Aufgaben:Exercise 1.09: BPSK and 4-QAM

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Phase diagrams of BPSK and 4-QAM

The diagram shows schematically the phase diagrams of the  "binary phase modulation"  $($abbreviated $\rm BPSK)$  and the  "quadrature amplitude modulation"  $($called  $\rm 4–QAM)$.

• The latter can be described by two BPSK systems with cosine and minus-sine carriers,  where for each of the subcomponents the transmission amplitude is reduced by a factor of  $\sqrt{2}$  compared to BPSK.

• The envelope of the total signal  $s(t)$  is thus also constant equal to  $s_{0}$.

• The error probability depending on the quotient  $E_{\rm B}/N_{0}$  is the same for BPSK and 4–QAM:

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

However,  the error probability of the BPSK system can also be expressed in the form

$$p_{\rm B,\hspace{0.04cm}BPSK} = {\rm Q}\left ( \frac{s_0}{\sigma_d } \right )\hspace{0.2cm}{\rm with}\hspace{0.2cm}\sigma_d = \sqrt{\frac{N_0}{T_{\rm B}}}$$

Correspondingly,  for the 4-QAM system:

$$p_{\rm B,\hspace{0.04cm}QAM} = {\rm Q}\left ( \frac{s_0/\sqrt{2}}{\sigma_d } \right )\hspace{0.2cm}{\rm with}\hspace{0.2cm}\sigma_d = \sqrt{\frac{N_0}{2 \cdot T_{\rm B}}}.$$

The equations are valid only under the condition of exact phase synchronization:

• If there is a phase offset  $\Delta\phi_{\rm T}$  between the transmitted and received carrier signals,  the error probability increases significantly,  with BPSK and QAM systems being degraded differently.

• In the phase diagram,  the phase offset is noticeable by a rotation of the point clouds. 

• In the diagram,  the centers of the point clouds for  $\Delta\phi_{\rm T} = 15^\circ$  are marked by yellow crosses,  while the red circles indicate the centers for  $\Delta\phi_{\rm T} = 0$. 

 $E_{\rm B}/N_{0} = 8$ always holds,  so the error probabilities of BPSK and QAM in the best case  (without phase shift)  are as follows   ⇒   Exercise 1.8Z:

$$p_{\rm B} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {\rm Q}(4)= 0.317 \cdot 10^{-4}.$$

Further remarks:

• If we denote the distance of the BPSK useful samples  (without noise)  from the  (vertical)  decision threshold by  $s_{0}$,  we get  $\sigma_{d} = s_{0}/4$  for the noise rms value.  The lighter circles in the diagram mark the contour lines with radius  $2\cdot \sigma_{d}$  and  $3\cdot \sigma_{d}$  of the two-dimensional Gaussian PDF.

• For the 4-QAM,  compared to the BPSK,  the distances of the noise-free samples drawn in red from the now two decision thresholds are each smaller by a factor of  $\sqrt{2}$,  but it also results in a noise rms value  $\sigma_{d}$noiseless smaller by the same factor.

Notes:

• The exercise belongs to the chapter  "Linear Digital Modulation - Coherent Demodulation".

• Reference is made in particular to the section  "Phase offset between transmitter and receiver".

• You can determine the values of the Q–function with the HTML5/JavaScript applet  "Complementary Gaussian Error Functions". 

Questions

1

What is the bit error probability for BPSK with  $\Delta\phi_{\rm T} = 15^\circ$?

$p_\text{B, BPSK} \ = \ $ $\ \% $

2

What is the bit error probability for BPSK with  $\Delta\phi_{\rm T} = 45^\circ$?

$p_\text{B, BPSK} \ = \ $ $\ \%$

3

What is the bit error probability for 4-QAM with  $\Delta\phi_{\rm T} = 15^\circ$?

$p_\text{B, 4-QAM} \ = \ $ $\ \%$

4

What is the bit error probability for 4-QAM with  $\Delta\phi_{\rm T} = 45^\circ$?

$p_\text{B, 4-QAM} \ = \ $ $\ \%$

Solution

Solution

(1)  Rotating the phase diagram by  $\Delta\phi_{\rm T} = 15^\circ$  decreases the distance of the useful samples from the threshold by  $\cos(15^\circ) \approx 0.966$.  It follows that:

$$p_{\rm B} = {\rm Q}(0.966 \cdot 4) \approx {\rm Q}(3.86)= 0.57 \cdot 10^{-4}\hspace{0.1cm}\underline {= 0.0057\, \%}.$$

(2)  Analogous to subtask  (1),  $\cos(45^\circ) \approx 0.707$  is obtained:

$$p_{\rm B} = {\rm Q}(0.707 \cdot 4) \approx {\rm Q}(2.83)\hspace{0.1cm}\underline {= 0.233 \, \%}.$$

(3)  For 4-QAM,  clockwise rotation by  $\Delta\phi_{\rm T}$  increases the distance

• from the horizontal threshold  (decision of the first bit)  equals  $s_{0} \cdot \cos(45^\circ + \Delta\phi_{\rm T})$,  i.e., smaller than without phase shift,

• from the vertical threshold  (decision of the second bit)  equal to $s_{0} \cdot \cos(45^\circ - \Delta\phi_{\rm T})$,  thus larger than without phase shift.

Thus,  we obtain for the average error probability:

$$p_{\rm B} = {1}/{2} \cdot {\rm Q}\left ( \frac{\cos(45^\circ+{\rm \Delta} \phi_{\rm T}) \cdot s_0}{0.25 \cdot s_0 / \sqrt{2}} \right ) + {1}/{2} \cdot {\rm Q}\left ( \frac{\cos(45^\circ-{\rm \Delta} \phi_{\rm T}) \cdot s_0}{0.25 \cdot s_0 / \sqrt{2}}\right ).$$

• This already takes into account the smaller noise rms value of the 4-QAM.

• As a check,  we calculate the error probability for  $\Delta\phi_{\rm T} = 0$:

$$p_{\rm B} = {1}/{2} \cdot {\rm Q}\left ( \frac{\cos(45^\circ) \cdot 4}{1 / \sqrt{2}} \right ) +{1}/{2} \cdot {\rm Q}\left ( \frac{\cos(45^\circ) \cdot 4}{1 / \sqrt{2}} \right )= {\rm Q}(4) = 0.317 \cdot 10^{-4}.$$

• On the other hand,  we obtain with  $\Delta\phi_{\rm T} = 15^\circ$:

$$p_{\rm B} = {1}/{2} \cdot {\rm Q}\left ( \frac{\cos(60^\circ) \cdot 4}{1 / \sqrt{2}} \right ) +{1}/{2} \cdot {\rm Q}\left ( \frac{\cos(30^\circ) \cdot 4}{1 / \sqrt{2}} \right )= {1}/{2} \cdot \left [{\rm Q}(2.83)+ {\rm Q}(4.90)\right]$$

$$\Rightarrow \hspace{0.3cm} p_{\rm B} \approx \frac{1}{2} \cdot \left [0.233 \cdot 10^{-2}+ 0.479 \cdot 10^{-6}\right] \hspace{0.1cm}\underline {= 0.117 \, \%}.$$

(4)  With a phase shift of  $45^\circ$,  one obtains from the equation generally derived above:

$$p_{\rm B} ={1}/{2} \cdot {\rm Q}\left ( \frac{\cos(90^\circ) \cdot 4}{1 / \sqrt{2}} \right ) +{1}/{2} \cdot {\rm Q}\left ( \frac{\cos(0^\circ) \cdot 4}{1 / \sqrt{2}} \right )= {1}/{2} \cdot \left [{\rm Q}(0)+ {\rm Q}(5.66)\right] \approx 0.25\hspace{0.1cm}\underline {= 25 \, \%}.$$

That is:

• The bit error rate for the first bit is  $50\%$.

• In contrast,  the second bit is decided almost error-free  $(\approx 10^{–8})$.

• Overall,  this results in a mean bit error probability of approx. $25\%$.