Aufgaben:Exercise 4.19: Orthogonal Multilevel FSK - Revision history

Hwang at 03:49, 11 September 2022

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Guenter at 12:41, 31 August 2022

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Guenter at 12:13, 31 August 2022

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Guenter at 11:59, 31 August 2022

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Hwang at 18:09, 16 August 2022

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Hwang at 18:08, 16 August 2022

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Guenter at 15:28, 15 August 2022

2022-08-15T15:28:50Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.19: Orthogonale mehrstufige FSK to Aufgaben:Exercise 4.19: Orthogonal Multilevel FSK

2022-08-15T15:25:43Z

Guenter moved page Aufgabe 4.19: Orthogonale mehrstufige FSK to Exercise 4.19: Orthogonal Multilevel FSK

Javier: Text replacement - "„" to """

2021-05-28T14:38:01Z

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2021-05-28T14:36:47Z

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@@ Line 111: / Line 111: @@
 '''(4)'''&nbsp; With &nbsp;$E_{\rm B} = E_{\rm S}/{\rm log}_2(M)$&nbsp; 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,&nbsp; because at constant&nbsp; $E_{\rm B}$&nbsp; the energy&nbsp; $E_{\rm S}$&nbsp; per symbol increases by the factor&nbsp; ${\rm log}_2 \, (M)$.
+*The error probability becomes smaller as the level number increases,&nbsp; because at constant&nbsp; $E_{\rm B}$&nbsp; the energy&nbsp; $E_{\rm S}$&nbsp; per symbol increases by the factor&nbsp; ${\rm log}_2 \, (M)$.
 *The factor&nbsp; $M-1$&nbsp; (considers the falsification possibilities of a signal space point)&nbsp; has less influence than the increase of the negative exponent:

@@ Line 8: / Line 8: @@
 :$$s_i(t)  = \sqrt{E_{\rm S}} \cdot \xi_i(t) \hspace{0.05cm}.$$
-*$\xi_i(t)$&nbsp; are complex basis functions for which in general&nbsp; $i = 1, \ \text{...} \ , N$.&nbsp; &nbsp; However,&nbsp; for orthogonal signaling,&nbsp; $M = N$ is always true.
+*$\xi_i(t)$&nbsp; are complex basis functions for which in general&nbsp; $i = 1, \ \text{...} \ , N$.&nbsp; &nbsp;
 *The diagram shows three different signal space constellations.&nbsp; However,&nbsp; not all three describe an orthogonal FSK.&nbsp; This is referred to in the subtask&nbsp; '''(1)'''.

@@ Line 69: / Line 69: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; <u>Solutions 1 and 3</u> are correct:
+'''(1)'''&nbsp; <u>Solutions 1 and 3</u>&nbsp; are correct:
-*In constellation &nbsp;$\rm B$&nbsp; the orthogonality is not given. Rather $M = 3$ and $N = 2$ are valid here.
+*In constellation &nbsp;$\rm B$&nbsp; the orthogonality is not given.&nbsp; Rather&nbsp; $M = 3$&nbsp; and&nbsp; $N = 2$&nbsp; are valid here.
-'''(2)'''&nbsp; For the binary FSK $(M = 2)$ applies with the abbreviation $x = E_{\rm S}/N_0 = 6$:
+'''(2)'''&nbsp; For the binary FSK&nbsp; $(M = 2)$&nbsp; applies with the abbreviation&nbsp; $x = E_{\rm S}/N_0 = 6$:
 :$$p_{\rm S} =  (-1)^{2} \cdot {1 \choose 1 }  \cdot {1}/{2} \cdot {\rm e }^{-x/2 } = {1}/{2} \cdot {\rm e }^{-3}  \hspace{0.15cm}\underline{\approx 2.49 \%}
    \hspace{0.05cm}.$$
-*Accordingly, for the ternary FSK $(M = 3)$ we obtain:
+*Accordingly,&nbsp; for the ternary FSK&nbsp; $(M = 3)$&nbsp; we obtain:
 :$$p_{\rm S} = (-1)^{2} \cdot {2 \choose 1 }  \cdot {1}/{2} \cdot {\rm e }^{-(1/2) \hspace{0.05cm} \cdot \hspace{0.05cm} x} +
   (-1)^{3} \cdot {2 \choose 2 }  \cdot {1}/{3}\cdot {\rm e }^{-(2/3) \hspace{0.05cm} \cdot \hspace{0.05cm} x}=
@@ Line 84: / Line 84: @@
    \hspace{0.05cm}.$$
-*Finally, for the quaternary FSK $(M = 4)$ we obtain:
+*Finally,&nbsp; for the quaternary FSK&nbsp; $(M = 4)$&nbsp; we obtain:
 :$$p_{\rm S} = (-1)^{2} \cdot {3 \choose 1 }  \cdot \frac{{\rm e }^{-x/2}}{2}   +
   (-1)^{3} \cdot {3 \choose 2 }  \cdot \frac{{\rm e }^{-2x/3}}{3}
@@ Line 93: / Line 93: @@
-'''(3)'''&nbsp; For equal $E_{\rm S}/N_0 = 6$, $p_{\rm S, \ max} &#8805; p_{\rm S}$ always holds:
+'''(3)'''&nbsp; For equal&nbsp; $E_{\rm S}/N_0 = 6$,&nbsp; $p_{\rm S, \ max} &#8805; p_{\rm S}$ always holds:
 :$$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 =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}.$$
-Analyzing the equation &nbsp;$p_{\rm S, \hspace{0.05cm}max} =  (M-1)/2 \cdot {\rm e }^{-E_{\rm S}/(2N_{\rm 0})}$&nbsp; 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 &nbsp; $p_{\rm S, \hspace{0.05cm}max} =  (M-1)/2 \cdot {\rm e }^{-E_{\rm S}/(2N_{\rm 0})}$ &nbsp; in more detail,&nbsp; 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, &nbsp; $1/2 \cdot  {\rm e }^{-E_{\rm S}/(2N_{\rm 0})}$ &nbsp; gives the falsification probability,&nbsp; for example from&nbsp; $\boldsymbol{s}_1$&nbsp; to&nbsp; $\boldsymbol{s}_2$&nbsp; or&nbsp; vice versa.
-* For the <i>M</i>&ndash;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&ndash;1$ other points by the factor $M -1$.
+* For the $M$&ndash;level system,&nbsp; the distance between&nbsp; $\boldsymbol{s}_1$&nbsp; and&nbsp; $\boldsymbol{s}_2$&nbsp; is exactly the same.&nbsp; But also the points&nbsp; $\boldsymbol{s}_3, \ \text{... ,} \, \boldsymbol{s}_M$&nbsp; are at the same distance from&nbsp; $\boldsymbol{s}_1$&nbsp; or&nbsp; from $\boldsymbol{s}_2$.
 '''(4)'''&nbsp; With &nbsp;$E_{\rm B} = E_{\rm S}/{\rm log}_2(M)$&nbsp; 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,&nbsp; because at constant&nbsp; $E_{\rm B}$&nbsp; the energy&nbsp; $E_{\rm S}$&nbsp; per symbol increases by the factor&nbsp; ${\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:
 :$$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},$$

@@ Line 3: / Line 3: @@
 [[File:P_ID2092__Dig_A_4_19.png|right|frame|Signal space constellations]]
-In this last exercise of this chapter we consider &nbsp;<i>Frequency Shift Keying</i>&nbsp; (FSK) with&nbsp; $M$&nbsp; waveforms and assume that they are orthogonal to each other in pairs. In this case, the equivalent low&ndash;pass signals&nbsp; $s_i(t)$ with $i = 1, \ \text{...} \ , M$&nbsp; can be represented in the following form:
+In this last exercise of this chapter we consider &nbsp;"Frequency Shift Keying"&nbsp; $\rm (FSK)$&nbsp; with&nbsp; $M$&nbsp; waveforms and assume that they are orthogonal to each other in pairs.&nbsp;
 :$$s_i(t)  = \sqrt{E_{\rm S}} \cdot \xi_i(t) \hspace{0.05cm}.$$
-$\xi_i(t)$&nbsp; are complex basis functions for which in general&nbsp; $i = 1, \ \text{...} \ , N$&nbsp; holds. However, for orthogonal signaling,&nbsp; $M = N$ is always true.
+*$\xi_i(t)$&nbsp; are complex basis functions for which in general&nbsp; $i = 1, \ \text{...} \ , N$.&nbsp; &nbsp; However,&nbsp; for orthogonal signaling,&nbsp; $M = N$ is always true.
 In the&nbsp; [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation| "theory part"]],&nbsp; the exact formula for the probability of a correct decision in the case of AWGN noise is given:
@@ Line 14: / Line 17: @@
    \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}
    \hspace{0.05cm}.$$
-An upper bound&nbsp; $(p_{\rm S, \ max} &#8805; p_{\rm S})$&nbsp; is obtained due to the alternating signs if we consider only the first term&nbsp; $(i=1)$&nbsp; of this sum:
+*An upper bound&nbsp; $(p_{\rm S, \ max} &#8805; p_{\rm S})$&nbsp; is obtained due to the alternating signs if we consider only the first term&nbsp; $(i=1)$&nbsp; of this sum:
 :$$p_{\rm S, \hspace{0.05cm}max} =  (M-1)/2 \cdot {\rm e }^{-E_{\rm S}/(2N_{\rm 0})}
   \hspace{0.05cm}.$$
-In subtask '''(4)''', this bound is to be evaluated for a given ratio&nbsp; $E_{\rm B}/N_0$,&nbsp; where&nbsp; $E_{\rm B}$&nbsp; is the average signal energy per bit:
+In subtask&nbsp; '''(4)''',&nbsp; this bound is to be evaluated for a given ratio&nbsp; $E_{\rm B}/N_0$,&nbsp; where&nbsp; $E_{\rm B}$&nbsp; is the average signal energy per bit:
 :$$E_{\rm B} = \frac{ E_{\rm S} }  { {\rm log_2}\hspace{0.1cm}(M)}
    \hspace{0.05cm}.$$
@@ Line 29: / Line 32: @@
 * Reference is made in particular to the section&nbsp;  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation#Non-coherent_demodulation_of_multi-level_FSK|"Non-coherent demodulation of multi-level FSK"]].
@@ Line 46: / Line 48: @@
 + Constellation &nbsp;$\rm C$.
-{For&nbsp; $E_{\rm S}/N_0 = 6$,&nbsp; calculate the error probability of binary, ternary and quaternary FSK. $E_{\rm S}$&nbsp; denotes the symbol energy.
+{For&nbsp; $E_{\rm S}/N_0 = 6$,&nbsp; calculate the error probability of binary, ternary and quaternary FSK.&nbsp; $E_{\rm S}$&nbsp; denotes the average symbol energy.
 |type="{}"}
 $M = 2 \text{:} \hspace{0.4cm} p_{\rm S} \ = \ $ { 2.49 3% } $\ \%$

← Older revision		Revision as of 18:09, 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 ${\rm log}_2 \, (M)$.

← Older revision		Revision as of 15:28, 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^]]

← Older revision	Revision as of 15:25, 15 August 2022
(No difference)

@@ Line 99: / Line 99: @@
 * Beim Binärsystem gibt $1/2 \cdot  {\rm e }^{-E_{\rm S}/(2N_{\rm 0})}$ die Verfälschungswahrscheinlichkeit an, zum Beispiel von $\boldsymbol{s}_1$ nach $\boldsymbol{s}_2$ oder umgekehrt.
 * Beim <i>M</i>&ndash;stufigen System ist der Abstand zwischen $\boldsymbol{s}_1$ und $\boldsymbol{s}_2$ genau so groß. Aber auch die Punkte $\boldsymbol{s}_1, \ \text{... ,} \, \boldsymbol{s}_M$ liegen im gleichen Abstand zu $\boldsymbol{s}_1$ bzw. zu $\boldsymbol{s}_2$.
-* Die &bdquo;Union&ndash;Bound" berücksichtigt die Verfälschungsmöglichkeiten eines Punktes zu jedem der allgemein $M&ndash;1$ anderen Punkte durch den Faktor $M -1$.
+* Die "Union&ndash;Bound" berücksichtigt die Verfälschungsmöglichkeiten eines Punktes zu jedem der allgemein $M&ndash;1$ anderen Punkte durch den Faktor $M -1$.