Exercise 4.12Z: 4-QAM Systems again

[[File:P_ID1724__Mod_Z_4_11.png|right|frame|Phase diagrams for 4–QAM, ideal and with degradations]

Graph $\rm (A)$ 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":

rectangular fundamental transmision pulse of symbol duration $T$,
rectangular impulse response of the matched filter of the same width $T$.

All phase diagrams presented here - both $\rm (A)$ and $\rm (B)$ and $\rm (C)$ - refer to the detection time points only. Thus, the transitions between the individual discrete-time points are not plotted in this phase diagram.

An AWGN channel with $10 · \lg E_{\rm B}/N_0 = 9 \ \rm dB$ is present.
Accordingly, for the bit error probability of the first system considered $\rm (A)$ :

$$p_{\rm B} = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{{E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right )\hspace{0.05cm}.$$

Phase diagrams $\rm (B)$ and $\rm (C)$ belong to two systems where the 4-QAM was not optimally realized. AWGN noise with $10 · \lg E_{\rm B}/N_0 = 9 \ \rm dB$ is also assumed in each of these.

Hints:

This exercise belongs to the chapter Quadrature Amplitude Modulation.
Particular reference is made to the page Phase offset between transmitter and receiver in the book "Digital Signal Transmission".
Causes and Effects of impulse interference are explained in the section with the same name of the book "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".

As a sufficiently goodapproximation for the complementary Gaussian error integral, you can use:

$${\rm erfc}(x) \approx \frac{1}{\sqrt{\pi}\cdot x} \cdot {\rm e}^{-x^2}.$$

Questions

$p_{\rm B} \ = \ $

$\ \cdot 10^{-5}$

	There is a phase offset between the transmitter and receiver.
	The receiver filter results in pulse interference.
	There is no degradation compared to system $\rm (A)$.

	There is a phase offset between the transmitter and receiver.
	The receiver filter results in pulse interference.
	There is no degradation compared to system $\rm (A)$.

	All three systems have the same bit error probability.
	The error probability of system $\rm (A)$ is the smallest.
	System $\rm (B)$ has a larger bit error probability than system $\rm (C)$.

Solution

(1) From $10 · \lg E_{\rm B}/N_0 = 9 \ \rm dB$ , ${E_{\rm B}}/{N_0} = 10^{0.9}\approx 7.95 \hspace{0.05cm} follows.$

With the given approximation, it further holds that:

$$p_{\rm B} = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{{E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) \approx \frac{1}{2 \cdot\sqrt{\pi \cdot{E_{\rm B}}/{N_0}} } \cdot {\rm e}^{-{E_{\rm B}}/{N_0}} = {1}/{2 \cdot\sqrt{7.95 \cdot \pi }} \cdot {\rm e}^{-7.95}\approx \hspace{0.15cm}\underline {3.5 \cdot 10^{-5}\hspace{0.05cm}}.$$

The exact value $p_{\rm B}\hspace{0.15cm}\underline { = 3.3 · 10^{–5}}$ is only slightly smaller.

(2) Answer 1 is correct:

Due to a phase shift of $Δϕ_{\rm T} = 30^\circ$ , the phase diagram was rotated, resulting in degradation.
The two components $\rm I$ and $\rm Q$ influence each other, but there is no pulse interference as in system $\rm (C)$.
A "Nyquist system" never leads to pulse interference.

(3) Answer 2 is correct:

In particular, the nine crosses in each quadrant of the phase diagram $\rm (C)$, which mark the noise-free case, show the influence of impulse interference.
Instead of the optimal receiver filter for a rectangular fundamental transmission pulse $g_s(t)$ ⇒ rectangular impulse response $h_{\rm E}(t)$ , a Gaussian low-pass filter with (normalized) cutoff frequency $f_{\rm G} · T = 0.6$ was used here.
This causes pulse interference. Even without noise, there are nine crosses in each quadrant indicating one leader and one follower per component.

(4) Answers 2 and 3 are correct:

Systems $\rm (B)$ and $\rm (C)$ are not optimal. This already shows that statement 1 is not correct.
In contrast, Answer 2 is right. 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 $\rm (A)$ und zu den Detektionszeitpunkten auch das gleiche Phasendiagramm. Die Übergänge zwischen den einzelnen Punkten sind jedoch unterschiedlich.
Auch die dritte Aussage ist zutreffend. Man erkennt bereits aus dem Phasendiagramm von System $\rm (B)$ Fehlentscheidungen und zwar immer dann, wenn Punkte farblich nicht zu den Quadranten passen.

Die Fehlerwahrscheinlichkeiten von System $\rm (B)$ und System $\rm (C)$ werden im Buch „Digitalsignalübertragung” hergeleitet. Die Ergebnisse einer Systemsimulation bestätigen die obigen Aussagen:

System $\rm (A)$: $p_{\rm B} ≈ 3.3 · 10^{–5}$ (siehe Teilaufgabe 1),
System $\rm (B)$: $p_{\rm B} ≈ 3.5 · 10^{–2}$,
System $\rm (C)$: $p_{\rm B} ≈ 2.4 · 10^{–4}$.