Aufgaben:Exercise 1.09: BPSK and 4-QAM - Revision history

Guenter at 07:30, 20 May 2022

2022-05-20T07:30:10Z

Hwang at 13:36, 16 May 2022

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Hwang at 15:13, 10 May 2022

2022-05-10T15:13:52Z

Hwang at 14:30, 10 May 2022

2022-05-10T14:30:42Z

Hwang at 14:26, 10 May 2022

2022-05-10T14:26:53Z

Hwang at 14:25, 10 May 2022

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Hwang at 14:23, 10 May 2022

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Guenter at 15:56, 6 May 2022

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Guenter at 15:43, 6 May 2022

2022-05-06T15:43:02Z

Hwang at 14:40, 29 March 2022

2022-03-29T14:40:27Z

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← Older revision		Revision as of 07:30, 20 May 2022
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	*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.		*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. ~~'''KORREKTUR: noiseless smaller?'''~~	+	*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  $\sigma_{d}$  smaller by the same factor.

← Older revision		Revision as of 13:36, 16 May 2022
Line 35:		Line 35:
	*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.		*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. '''KORREKTUR: noiseless'''	+	*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. '''KORREKTUR: noiseless smaller?'''

← Older revision		Revision as of 15:13, 10 May 2022
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−	$E_{\rm B}/N_{0} = 8$ always holds,  so the error probabilities of BPSK and QAM in the best case  (without phase shift)  are as follows   ⇒   [[Aufgaben:Exercise_1.08Z:_BPSK_Error_Probability\|Exercise 1.8Z]]:	+	$E_{\rm B}/N_{0} = 8$ always holds,  so the error probabilities of BPSK and QAM in the best case  (without phase shift)  are as follows   ⇒   [[Aufgaben:Exercise_1.08Z:_BPSK_Error_Probability\|"Exercise 1.8Z"]]:
	:$$p_{\rm B} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {\rm Q}(4)= 0.317 \cdot 10^{-4}.$$		:$$p_{\rm B} = {\rm Q}\left ( \sqrt{{2 \cdot E_{\rm B}}/{N_0 }} \hspace{0.1cm}\right ) = {\rm Q}(4)= 0.317 \cdot 10^{-4}.$$

← Older revision		Revision as of 14:30, 10 May 2022
Line 35:		Line 35:
	*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.		*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.	+	*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. '''KORREKTUR: noiseless'''

← Older revision		Revision as of 14:26, 10 May 2022
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	*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.		*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.	+	*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.

@@ Line 33: / Line 33: @@
 <u>Further remarks:</u>
-*If we denote the distance of the BPSK useful samples&nbsp; (without noise)&nbsp; from the&nbsp; (vertical)&nbsp; decision threshold by &nbsp;$s_{0}$,&nbsp; we get &nbsp;$\sigma_{d} = s_{0}/4$&nbsp; for the noise rms value.&nbsp; The lighter circles in the diagram mark the contour lines with radius &nbsp;$2\cdot \sigma_{d}$&nbsp; and &nbsp;$3\cdot \sigma_{d}$&nbsp; of the two-dimensionl Gaussian PDF.
+*If we denote the distance of the BPSK useful samples&nbsp; (without noise)&nbsp; from the&nbsp; (vertical)&nbsp; decision threshold by &nbsp;$s_{0}$,&nbsp; we get &nbsp;$\sigma_{d} = s_{0}/4$&nbsp; for the noise rms value.&nbsp; The lighter circles in the diagram mark the contour lines with radius &nbsp;$2\cdot \sigma_{d}$&nbsp; and &nbsp;$3\cdot \sigma_{d}$&nbsp; of the two-dimensional Gaussian PDF.
 *For the 4-QAM,&nbsp; compared to the BPSK,&nbsp; the distances of the noise-free samples drawn in red from the now two decision thresholds are each smaller by a factor of &nbsp;$\sqrt{2}$,&nbsp; but it also results in a noise rms value &nbsp;$\sigma_{d}$noiseless smaller by the same factor.

@@ Line 18: / Line 18: @@
 Correspondingly,&nbsp; for the 4-QAM system:
 :$$p_{\rm B,\hspace{0.04cm}QAM} = {\rm Q}\left ( \frac{s_0/\sqrt{2}}{\sigma_d } \right
-  )\hspace{0.2cm}{\rm mit}\hspace{0.2cm}\sigma_d = \sqrt{\frac{N_0}{2 \cdot T_{\rm
+  )\hspace{0.2cm}{\rm with}\hspace{0.2cm}\sigma_d = \sqrt{\frac{N_0}{2 \cdot T_{\rm
   B}}}.$$

@@ Line 75: / Line 75: @@
 {{ML-Kopf}}
-'''(1)'''&nbsp; 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:
+'''(1)'''&nbsp; Rotating the phase diagram by&nbsp; $\Delta\phi_{\rm T} = 15^\circ$&nbsp; decreases the distance of the useful samples from the threshold by&nbsp; $\cos(15^\circ) \approx 0.966$.&nbsp; 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)'''&nbsp; Analogous to subtask '''(1)''', $\cos(45^\circ) \approx 0.707$ is obtained:
+'''(2)'''&nbsp; Analogous to subtask&nbsp; '''(1)''',&nbsp; $\cos(45^\circ) \approx 0.707$&nbsp; is obtained:
 :$$p_{\rm B} = {\rm Q}(0.707 \cdot 4) \approx {\rm Q}(2.83)\hspace{0.1cm}\underline {= 0.233 \, \%}.$$
-'''(3)'''&nbsp; For 4-QAM, clockwise rotation by $\Delta\phi_{\rm T}$ increases the distance
+'''(3)'''&nbsp; For 4-QAM,&nbsp; clockwise rotation by&nbsp; $\Delta\phi_{\rm T}$&nbsp; 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 horizontal threshold&nbsp; (decision of the first bit)&nbsp; equals&nbsp; $s_{0} \cdot \cos(45^\circ + \Delta\phi_{\rm T})$,&nbsp; i.e., smaller than without phase shift,
 Thus, we obtain for the average error probability:
 :$$p_{\rm B} = {1}/{2} \cdot {\rm Q}\left ( \frac{\cos(45^\circ+{\rm \Delta} \phi_{\rm
   T}) \cdot s_0}{0.25 \cdot s_0 / \sqrt{2}} \right
@@ Line 95: / Line 96: @@
   ).$$
 *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 $\Delta\phi_{\rm T} = 15^\circ$:
+*On the other hand,&nbsp; we obtain with&nbsp; $\Delta\phi_{\rm T} = 15^\circ$:
 :$$p_{\rm B}  = {1}/{2} \cdot {\rm Q}\left ( \frac{\cos(60^\circ) \cdot 4}{1 / \sqrt{2}} \right
   ) +{1}/{2} \cdot {\rm Q}\left ( \frac{\cos(30^\circ) \cdot 4}{1 / \sqrt{2}} \right
@@ Line 108: / Line 110: @@
-'''(4)'''&nbsp; With a phase shift of $45^\circ$, one obtains from the equation generally derived above:
+'''(4)'''&nbsp; With a phase shift of&nbsp; $45^\circ$,&nbsp; one obtains from the equation generally derived above:
 :$$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
@@ Line 114: / Line 116: @@
 That is:
-*The error rate for the first bit is $50\%$.
+*The bit error rate for the first bit is&nbsp; $50\%$.
 *In contrast, the second bit is decided almost error-free $(\approx 10^{–8})$.
-*Overall, this results in a mean error probability of approx. $25\%$.
+*In contrast,&nbsp; the second bit is decided almost error-free&nbsp; $(\approx 10^{–8})$.

@@ Line 4: / Line 4: @@
 [[File:P_ID1682__Dig_A_4_2.png|right|frame|Phase diagrams of BPSK and 4-QAM]]
-The diagram shows schematically the phase diagrams of the ''binary phase modulation''&nbsp; (abbreviated '''BPSK''') and the ''quadrature amplitude modulation''&nbsp; (called '''4–QAM''').
+The diagram shows schematically the phase diagrams of the&nbsp; "binary phase modulation"&nbsp; $($abbreviated $\rm BPSK)$&nbsp; and the&nbsp; "quadrature amplitude modulation"&nbsp; $($called&nbsp; $\rm 4–QAM)$.
-*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 &nbsp;$\sqrt{2}$&nbsp; compared to BPSK.
+*The latter can be described by two BPSK systems with cosine and minus-sine carriers,&nbsp; where for each of the subcomponents the transmission amplitude is reduced by a factor of &nbsp;$\sqrt{2}$&nbsp; compared to BPSK.
 *The envelope of the total signal &nbsp;$s(t)$&nbsp; is thus also constant equal to &nbsp;$s_{0}$.
 *The error probability depending on the quotient &nbsp;$E_{\rm B}/N_{0}$&nbsp; 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:
+Correspondingly,&nbsp; for the 4-QAM system:
 :$$p_{\rm B,\hspace{0.04cm}QAM} = {\rm Q}\left ( \frac{s_0/\sqrt{2}}{\sigma_d } \right
   )\hspace{0.2cm}{\rm mit}\hspace{0.2cm}\sigma_d = \sqrt{\frac{N_0}{2 \cdot T_{\rm
@@ Line 19: / Line 22: @@
 The equations are valid only under the condition of exact phase synchronization:
-*If there is a phase offset &nbsp;$\Delta\phi_{\rm T}$&nbsp; between the transmitted and received carrier signals, the error probability increases significantly, with BPSK and QAM systems being degraded differently.
+*If there is a phase offset &nbsp;$\Delta\phi_{\rm T}$&nbsp; between the transmitted and received carrier signals,&nbsp; the error probability increases significantly,&nbsp; with BPSK and QAM systems being degraded differently.
-*In the phase diagram, the phase offset is noticeable by a rotation of the point clouds. In the diagram, the centers of the point clouds for &nbsp;$\Delta\phi_{\rm T} = 15^\circ$&nbsp; are marked by yellow crosses, while the red circles indicate the centers for &nbsp;$\Delta\phi_{\rm T} = 0$.&nbsp;
-&nbsp;$E_{\rm B}/N_{0} = 8$ always holds, so the error probabilities of BPSK and QAM in the best case (without phase shift) are respectively as follows  &nbsp; &rArr; &nbsp;  [[Aufgaben:Exercise_1.08Z:_BPSK_Error_Probability|Exercise 1.8Z]]:
 :$$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:''
+<u>Further remarks:</u>
-*If we denote the distance of the BPSK useful samples from the (vertical) decision threshold by &nbsp;$s_{0}$, we get &nbsp;$\sigma_{d} = s_{0}/4$ for the noise rms value. The lighter circles in the diagram mark the contour lines with radius &nbsp;$2\cdot \sigma_{d}$&nbsp; and &nbsp;$3\cdot \sigma_{d}$&nbsp; of the Gaussian 2D PDF.
+*If we denote the distance of the BPSK useful samples&nbsp; (without noise)&nbsp; from the&nbsp; (vertical)&nbsp; decision threshold by &nbsp;$s_{0}$,&nbsp; we get &nbsp;$\sigma_{d} = s_{0}/4$&nbsp; for the noise rms value.&nbsp; The lighter circles in the diagram mark the contour lines with radius &nbsp;$2\cdot \sigma_{d}$&nbsp; and &nbsp;$3\cdot \sigma_{d}$&nbsp; of the two-dimensionl Gaussian PDF.
-''Notes:''
+*Reference is made in particular to the section&nbsp; [[Digital_Signal_Transmission/Linear_Digital_Modulation_-_Coherent_Demodulation#Phase_offset_between_transmitter_and_receiver|"Phase offset between transmitter and receiver"]].
-*You can determine the values of the Q function with the applet &nbsp;[[Applets:Komplementäre_Gaußsche_Fehlerfunktionen|Complementary Gaussian Error Functions]].&nbsp;
+*You can determine the values of the Q&ndash;function with the HTML5/JavaScript applet &nbsp;[[Applets:Komplementäre_Gaußsche_Fehlerfunktionen|"Complementary Gaussian Error Functions"]].&nbsp;
@@ Line 60: / Line 65: @@
 $p_\text{B, 4-QAM} \ = \ $ { 0.117 3% } $\ \%$
-{What is the error probability for 4-QAM with &nbsp;$\Delta\phi_{\rm T} = 45^\circ$?
+{What is the bit error probability for 4-QAM with &nbsp;$\Delta\phi_{\rm T} = 45^\circ$?
 |type="{}"}
 $p_\text{B, 4-QAM} \ = \ $ { 25 3% } $\ \%$