View source for Aufgaben:Exercise 4.10Z: Signal Space Constellation of the 16-QAM


{{quiz-Header|Buchseite=Modulationsverfahren/Quadratur–Amplitudenmodulation
}}

[[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.&nbsp;  The diagram shows the possible complex amplitude coefficients &nbsp;$a = a_{\rm I} + {\rm j} · a_{\rm Q}$.

As in&nbsp; [[Aufgaben:Exercise_4.10:_Signal_Waveforms_of_the_16-QAM|Exercise 4.10]],&nbsp; the following should be assumed:
* The possible amplitude coefficients &nbsp;$a_{\rm I}$&nbsp; and &nbsp;$a_{\rm Q}$&nbsp; of the two component signals are&nbsp;$ ±1$&nbsp; and &nbsp;$±1/3$, respectively. 
* The basic transmission pulse &nbsp;$g_s(t)$&nbsp; is rectangular with amplitude &nbsp;$g_0 = 1\ \rm  V$&nbsp; and duration &nbsp;$T = 1 \ \rm &micro; s$.
*  The source signal &nbsp;$q(t)$&nbsp; before the serial-to-parallel converter is binary and redundancy-free.





Hints: 
*This exercise belongs to the chapter &nbsp; [[Modulation_Methods/Quadrature_Amplitude_Modulation|"Quadrature Amplitude Modulation"]].
*The page&nbsp; [[Modulation_Methods/Quadrature_Amplitude_Modulation#Quadratic_QAM_signal_space_constellations|"Quadratic QAM signal space constellations"]]&nbsp; is helpful for completing this exercise. 
*The signals belonging to the colored points are shown in the same colors as in&nbsp; [[Aufgaben:Exercise_4.10:_Signal_Waveforms_of_the_16-QAM|Exercise 4.10]].
 


===Questions===

<quiz display=simple>
{What is the bit rate &nbsp;$R_{\rm B}$&nbsp; of the binary source signal &nbsp;$q(t)$?
|type="{}"}
$R_{\rm B}\ = \ $  { 4 3% } $\ \rm Mbit/s$


{Give the magnitude and the phase&nbsp; $($between &nbsp;$±180^\circ)$&nbsp; for the red symbol &nbsp; &rArr; &nbsp; $a = 1 +{\rm j}$.
|type="{}"}
$|a| \ = \ $ { 1.414 3% }
${\rm arc} \ a \ = \ $ { 45 3% } $\ \rm degrees$

{Give the magnitude and the phase for the blue symbol &nbsp; &rArr; &nbsp; $a = 1/3 +{\rm j}/3$.
|type="{}"}
$|a| \ = \ $ { 0.471 3% } 
${\rm arc} \ a \ = \ $ { 45 3% } $\ \rm degrees$

{Give the magnitude and the phase for the green symbol&nbsp; &rArr; &nbsp; $a = -1 +{\rm j}/3$.
|type="{}"}
$|a| \ = \ $ { 1.054 3% } 
${\rm arc} \ a \ = \ $ { 161.57 } $\ \rm degrees$

{Give the magnitude and the phase for the purple symbol&nbsp; &rArr; &nbsp; $a = -1 -{\rm j}/3$.
|type="{}"}
$|a| \ = \ $ { 1.054 3% } 
${\rm arc} \ a \ = \ ${ -166.57--156.57 } $\ \rm degrees$

{How many different magnitudes &nbsp; &rArr; &nbsp; $N_{|a|}$&nbsp; and phase positions &nbsp; &rArr; &nbsp; $N_{arc}$ are possible?
|type="{}"}
$N_{|a|}\ = \ $ { 3 } 
$N_{\rm arc}\ = \ $ { 12 } 
</quiz>

===Solution===
{{ML-Kopf}}
'''(1)'''&nbsp; Each&nbsp; $\log_2 \ 16 = 4$&nbsp; bits of the source signal are represented by one symbol,&nbsp;   two bits by the four-level coefficient&nbsp; $a_{\rm I}$&nbsp; and two more by&nbsp; $a_{\rm Q}$. 
*Thus,&nbsp; the bit duration is&nbsp; $T_{\rm B} = T/4 = 0.25 \ \rm &micro; s$. 
*And the bit rate is&nbsp; $R_{\rm B} = 1/T_{\rm B}\hspace{0.15cm}\underline { = 4 \ \rm  Mbit/s}$.



'''(2)'''&nbsp; From geometry,&nbsp; for &nbsp; $a = 1 + {\rm j}$&nbsp; 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)'''&nbsp; The angle is obtained as in subtask&nbsp; '''(2)''',&nbsp; the magnitude is smaller by a factor of &nbsp; $3$: 
:$$|a| = \sqrt{(1/3)^2 + (1/3)^2}= \sqrt{2}\hspace{0.15cm}\underline { =0.471}\hspace{0.05cm}, \hspace{0.5cm} {\rm arc}\hspace{0.15cm} a  \hspace{0.15cm}\underline {= 45^{\circ}}\hspace{0.05cm}.$$



'''(4)'''&nbsp; For the complex amplitude coefficient&nbsp; $a = -1 + {\rm j}/3$,&nbsp; 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)'''&nbsp; The purple symbol&nbsp; $a = -1 - {\rm j}/3$&nbsp; has the same magnitude as the green symbol according to subtask&nbsp; '''(4)''',&nbsp; 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)'''&nbsp; For the magnitude: &nbsp; $N_{|a|}\hspace{0.15cm}\underline { = 3}$&nbsp; different results are possible: &nbsp;$1.414$, &nbsp;$1.054$&nbsp; and &nbsp;$0.471$.

*In contrast,&nbsp; there are &nbsp; $N_{\rm arc}\hspace{0.15cm}\underline { = 12}$&nbsp; possible phase positions,&nbsp; namely:
:$$ \pm \arctan (1/3) = \pm 18.43^{\circ}, \hspace{0.2cm}\pm \arctan (1) = \pm 45^{\circ}, \hspace{0.2cm}\pm \arctan (3) = \pm 71.57^{\circ}\hspace{0.05cm},$$
:$$\pm (180^{\circ}-71.57^{\circ}) = \pm 108.43^{\circ}, \hspace{0.2cm}\pm (180^{\circ}-45^{\circ}) = \pm 135^{\circ}, \hspace{0.2cm}\pm 161.57^{\circ} \hspace{0.05cm}.$$

{{ML-Fuß}}



[[Category:Modulation Methods: Exercises|^4.3 Quadrature Amplitude Modulation^]]