Aufgaben:Exercise 4.12Z: 4-QAM Systems again

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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

1

Using the given approximation,  calculate the bit error probability of system  $\rm (A)$.

System  $\rm (A):\ \ p_{\rm B} \ = \ $ $\ \cdot 10^{-5}$

2

What are the properties of system  $\rm (B)$ ?

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

3

What are the properties of system  $\rm (C)$ ?

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

4

Which statements about the error probabilities are correct ?

	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

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