Exercise 4.10: Signal Waveforms of the 16-QAM

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Signal waveforms of 16–QAM for
four typical symbols

Let us consider the 16–QAM method according to the block diagram  in the theory section.

Very briefly,  it can be described as follows:

• After serial-parallel conversion and subsequent signal space assignment,  four bits of the binary redundancy-free source signal  $q(t)$  at the input each result in a complex-valued amplitude coefficient

$$a = a_{\rm I} +{\rm j} · a_{\rm Q}.$$

• With the rectangular transmission pulse  $g_s(t)$  in the range from  $0$  to  $T$  and of height  $g_0$,  after multiplication with the cosine function or minus-sine function in the given time interval,  we obtain:

$$s_{\rm cos}(t) = a_{\rm I}\cdot g_0 \cdot \cos(2 \pi f_{\rm T} t)\hspace{0.05cm},$$

$$s_{\rm -sin}(t) = -a_{\rm Q} \cdot g_0 \cdot \sin(2 \pi f_{\rm T} t)\hspace{0.05cm}.$$

• The 16-QAM transmitted signal is then the sum of these two component signals:

$$s(t) = s_{\rm cos}(t)+ s_{\rm -sin}(t) \hspace{0.05cm}.$$

The graph shows the signals  $s_{\rm cos}(t)$,  $s_{\rm –sin}(t)$  and  $s(t)$ for four selected symbols.  Using these,  the amplitude coefficients are to be determined.

Hints:

• This exercise belongs to the chapter  "Quadrature Amplitude Modulation".
• The page  "Signal waveforms for 4–QAM"  is helpful for completing this exercise.
• The signal space allocation considered can be seen in  Exercise 4.10Z.  The color highlights also correspond.
• From question   (6)  onwards,  use the parameter values $g_0 = 1 \ \rm V$  and  $T = 1 \ \rm µ s$.
• Energy values are to be given in  $\rm V^2s$ ; like this,  they refer to the reference resistance  $R = 1 \ \rm \Omega$.

Questions

1

What are the real and imaginary part of the amplitude coefficient  $a$  in the red time interval  $(0 < t < T)$?

$a_{\rm I} \ = \ $

$a_{\rm Q} \ = \ $

2

What is the relationship between  $s_0$  (maximum envelope of the transmitted signal) and  $g_0$  (maximum envelope of the partial signals)?

$s_0/g_0 \ = \ $

3

What is the amplitude coefficient  $a$  in the blue time interval  $(T < t < 2T)$?

$a_{\rm I} \ = \ $

$a_{\rm Q} \ = \ $

4

What is the amplitude coefficient  $a$  in the green time interval  $(2T < t < 3T)$?  Determine also its magnitude and phase.

$a_{\rm I} \ = \ $

$a_{\rm Q} \ = \ $

5

What is the amplitude coefficient  $a$  in the purple time interval  $(3T < t < 4T)$?

$a_{\rm I} \ = \ $

$a_{\rm Q} \ = \ $

6

What is the maximum energy  $E_\text{S, max}$  expended per symbol? Under what condition is the average energy per symbol equal to $E_\text{S, max}$?

$E_\text{S, max} \ = \ $ $\ \cdot 10^{-6} \ \rm V^2 s$

7

What is the maximum energy  $E_\text{B, max}$  per bit?

$E_\text{B, max} \ = \ $ $\ \cdot 10^{-6} \ \rm V^2 s$

8

What is the minimum energy  $E_\text{B, min}$  per bit?

$E_\text{B, min} \ = \ $ $\ \cdot 10^{-6} \ \rm V^2 s$

Solution

Solution

(1)  From the  (red)  inphase signal   ⇒   the real part follows  (first equation according to the definition,  second equation according to the graph):

$$s_{\rm cos}(t)= a_{\rm I}\cdot g_0 \cdot \cos(2 \pi f_{\rm T} t)= g_0 \cdot \cos(2 \pi f_{\rm T} t)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}a_{\rm I}\hspace{0.15cm}\underline {= +1} \hspace{0.05cm}.$$

• Accordingly,  we recognize from the quadrature signal   ⇒   imaginary part:

$$s_{\rm -sin}(t)= -a_{\rm Q}\cdot g_0 \cdot \sin(2 \pi f_{\rm T} t)= -g_0 \cdot \sin(2 \pi f_{\rm T} t)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}a_{\rm Q}\hspace{0.15cm}\underline {= +1} \hspace{0.05cm}.$$

(2)  The two partial signals each have the  (maximum)  envelope  $g_0$,  while  $s_0$ characterizes the transmitted signal  $s(t)$.

• As can be seen from the signal space allocation  (see Exercise 4.10Z) :

$${s_0}/{ g_0 }= \sqrt{2}\hspace{0.15cm}\underline { = 1.414} \hspace{0.05cm}.$$

(3)  The amplitude coefficients  $a_{\rm I}$  and  $a_{\rm Q}$  have the same signs as in subtask  (1),  but with smaller magnitude:

$$a_{\rm I} = + 1/3\hspace{0.15cm}\underline {= +0.333} \hspace{0.05cm},\hspace{0.5cm}a_{\rm Q} = + 1/3\hspace{0.15cm}\underline {= +0.333} \hspace{0.05cm}.$$

(4)  In the third  (green)  interval,  we can see a minus-cosine signal with amplitude  $g_0$  and a minus-sine signal with amplitude  $g_0/3$:

$$a_{\rm I} = \hspace{0.15cm}\underline {= -1} \hspace{0.05cm},\hspace{0.5cm}a_{\rm Q} = + 1/3\hspace{0.15cm}\underline {= +0.333} \hspace{0.05cm}.$$

• As is yet to be calculated in sub-task  (4)  of Exercise 4.10Z,  here the magnitude is equal to  $|a| =1.054$  and the phase angle is   ${\rm arc} \ a \approx 161^\circ$.

(5)  The violet signal does not differ from the green interval in the in-phase component except in the sign of the quadrature component:

$$a_{\rm I} = \hspace{0.15cm}\underline {= -1} \hspace{0.05cm},\hspace{0.5cm}a_{\rm Q} = - 1/3\hspace{0.15cm}\underline {= -0.333} \hspace{0.05cm}.$$

(6)  The maximum signal energy occurs when one of the four outer vertices is occupied. In that case:

$$E_{\rm S, \hspace{0.05cm}max} = {1}/{2}\cdot s_0^2 \cdot T = {1}/{2}\cdot \left (\sqrt{2} \cdot g_0 \right )^2 \cdot T = g_0^2 \cdot T = (1\,{\rm V})^2 \cdot (1\,{\rm \mu s}) \hspace{0.15cm}\underline {= 10^{-6}\,{\rm V^2s}}\hspace{0.05cm}.$$

• The average signal energy is equal to the maximum value when only the corner points of the signal space mapping are occupied and the  "inner symbols"  are excluded from the encoding.

(7)  Four bits are transmitted per symbol.  From this it follows that:

$$E_{\rm B, \hspace{0.05cm}max} = {E_{\rm S, \hspace{0.05cm}max}}/{4}\hspace{0.15cm}\underline {= 0.25 \cdot 10^{-6}\,{\rm V^2s}}\hspace{0.05cm}.$$

(8)  The minimum signal energy is obtained at one of the inner signal space points and is smaller by a factor of $9$  than in subtask  (7):

$$E_{\rm B, \hspace{0.05cm}min} = \frac{E_{\rm B, \hspace{0.05cm}max}}{9} = \frac{g_0^2 \cdot T}{36} \hspace{0.15cm}\underline { \approx 0.028 \cdot 10^{-6}\,{\rm V^2s}}\hspace{0.05cm}.$$

• In the theory section,  it was shown that,  assuming all symbols are equally likely,  the average signal energy per bit for 16-QAM is approximately:

$$E_{\rm B} \approx 0.139 · g_0^2 \cdot T = 0.035 \cdot 10^{-6}\,{\rm V^2s}.$$