Aufgaben:Exercise 4.12Z: 4-QAM Systems again - Revision history

Guenter at 14:53, 19 April 2022

2022-04-19T14:53:03Z

Guenter at 14:40, 19 April 2022

2022-04-19T14:40:46Z

Guenter at 13:46, 6 April 2022

2022-04-06T13:46:03Z

Guenter: Guenter moved page Aufgaben:Exercise 4.12Z: 4-QAM systems again to Aufgaben:Exercise 4.12Z: 4-QAM Systems again

2022-04-06T13:42:46Z

Guenter moved page Exercise 4.12Z: 4-QAM systems again to Exercise 4.12Z: 4-QAM Systems again

Reed at 13:44, 20 March 2022

2022-03-20T13:44:36Z

Reed at 13:40, 20 March 2022

2022-03-20T13:40:11Z

Reed: /* Solution */

2022-03-19T19:04:46Z

Solution

Reed at 18:57, 19 March 2022

2022-03-19T18:57:27Z

Reed: /* Questions */

2022-03-19T18:57:12Z

Questions

Reed at 18:54, 19 March 2022

2022-03-19T18:54:18Z

@@ Line 65: / Line 65: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp;  From &nbsp; $10 · \lg E_{\rm B}/N_0 = 9 \ \rm dB$&nbsp;, &nbsp; ${E_{\rm B}}/{N_0} = 10^{0.9}\approx 7.95 \hspace{0.05cm} follows.$&nbsp;
+'''(1)'''&nbsp;  From &nbsp; $10 · \lg E_{\rm B}/N_0 = 9 \ \rm dB$ &nbsp; &rArr;  &nbsp; ${E_{\rm B}}/{N_0} = 10^{0.9}\approx 7.95 \hspace{0.05cm}$&nbsp; follows:$&nbsp;
-*With the given approximation, it further holds that:
+*With the given approximation,&nbsp; it further holds:
 :$$p_{\rm B}  =  {1}/{2}\cdot {\rm erfc}\left ( \sqrt{{E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) \approx \frac{1}{2 \cdot\sqrt{\pi \cdot{E_{\rm B}}/{N_0}} } \cdot {\rm e}^{-{E_{\rm B}}/{N_0}}  =  {1}/{2 \cdot\sqrt{7.95 \cdot \pi }} \cdot {\rm e}^{-7.95}\approx \hspace{0.15cm}\underline {3.5 \cdot 10^{-5}\hspace{0.05cm}}.$$
 *The exact value&nbsp; $p_{\rm B}\hspace{0.15cm}\underline { = 3.3 · 10^{–5}}$&nbsp; is only slightly smaller.
@@ Line 72: / Line 72: @@
-'''(2)'''&nbsp;  <u>Answer 1</u> is correct:
+'''(2)'''&nbsp;  <u>Answer 1</u>&nbsp; is correct:
-*Due to a phase shift of &nbsp; $Δϕ_{\rm T} = 30^\circ$&nbsp;, the phase diagram was rotated, resulting in degradation.
+*Due to a phase shift of &nbsp; $Δϕ_{\rm T} = 30^\circ$,&nbsp; the phase diagram was rotated,&nbsp; resulting in degradation.
-*The two components &nbsp; $\rm I$&nbsp; and&nbsp; $\rm Q$&nbsp; influence each other, but there is no pulse interference as in system &nbsp;$\rm (C)$.
+*The two components &nbsp; $\rm I$&nbsp; and&nbsp; $\rm Q$&nbsp; influence each other,&nbsp; but there is no intersymbol interference as in system &nbsp;$\rm (C)$.&nbsp;
-*A "Nyquist system" never leads to pulse interference.
+*A&nbsp; "Nyquist system"&nbsp; never leads to intersymbol interference.
 '''(4)'''&nbsp;  <u>Answers 2 and 3</u> are correct:
 *Systems &nbsp;$\rm (B)$&nbsp; and &nbsp;$\rm (C)$&nbsp; are not optimal.&nbsp; This already shows that statement 1 is not correct.
-* In contrast, Answer 2 is right.&nbsp; Every 4-QAM system, which follows the matched filter principle and additionally fulfills the first Nyquist criterion, has the error probability given above
+* In contrast,&nbsp; Answer 2 is right.&nbsp; Every 4-QAM system,&nbsp; which follows the matched filter principle and additionally fulfills the first Nyquist criterion,&nbsp; has the error probability given above:
 :$$p_{\rm B} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{{E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ).$$
-*Thus, the so-called "root-Nyquist configuration", which was treated for example in Exercise 4.12, has exactly the same error probability as system &nbsp;$\rm (A)$&nbsp; and also the same phase diagram at the detection times.  The transitions between the individual points are nevertheless different.
+*Thus,&nbsp; the so-called&nbsp; "root-Nyquist configuration",&nbsp; which was treated for example in Exercise 4.12,&nbsp; has exactly the same error probability as system &nbsp;$\rm (A)$&nbsp; and also the same phase diagram at the detection times.&nbsp;  The transitions between the individual points are nevertheless different.
-*The third statement is also true.  One can already recognize incorrect decisions from the phase diagram of system &nbsp;$\rm (B)$&nbsp;, and this will always be the case when the points do not match the quadrants in terms of color.
+*The third statement is also true.&nbsp;  One can already recognize incorrect decisions from the phase diagram of system &nbsp;$\rm (B)$,&nbsp; and this will always be the case when the points do not match the quadrants in terms of color.

@@ Line 5: / Line 5: @@
 [[File:P_ID1724__Mod_Z_4_11.png|right|frame|Phase diagrams for 4–QAM, ideal and with degradations]]
-Graph &nbsp;$\rm (A)$&nbsp; shows the phase diagram of the 4-QAM after the matched filter, where an optimal realization form was chosen in the case of AWGN noise under the constraint of "peak limiting":
+Graph &nbsp;$\rm (A)$&nbsp; shows the phase diagram of the 4-QAM after the matched filter,&nbsp; where an optimal realization form was chosen in the case of AWGN noise under the constraint of&nbsp; "peak limiting":
-* rectangular fundamental transmision pulse of symbol duration &nbsp;$T$,
+* rectangular basic transmision pulse of symbol duration &nbsp;$T$,
 * rectangular impulse response of the matched filter of the same width &nbsp;$T$.
-All phase diagrams presented here - both &nbsp;$\rm (A)$&nbsp; and  &nbsp;$\rm (B)$&nbsp; and &nbsp;$\rm (C)$&nbsp; - refer to the detection time points only. Thus, the transitions between the individual discrete-time points are not plotted in this phase diagram.
+All phase diagrams presented here &ndash; &nbsp;$\rm (A)$&nbsp; and  &nbsp;$\rm (B)$&nbsp; and &nbsp;$\rm (C)$&nbsp; &ndash; refer to the detection time points only.&nbsp; Thus,&nbsp; the transitions between the individual discrete-time points are not plotted in this phase diagram.
-*An AWGN channel with &nbsp;$10 · \lg E_{\rm B}/N_0 = 9 \ \rm dB$&nbsp; is present.
+*An AWGN channel with &nbsp; $10 · \lg E_{\rm B}/N_0 = 9 \ \rm dB$ &nbsp; is present.
-*Accordingly, for the bit error probability of the first system considered &nbsp;$\rm (A)$&nbsp;:
+*Accordingly,&nbsp; for the bit error probability of the first system considered &nbsp;$\rm (A)$&nbsp;:
 :$$p_{\rm B} = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{{E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right )\hspace{0.05cm}.$$
-Phase diagrams &nbsp;$\rm (B)$&nbsp; and &nbsp;$\rm (C)$&nbsp;  belong to two systems where the 4-QAM was not optimally realized. AWGN noise with &nbsp;$10 · \lg E_{\rm B}/N_0 = 9 \ \rm dB$&nbsp; is also assumed in each of these.
+The phase diagrams &nbsp;$\rm (B)$&nbsp; and &nbsp;$\rm (C)$&nbsp;  belong to two systems where the 4-QAM was not optimally realized.&nbsp; AWGN noise with &nbsp;$10 · \lg E_{\rm B}/N_0 = 9 \ \rm dB$&nbsp; is also assumed in each of these.
 Hints:
+*This exercise belongs to the chapter&nbsp; [[Modulation_Methods/Quadrature_Amplitude_Modulation|"Quadrature Amplitude Modulation"]].
+*Reference is also made to the page&nbsp; [[Digital_Signal_Transmission/Linear_Digital_Modulation_-_Coherent_Demodulation#Phase_offset_between_transmitter_and_receiver|"Phase offset between transmitter and receiver"]] in the book&nbsp; "Digital Signal Transmission".
+*Causes and Effects of intersymbol interference are explained in the &nbsp;[[Digital_Signal_Transmission/Causes_and_Effects_of_Intersymbol_Interference|section with the same name]]&nbsp; of the book&nbsp; "Digital Signal Transmission".
 *The crosses in the graphs mark possible points in the phase diagrams if no AWGN noise were present.
-*The point clouds due to the AWGN noise all have the same diameter. The red cloud appears slightly smaller than the others only because "red" is harder to see on "black".
+*The point clouds due to the AWGN noise all have the same diameter.&nbsp; The red cloud appears slightly smaller than the others only because&nbsp; "red"&nbsp; is harder to see on a black background.
+*As a sufficiently good approximation for the complementary Gaussian error integral,&nbsp; you can use:
 :$${\rm erfc}(x) \approx \frac{1}{\sqrt{\pi}\cdot x} \cdot {\rm e}^{-x^2}.$$
@@ Line 42: / Line 35: @@
 <quiz display=simple>
-{Using the given approximation, calculate the bit error probability of system &nbsp;$\rm (A)$.
+{Using the given approximation,&nbsp; calculate the bit error probability of system &nbsp;$\rm (A)$.
 |type="{}"}
-$p_{\rm B} \ = \ $ { 3.5 3% } $\ \cdot 10^{-5}$
+System &nbsp;$\rm (A):\ \ p_{\rm B} \ = \ $ { 3.5 3% } $\ \cdot 10^{-5}$
@@ Line 50: / Line 43: @@
 {What are the properties of system &nbsp;$\rm (B)$&nbsp;?
 |type="[]"}
-+ There is a phase offset between the transmitter and receiver.
++ There is a phase offset between transmitter and receiver.
-- The receiver filter results in pulse interference.
+- The receiver filter results in intersymbol interference.
 - There is no degradation compared to system &nbsp;$\rm (A)$.
-{ What are the properties of system&nbsp;$\rm (C)$&nbsp;?
+{ What are the properties of system &nbsp;$\rm (C)$&nbsp;?
 |type="[]"}
-- There is a phase offset between the transmitter and receiver.
+- There is a phase offset between transmitter and receiver.
-+ The receiver filter results in pulse interference.
++ The receiver filter results in intersymbol interference.
 - There is no degradation compared to system &nbsp;$\rm (A)$.
@@ Line 64: / Line 57: @@
 - All three systems have the same bit error probability.
 + The error probability of system &nbsp;$\rm (A)$&nbsp; is the smallest.
-+ System &nbsp;$\rm (B)$&nbsp; has a larger bit error probability than system&nbsp;$\rm (C)$.
++ System &nbsp;$\rm (B)$&nbsp; has a larger bit error probability than system&nbsp; $\rm (C)$.

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID1724__Mod_Z_4_11.png|right|frame|Phase diagrams for 4–QAM, ideal and with degradations]
+[[File:P_ID1724__Mod_Z_4_11.png|right|frame|Phase diagrams for 4–QAM, ideal and with degradations]]
 Graph &nbsp;$\rm (A)$&nbsp; shows the phase diagram of the 4-QAM after the matched filter, where an optimal realization form was chosen in the case of AWGN noise under the constraint of "peak limiting":

@@ Line 97: / Line 97: @@
 * In contrast, Answer 2 is right.&nbsp; Every 4-QAM system, which follows the matched filter principle and additionally fulfills the first Nyquist criterion, has the error probability given above
 :$$p_{\rm B} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{{E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ).$$
-*Die so genannte „Wurzel–Nyquist–Konfiguration”, die zum Beispiel in der Aufgabe 4.12 behandelt wurde, hat somit die genau gleiche Fehlerwahrscheinlichkeit wie das System &nbsp;$\rm (A)$&nbsp; und zu den Detektionszeitpunkten auch das gleiche Phasendiagramm.&nbsp; Die Übergänge zwischen den einzelnen Punkten sind jedoch unterschiedlich.
+*Thus, the so-called "root-Nyquist configuration", which was treated for example in Exercise 4.12, has exactly the same error probability as system &nbsp;$\rm (A)$&nbsp; and also the same phase diagram at the detection times.  The transitions between the individual points are nevertheless different.
-*Auch die dritte Aussage ist zutreffend.&nbsp; Man erkennt bereits aus dem Phasendiagramm von System &nbsp;$\rm (B)$&nbsp; Fehlentscheidungen und zwar immer dann, wenn Punkte farblich nicht zu den Quadranten passen.
+*The third statement is also true.  One can already recognize incorrect decisions from the phase diagram of system &nbsp;$\rm (B)$&nbsp;, and this will always be the case when the points do not match the quadrants in terms of color.
-Die Fehlerwahrscheinlichkeiten von System &nbsp;$\rm (B)$&nbsp; und System &nbsp;$\rm (C)$&nbsp; werden im Buch „Digitalsignalübertragung” hergeleitet. Die Ergebnisse einer Systemsimulation bestätigen die obigen Aussagen:
+The error probabilities of system &nbsp;$\rm (B)$&nbsp; and system &nbsp;$\rm (C)$&nbsp; are derived in the book "Digital Signal Transmission". The results of a system simulation confirm the above statements:
-* System &nbsp;$\rm (A)$: &nbsp; &nbsp; $p_{\rm B} ≈ 3.3 · 10^{–5}$ (siehe Teilaufgabe 1),
+* System &nbsp;$\rm (A)$: &nbsp; &nbsp; $p_{\rm B} ≈ 3.3 · 10^{–5}$ (see Question 1),
 * System &nbsp;$\rm (B)$: &nbsp; &nbsp; $p_{\rm B} ≈ 3.5 · 10^{–2}$,
 * System &nbsp;$\rm (C)$: &nbsp; &nbsp; $p_{\rm B} ≈ 2.4 · 10^{–4}$.

@@ Line 89: / Line 89: @@
 *In particular, the nine crosses in each quadrant of the phase diagram &nbsp;$\rm (C)$, which mark the noise-free case, show the influence of impulse interference.
 *Instead of the optimal receiver filter for a rectangular fundamental transmission pulse&nbsp; $g_s(t)$ &nbsp; &rArr; &nbsp; rectangular impulse response &nbsp; $h_{\rm E}(t)$&nbsp;, a [[Signal_Representation/Special_Cases_of_Pulses#Gaussian_pulse|Gaussian low-pass filter]] with (normalized) cutoff frequency &nbsp; $f_{\rm G} · T = 0.6$&nbsp; was used here.
-*Dieser bewirkt Impulsinterferenzen.&nbsp; Auch ohne Rauschen gibt es in jedem Quadranten neun Kreuze, die auf je einen Vor&ndash; und Nachläufer pro Komponente hinweisen.
+*This causes pulse interference.  Even without noise, there are nine crosses in each quadrant indicating one leader and one follower per component.
-'''(4)'''&nbsp;  Richtig sind die <u>Lösungsvorschläge 2 und 3</u>:
+'''(4)'''&nbsp;  <u>Answers 2 and 3</u> are correct:
-*Die Systeme &nbsp;$\rm (B)$&nbsp; und &nbsp;$\rm (C)$&nbsp; sind nicht optimal.&nbsp; Daraus ist bereits ersichtlich, dass die Aussage 1 nicht zutrifft.
+*Systems &nbsp;$\rm (B)$&nbsp; and &nbsp;$\rm (C)$&nbsp; are not optimal.&nbsp; This already shows that statement 1 is not correct.
-* Dagegen ist  die Aussage 2 richtig.&nbsp; Jedes 4–QAM–System, das dem Matched–Filter–Prinzip folgt und zusätzlich die erste Nyquistbedingung erfüllt, besitzt die vorne angegebene Fehlerwahrscheinlichkeit
+* In contrast, Answer 2 is right.&nbsp; Every 4-QAM system, which follows the matched filter principle and additionally fulfills the first Nyquist criterion, has the error probability given above
 :$$p_{\rm B} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{{E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ).$$
 *Die so genannte „Wurzel–Nyquist–Konfiguration”, die zum Beispiel in der Aufgabe 4.12 behandelt wurde, hat somit die genau gleiche Fehlerwahrscheinlichkeit wie das System &nbsp;$\rm (A)$&nbsp; und zu den Detektionszeitpunkten auch das gleiche Phasendiagramm.&nbsp; Die Übergänge zwischen den einzelnen Punkten sind jedoch unterschiedlich.

@@ Line 70: / Line 70: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp;  Aus der Angabe&nbsp; $10 · \lg E_{\rm B}/N_0 = 9 \ \rm dB$&nbsp; folgt &nbsp; ${E_{\rm B}}/{N_0} = 10^{0.9}\approx 7.95 \hspace{0.05cm}.$&nbsp;

@@ Line 30: / Line 30: @@
 *Particular reference is made to the page&nbsp; [[Digital_Signal_Transmission/Linear_Digital_Modulation_-_Coherent_Demodulation#Phase_offset_between_transmitter_and_receiver|Phase offset between transmitter and receiver]] in the book "Digital Signal Transmission".
 *Causes and Effects of impulse interference are explained in the &nbsp;[[Digital_Signal_Transmission/Causes_and_Effects_of_Intersymbol_Interference|section with the same name]]&nbsp; of the book "Digital Signal Transmission".
-*Die Kreuze in den Grafiken markieren mögliche Punkte in den Phasendiagrammen, wenn kein AWGN–Rauschen vorhanden wäre.
+*The crosses in the graphs mark possible points in the phase diagrams if no AWGN noise were present.
-*Die Punktwolken aufgrund des AWGN–Rauschens haben alle gleichen Durchmesser. Die rote Wolke erscheint nur deshalb etwas kleiner als die anderen, da "Rot" auf "Schwarz" schlechter zu erkennen ist.
+*The point clouds due to the AWGN noise all have the same diameter. The red cloud appears slightly smaller than the others only because "red" is harder to see on "black".
-*Als eine hinreichend gute Näherung für das komplementäre Gaußsche Fehlerintegral können Sie verwenden:
 :$${\rm erfc}(x) \approx \frac{1}{\sqrt{\pi}\cdot x} \cdot {\rm e}^{-x^2}.$$
-===Fragebogen===
+===Questions===
 <quiz display=simple>