Aufgaben:Exercise 4.19: Orthogonal Multilevel FSK - Revision history

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	[[Category:Digital Signal Transmission: Exercises\|^4.5 Non-Coherent Demodulation^]]		[[Category:Digital Signal Transmission: Exercises\|^4.5 Non-Coherent Demodulation^]]
	[[de:Aufgaben:4.~~19_Orthogonale_mehrstufige_FSK~~]]		[[de:Aufgaben:Aufgabe 4.19: Orthogonale mehrstufige FSK]]

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	In the  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation\| "theory part"]],  the exact formula for the probability of a correct decision in the case of AWGN noise is given:		In the  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation\| "theory part"]],  the exact formula for the probability of a correct decision in the case of AWGN noise is given:
	:$${\rm Pr}({\cal{C}}) =\sum_{i = 0}^{M-1} (-1)^i \cdot {M-1 \choose i } \cdot \frac{1}{i+1} \cdot {\rm e }^{ - i/(i+1) \hspace{0.05cm}\cdot \hspace{0.05cm}E_{\rm S}/ N_0}		:$${\rm Pr}({\cal{C}}) =\sum_{i = 0}^{M-1} (-1)^i \cdot {M-1 \choose i } \cdot \frac{1}{i+1} \cdot {\rm e }^{ - i/(i+1) \hspace{0.05cm}\cdot \hspace{0.05cm}E_{\rm S}/ N_0}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	*From this it is very easy to calculate the symbol error probability:		*From this it is very easy to calculate the symbol error probability:
	:$$p_{\rm S} = {\rm Pr}({\cal{E}}) = 1 - {\rm Pr}({\cal{C}}) = \sum_{i = 1}^{M-1} (-1)^{i+1} \cdot {M-1 \choose i } \cdot \frac{1}{i+1} \cdot {\rm e }^{ - i/(i+1) \hspace{0.05cm}\cdot \hspace{0.05cm}E_{\rm S}/ N_0}		:$$p_{\rm S} = {\rm Pr}({\cal{E}}) = 1 - {\rm Pr}({\cal{C}}) = \sum_{i = 1}^{M-1} (-1)^{i+1} \cdot {M-1 \choose i } \cdot \frac{1}{i+1} \cdot {\rm e }^{ - i/(i+1) \hspace{0.05cm}\cdot \hspace{0.05cm}E_{\rm S}/ N_0}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	*An upper bound  $(p_{\rm S, \ max} ≥ p_{\rm S})$  is obtained due to the alternating signs if we consider only the first term  $(i=1)$  of this sum:		*An upper bound  $(p_{\rm S, \ max} ≥ p_{\rm S})$  is obtained due to the alternating signs if we consider only the first term  $(i=1)$  of this sum:
	:$$p_{\rm S, \hspace{0.05cm}max} = (M-1)/2 \cdot {\rm e }^{-E_{\rm S}/(2N_{\rm 0})}		:$$p_{\rm S, \hspace{0.05cm}max} = (M-1)/2 \cdot {\rm e }^{-E_{\rm S}/(2N_{\rm 0})}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	In subtask  '''(4)''',  this bound is to be evaluated for a given ratio  $E_{\rm B}/N_0$,  where  $E_{\rm B}$  is the average signal energy per bit:		In subtask  '''(4)''',  this bound is to be evaluated for a given ratio  $E_{\rm B}/N_0$,  where  $E_{\rm B}$  is the average signal energy per bit:
	:$$E_{\rm B} = \frac{ E_{\rm S} } { {\rm log_2}\hspace{0.1cm}(M)}		:$$E_{\rm B} = \frac{ E_{\rm S} } { {\rm log_2}\hspace{0.1cm}(M)}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$


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	'''(2)'''  For the binary FSK  $(M = 2)$  applies with the abbreviation  $x = E_{\rm S}/N_0 = 6$:		'''(2)'''  For the binary FSK  $(M = 2)$  applies with the abbreviation  $x = E_{\rm S}/N_0 = 6$:
	:$$p_{\rm S} = (-1)^{2} \cdot {1 \choose 1 } \cdot {1}/{2} \cdot {\rm e }^{-x/2 } = {1}/{2} \cdot {\rm e }^{-3} \hspace{0.15cm}\underline{\approx 2.49 \%}		:$$p_{\rm S} = (-1)^{2} \cdot {1 \choose 1 } \cdot {1}/{2} \cdot {\rm e }^{-x/2 } = {1}/{2} \cdot {\rm e }^{-3} \hspace{0.15cm}\underline{\approx 2.49 \%}\hspace{0.05cm}.$$
	\hspace{0.05cm}.$$

	*Accordingly,  for the ternary FSK  $(M = 3)$  we obtain:		*Accordingly,  for the ternary FSK  $(M = 3)$  we obtain:
	:$$p_{\rm S} = (-1)^{2} \cdot {2 \choose 1 } \cdot {1}/{2} \cdot {\rm e }^{-(1/2) \hspace{0.05cm} \cdot \hspace{0.05cm} x} +		:$$p_{\rm S} = (-1)^{2} \cdot {2 \choose 1 } \cdot {1}/{2} \cdot {\rm e }^{-(1/2) \hspace{0.05cm} \cdot \hspace{0.05cm} x} +(-1)^{3} \cdot {2 \choose 2 } \cdot {1}/{3}\cdot {\rm e }^{-(2/3) \hspace{0.05cm} \cdot \hspace{0.05cm} x}={\rm e }^{-3} - {1}/{3} \cdot {\rm e }^{-4} \approx 0.0498 - 0.0061 \hspace{0.15cm}\underline{ =4.37\%}\hspace{0.05cm}.$$
	(-1)^{3} \cdot {2 \choose 2 } \cdot {1}/{3}\cdot {\rm e }^{-(2/3) \hspace{0.05cm} \cdot \hspace{0.05cm} x}=
	{\rm e }^{-3} - {1}/{3} \cdot {\rm e }^{-4} \approx 0.0498 - 0.0061 \hspace{0.15cm}\underline{ =4.37\%}
	\hspace{0.05cm}.$$

	*Finally,  for the quaternary FSK  $(M = 4)$  we obtain:		*Finally,  for the quaternary FSK  $(M = 4)$  we obtain:
	:$$p_{\rm S} = (-1)^{2} \cdot {3 \choose 1 } \cdot \frac{{\rm e }^{-x/2}}{2} +		:$$p_{\rm S} = (-1)^{2} \cdot {3 \choose 1 } \cdot \frac{{\rm e }^{-x/2}}{2} +(-1)^{3} \cdot {3 \choose 2 } \cdot \frac{{\rm e }^{-2x/3}}{3}+ (-1)^{4} \cdot {4 \choose 3 } \cdot \frac{{\rm e }^{-3x/4 }}{4} ={3}/ {2} \cdot{\rm e }^{-3} - {\rm e }^{-4} + {\rm e }^{-4.5} \hspace{0.15cm}\underline{\approx 6.75\%}\hspace{0.05cm}.$$
	(-1)^{3} \cdot {3 \choose 2 } \cdot \frac{{\rm e }^{-2x/3}}{3}
	+ (-1)^{4} \cdot {4 \choose 3 } \cdot \frac{{\rm e }^{-3x/4 }}{4} =
	{3}/ {2} \cdot{\rm e }^{-3} - {\rm e }^{-4} + {\rm e }^{-4.5} \hspace{0.15cm}\underline{\approx 6.75\%}
	\hspace{0.05cm}.$$

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	[[Category:Digital Signal Transmission: Exercises\|^4.5 Non-Coherent Demodulation^]]		[[Category:Digital Signal Transmission: Exercises\|^4.5 Non-Coherent Demodulation^]]
			[[de:Aufgaben:4.19_Orthogonale_mehrstufige_FSK]]

Hwang at 03:49, 11 September 2022

2022-09-11T03:49:34Z

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	'''(4)'''  With  $E_{\rm B} = E_{\rm S}/{\rm log}_2(M)$  one obtains $p_{\rm S, \hspace{0.05cm}max} = (M-1)/2 \cdot {\rm e }^{-\log_2 \ (M) E_{\rm B}/(2N_{\rm 0})}$.		'''(4)'''  With  $E_{\rm B} = E_{\rm S}/{\rm log}_2(M)$  one obtains $p_{\rm S, \hspace{0.05cm}max} = (M-1)/2 \cdot {\rm e }^{-\log_2 \ (M) E_{\rm B}/(2N_{\rm 0})}$.

	*The error probability becomes smaller as the level-number increases,  because at constant  $E_{\rm B}$  the energy  $E_{\rm S}$  per symbol increases by the factor  ${\rm log}_2 \, (M)$.		*The error probability becomes smaller as the level number increases,  because at constant  $E_{\rm B}$  the energy  $E_{\rm S}$  per symbol increases by the factor  ${\rm log}_2 \, (M)$.

	*The factor  $M-1$  (considers the falsification possibilities of a signal space point)  has less influence than the increase of the negative exponent:		*The factor  $M-1$  (considers the falsification possibilities of a signal space point)  has less influence than the increase of the negative exponent:

Guenter at 12:41, 31 August 2022

2022-08-31T12:41:44Z

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	:$$s_i(t) = \sqrt{E_{\rm S}} \cdot \xi_i(t) \hspace{0.05cm}.$$		:$$s_i(t) = \sqrt{E_{\rm S}} \cdot \xi_i(t) \hspace{0.05cm}.$$

	*$\xi_i(t)$  are complex basis functions for which in general  $i = 1, \ \text{...} \ , N$.    However,  for orthogonal signaling,  $M = N$ is always true.		*$\xi_i(t)$  are complex basis functions for which in general  $i = 1, \ \text{...} \ , N$.

			*However,  for orthogonal signaling,  $M = N$ is always true.

	*The diagram shows three different signal space constellations.  However,  not all three describe an orthogonal FSK.  This is referred to in the subtask  '''(1)'''.		*The diagram shows three different signal space constellations.  However,  not all three describe an orthogonal FSK.  This is referred to in the subtask  '''(1)'''.

Guenter at 12:13, 31 August 2022

2022-08-31T12:13:01Z

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  <u>Solutions 1 and 3</u> are correct:		'''(1)'''  <u>Solutions 1 and 3</u>  are correct:
	*In constellation  $\rm B$  the orthogonality is not given. Rather $M = 3$ and $N = 2$ are valid here.		*In constellation  $\rm B$  the orthogonality is not given.  Rather  $M = 3$  and  $N = 2$  are valid here.



	'''(2)'''  For the binary FSK $(M = 2)$ applies with the abbreviation $x = E_{\rm S}/N_0 = 6$:		'''(2)'''  For the binary FSK  $(M = 2)$  applies with the abbreviation  $x = E_{\rm S}/N_0 = 6$:
	:$$p_{\rm S} = (-1)^{2} \cdot {1 \choose 1 } \cdot {1}/{2} \cdot {\rm e }^{-x/2 } = {1}/{2} \cdot {\rm e }^{-3} \hspace{0.15cm}\underline{\approx 2.49 \%}		:$$p_{\rm S} = (-1)^{2} \cdot {1 \choose 1 } \cdot {1}/{2} \cdot {\rm e }^{-x/2 } = {1}/{2} \cdot {\rm e }^{-3} \hspace{0.15cm}\underline{\approx 2.49 \%}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	*Accordingly, for the ternary FSK $(M = 3)$ we obtain:		*Accordingly,  for the ternary FSK  $(M = 3)$  we obtain:
	:$$p_{\rm S} = (-1)^{2} \cdot {2 \choose 1 } \cdot {1}/{2} \cdot {\rm e }^{-(1/2) \hspace{0.05cm} \cdot \hspace{0.05cm} x} +		:$$p_{\rm S} = (-1)^{2} \cdot {2 \choose 1 } \cdot {1}/{2} \cdot {\rm e }^{-(1/2) \hspace{0.05cm} \cdot \hspace{0.05cm} x} +
	(-1)^{3} \cdot {2 \choose 2 } \cdot {1}/{3}\cdot {\rm e }^{-(2/3) \hspace{0.05cm} \cdot \hspace{0.05cm} x}=		(-1)^{3} \cdot {2 \choose 2 } \cdot {1}/{3}\cdot {\rm e }^{-(2/3) \hspace{0.05cm} \cdot \hspace{0.05cm} x}=
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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	*Finally, for the quaternary FSK $(M = 4)$ we obtain:		*Finally,  for the quaternary FSK  $(M = 4)$  we obtain:
	:$$p_{\rm S} = (-1)^{2} \cdot {3 \choose 1 } \cdot \frac{{\rm e }^{-x/2}}{2} +		:$$p_{\rm S} = (-1)^{2} \cdot {3 \choose 1 } \cdot \frac{{\rm e }^{-x/2}}{2} +
	(-1)^{3} \cdot {3 \choose 2 } \cdot \frac{{\rm e }^{-2x/3}}{3}		(-1)^{3} \cdot {3 \choose 2 } \cdot \frac{{\rm e }^{-2x/3}}{3}
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	'''(3)'''  For equal $E_{\rm S}/N_0 = 6$, $p_{\rm S, \ max} ≥ p_{\rm S}$ always holds:		'''(3)'''  For equal  $E_{\rm S}/N_0 = 6$,  $p_{\rm S, \ max} ≥ p_{\rm S}$ always holds:
	:$$M =2\text{:} \hspace{0.2cm} p_{\rm S, \hspace{0.05cm}max} \hspace{0.15cm}\underline{=2.49\%} = p_{\rm S} \hspace{0.05cm},$$		:$$M =2\text{:} \hspace{0.2cm} p_{\rm S, \hspace{0.05cm}max} \hspace{0.15cm}\underline{=2.49\%} = p_{\rm S} \hspace{0.05cm},$$
	:$$M =3\text{:} \hspace{0.2cm} p_{\rm S, \hspace{0.05cm}max} \hspace{0.15cm}\underline{=4.98\%} > 4.37\% = p_{\rm S} \hspace{0.05cm},$$		:$$M =3\text{:} \hspace{0.2cm} p_{\rm S, \hspace{0.05cm}max} \hspace{0.15cm}\underline{=4.98\%} > 4.37\% = p_{\rm S} \hspace{0.05cm},$$
	:$$M =4\text{:} \hspace{0.2cm} p_{\rm S, \hspace{0.05cm}max} \hspace{0.15cm}\underline{=7.47\%} > {6.75\%} = p_{\rm S} \hspace{0.05cm}.$$		:$$M =4\text{:} \hspace{0.2cm} p_{\rm S, \hspace{0.05cm}max} \hspace{0.15cm}\underline{=7.47\%} > {6.75\%} = p_{\rm S} \hspace{0.05cm}.$$

	Analyzing the equation  $p_{\rm S, \hspace{0.05cm}max} = (M-1)/2 \cdot {\rm e }^{-E_{\rm S}/(2N_{\rm 0})}$  in more detail, we see that this bound exactly specifies the [[Digital_Signal_Transmission/Approximation_of_the_Error_Probability#Union_Bound_-_Upper_bound_for_the_error_probability\| "Union Bound"]]:		Analyzing the equation   $p_{\rm S, \hspace{0.05cm}max} = (M-1)/2 \cdot {\rm e }^{-E_{\rm S}/(2N_{\rm 0})}$   in more detail,  we see that this bound exactly specifies the [[Digital_Signal_Transmission/Approximation_of_the_Error_Probability#Union_Bound_-_Upper_bound_for_the_error_probability\| "Union Bound"]]:
	* For the binary system, $1/2 \cdot {\rm e }^{-E_{\rm S}/(2N_{\rm 0})}$ gives the falsification probability, for example from $\boldsymbol{s}_1$ to $\boldsymbol{s}_2$ or vice versa.		* For the binary system,   $1/2 \cdot {\rm e }^{-E_{\rm S}/(2N_{\rm 0})}$   gives the falsification probability,  for example from  $\boldsymbol{s}_1$  to  $\boldsymbol{s}_2$  or  vice versa.
	* For the ~~<i>~~M~~</i>~~–level system, the distance between $\boldsymbol{s}_1$ and $\boldsymbol{s}_2$ is exactly the same. But also the points $\boldsymbol{s}_1, \ \text{... ,} \, \boldsymbol{s}_M$ are at the same distance from $\boldsymbol{s}_1$ or from $\boldsymbol{s}_2$.
	* The "Union Bound" considers the distortion possibilities of a point to each of the generally $M–1$ other points by the factor $M -1$.		* For the $M$–level system,  the distance between  $\boldsymbol{s}_1$  and  $\boldsymbol{s}_2$  is exactly the same.  But also the points  $\boldsymbol{s}_3, \ \text{... ,} \, \boldsymbol{s}_M$  are at the same distance from  $\boldsymbol{s}_1$  or  from $\boldsymbol{s}_2$.

			* The  "Union Bound"  considers the distortion possibilities of a point to each of the generally  $M–1$  other points by the factor  $M -1$.



	'''(4)'''  With  $E_{\rm B} = E_{\rm S}/{\rm log}_2(M)$  one obtains $p_{\rm S, \hspace{0.05cm}max} = (M-1)/2 \cdot {\rm e }^{-\log_2 \ (M) E_{\rm B}/(2N_{\rm 0})}$.		'''(4)'''  With  $E_{\rm B} = E_{\rm S}/{\rm log}_2(M)$  one obtains $p_{\rm S, \hspace{0.05cm}max} = (M-1)/2 \cdot {\rm e }^{-\log_2 \ (M) E_{\rm B}/(2N_{\rm 0})}$.

	*The error probability becomes smaller as the number ~~of levels~~ increases, because at constant $E_{\rm B}$ the energy $E_{\rm S}$ per symbol increases by the factor ${\rm log}_2 \, (M)$.		*The error probability becomes smaller as the level-number increases,  because at constant  $E_{\rm B}$  the energy  $E_{\rm S}$  per symbol increases by the factor  ${\rm log}_2 \, (M)$.
	*The factor $M-1$ (considers the falsification possibilities of a signal space point) has less influence than the increase of the negative exponent:
			*The factor  $M-1$  (considers the falsification possibilities of a signal space point)  has less influence than the increase of the negative exponent:
	:$$M =2\hspace{-0.1cm}: \hspace{0.2cm} p_{\rm S, \hspace{0.05cm}max} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {1}/{ 2} \cdot {\rm e }^{-3} \hspace{0.15cm} \underline{= 2.49\%} \hspace{0.05cm},$$		:$$M =2\hspace{-0.1cm}: \hspace{0.2cm} p_{\rm S, \hspace{0.05cm}max} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {1}/{ 2} \cdot {\rm e }^{-3} \hspace{0.15cm} \underline{= 2.49\%} \hspace{0.05cm},$$
	:$$M =3\hspace{-0.1cm}: \hspace{0.2cm} p_{\rm S, \hspace{0.05cm}max} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm e }^{-4.755} \hspace{0.5cm} \underline{= 0.86\%} \hspace{0.05cm},$$		:$$M =3\hspace{-0.1cm}: \hspace{0.2cm} p_{\rm S, \hspace{0.05cm}max} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\rm e }^{-4.755} \hspace{0.5cm} \underline{= 0.86\%} \hspace{0.05cm},$$

Guenter at 11:59, 31 August 2022

2022-08-31T11:59:22Z

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	[[File:P_ID2092__Dig_A_4_19.png\|right\|frame\|Signal space constellations]]		[[File:P_ID2092__Dig_A_4_19.png\|right\|frame\|Signal space constellations]]
	In this last exercise of this chapter we consider  ~~<i>~~Frequency Shift Keying~~</i>~~  (FSK) with  $M$  waveforms and assume that they are orthogonal to each other in pairs. In this case, the equivalent low–pass signals  $s_i(t)$ with $i = 1, \ \text{...} \ , M$  can be represented in the following form:		In this last exercise of this chapter we consider  "Frequency Shift Keying"  $\rm (FSK)$  with  $M$  waveforms and assume that they are orthogonal to each other in pairs.

			*In this case,  the equivalent low–pass signals  $s_i(t)$  with  $i = 1, \ \text{...} \ , M$  can be represented in the following form:
	:$$s_i(t) = \sqrt{E_{\rm S}} \cdot \xi_i(t) \hspace{0.05cm}.$$		:$$s_i(t) = \sqrt{E_{\rm S}} \cdot \xi_i(t) \hspace{0.05cm}.$$

	$\xi_i(t)$  are complex basis functions for which in general  $i = 1, \ \text{...} \ , N$  ~~holds.~~ However, for orthogonal signaling,  $M = N$ is always true.		*$\xi_i(t)$  are complex basis functions for which in general  $i = 1, \ \text{...} \ , N$.    However,  for orthogonal signaling,  $M = N$ is always true.

			*The diagram shows three different signal space constellations.  However,  not all three describe an orthogonal FSK.  This is referred to in the subtask  '''(1)'''.

	~~The diagram shows three different signal space constellations. However, not all three describe an orthogonal FSK. This is referred to in the subtask '''(1)'''.~~

	In the  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation\| "theory part"]],  the exact formula for the probability of a correct decision in the case of AWGN noise is given:		In the  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation\| "theory part"]],  the exact formula for the probability of a correct decision in the case of AWGN noise is given:
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	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	From this it is very easy to calculate the symbol error probability:		*From this it is very easy to calculate the symbol error probability:
	:$$p_{\rm S} = {\rm Pr}({\cal{E}}) = 1 - {\rm Pr}({\cal{C}}) = \sum_{i = 1}^{M-1} (-1)^{i+1} \cdot {M-1 \choose i } \cdot \frac{1}{i+1} \cdot {\rm e }^{ - i/(i+1) \hspace{0.05cm}\cdot \hspace{0.05cm}E_{\rm S}/ N_0}		:$$p_{\rm S} = {\rm Pr}({\cal{E}}) = 1 - {\rm Pr}({\cal{C}}) = \sum_{i = 1}^{M-1} (-1)^{i+1} \cdot {M-1 \choose i } \cdot \frac{1}{i+1} \cdot {\rm e }^{ - i/(i+1) \hspace{0.05cm}\cdot \hspace{0.05cm}E_{\rm S}/ N_0}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	An upper bound  $(p_{\rm S, \ max} ≥ p_{\rm S})$  is obtained due to the alternating signs if we consider only the first term  $(i=1)$  of this sum:		*An upper bound  $(p_{\rm S, \ max} ≥ p_{\rm S})$  is obtained due to the alternating signs if we consider only the first term  $(i=1)$  of this sum:
	:$$p_{\rm S, \hspace{0.05cm}max} = (M-1)/2 \cdot {\rm e }^{-E_{\rm S}/(2N_{\rm 0})}		:$$p_{\rm S, \hspace{0.05cm}max} = (M-1)/2 \cdot {\rm e }^{-E_{\rm S}/(2N_{\rm 0})}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$

	In subtask '''(4)''', this bound is to be evaluated for a given ratio  $E_{\rm B}/N_0$,  where  $E_{\rm B}$  is the average signal energy per bit:		In subtask  '''(4)''',  this bound is to be evaluated for a given ratio  $E_{\rm B}/N_0$,  where  $E_{\rm B}$  is the average signal energy per bit:
	:$$E_{\rm B} = \frac{ E_{\rm S} } { {\rm log_2}\hspace{0.1cm}(M)}		:$$E_{\rm B} = \frac{ E_{\rm S} } { {\rm log_2}\hspace{0.1cm}(M)}
	\hspace{0.05cm}.$$		\hspace{0.05cm}.$$
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			Notes:
			* The exercise belongs to the chapter  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation\| "Carrier Frequency Systems with Non-Coherent Demodulation"]].


	~~''Notes:''~~
	* The exercise belongs to the chapter  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation\| "Carrier Frequency Systems with Non-Coherent Demodulation"]].
	* Reference is made in particular to the section  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation#Non-coherent_demodulation_of_multi-level_FSK\|"Non-coherent demodulation of multi-level FSK"]].		* Reference is made in particular to the section  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation#Non-coherent_demodulation_of_multi-level_FSK\|"Non-coherent demodulation of multi-level FSK"]].

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	+ Constellation  $\rm C$.		+ Constellation  $\rm C$.

	{For  $E_{\rm S}/N_0 = 6$,  calculate the error probability of binary, ternary and quaternary FSK. $E_{\rm S}$  denotes the symbol energy.		{For  $E_{\rm S}/N_0 = 6$,  calculate the error probability of binary, ternary and quaternary FSK.  $E_{\rm S}$  denotes the average symbol energy.
	\|type="{}"}		\|type="{}"}
	$M = 2 \text{:} \hspace{0.4cm} p_{\rm S} \ = \ $ { 2.49 3% } $\ \%$		$M = 2 \text{:} \hspace{0.4cm} p_{\rm S} \ = \ $ { 2.49 3% } $\ \%$

Hwang at 18:09, 16 August 2022

2022-08-16T18:09:36Z

← Older revision		Revision as of 20:09, 16 August 2022
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	'''(4)'''  With  $E_{\rm B} = E_{\rm S}/{\rm log}_2(M)$  one obtains$p_{\rm S, \hspace{0.05cm}max} = (M-1)/2 \cdot {\rm e }^{-\log_2 \ (M) E_{\rm B}/(2N_{\rm 0})}$.		'''(4)'''  With  $E_{\rm B} = E_{\rm S}/{\rm log}_2(M)$  one obtains $p_{\rm S, \hspace{0.05cm}max} = (M-1)/2 \cdot {\rm e }^{-\log_2 \ (M) E_{\rm B}/(2N_{\rm 0})}$.

	*The error probability becomes smaller as the number of levels increases, because at constant $E_{\rm B}$ the energy $E_{\rm S}$ per symbol increases by the factor ${\rm log}_2 \, (M)$.		*The error probability becomes smaller as the number of levels increases, because at constant $E_{\rm B}$ the energy $E_{\rm S}$ per symbol increases by the factor ${\rm log}_2 \, (M)$.

Hwang at 18:08, 16 August 2022

2022-08-16T18:08:55Z

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Guenter at 15:28, 15 August 2022

2022-08-15T15:28:50Z

← Older revision		Revision as of 17:28, 15 August 2022
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	[[Category:Digital Signal Transmission: Exercises\|^4.5 ~~Inkohärente~~ Demodulation^]]		[[Category:Digital Signal Transmission: Exercises\|^4.5 Non-Coherent Demodulation^]]