Difference between revisions of "Aufgaben:Exercise 1.09: BPSK and 4-QAM"

Latest revision as of 08:30, 20 May 2022

Phase diagrams of BPSK and 4-QAM

The diagram shows schematically the phase diagrams of the "binary phase modulation" $($abbreviated $\rm BPSK)$ and the "quadrature amplitude modulation" $($called $\rm 4–QAM)$.

The latter can be described by two BPSK systems with cosine and minus-sine carriers, where for each of the subcomponents the transmission amplitude is reduced by a factor of $\sqrt{2}$ compared to BPSK.

The envelope of the total signal $s(t)$ is thus also constant equal to $s_{0}$.

The error probability depending on the quotient $E_{\rm B}/N_{0}$ is the same for BPSK and 4–QAM:

$$p_{\rm B} = \ {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = \ {1}/{2}\cdot {\rm erfc}\left ( \sqrt{E_{\rm B}/{ N_0 }} \right ).$$

However, the error probability of the BPSK system can also be expressed in the form

$$p_{\rm B,\hspace{0.04cm}BPSK} = {\rm Q}\left ( \frac{s_0}{\sigma_d } \right )\hspace{0.2cm}{\rm with}\hspace{0.2cm}\sigma_d = \sqrt{\frac{N_0}{T_{\rm B}}}$$

Correspondingly, for the 4-QAM system:

$$p_{\rm B,\hspace{0.04cm}QAM} = {\rm Q}\left ( \frac{s_0/\sqrt{2}}{\sigma_d } \right )\hspace{0.2cm}{\rm with}\hspace{0.2cm}\sigma_d = \sqrt{\frac{N_0}{2 \cdot T_{\rm B}}}.$$

The equations are valid only under the condition of exact phase synchronization:

If there is a phase offset $\Delta\phi_{\rm T}$ between the transmitted and received carrier signals, the error probability increases significantly, with BPSK and QAM systems being degraded differently.

In the phase diagram, the phase offset is noticeable by a rotation of the point clouds.

In the diagram, the centers of the point clouds for $\Delta\phi_{\rm T} = 15^\circ$ are marked by yellow crosses, while the red circles indicate the centers for $\Delta\phi_{\rm T} = 0$.

$E_{\rm B}/N_{0} = 8$ always holds, so the error probabilities of BPSK and QAM in the best case (without phase shift) are as follows ⇒ "Exercise 1.8Z":

$$p_{\rm B} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {\rm Q}(4)= 0.317 \cdot 10^{-4}.$$

Further remarks:

If we denote the distance of the BPSK useful samples (without noise) from the (vertical) decision threshold by $s_{0}$, we get $\sigma_{d} = s_{0}/4$ for the noise rms value. The lighter circles in the diagram mark the contour lines with radius $2\cdot \sigma_{d}$ and $3\cdot \sigma_{d}$ of the two-dimensional Gaussian PDF.

For the 4-QAM, compared to the BPSK, the distances of the noise-free samples drawn in red from the now two decision thresholds are each smaller by a factor of $\sqrt{2}$, but it also results in a noise rms value $\sigma_{d}$ smaller by the same factor.

Notes:

The exercise belongs to the chapter "Linear Digital Modulation - Coherent Demodulation".

Reference is made in particular to the section "Phase offset between transmitter and receiver".

You can determine the values of the Q–function with the HTML5/JavaScript applet "Complementary Gaussian Error Functions".

Questions

Solution

(1) Rotating the phase diagram by $\Delta\phi_{\rm T} = 15^\circ$ decreases the distance of the useful samples from the threshold by $\cos(15^\circ) \approx 0.966$. It follows that:

$$p_{\rm B} = {\rm Q}(0.966 \cdot 4) \approx {\rm Q}(3.86)= 0.57 \cdot 10^{-4}\hspace{0.1cm}\underline {= 0.0057\, \%}.$$

(2) Analogous to subtask (1), $\cos(45^\circ) \approx 0.707$ is obtained:

$$p_{\rm B} = {\rm Q}(0.707 \cdot 4) \approx {\rm Q}(2.83)\hspace{0.1cm}\underline {= 0.233 \, \%}.$$

(3) For 4-QAM, clockwise rotation by $\Delta\phi_{\rm T}$ increases the distance

from the horizontal threshold (decision of the first bit) equals $s_{0} \cdot \cos(45^\circ + \Delta\phi_{\rm T})$, i.e., smaller than without phase shift,

from the vertical threshold (decision of the second bit) equal to $s_{0} \cdot \cos(45^\circ - \Delta\phi_{\rm T})$, thus larger than without phase shift.

Thus, we obtain for the average error probability:

$$p_{\rm B} = {1}/{2} \cdot {\rm Q}\left ( \frac{\cos(45^\circ+{\rm \Delta} \phi_{\rm T}) \cdot s_0}{0.25 \cdot s_0 / \sqrt{2}} \right ) + {1}/{2} \cdot {\rm Q}\left ( \frac{\cos(45^\circ-{\rm \Delta} \phi_{\rm T}) \cdot s_0}{0.25 \cdot s_0 / \sqrt{2}}\right ).$$

This already takes into account the smaller noise rms value of the 4-QAM.

As a check, we calculate the error probability for $\Delta\phi_{\rm T} = 0$:

$$p_{\rm B} = {1}/{2} \cdot {\rm Q}\left ( \frac{\cos(45^\circ) \cdot 4}{1 / \sqrt{2}} \right ) +{1}/{2} \cdot {\rm Q}\left ( \frac{\cos(45^\circ) \cdot 4}{1 / \sqrt{2}} \right )= {\rm Q}(4) = 0.317 \cdot 10^{-4}.$$

On the other hand, we obtain with $\Delta\phi_{\rm T} = 15^\circ$:

$$p_{\rm B} = {1}/{2} \cdot {\rm Q}\left ( \frac{\cos(60^\circ) \cdot 4}{1 / \sqrt{2}} \right ) +{1}/{2} \cdot {\rm Q}\left ( \frac{\cos(30^\circ) \cdot 4}{1 / \sqrt{2}} \right )= {1}/{2} \cdot \left [{\rm Q}(2.83)+ {\rm Q}(4.90)\right]$$

$$\Rightarrow \hspace{0.3cm} p_{\rm B} \approx \frac{1}{2} \cdot \left [0.233 \cdot 10^{-2}+ 0.479 \cdot 10^{-6}\right] \hspace{0.1cm}\underline {= 0.117 \, \%}.$$

(4) With a phase shift of $45^\circ$, one obtains from the equation generally derived above:

$$p_{\rm B} ={1}/{2} \cdot {\rm Q}\left ( \frac{\cos(90^\circ) \cdot 4}{1 / \sqrt{2}} \right ) +{1}/{2} \cdot {\rm Q}\left ( \frac{\cos(0^\circ) \cdot 4}{1 / \sqrt{2}} \right )= {1}/{2} \cdot \left [{\rm Q}(0)+ {\rm Q}(5.66)\right] \approx 0.25\hspace{0.1cm}\underline {= 25 \, \%}.$$

That is:

The bit error rate for the first bit is $50\%$.

In contrast, the second bit is decided almost error-free $(\approx 10^{–8})$.

Overall, this results in a mean bit error probability of approx. $25\%$.

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Digitalsignalübertragung/Lineare digitale Modulation – Kohärente Demodulation
+{{quiz-Header|Buchseite=Digital_Signal_Transmission/Linear_Digital_Modulation_-_Coherent_Demodulation
 }}
-[[File:P_ID1682__Dig_A_4_2.png|right|frame|Phasendiagramme von BPSK und 4–QAM]]
+[[File:P_ID1682__Dig_A_4_2.png|right|frame|Phase diagrams of BPSK and 4-QAM]]
-Die Grafik zeigt schematisch die Phasendiagramme der binären Phasenmodulation (abgekürzt BPSK) und der Quadraturamplitudenmodulation (4–QAM genannt). Die letztere lässt sich durch zwei BPSK–Systeme mit Cosinus– und Minus–Sinus–Träger beschreiben, wobei bei jedem der Teilkomponenten die Sendeamplitude gegenüber der BPSK um den Faktor „Wurzel aus 2” reduziert ist. Die Hüllkurve des Gesamtsignals $s(t)$ ist somit ebenfalls konstant gleich $s_{0}$.
+The diagram shows schematically the phase diagrams of the&nbsp; "binary phase modulation"&nbsp; $($abbreviated $\rm BPSK)$&nbsp; and the&nbsp; "quadrature amplitude modulation"&nbsp; $($called&nbsp; $\rm 4–QAM)$.
-Die Fehlerwahrscheinlichkeit abhängig vom Quotienten $E_{\rm B}/N_{0}$ lautet bei BPSK und 4–QAM gleichermaßen:
+*The latter can be described by two BPSK systems with cosine and minus-sine carriers,&nbsp; where for each of the subcomponents the transmission amplitude is reduced by a factor of &nbsp;$\sqrt{2}$&nbsp; compared to BPSK.
+*The envelope of the total signal &nbsp;$s(t)$&nbsp; is thus also constant equal to &nbsp;$s_{0}$.
+*The error probability depending on the quotient &nbsp;$E_{\rm B}/N_{0}$&nbsp; is the same for BPSK and 4–QAM:
 :$$p_{\rm B}  = \ {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right
   )  = \ {1}/{2}\cdot {\rm erfc}\left ( \sqrt{E_{\rm B}/{ N_0 }} \right ).$$
-Die Fehlerwahrscheinlichkeit des BPSK–Systems kann aber auch in der Form
+However,&nbsp; the error probability of the BPSK system can also be expressed in the form
 :$$p_{\rm B,\hspace{0.04cm}BPSK} = {\rm Q}\left ( \frac{s_0}{\sigma_d } \right
-  )\hspace{0.2cm}{\rm mit}\hspace{0.2cm}\sigma_d = \sqrt{\frac{N_0}{T_{\rm B}}}$$
+  )\hspace{0.2cm}{\rm with}\hspace{0.2cm}\sigma_d = \sqrt{\frac{N_0}{T_{\rm B}}}$$
-dargestellt werden. Entsprechend gilt für das 4–QAM–System:
+Correspondingly,&nbsp; for the 4-QAM system:
 :$$p_{\rm B,\hspace{0.04cm}QAM} = {\rm Q}\left ( \frac{s_0/\sqrt{2}}{\sigma_d } \right
-  )\hspace{0.2cm}{\rm mit}\hspace{0.2cm}\sigma_d = \sqrt{\frac{N_0}{2 \cdot T_{\rm
+  )\hspace{0.2cm}{\rm with}\hspace{0.2cm}\sigma_d = \sqrt{\frac{N_0}{2 \cdot T_{\rm
   B}}}.$$
-Beide Gleichungen gelten allerdings nur unter der Voraussetzung einer exakten Phasensynchronisation. Bei einem Phasenversatz $\Delta\phi_{\rm T}$ zwischen sender– und empfangsseitigem Trägersignal erhöht sich die Fehlerwahrscheinlichkeit signifikant, wobei BPSK– und QAM–System unterschiedlich degradiert werden. Im Phasendiagramm macht sich der Phasenversatz durch eine Rotation der Punktwolken bemerkbar. In der Grafik sind die Mittelpunkte der Punktwolken für $\Delta\phi_{\rm T} = 15°$ durch gelbe Kreuze markiert, während die roten Kreise die Mittelpunkte für $\Delta\phi_{\rm T} = 0$ angeben.
-Es gilt stets $E_{\rm B}/N_{0} = 8$, so dass sich die Fehlerwahrscheinlichkeiten von BPSK und QAM im günstigsten Fall (ohne Phasenversatz) jeweils wie folgt ergeben  $\Rightarrow$ [[Aufgaben:1.08Z_BPSK-Fehlerwahrscheinlichkeit|Aufgabe Z1.8:]]
+The equations are valid only under the condition of exact phase synchronization:
+*If there is a phase offset &nbsp;$\Delta\phi_{\rm T}$&nbsp; between the transmitted and received carrier signals,&nbsp; the error probability increases significantly,&nbsp; with BPSK and QAM systems being degraded differently.
+*In the phase diagram,&nbsp; the phase offset is noticeable by a rotation of the point clouds.&nbsp;
+*In the diagram,&nbsp; the centers of the point clouds for &nbsp;$\Delta\phi_{\rm T} = 15^\circ$&nbsp; are marked by yellow crosses,&nbsp; while the red circles indicate the centers for &nbsp;$\Delta\phi_{\rm T} = 0$.&nbsp;
+&nbsp;$E_{\rm B}/N_{0} = 8$ always holds,&nbsp; so the error probabilities of BPSK and QAM in the best case&nbsp; (without phase shift)&nbsp; are as follows  &nbsp; &rArr; &nbsp;  [[Aufgaben:Exercise_1.08Z:_BPSK_Error_Probability|"Exercise 1.8Z"]]:
 :$$p_{\rm B} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {\rm Q}(4)= 0.317 \cdot 10^{-4}.$$
-Bezeichnet man den Abstand der BPSK–Nutzabtastwerte von der (vertikalen) Entscheiderschwelle mit $s_{0}$, so ergibt sich für den Rauscheffektivwert $\sigma_{d} = s_{0}/4$. Die helleren Kreise in der Grafik markieren die Höhenlinien mit dem Radius $2\sigma_{d}$ bzw. $3\sigma_{d}$ der Gaußschen 2D–WDF.
+<u>Further remarks:</u>
+*If we denote the distance of the BPSK useful samples&nbsp; (without noise)&nbsp; from the&nbsp; (vertical)&nbsp; decision threshold by &nbsp;$s_{0}$,&nbsp; we get &nbsp;$\sigma_{d} = s_{0}/4$&nbsp; for the noise rms value.&nbsp; The lighter circles in the diagram mark the contour lines with radius &nbsp;$2\cdot \sigma_{d}$&nbsp; and &nbsp;$3\cdot \sigma_{d}$&nbsp; of the two-dimensional Gaussian PDF.
+*For the 4-QAM,&nbsp; compared to the BPSK,&nbsp; the distances of the noise-free samples drawn in red from the now two decision thresholds are each smaller by a factor of &nbsp;$\sqrt{2}$,&nbsp; but it also results in a noise rms value &nbsp;$\sigma_{d}$&nbsp; smaller by the same factor.
-Bei der 4–QAM sind gegenüber der BPSK die Abstände der rot eingezeichneten Nutzabtastwerte von den nun zwei Entscheiderschwellen jeweils um den Faktor „Wurzel aus 2” geringer, aber es ergibt sich auch ein um den gleichen Faktor kleinerer Rauscheffektivwert $\sigma_{d}$.
-''Hinweis:''
-*Die Aufgabe bezieht sich auf die Seite [[Digitalsignalübertragung/Lineare_digitale_Modulation_–_Kohärente_Demodulation#Phasenversatz_zwischen_Sender_und_Empf.C3.A4nger|Phasenversatz zwischen Sender und Empfänger]] im [[Digitalsignalübertragung/Lineare_digitale_Modulation_–_Kohärente_Demodulation|Lineare digitale Modulation – Kohärente Demodulation]].
+Notes:
-*Die Werte der Q–Funktion können Sie mit folgenden Interaktionsmodul ermitteln: [[Komplementäre Gaußsche Fehlerfunktionen]]
+*The exercise belongs to the chapter&nbsp;  [[Digital_Signal_Transmission/Linear_Digital_Modulation_-_Coherent_Demodulation|"Linear Digital Modulation - Coherent Demodulation"]].
-===Fragebogen===
+*Reference is made in particular to the section&nbsp; [[Digital_Signal_Transmission/Linear_Digital_Modulation_-_Coherent_Demodulation#Phase_offset_between_transmitter_and_receiver|"Phase offset between transmitter and receiver"]].
+*You can determine the values of the Q&ndash;function with the HTML5/JavaScript applet &nbsp;[[Applets:Komplementäre_Gaußsche_Fehlerfunktionen|"Complementary Gaussian Error Functions"]].&nbsp;
+===Questions===
 <quiz display=simple>
-{Wie groß ist die Bitfehlerwahrscheinlichkeit bei BPSK mit $\Delta\phi_{\rm T} = 15°$?
+{What is the bit error probability for BPSK with &nbsp;$\Delta\phi_{\rm T} = 15^\circ$?
 |type="{}"}
-$p_{\rm B} \ = \ $ { 0.57 3% } $\ \cdot 10^{-4} $
+$p_\text{B, BPSK} \ = \ $ { 0.0057 3% } $\ \% $
-{Welche Bitfehlerwahrscheinlichkeit ergibt sich bei BPSK mit $\Delta\phi_{\rm T} = 45°$?
+{What is the bit error probability for BPSK with &nbsp;$\Delta\phi_{\rm T} = 45^\circ$?
 |type="{}"}
-$p_{\rm B} \ = \ $ { 0.233 3% } $\ \cdot 10^{-2} $
+$p_\text{B, BPSK} \ = \ $ { 0.233 3% } $\ \%$
-{Wie groß ist die Bitfehlerwahrscheinlichkeit bei 4–QAM mit $\Delta\phi_{\rm T} = 15°$°?
+{What is the bit error probability for 4-QAM with &nbsp;$\Delta\phi_{\rm T} = 15^\circ$?
 |type="{}"}
-$p_{\rm B} \ = \ $ { 0.117 3% } $\ \cdot 10^{-2} $
+$p_\text{B, 4-QAM} \ = \ $ { 0.117 3% } $\ \%$
-{Welche Fehlerwahrscheinlichkeit ergibt sich bei 4–QAM mit $\Delta\phi_{\rm T} = 45°$?
+{What is the bit error probability for 4-QAM with &nbsp;$\Delta\phi_{\rm T} = 45^\circ$?
 |type="{}"}
-$p_{\rm B} \ = \ $ { 0.25 3% }
+$p_\text{B, 4-QAM} \ = \ $ { 25 3% } $\ \%$
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp;
-'''(2)'''&nbsp;
+'''(1)'''&nbsp; Rotating the phase diagram by&nbsp; $\Delta\phi_{\rm T} = 15^\circ$&nbsp; decreases the distance of the useful samples from the threshold by&nbsp; $\cos(15^\circ) \approx 0.966$.&nbsp; It follows that:
-'''(3)'''&nbsp;
+:$$p_{\rm B} = {\rm Q}(0.966 \cdot 4) \approx {\rm Q}(3.86)= 0.57 \cdot 10^{-4}\hspace{0.1cm}\underline {= 0.0057\, \%}.$$
-'''(4)'''&nbsp;
-'''(5)'''&nbsp;
-'''(6)'''&nbsp;
+'''(2)'''&nbsp; Analogous to subtask&nbsp; '''(1)''',&nbsp; $\cos(45^\circ) \approx 0.707$&nbsp; is obtained:
+:$$p_{\rm B} = {\rm Q}(0.707 \cdot 4) \approx {\rm Q}(2.83)\hspace{0.1cm}\underline {= 0.233 \, \%}.$$
+'''(3)'''&nbsp; For 4-QAM,&nbsp; clockwise rotation by&nbsp; $\Delta\phi_{\rm T}$&nbsp; increases the distance
+*from the horizontal threshold&nbsp; (decision of the first bit)&nbsp; equals&nbsp; $s_{0} \cdot \cos(45^\circ + \Delta\phi_{\rm T})$,&nbsp; i.e., smaller than without phase shift,
+*from the vertical threshold&nbsp; (decision of the second bit)&nbsp; equal to $s_{0} \cdot \cos(45^\circ - \Delta\phi_{\rm T})$,&nbsp; thus larger than without phase shift.
+Thus,&nbsp; we obtain for the average error probability:
+:$$p_{\rm B} = {1}/{2} \cdot {\rm Q}\left ( \frac{\cos(45^\circ+{\rm \Delta} \phi_{\rm
+ T}) \cdot s_0}{0.25 \cdot s_0 / \sqrt{2}} \right
+ ) + {1}/{2} \cdot {\rm Q}\left ( \frac{\cos(45^\circ-{\rm \Delta} \phi_{\rm
+ T}) \cdot s_0}{0.25 \cdot s_0 / \sqrt{2}}\right
+ ).$$
+*This already takes into account the smaller noise rms value of the 4-QAM.
+*As a check,&nbsp; we calculate the error probability for&nbsp; $\Delta\phi_{\rm T} = 0$:
+:$$p_{\rm B} = {1}/{2} \cdot {\rm Q}\left ( \frac{\cos(45^\circ) \cdot 4}{1 / \sqrt{2}} \right
+ ) +{1}/{2} \cdot {\rm Q}\left ( \frac{\cos(45^\circ) \cdot 4}{1 / \sqrt{2}} \right
+ )= {\rm Q}(4) = 0.317 \cdot 10^{-4}.$$
+*On the other hand,&nbsp; we obtain with&nbsp; $\Delta\phi_{\rm T} = 15^\circ$:
+:$$p_{\rm B}  = {1}/{2} \cdot {\rm Q}\left ( \frac{\cos(60^\circ) \cdot 4}{1 / \sqrt{2}} \right
+ ) +{1}/{2} \cdot {\rm Q}\left ( \frac{\cos(30^\circ) \cdot 4}{1 / \sqrt{2}} \right
+ )= {1}/{2} \cdot \left [{\rm Q}(2.83)+ {\rm Q}(4.90)\right]$$
+:$$\Rightarrow \hspace{0.3cm} p_{\rm B} \approx
+ \frac{1}{2} \cdot \left [0.233 \cdot 10^{-2}+ 0.479 \cdot 10^{-6}\right] \hspace{0.1cm}\underline {= 0.117 \, \%}.$$
+'''(4)'''&nbsp; With a phase shift of&nbsp; $45^\circ$,&nbsp; one obtains from the equation generally derived above:
+:$$p_{\rm B}  ={1}/{2} \cdot {\rm Q}\left ( \frac{\cos(90^\circ) \cdot 4}{1 / \sqrt{2}} \right
+ ) +{1}/{2} \cdot {\rm Q}\left ( \frac{\cos(0^\circ) \cdot 4}{1 / \sqrt{2}} \right
+ )= {1}/{2} \cdot \left [{\rm Q}(0)+ {\rm Q}(5.66)\right] \approx  0.25\hspace{0.1cm}\underline {=  25 \, \%}.$$
+That is:
+*The bit error rate for the first bit is&nbsp; $50\%$.
+*In contrast,&nbsp; the second bit is decided almost error-free&nbsp; $(\approx 10^{–8})$.
+*Overall,&nbsp; this results in a mean bit error probability of approx. $25\%$.
 {{ML-Fuß}}
@@ Line 64: / Line 128: @@
-[[Category:Aufgaben zu Digitalsignalübertragung|^1.5 Lineare digitale Modulation^]]
+[[Category:Digital Signal Transmission: Exercises|^1.5 Linear Digital Modulation^]]