Difference between revisions of "Aufgaben:Exercise 4.16: Binary Frequency Shift Keying"

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{{quiz-Header|Buchseite=Digitalsignalübertragung/Trägerfrequenzsysteme mit kohärenter Demodulation}}
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{{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|FSK band-pass signals]]
Bei der binären FSK werden die beiden Nachrichten $m_0$ und $m_1$ durch zwei unterschiedliche Frequenzen dargestellt. Für die beiden möglichen Bandpass–Signale gilt dann jeweils im Bereich $0 ≤ t ≤ T$ mit $f_0 = f_{\rm T} + \Delta f_{\rm A}$ sowie $f_1 = f_{\rm T} \, – \Delta f_{\rm A}$:
+
In binary FSK,  the two messages  $m_0$  and  $m_1$  are represented by two different frequencies.  For the two possible band–pass signals then applies in each case in the range   $0 ≤ t ≤ T$   with   $f_0 = f_{\rm T} + \Delta f_{\rm A}$   as well as  $f_1 = f_{\rm T} \, – \Delta f_{\rm A}$:
 
:$$s_{\rm BP0}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  \sqrt{2E/T} \cdot \cos( 2\pi  f_0  t)\hspace{0.05cm},$$
 
:$$s_{\rm BP0}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  \sqrt{2E/T} \cdot \cos( 2\pi  f_0  t)\hspace{0.05cm},$$
 
:$$ s_{\rm BP1}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  \sqrt{2E/T} \cdot \cos( 2\pi  f_1  t)\hspace{0.05cm}.$$
 
:$$ s_{\rm BP1}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  \sqrt{2E/T} \cdot \cos( 2\pi  f_1  t)\hspace{0.05cm}.$$
  
Die Grafik zeigt beispielhafte Signale. In obiger Gleichung gibt $f_{\rm T}$ die <i>Trägerfrequenz</i> an und $\Delta f_{\rm A}$ den <i>Frequenzhub</i> als die maximale Abweichung der [[Modulationsverfahren/Frequenzmodulation_(FM)#Augenblicksfrequenz|Augenblicksfrequenz]] von der Trägerfrequenz an. $T$ ist die Symboldauer und $E$ die Signalenergie. Dabei gilt gleichermaßen für die mittlere Symbolenergie und die mittlere Bitenergie:
+
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&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.&nbsp; Here,&nbsp; it is equally true for the&nbsp; "average symbol energy"&nbsp; and the&nbsp; "average bit energy":
 
:$$E_{\rm S} = E_{\rm B} = E\hspace{0.05cm}.$$
 
:$$E_{\rm S} = E_{\rm B} = E\hspace{0.05cm}.$$
  
Meist arbeitet man mit dem <i>Modulationsindex</i>, der als das Verhältnis von Gesamtfrequenzhub und Symbolrate definiert ist:
+
Mostly one works with the&nbsp; "modulation index",&nbsp; which is defined as the ratio of total frequency deviation and symbol rate:
 
:$$h = \frac{2 \cdot \Delta f_{\rm A}}{1/T} = 2 \cdot \Delta f_{\rm A} \cdot T \hspace{0.05cm}.$$
 
:$$h = \frac{2 \cdot \Delta f_{\rm A}}{1/T} = 2 \cdot \Delta f_{\rm A} \cdot T \hspace{0.05cm}.$$
  
Die äquivalente Tiefpassdarstellung führt unter Verwendung von $h$ zu den beiden komplexen Signalen
+
Using&nbsp; $h$,&nbsp; the equivalent low-pass representation leads to the two complex signals
 
:$$ s_{\rm TP0}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  \sqrt{E/T} \cdot {\rm e}^{\hspace{0.05cm}+{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} \pi \hspace{0.03cm}\cdot \hspace{0.03cm} h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T}\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T\hspace{0.05cm},$$
 
:$$ s_{\rm TP0}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  \sqrt{E/T} \cdot {\rm e}^{\hspace{0.05cm}+{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} \pi \hspace{0.03cm}\cdot \hspace{0.03cm} h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T}\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T\hspace{0.05cm},$$
 
:$$ s_{\rm TP1}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  \sqrt{E/T} \cdot {\rm e}^{\hspace{0.05cm}-{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} \pi \hspace{0.03cm}\cdot \hspace{0.03cm} h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T}\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T\hspace{0.05cm}.$$
 
:$$ s_{\rm TP1}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  \sqrt{E/T} \cdot {\rm e}^{\hspace{0.05cm}-{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} \pi \hspace{0.03cm}\cdot \hspace{0.03cm} h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T}\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T\hspace{0.05cm}.$$
  
Eine orthogonale FSK liegt vor, wenn das innere Produkt den Wert $0$ ergibt:
+
An orthogonal FSK exists when the inner product gives the value &nbsp;$0$:&nbsp;
 
:$$<  \hspace{-0.05cm}s_{\rm TP0}(t) \cdot s_{\rm TP1}(t) \hspace{-0.05cm}> \hspace{0.2cm}=   
 
:$$<  \hspace{-0.05cm}s_{\rm TP0}(t) \cdot s_{\rm TP1}(t) \hspace{-0.05cm}> \hspace{0.2cm}=   
 
  \int_{0}^{T} s_{\rm TP0}(t) \cdot s_{\rm TP1}^{\star}(t) \,{\rm d} t =0 \hspace{0.05cm}.$$
 
  \int_{0}^{T} s_{\rm TP0}(t) \cdot s_{\rm TP1}^{\star}(t) \,{\rm d} t =0 \hspace{0.05cm}.$$
  
In diesem Fall ist auch eine nichtkohärente Demodulation wie im Kapitel [[Digitalsignal%C3%BCbertragung/Tr%C3%A4gerfrequenzsysteme_mit_nichtkoh%C3%A4renter_Demodulation| Trägerfrequenzensysteme mit nichtkohärenter Demodulation]] beschrieben möglich.
+
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.
  
Das innere Produkt der BP&ndash;Signale kann aus dem inneren Produkt der TP&ndash;Signale durch Realteilbildung ermittelt werden:
+
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:
 
:$$<  \hspace{-0.05cm}s_{\rm BP0}(t) \cdot s_{\rm BP1}(t) \hspace{-0.05cm}> \hspace{0.2cm}=   
 
:$$<  \hspace{-0.05cm}s_{\rm BP0}(t) \cdot 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{0.15cm} \right ]\hspace{0.05cm}.$$
 
  {\rm Re}\left [ \hspace{0.1cm}<  \hspace{-0.05cm}s_{\rm TP0}(t) \cdot s_{\rm TP1}(t) \hspace{-0.05cm}> \hspace{0.15cm} \right ]\hspace{0.05cm}.$$
  
Gilt $&#9001; s_{\rm BP0}(t) \cdot s_{\rm BP1}(t)&#9002; = 0$, aber gleichzeitig auch $&#9001; s_{\rm TP0}(t) \cdot s_{\rm TP1}(t)&#9002; &ne; 0$, so ist zwar eine kohärente Demodulation möglich, aber keine nichtkohärente.
+
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.
 +
 
  
  
''Hinweise:''
 
* Die Aufgabe gehört zum Kapitel  [[Digitalsignal%C3%BCbertragung/Tr%C3%A4gerfrequenzsysteme_mit_koh%C3%A4renter_Demodulation| Trägerfrequenzsysteme mit kohärenter Demodulation]].
 
* Bezug genommen wird insbesondere auf die Seiten [[Digitalsignalübertragung/Trägerfrequenzsysteme_mit_kohärenter_Demodulation#Binary_Frequency_Shift_Keying_.282.E2.80.93FSK.29|'' Binary_Frequency_Shift_Keying'']](BPSK) und  [[Digitalsignal%C3%BCbertragung/Tr%C3%A4gerfrequenzsysteme_mit_koh%C3%A4renter_Demodulation#Minimum_Shift_Keying_.28MSK.29|''Minimum Shift Keying'']] (MSK).
 
* Sollte die Eingabe des Zahlenwertes „0” erforderlich sein, so geben Sie bitte „0.” ein.
 
  
  
 +
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;$\rm (BFSK)$&nbsp; and &nbsp;[[Digital_Signal_Transmission/Carrier_Frequency_Systems_with_Coherent_Demodulation#Minimum_Shift_Keying_.28MSK.29|"Minimum Shift Keying"]]&nbsp; $\rm (MSK)$.
 +
  
===Fragebogen===
+
 
 +
 
 +
===Questions===
 
<quiz display=simple>
 
<quiz display=simple>
{Welche Trägerfrequenz $f_{\rm T}$ und welcher Frequenzhub $\Delta f_{\rm A}$ liegen der Grafik auf der Angabenseite zugrunde?
+
{Which carrier frequency&nbsp; $f_{\rm T}$&nbsp; and which frequency deviation&nbsp; $\Delta f_{\rm A}$&nbsp; are the basis of the graph in the information section?
 
|type="{}"}
 
|type="{}"}
 
$f_{\rm T}\hspace{0.48cm} = \ $  { 4 3% } $\ \cdot 1/T$
 
$f_{\rm T}\hspace{0.48cm} = \ $  { 4 3% } $\ \cdot 1/T$
 
$\Delta f_{\rm A}\ = \ $ { 0.5 3% } $\ \cdot 1/T$
 
$\Delta f_{\rm A}\ = \ $ { 0.5 3% } $\ \cdot 1/T$
  
{Welchem Modulationsindex $h$ entspricht dieser Frequenzhub?
+
{To what modulation index&nbsp; $h$&nbsp; does this frequency deviation correspond?
 
|type="{}"}
 
|type="{}"}
 
$h\ = \ ${ 1 3% }
 
$h\ = \ ${ 1 3% }
  
{Für welche Werte von $h$ ist die Orthogonalität der Tiefpass&ndash;Signale gegeben?
+
{For which values of&nbsp; $h$&nbsp; is the orthogonality of the low&ndash;pass signals given?
 
|type="[]"}
 
|type="[]"}
 
- $h = 0.5$,
 
- $h = 0.5$,
Line 55: Line 65:
 
+ $h = 2$.
 
+ $h = 2$.
  
{Für welche Werte von $h$ ist die Orthogonalität der Bandpass&ndash;Signale gegeben?
+
{For which values of&nbsp; $h$&nbsp; is the orthogonality of the band&ndash;pass signals given?
 
|type="[]"}
 
|type="[]"}
 
+ $h = 0.5$,
 
+ $h = 0.5$,
Line 62: Line 72:
 
+ $h = 2$.
 
+ $h = 2$.
  
{Welche Aussagen gelten hinsichtlich kohärenter bzw. nichtkohärenter Demodulation?
+
{Which statements are true regarding coherent or non-coherent demodulation?
 
|type="[]"}
 
|type="[]"}
- Kohärente Demodulation ist immer möglich.
+
- Coherent demodulation is always possible.
+ Ist nichtkohärente Demodulation möglich, so geht auch eine kohärente Demodulation.
+
+ If non-coherent demodulation is possible, coherent demodulation is also possible.
- Ist kohärente Demodulation möglich, so geht auch eine nichtkohärente Demodulation.
+
- If coherent demodulation is possible, non-coherent demodulation is also possible.
 
</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Durch Abzählen der Schwingungen innerhalb einer Symboldauer $T$ kommt man zu den beiden Frequenzen $f_0 = 4.5/T$ und $f_1 = 3.5/T$. Daraus berechnen sich Trägerfrequenzen und Frequenzhub zu
+
'''(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 deviations are calculated to
 
:$$f_{\rm T} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  {1}/{2}\cdot (f_0 + f_1) = \underline{4 \cdot 1/T}\hspace{0.05cm},$$
 
:$$f_{\rm T} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  {1}/{2}\cdot (f_0 + f_1) = \underline{4 \cdot 1/T}\hspace{0.05cm},$$
 
:$$ \Delta f_{\rm A} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  {1}/{2}\cdot (f_0 - f_1)= \underline{0.5 \cdot 1/T }\hspace{0.05cm}.$$
 
:$$ \Delta f_{\rm A} \hspace{-0.1cm} \ = \ \hspace{-0.1cm}  {1}/{2}\cdot (f_0 - f_1)= \underline{0.5 \cdot 1/T }\hspace{0.05cm}.$$
  
  
'''(2)'''&nbsp; Mit der angegebenen Gleichung gilt für den Modulationsindex:
+
'''(2)'''&nbsp; With the given equation,&nbsp; the modulation index is:
 
:$$h = 2 \cdot \Delta f_{\rm A} \cdot T  = 2 \cdot 0.5 \cdot 1/T \cdot T \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \underline{h= 1}\hspace{0.05cm}. $$
 
:$$h = 2 \cdot \Delta f_{\rm A} \cdot T  = 2 \cdot 0.5 \cdot 1/T \cdot T \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \underline{h= 1}\hspace{0.05cm}. $$
  
  
'''(3)'''&nbsp; Das innere Produkt der TP&ndash;Signale lautet:
+
'''(3)'''&nbsp; The inner product of the low&ndash;pass signals is:
 
:$$<  \hspace{-0.05cm} s_{\rm TP0}(t) \hspace{0.01cm} \ \cdot \ \hspace{0.01cm} s_{\rm TP1}(t) \hspace{-0.05cm} > \hspace{0.2cm} =   
 
:$$<  \hspace{-0.05cm} s_{\rm TP0}(t) \hspace{0.01cm} \ \cdot \ \hspace{0.01cm} s_{\rm TP1}(t) \hspace{-0.05cm} > \hspace{0.2cm} =   
  \int_{0}^{T} s_{\rm TP0}(t) \cdot s_{\rm TP1}^{\star}(t) \,{\rm d} t =$$
+
  \int_{0}^{T} s_{\rm TP0}(t) \cdot s_{\rm TP1}^{\star}(t) \,{\rm d} t = \frac{E}{T}  \cdot \int_{0}^{T} {\rm e}^{\hspace{0.05cm}{\rm j}  2\pi  h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T} \,{\rm d} t = \frac{E}{{\rm j}2\pi  h} \cdot \left [ {\rm e}^{\hspace{0.05cm}{\rm j}  2\pi  h} - 1 \right ]  
:$$ \hspace{-1.1cm} \ = \ \hspace{-0.1cm} \frac{E}{T}  \cdot \int_{0}^{T} {\rm e}^{\hspace{0.05cm}{\rm j}  2\pi  h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T} \,{\rm d} t = \frac{E}{{\rm j}2\pi  h} \cdot \left [ {\rm e}^{\hspace{0.05cm}{\rm j}  2\pi  h} - 1 \right ]  
 
 
   \hspace{0.05cm}.$$
 
   \hspace{0.05cm}.$$
  
Orthogonalität bedeutet, dass dieses innere Produkt $0$ sein muss:
+
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} =   
 
:$$<  \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
 
   \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, ...$$
 
\hspace{0.3cm}\Rightarrow \hspace{0.3cm} h = 1, 2, 3, ...$$
  
Richtig sind demzufolge die <u>Lösungsvorschläge 3 und 4</u>. Ist der Modulationsindex $h$ ganzzahlig, so kann nichtkohärent demoduliert werden, ohne dass die Orthogonalität verletzt wird.
+
Consequently,&nbsp; <u>solutions 3 and 4</u>&nbsp; are correct:
 +
*If the modulation index $h$&nbsp; is integer,&nbsp; non-coherent demodulation can be performed without violating orthogonality.
 +
 
  
  
'''(4)'''&nbsp; Für das innere Produkt der Bandpass&ndash;Signale kann nach den Erläuterungen auf der Angabenseite geschrieben werden:
+
'''(4)'''&nbsp; <u>Solutions 1, 3 and 4</u> are correct:
 +
*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}=   
 
:$$<  \hspace{-0.05cm}s_{\rm BP0}(t) \hspace{0.01cm} \ \cdot \ \hspace{0.01cm} s_{\rm BP1}(t) \hspace{-0.05cm}> \hspace{0.2cm}=   
  {\rm Re}\left [ \hspace{0.1cm}<  \hspace{-0.05cm}s_{\rm TP0}(t) \cdot s_{\rm TP1}(t) \hspace{-0.05cm}> \hspace{0.2cm} \right ] = {\rm Re}\left [ \frac{E}{{\rm j}2\pi  h} \cdot \left ( {\rm e}^{\hspace{0.05cm}{\rm j}  2\pi  h} - 1 \right ) \right ] =$$
+
  {\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 ]$$
:$$ \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {\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}.$$
+
:$$ \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&nbsp; $0$&nbsp; whenever the modulation index&nbsp; $h$&nbsp; is an integer multiple of&nbsp; $0.5$.
  
Dieses Ergebnis ist immer dann $0$, wenn der Modulationsindex $h$ ein ganzzahliges Vielfaches von $0.5$ ist. Richtig sind also die <u>Lösungsvorschläge 1, 3 und 4</u>.
 
  
 +
'''(5)'''&nbsp; <u>Solution 2</u>&nbsp; is correct.
 +
*For coherent demodulation,&nbsp; $h$&nbsp; must be a multiple of&nbsp; $0.5$.
 +
 +
*If non-coherent demodulation is possible,&nbsp; as for example in the case considered here&nbsp; ($h = 1$),&nbsp; coherent demodulation is also applicable.
  
'''(5)'''&nbsp; Richtig ist hier nur der <u>Lösungsvorschlag 2</u>. Für kohärente Demodulation muss $h$ ein Vielfaches von $0.5$ sein. Ist nichtkohärente Demodulation möglich, wie zum Beispiel im hier betrachteten Fall ($h = 1$), so ist auch kohärente Demodulation anwendbar. Dagegen kann für $h = 0.5$ zwar kohärent demoduliert werden, aber eine nichtkohärente Demodulation (die auf die Hüllkurve angewiesen ist) versagt.
+
*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}}
 
{{ML-Fuß}}
  
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[[Category:Aufgaben zu Digitalsignalübertragung|^4.4 Kohärente Demodulation^]]
+
[[Category:Digital Signal Transmission: Exercises|^4.4 Coherent Demodulation^]]

Latest revision as of 20:51, 1 September 2022

FSK band-pass signals

In binary FSK,  the two messages  $m_0$  and  $m_1$  are represented by two different frequencies.  For the two possible band–pass signals then applies in each case in the range   $0 ≤ t ≤ T$   with   $f_0 = f_{\rm T} + \Delta f_{\rm A}$   as well as  $f_1 = f_{\rm T} \, – \Delta f_{\rm A}$:

$$s_{\rm BP0}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{2E/T} \cdot \cos( 2\pi f_0 t)\hspace{0.05cm},$$
$$ s_{\rm BP1}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{2E/T} \cdot \cos( 2\pi f_1 t)\hspace{0.05cm}.$$

The graph shows example signals  $($in each case only of one symbol duration  $T)$.

  • In the above equation,  $f_{\rm T}$  indicates the  "carrier frequency"  and  $\Delta f_{\rm A}$  indicates the  "frequency deviation"  as the maximum deviation of the  "instantaneous frequency"  from the carrier frequency.
  • $E$  is the signal energy.  Here,  it is equally true for the  "average symbol energy"  and the  "average bit energy":
$$E_{\rm S} = E_{\rm B} = E\hspace{0.05cm}.$$

Mostly one works with the  "modulation index",  which is defined as the ratio of total frequency deviation and symbol rate:

$$h = \frac{2 \cdot \Delta f_{\rm A}}{1/T} = 2 \cdot \Delta f_{\rm A} \cdot T \hspace{0.05cm}.$$

Using  $h$,  the equivalent low-pass representation leads to the two complex signals

$$ s_{\rm TP0}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{E/T} \cdot {\rm e}^{\hspace{0.05cm}+{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} \pi \hspace{0.03cm}\cdot \hspace{0.03cm} h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T}\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T\hspace{0.05cm},$$
$$ s_{\rm TP1}(t) \hspace{-0.1cm} \ = \ \hspace{-0.1cm} \sqrt{E/T} \cdot {\rm e}^{\hspace{0.05cm}-{\rm j} \hspace{0.03cm}\cdot \hspace{0.03cm} \pi \hspace{0.03cm}\cdot \hspace{0.03cm} h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T}\hspace{0.05cm},\hspace{0.2cm} 0 \le t \le T\hspace{0.05cm}.$$

An orthogonal FSK exists when the inner product gives the value  $0$: 

$$< \hspace{-0.05cm}s_{\rm TP0}(t) \cdot s_{\rm TP1}(t) \hspace{-0.05cm}> \hspace{0.2cm}= \int_{0}^{T} s_{\rm TP0}(t) \cdot s_{\rm TP1}^{\star}(t) \,{\rm d} t =0 \hspace{0.05cm}.$$

In this case,  non-coherent demodulation as described in the chapter  "Carrier Frequency Systems with Non-Coherent Demodulation"  is also possible.

The inner product of the band–pass signals can be determined from the inner product of the low–pass signals by real partitioning:

$$< \hspace{-0.05cm}s_{\rm BP0}(t) \cdot s_{\rm BP1}(t) \hspace{-0.05cm}> \hspace{0.2cm}= {\rm Re}\left [ \hspace{0.1cm}< \hspace{-0.05cm}s_{\rm TP0}(t) \cdot s_{\rm TP1}(t) \hspace{-0.05cm}> \hspace{0.15cm} \right ]\hspace{0.05cm}.$$

If  $〈 s_{\rm BP0}(t) \cdot s_{\rm BP1}(t)〉 = 0$, but at the same time  $〈 s_{\rm TP0}(t) \cdot s_{\rm TP1}(t)〉 ≠ 0$,

  • so coherent demodulation is possible,
  • but no non-coherent demodulation.



Notes:



Questions

1

Which carrier frequency  $f_{\rm T}$  and which frequency deviation  $\Delta f_{\rm A}$  are the basis of the graph in the information section?

$f_{\rm T}\hspace{0.48cm} = \ $

$\ \cdot 1/T$
$\Delta f_{\rm A}\ = \ $

$\ \cdot 1/T$

2

To what modulation index  $h$  does this frequency deviation correspond?

$h\ = \ $

3

For which values of  $h$  is the orthogonality of the low–pass signals given?

$h = 0.5$,
$h = \pi/4$,
$h = 1$,
$h = 2$.

4

For which values of  $h$  is the orthogonality of the band–pass signals given?

$h = 0.5$,
$h = \pi/4$,
$h = 1$,
$h = 2$.

5

Which statements are true regarding coherent or non-coherent demodulation?

Coherent demodulation is always possible.
If non-coherent demodulation is possible, coherent demodulation is also possible.
If coherent demodulation is possible, non-coherent demodulation is also possible.


Solution

(1)  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.

  • From this carrier frequencies and frequency deviations are calculated to
$$f_{\rm T} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {1}/{2}\cdot (f_0 + f_1) = \underline{4 \cdot 1/T}\hspace{0.05cm},$$
$$ \Delta f_{\rm A} \hspace{-0.1cm} \ = \ \hspace{-0.1cm} {1}/{2}\cdot (f_0 - f_1)= \underline{0.5 \cdot 1/T }\hspace{0.05cm}.$$


(2)  With the given equation,  the modulation index is:

$$h = 2 \cdot \Delta f_{\rm A} \cdot T = 2 \cdot 0.5 \cdot 1/T \cdot T \hspace{0.3cm} \Rightarrow \hspace{0.3cm} \underline{h= 1}\hspace{0.05cm}. $$


(3)  The inner product of the low–pass signals is:

$$< \hspace{-0.05cm} s_{\rm TP0}(t) \hspace{0.01cm} \ \cdot \ \hspace{0.01cm} s_{\rm TP1}(t) \hspace{-0.05cm} > \hspace{0.2cm} = \int_{0}^{T} s_{\rm TP0}(t) \cdot s_{\rm TP1}^{\star}(t) \,{\rm d} t = \frac{E}{T} \cdot \int_{0}^{T} {\rm e}^{\hspace{0.05cm}{\rm j} 2\pi h \hspace{0.03cm}\cdot \hspace{0.03cm}t/T} \,{\rm d} t = \frac{E}{{\rm j}2\pi h} \cdot \left [ {\rm e}^{\hspace{0.05cm}{\rm j} 2\pi h} - 1 \right ] \hspace{0.05cm}.$$

Orthogonality means that this inner product must be  $0$:

$$< \hspace{-0.05cm} s_{\rm TP0}(t) \cdot s_{\rm TP1}(t) \hspace{-0.05cm} > \hspace{0.2cm} = \frac{E}{{\rm j}2\pi h} \cdot \left [ {\rm e}^{\hspace{0.05cm}{\rm j} 2\pi h} - 1 \right ] = 0 \hspace{0.3cm}\Rightarrow \hspace{0.3cm} h = 1, 2, 3, ...$$

Consequently,  solutions 3 and 4  are correct:

  • If the modulation index $h$  is integer,  non-coherent demodulation can be performed without violating orthogonality.


(4)  Solutions 1, 3 and 4 are correct:

  • For the inner product of the band–pass signals,  according to the explanations in the information section,  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  $0.5$.


(5)  Solution 2  is correct.

  • For coherent demodulation,  $h$  must be a multiple of  $0.5$.
  • If non-coherent demodulation is possible,  as for example in the case considered here  ($h = 1$),  coherent demodulation is also applicable.
  • In contrast,  for $h = 0.5$,  coherent demodulation can be applied,  but non-coherent demodulation  (which relies on the envelope) fails.