Exercise 4.14Z: 4-QAM and 4-PSK

From LNTwww

Signal space constellation of the  "4-QAM"  and  "4-PSK"

For  "quadrature amplitude modulation"  $\rm (M–QAM)$,  an upper bound  ("Union–Bound")  on the symbol error probability was given in the theory section for  $M ≥ 16$: 

$$ p_{\rm UB} = 4 \cdot {\rm Q} \left [ \sqrt{ { E_{\rm S}}/{ N_0}} \hspace{0.05cm}\right ] \ge p_{\rm S} \hspace{0.05cm}.$$

In the theory section,  one can also find the  "Union–Bound" for  "M–level phase modulation"    $\rm (M–PSK)$, 

$$ p_{\rm UB} = 2 \cdot {\rm Q} \left [ \sin ({ \pi}/{ M}) \cdot \sqrt{ { 2E_{\rm S}}/{ N_0}} \hspace{0.05cm}\right ] \ge p_{\rm S} \hspace{0.05cm}.$$

In both methods,  each signal space point has exactly the same energy,  namely  $E_{\rm S}$.

From the graph,  one can see that for the special case  $M = 4$,  the two modulation processes should actually be identical,  which is not directly evident from the above equations.

The 4–PSK is shown here with the phase offset  $\phi_{\rm off} = 0$.  With a general phase offset,  on the other hand,  the in-phase and quadrature components of the signal space points are generally:  $(i = 0, \ ... \ , M = 1)$:

$$s_{{\rm I}i} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \cos \left ( { 2\pi i}/{ M} + \phi_{\rm off} \right ) \hspace{0.05cm},$$
$$ s_{{\rm Q}i} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sin \left ( { 2\pi i}/{ M} + \phi_{\rm off} \right ) \hspace{0.05cm}.$$


  • In the above diagram the Gray mapping of the symbols to bit-duples is shown in red.



For which phase offset do the 4–QAM and the 4–PSK match exactly?

$\phi_{\rm off}\ = \ $

$\ \rm degree$


What is the upper bound  $($Union Bound,  $p_{\rm UB} ≥ p_{\rm S})$  for the 4–PSK?

$p_{\rm UB} = 4 \cdot {\rm Q}[\sqrt{E_{\rm S}/N_0}\hspace{0.05cm}]$,
$p_{\rm UB} = 2 \cdot {\rm Q}[\sqrt{E_{\rm S}/N_0}\hspace{0.05cm}]$,
$p_{\rm UB} = 2 \cdot {\rm Q}[\sqrt{2E_{\rm S}/N_0}\hspace{0.05cm}]$.


Specify a closer upper bound for the 4–QAM.

$p_{\rm S} ≤ 4 \cdot {\rm Q}[\sqrt{E_{\rm S}/N_0}\hspace{0.05cm}]$,
$p_{\rm S} ≤ 2 \cdot {\rm Q}[\sqrt{E_{\rm S}/N_0}\hspace{0.05cm}]$,
$p_{\rm S} ≤ 2 \cdot {\rm Q}[\sqrt{2E_{\rm S}/N_0}\hspace{0.05cm}]$.


What is the bit error probability bound for the 4–QAM,  assuming Gray coding?

$p_{\rm B} ≤ 2 \cdot {\rm Q}[\sqrt{2E_{\rm B}/N_0}\hspace{0.05cm}]$,
$p_{\rm B} ≤ {\rm Q}[\sqrt{2E_{\rm B}/N_0}\hspace{0.05cm}]$,
$p_{\rm B} ≤ {\rm Q}[\sqrt{E_{\rm B}/N_0}\hspace{0.05cm}]$.


(1)  With  $M = 4$,  the signal space points are  $\boldsymbol{s}_i = (s_{\rm I \it i}, s_{\rm Q \it i})$  of digital phase modulation  $(i = 0, \ \text{...} \ , 3)$:

$$s_{{\rm I}i} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \cos \left ( { 2\pi i}/{ M} + \phi_{\rm off} \right ) \hspace{0.05cm},$$
$$ s_{{\rm Q}i} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sin \left ( { 2\pi i}/{ M} + \phi_{\rm off} \right ) \hspace{0.05cm}.$$
  • With  $\phi_{\rm off} \ \underline {= \pi/2 \ (45^°)}$,  we obtain exactly the signal space points of the 4–QAM:
$$\boldsymbol{ s}_{\rm 0} = (+\sqrt{2}, +\sqrt{2})\hspace{0.05cm},\hspace{0.2cm}\boldsymbol{ s}_{\rm 1} = (-\sqrt{2}, +\sqrt{2})\hspace{0.05cm},\hspace{0.2cm} \boldsymbol{ s}_{\rm 3} = (-\sqrt{2}, -\sqrt{2})\hspace{0.05cm},\hspace{0.2cm}\boldsymbol{ s}_{\rm 4} = (+\sqrt{2}, -\sqrt{2}) \hspace{0.05cm}.$$

(2)  Solution 2  is correct: For the  "4–PSK"  holds:

$$p_{\rm S} \le p_{\rm UB} \hspace{-0.15cm} \ = \ \hspace{-0.15cm}2 \cdot {\rm Q} \left [ \sin ({ \pi}/{ M}) \cdot \sqrt{ { 2E_{\rm S}}/{ N_0}} \right ] = 2 \cdot {\rm Q} \left [ { 1}/{ \sqrt{2}} \cdot \sqrt{ { 2E_{\rm S}}/{ N_0}} \right ]= 2 \cdot {\rm Q} \left [ \sqrt{ { E_{\rm S}}/{ N_0}} \right ] \hspace{0.05cm}.$$

(3)  Solution 2  is correct:

  • The  "4–QAM"  is identical with the  "4–PSK"  (regarding error probability even independent of the phase offset).
  • Solution 1,  on the other hand,  gives the  "Union Bound"  of the  "M–QAM"  in general,  where  $M = 4$  is used.
  • However,  since there are no inner symbols in  "4–QAM",  this bound is too pessimistic.
  • The resulting  "Union Bound"  is then twice as large as the 4–PSK bound.

(4)  Here again the  second solution  is correct:

  • In Gray coding,  each symbol error results in a bit error if only adjacent regions are considered:   $p_{\rm B} \approx p_{\rm S}/2$.
  • Furthermore,  $E_{\rm S} = 2 \ E_{\rm B}$  is valid.  It follows that
$$p_{\rm B} = \frac{p_{\rm S}}{2} \le {\rm Q} \left [ \sqrt{ { E_{\rm S}}/{ N_0}} \right ] = {\rm Q} \left [ \sqrt{ { 2E_{\rm B}}/{ N_0}} \right ] \hspace{0.05cm}.$$
  • As derived in the solution to  "Exercise 4.13",  it is even exactly valid:
$$p_{\rm B} = {\rm Q} \left [ \sqrt{ { 2E_{\rm B}}/{ N_0}} \right ] \hspace{0.05cm}.$$
  • In this derivation,  it was used that the  "4–QAM"  can be represented by two orthogonal BPSK modulations  (with cosine and minus sinusoidal carriers,  respectively).
  • Thus,  the bit error probability of the  "4–QAM"  and thus also of the  "4–PSK" as a function of  $E_{\rm B}/N_0$  is the same as for BPSK.