# Exercise 4.12Z: 4-QAM Systems again

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

### 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 $\rm (C)$: $p_{\rm B} ≈ 2.4 · 10^{–4}$.