Exercise 4.11Z: Error Probability with QAM

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Table with two different  "Gaussian error functions"

We now make the following assumptions:

• binary bipolar amplitude coefficients  $a_ν ∈ \{±1\}$,
• rectangular basic transmission pulse with amplitude  $s_0$  and bit duration  $T_{\rm B}$,
• AWGN noise with noise power density  $N_0$,
• a receiver according to the matched-filter principle,
• the best possible demodulation and detection.

The bit error probability of binary phase modulation   $\rm (BPSK)$  under these conditions can be calculated using the following equations:

$$p_{\rm B, \hspace{0.05cm}BPSK} = {\rm Q}\left ({s_0}/{\sigma_d } \right ), \hspace{0.2cm} E_{\rm B} = {1}/{2} \cdot s_0^2 \cdot T_{\rm B} ,\hspace{0.2cm} \sigma_d^2 = {N_0}/{T_{\rm B} }$$

$$\Rightarrow \hspace{0.3cm} p_{\rm B, \hspace{0.05cm}BPSK} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{{E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ).$$

• The corresponding equations of  $\rm 4–QAM$  are:

$$p_{\rm B, \hspace{0.05cm}4-QAM} = {\rm Q}\left ( {g_0}/{\sigma_d } \right ), \hspace{0.2cm}g_{0} = {s_0}/{\sqrt{2}}, \hspace{0.2cm}E_{\rm B} = {1}/{2} \cdot s_0^2 \cdot T_{\rm B} ,\hspace{0.2cm} \sigma_d^2 = {N_0}/({2 \cdot T_{\rm B} }).$$

• Here it is taken into account that – in order to achieve the same transmission energy per bit as with BPSK – one must reduce the amplitude $g_0$  of the rectangular pulses in the two sub-branches of 4-QAM by a factor of  $\sqrt{2}$ .
• The envelope is then equal to  $s_0$ for both systems.

Hints:

• This exercise belongs to the chapter  "Quadrature Amplitude Modulation".
• Reference is also made to the page  "Error probabilities – a brief overview"  in the previous chapter.
• Always assume the following numerical values:   $s_0 = 2\,{\rm V}, \hspace{0.05cm} N_0 = 0.25 \cdot 10^{-6}\,{\rm V^2/Hz}\hspace{0.05cm}.$
• The bit duration is  $T_{\rm B} = 1 \ \rm µ s$  (question 1)  and  $T_{\rm B} = 2 \ \rm µ s$  (from question 2 onwards).
• In the table,  the two common Gaussian error functions  ${\rm Q}(x)$  and  $1/2 \cdot {\rm erfc}(x)$  are given.
• Energies are to be given in  $\rm V^2s$;  thus, they refer  to the reference resistance  $R = 1 \ \rm \Omega$.

Questions

1

What error probability  $p_\text{B, BPSK}$  results for  BPSK  when $T_{\rm B} = 1 \ \rm µ s$?

$p_\text{B, BPSK} \ = \ $ $\ \rm 10^{-4}$

2

What error probability  $p_\text{B, BPSK}$  results for  BPSK  when $T_{\rm B} = 2 \ \rm µ s$?

$p_\text{B, BPSK} \ = \ $ $\ \rm 10^{-8}$

3

What error probability  $p_\text{B, 4-QAM}$  is obtained for  4-QAM  when  $E_{\rm B} = 4 · 10^{–6} \ \rm V^2s$?

$p_\text{B, 4-QAM} \ = \ $ $\ \rm 10^{-8}$

4

Which statements apply if we consider only one branch   $\rm (I$  or  $\rm Q)$  of the 4–QAM?

	The same result is obtained as for the entire 4-QAM.
	The distance of the noisless samples from the threshold  $E=0$  is the same  $(s_0)$  as in BPSK.
	The same result is obtained for the noise power as in BPSK.

Solution

Solution

(1)  With the values given, for  Binary Phase Shift Keying  (BPSK), one gets:

$$E_{\rm B} = {1}/{2} \cdot s_0^2 \cdot T_{\rm B} = \frac{1}{2}\cdot (2\,{\rm V})^2 \cdot 1\,{\rm µ s} = 2 \cdot 10^{-6}\,{\rm V^2s} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {E_{\rm B}}/{N_0} = \frac {2 \cdot 10^{-6}\,{\rm V^2s}}{0.25 \cdot 10^{-6}\,{\rm V^2/Hz}} = 8$$

$$\Rightarrow \hspace{0.3cm} p_\text{B, BPSK} = {\rm Q}\left ( \sqrt{16} \right ) = {\rm Q}\left ( 4 \right ) = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{8}\right )\hspace{0.05cm}.$$

• Based on the given   $x$–values in the table, it is convenient to use the first equation in this subtask:

$$p_\text{B, BPSK} = {\rm Q}\left ( 4 \right ) \hspace{0.15cm}\underline {= 0.317 \cdot 10^{-4} }\hspace{0.05cm}.$$

(2)  With twice the bit duration, the energy is also twice as large:  $E_{\rm B} = 4 · 10^{–6} \ \rm V^2s$ ⇒ $E_{\rm B}/N_0 = 16$.

• It follows that:

$$p_\text{B, BPSK} = {\rm Q}\left ( \sqrt{32} \right ) = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{16}\right ) ={1}/{2}\cdot {\rm erfc}\left ( 4\right ) \hspace{0.15cm}\underline {= 0.771 \cdot 10^{-8}}\hspace{0.05cm}.$$

• For pragmatic reasons, the last column of the table was used here.

(3)  Substituting the equations given for the 4-QAM into each other, we get the same result as for the BPSK:

$$p_{\rm B, \hspace{0.05cm}4-QAM} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {1}/{2}\cdot {\rm erfc}\left ( \sqrt{{E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) \equiv p_\text{B, BPSK}.$$

• Also, since the energy per bit has not changed from subtask  (2) , the same error probability will arise:

$$p_{\rm B, \hspace{0.05cm}4-QAM}= {\rm Q}\left ( \sqrt{32} \right ) = {1}/{2}\cdot {\rm erfc}\left ( 4\right ) \hspace{0.15cm}\underline {= 0.771 \cdot 10^{-8}}\hspace{0.05cm}.$$

(4)  Only Answer 1 is correct:

• Of course the error probability of error is the same in the two branches. Why would it not be?
• This would no longer be true with a phase offset between the transmitter and receiver though.
• However, the distance of the useful samples from the threshold is  $g_0$  here and thus smaller than the envelope   $s_0$  of the entire 4-QAM by a factor of $\sqrt{2}$ .
• However, if the inphase branch (or the quadrature branch) is considered as a stand-alone BPSK, the noise power is also half that of BPSK because of the lower symbol rate. Therefore, the error probability remains the same.