Exercise 4.12Z: 4-QAM Systems again

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

All phase diagrams presented here – $\rm (A)$ 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 $\rm (A)$ :

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

The 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".
Reference is also made to the page "Phase offset between transmitter and receiver" in the book "Digital Signal Transmission".
Causes and Effects of intersymbol 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 a black background.
As a sufficiently good approximation 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

System $\rm (A):\ \ p_{\rm B} \ = \ $

$\ \cdot 10^{-5}$

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

	There is a phase offset between transmitter and receiver.
	The receiver filter results in intersymbol 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:

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

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 intersymbol interference as in system $\rm (C)$.
A "Nyquist system" never leads to intersymbol 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 intersymbol interference.
Instead of the optimal receiver filter for a rectangular basic 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 intersymbol 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:

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

Thus, the so-called "root-Nyquist configuration", which was treated for example in Exercise 4.12, has exactly the same error probability as system $\rm (A)$ and also the same phase diagram at the detection times. 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 $\rm (B)$, and this will always be the case when the points do not match the quadrants in terms of color.

The error probabilities of system $\rm (B)$ and system $\rm (C)$ are derived in the book "Digital Signal Transmission". The results of a system simulation confirm the above statements:

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