Aufgaben:Exercise 4.16: Binary Frequency Shift Keying - Revision history

Hwang at 19:51, 1 September 2022

2022-09-01T19:51:04Z

Guenter at 10:06, 24 August 2022

2022-08-24T10:06:42Z

Guenter at 08:30, 24 August 2022

2022-08-24T08:30:36Z

Hwang at 10:55, 16 August 2022

2022-08-16T10:55:01Z

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Hwang at 21:01, 15 August 2022

2022-08-15T21:01:19Z

Guenter at 12:00, 13 July 2022

2022-07-13T12:00:03Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.16: Binary Frequency Shift Keying to Aufgaben:Exercise 4.16: Binary Frequency Shift Keying

2022-07-13T11:54:48Z

Guenter moved page Aufgabe 4.16: Binary Frequency Shift Keying to Exercise 4.16: Binary Frequency Shift Keying

Javier: Text replacement - "Category:Aufgaben zu Digitalsignalübertragung" to "Category:Digital Signal Transmission: Exercises"

2021-03-23T12:47:39Z

Text replacement - "Category:Aufgaben zu Digitalsignalübertragung" to "Category:Digital Signal Transmission: Exercises"

Javier: Text replacement - "[[Modulationsverfahren" to "[[Modulation_Methods"

2020-07-09T12:54:08Z

Text replacement - "[[Modulationsverfahren" to "[[Modulation_Methods"

Javier: Text replacement - "[[Digitalsignalübertragung" to "[[Digital_Signal_Transmission"

2020-07-09T12:44:21Z

Text replacement - "[[Digitalsignalübertragung" to "[[Digital_Signal_Transmission"

@@ Line 83: / Line 83: @@
 '''(1)'''&nbsp; By counting the oscillations within a symbol duration&nbsp; $T$,&nbsp; the two frequencies&nbsp; $f_0 = 4.5/T$&nbsp; and&nbsp; $f_1 = 3.5/T$&nbsp; are obtained.
-*From thes carriere frequencies and frequency deviations are calculated to
+*From this carrier frequencies and frequency deviations are calculated to
 :$$f_{\rm T} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  {1}/{2}\cdot (f_0 + f_1) = \underline{4 \cdot 1/T}\hspace{0.05cm},$$
 :$$ \Delta f_{\rm A} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  {1}/{2}\cdot (f_0 - f_1)= \underline{0.5 \cdot 1/T }\hspace{0.05cm}.$$

@@ Line 81: / Line 81: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; By counting the oscillations within a symbol duration $T$, the two frequencies $f_0 = 4.5/T$ and $f_1 = 3.5/T$ are obtained.
+'''(1)'''&nbsp; By counting the oscillations within a symbol duration&nbsp; $T$,&nbsp; the two frequencies&nbsp; $f_0 = 4.5/T$&nbsp; and&nbsp; $f_1 = 3.5/T$&nbsp; are obtained.
-*From this carrier frequencies and frequency deviation are calculated to
 :$$f_{\rm T} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  {1}/{2}\cdot (f_0 + f_1) = \underline{4 \cdot 1/T}\hspace{0.05cm},$$
 :$$ \Delta f_{\rm A} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  {1}/{2}\cdot (f_0 - f_1)= \underline{0.5 \cdot 1/T }\hspace{0.05cm}.$$
-'''(2)'''&nbsp; With the given equation, the modulation index is:
+'''(2)'''&nbsp; With the given equation,&nbsp; the modulation index is:
 :$$h = 2 \cdot \Delta f_{\rm A} \cdot T  = 2 \cdot 0.5 \cdot 1/T \cdot T \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \underline{h= 1}\hspace{0.05cm}. $$
@@ Line 96: / Line 97: @@
    \hspace{0.05cm}.$$
-Orthogonality means that this inner product must be $0$:
+Orthogonality means that this inner product must be&nbsp; $0$:
 :$$<  \hspace{-0.05cm} s_{\rm TP0}(t)  \cdot   s_{\rm TP1}(t) \hspace{-0.05cm} > \hspace{0.2cm} =
    \frac{E}{{\rm j}2\pi   h} \cdot \left [ {\rm e}^{\hspace{0.05cm}{\rm j}  2\pi  h} - 1 \right ] = 0
 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} h = 1, 2, 3, ...$$
-Consequently, <u>solutions 3 and 4</u> are correct:
+Consequently,&nbsp; <u>solutions 3 and 4</u>&nbsp; are correct:
-*If the modulation index $h$ is integer, non-coherent demodulation can be performed without violating orthogonality.
+*If the modulation index $h$&nbsp; is integer,&nbsp; non-coherent demodulation can be performed without violating orthogonality.
 '''(4)'''&nbsp; <u>Solutions 1, 3 and 4</u> are correct:
-*For the inner product of the band&ndash;pass signals, according to the explanations in the information section, it can be written:
+*For the inner product of the band&ndash;pass signals,&nbsp; according to the explanations in the information section,&nbsp; it can be written:
 :$$<  \hspace{-0.05cm}s_{\rm BP0}(t) \hspace{0.01cm} \ \cdot \ \hspace{0.01cm} s_{\rm BP1}(t) \hspace{-0.05cm}> \hspace{0.2cm}=
   {\rm Re}\left [ \hspace{0.1cm}<  \hspace{-0.05cm}s_{\rm TP0}(t) \cdot s_{\rm TP1}(t) \hspace{-0.05cm}> \hspace{0.2cm} \right ] = {\rm Re}\left [ \frac{E}{{\rm j}2\pi   h} \cdot \left ( {\rm e}^{\hspace{0.05cm}{\rm j}  2\pi  h} - 1 \right ) \right ]$$
 :$$ \Rightarrow\hspace{0.3cm}<  \hspace{-0.05cm}s_{\rm BP0}(t) \hspace{0.01cm} \ \cdot \ \hspace{0.01cm} s_{\rm BP1}(t) \hspace{-0.05cm}> \hspace{0.2cm}= {\rm Re}\left [ \frac{E}{2\pi   h} \cdot \left ( \sin( 2\pi  h) - {\rm j} \cdot [\cos( 2\pi  h) - 1 ]\right ) \right ] = \frac{E \cdot \sin( 2\pi  h)}{2\pi   h} \hspace{0.05cm}.$$
-*This result is $0$ whenever the modulation index $h$ is an integer multiple of $0.5$.
+*This result is&nbsp; $0$&nbsp; whenever the modulation index&nbsp; $h$&nbsp; is an integer multiple of&nbsp; $0.5$.
-'''(5)'''&nbsp; <u>Solution 2</u> is correct.
+*In contrast,&nbsp; for $h = 0.5$,&nbsp; coherent demodulation can be applied,&nbsp; but non-coherent demodulation&nbsp; (which relies on the envelope) fails.
 {{ML-Fuß}}

@@ Line 2: / Line 2: @@
 {{quiz-Header|Buchseite=Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation}}
-[[File:P_ID2075__Dig_A_4_16.png|right|frame|Band-pass signals of the FSK]]
+[[File:P_ID2075__Dig_A_4_16.png|right|frame|FSK band-pass signals]]
-In binary FSK, the two messages&nbsp; $m_0$&nbsp; and&nbsp; $m_1$&nbsp; are represented by two different frequencies. For the two possible band&ndash;pass signals then applies in each case in the range&nbsp; $0 &#8804; t &#8804; T$&nbsp; with &nbsp;$f_0 = f_{\rm T} + \Delta f_{\rm A}$&nbsp; as well as &nbsp;$f_1 = f_{\rm T} \, &ndash; \Delta f_{\rm A}$:
+In binary FSK,&nbsp; the two messages&nbsp; $m_0$&nbsp; and&nbsp; $m_1$&nbsp; are represented by two different frequencies.&nbsp; For the two possible band&ndash;pass signals then applies in each case in the range &nbsp; $0 &#8804; t &#8804; T$ &nbsp; with &nbsp; $f_0 = f_{\rm T} + \Delta f_{\rm A}$ &nbsp; as well as &nbsp;$f_1 = f_{\rm T} \, &ndash; \Delta f_{\rm A}$:
 :$$s_{\rm BP0}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  \sqrt{2E/T} \cdot \cos( 2\pi  f_0  t)\hspace{0.05cm},$$
 :$$ s_{\rm BP1}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  \sqrt{2E/T} \cdot \cos( 2\pi  f_1  t)\hspace{0.05cm}.$$
-The graph shows example signals $($in each case only one symbol duration &nbsp;$T)$.
+The graph shows example signals&nbsp; $($in each case only of one symbol duration &nbsp;$T)$.
-*In the above equation,&nbsp; $f_{\rm T}$&nbsp; indicates the <i>carrier frequency</i>&nbsp; and&nbsp; $\Delta f_{\rm A}$&nbsp; indicates the <i>frequency deviation</i>&nbsp; as the maximum deviation of the&nbsp; [[Modulation_Methods/Frequency_Modulation_(FM)#Instantaneous_frequency|"instantaneous frequency"]]&nbsp; from the carrier frequency.
+*In the above equation,&nbsp; $f_{\rm T}$&nbsp; indicates the&nbsp; "carrier frequency"&nbsp; and&nbsp; $\Delta f_{\rm A}$&nbsp; indicates the&nbsp; "frequency deviation"&nbsp; as the maximum deviation of the&nbsp; [[Modulation_Methods/Frequency_Modulation_(FM)#Instantaneous_frequency|"instantaneous frequency"]]&nbsp; from the carrier frequency.
-* $E$&nbsp; is the signal energy. Here, it is equally true for the mean symbol energy and the mean bit energy:
 :$$E_{\rm S} = E_{\rm B} = E\hspace{0.05cm}.$$
-Mostly, however, one works with the <i>modulation index</i>, which is defined as the ratio of total frequency deviation and symbol rate:
+Mostly one works with the&nbsp; "modulation index",&nbsp; which is defined as the ratio of total frequency deviation and symbol rate:
 :$$h = \frac{2 \cdot \Delta f_{\rm A}}{1/T} = 2 \cdot \Delta f_{\rm A} \cdot T \hspace{0.05cm}.$$
@@ Line 23: / Line 24: @@
   \int_{0}^{T} s_{\rm TP0}(t) \cdot s_{\rm TP1}^{\star}(t) \,{\rm d} t =0 \hspace{0.05cm}.$$
-In this case, non-coherent demodulation as described in the chapter&nbsp; [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation| "Carrier Frequency Systems with Non-Coherent Demodulation"]]&nbsp; is also possible.
+In this case,&nbsp; non-coherent demodulation as described in the chapter&nbsp; [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Non-Coherent_Demodulation| "Carrier Frequency Systems with Non-Coherent Demodulation"]]&nbsp; is also possible.
 The inner product of the band&ndash;pass signals can be determined from the inner product of the low&ndash;pass signals by real partitioning:
@@ Line 31: / Line 32: @@
 If&nbsp; $&#9001; s_{\rm BP0}(t) \cdot s_{\rm BP1}(t)&#9002; = 0$, but at the same time&nbsp; $&#9001; s_{\rm TP0}(t) \cdot s_{\rm TP1}(t)&#9002; &ne; 0$,
 *so coherent demodulation is possible,
 *but no non-coherent demodulation.
@@ Line 37: / Line 39: @@
-''Notes:''
+Notes:
 * The exercise belongs to the chapter&nbsp;  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation| "Carrier Frequency Systems with Coherent Demodulation"]].
 * Reference is made in particular to the sections&nbsp; [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#Binary_frequency_shift_keying_.282.E2.80.93FSK.29|"Binary Frequency Shift Keying"]]&nbsp;(BFSK) and &nbsp;[[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#Minimum_Shift_Keying_.28MSK.29|"Minimum Shift Keying"]]&nbsp; (MSK).

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Digitalsignalübertragung/Trägerfrequenzsysteme mit kohärenter Demodulation}}
+{{quiz-Header|Buchseite=Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation}}
-[[File:P_ID2075__Dig_A_4_16.png|right|frame|Bandpass-Signale der FSK]]
+[[File:P_ID2075__Dig_A_4_16.png|right|frame|Band-pass signals of the FSK]]
-Bei der binären FSK werden die beiden Nachrichten&nbsp; $m_0$&nbsp; und&nbsp; $m_1$&nbsp; durch zwei unterschiedliche Frequenzen dargestellt. Für die beiden möglichen Bandpass&ndash;Signale gilt dann jeweils im Bereich&nbsp; $0 &#8804; t &#8804; T$&nbsp; mit &nbsp;$f_0 = f_{\rm T} + \Delta f_{\rm A}$&nbsp; sowie &nbsp;$f_1 = f_{\rm T} \, &ndash; \Delta f_{\rm A}$:
+In binary FSK, the two messages&nbsp; $m_0$&nbsp; and&nbsp; $m_1$&nbsp; are represented by two different frequencies. For the two possible band&ndash;pass signals then applies in each case in the range&nbsp; $0 &#8804; t &#8804; T$&nbsp; with &nbsp;$f_0 = f_{\rm T} + \Delta f_{\rm A}$&nbsp; as well as &nbsp;$f_1 = f_{\rm T} \, &ndash; \Delta f_{\rm A}$:
 :$$s_{\rm BP0}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  \sqrt{2E/T} \cdot \cos( 2\pi  f_0  t)\hspace{0.05cm},$$
 :$$ s_{\rm BP1}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  \sqrt{2E/T} \cdot \cos( 2\pi  f_1  t)\hspace{0.05cm}.$$
-Die Grafik zeigt beispielhafte Signale $($jeweils nur eine Symboldauer &nbsp;$T)$.
+The graph shows example signals $($in each case only one symbol duration &nbsp;$T)$.
 *In obiger Gleichung gibt&nbsp; $f_{\rm T}$&nbsp; die <i>Trägerfrequenz</i>&nbsp; an und&nbsp; $\Delta f_{\rm A}$&nbsp; den <i>Frequenzhub</i>&nbsp; als die maximale Abweichung der&nbsp; [[Modulation_Methods/Frequenzmodulation_(FM)#Augenblicksfrequenz|Augenblicksfrequenz]]&nbsp; von der Trägerfrequenz an.
 * $E$&nbsp; ist die Signalenergie. Dabei gilt gleichermaßen für die mittlere Symbolenergie und die mittlere Bitenergie:
@@ Line 38: / Line 38: @@
 ''Hinweise:''
-* Die Aufgabe gehört zum Kapitel&nbsp;  [[Digitalsignal%C3%BCbertragung/Tr%C3%A4gerfrequenzsysteme_mit_koh%C3%A4renter_Demodulation| Trägerfrequenzsysteme mit kohärenter Demodulation]].
+* Die Aufgabe gehört zum Kapitel&nbsp;  [[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation| "Carrier Frequency Systems with Coherent Demodulation"]].
 * Bezug genommen wird insbesondere auf die Seiten&nbsp; [[Digital_Signal_Transmission/Trägerfrequenzsysteme_mit_kohärenter_Demodulation#Binary_Frequency_Shift_Keying_.282.E2.80.93FSK.29| Binary Frequency Shift Keying]]&nbsp;(BFSK) und  &nbsp;[[Digitalsignal%C3%BCbertragung/Tr%C3%A4gerfrequenzsysteme_mit_koh%C3%A4renter_Demodulation#Minimum_Shift_Keying_.28MSK.29|Minimum Shift Keying]]&nbsp; (MSK).

← Older revision		Revision as of 12:00, 13 July 2022
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−	[[Category:Digital Signal Transmission: Exercises\|^4.4 ~~Kohärente~~ Demodulation^]]	+	[[Category:Digital Signal Transmission: Exercises\|^4.4 Coherent Demodulation^]]

← Older revision		Revision as of 12:47, 23 March 2021
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−	[[Category:~~Aufgaben zu Digitalsignalübertragung~~\|^4.4 Kohärente Demodulation^]]	+	[[Category:Digital Signal Transmission: Exercises\|^4.4 Kohärente Demodulation^]]