle changes sign:
- $$|a| \hspace{0.15cm}\underline {= 1.054}\hspace{0.05cm},\hspace{0.5cm} {\rm arc}\hspace{0.15cm} a \hspace{0.15cm}\underline {= -161.57^{\circ}}\hspace{0.05cm}.$$
(6) For the magnitude: $N_{|a|}\hspace{0.15cm}\underline { = 3}$ different results are possible: $1.414$, $1.054$ and $0.471$.
- In contrast, there are $N_{\rm arc}\hspace{0.15cm}\underline { = 12}$ possible phase positions, namely:
- $$ \pm \arctan (1/3) = \pm 18.43^{\circ}, \hspace{0.2cm}\pm \arctan (1) = \pm 45^{\circ}, \hspace{0.2cm}\pm \arctan (3) = \pm 71.57^{\circ}\hspace{0.05cm},$$
- $$\pm (180^{\circ}-71.57^{\circ}) = \pm 108.43^{\circ}, \hspace{0.2cm}\pm (180^{\circ}-45^{\circ}) = \pm 135^{\circ}, \hspace{0.2cm}\pm 161.57^{\circ} \hspace{0.05cm}.$$