Difference between revisions of "Aufgaben:Exercise 5.7: OFDM Transmitter using IDFT"

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{{quiz-Header|Buchseite=Modulationsverfahren/Realisierung von OFDM-Systemen
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{{quiz-Header|Buchseite=Modulation_Methods/Implementation_of_OFDM_Systems
 
}}
 
}}
  
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[[File:P_ID1662__A_5_7.png|right|]]
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[[File:P_ID1662__A_5_7.png|right|frame|Block diagram of the IDFT ]]
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In dieser Aufgabe wird ein OFDM–Sender genauer betrachtet, der mit Hilfe der Inversen Diskreten Fouriertransformation (IDFT) realisiert wird. Dabei gelte:
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In this exercise,  we take a closer look at an OFDM transmitter implemented using the  "Inverse Discrete Fourier Transform"  $\rm (IDFT)$.    Thereby it is valid:
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:* Das System habe N = 4 Träger.
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* The system has  $N = 4$  carriers.
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:* Die Rahmendauer sei $T_R = 0.25 ms$.
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* The frame duration is  $T_{\ \rm R} = 0.25 \ \rm ms$.
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:* Ein Guard–Intervall wird nicht verwendet.
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* A guard interval is not used.
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:* In einem Rahmen werden 16 Bit übertragen.  
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* In each frame  $16$  bits are transmitted.
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*The upper right diagram shows the block  "IDFT"  of the OFDM transmitter structure.
 +
*Here,  four bits each result in a complex symbol according to the  $\rm16–QAM$  signal space allocation sketched below left.
  
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Die Grafik zeigt den Block IDFT der OFDM–Senderstruktur. Jeweils vier Bit ergeben hierbei ein komplexes Symbol gemäß der unten gegebenen 16–QAM–Signalraumzuordung.
 
  
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'''Hinweis:''' Die Aufgabe bezieht sich auf das [http://en.lntwww.de/Modulationsverfahren/Realisierung_von_OFDM-Systemen Kapitel 5.6] dieses Buches sowie auf [http://en.lntwww.de/Signaldarstellung/Diskrete_Fouriertransformation_(DFT) Kapitel 5.2] des Buches „Signaldarstellung”. Die Gleichung der IDFT lautet mit ν = 0, ... , N–1:
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[[File:P_ID1666__A_5_7_Signalraum.png|left|frame|Suggested 16–QAM signal space allocation]]
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$$d_{\nu ,k} = \sum\limits_{\mu = 0}^{N - 1} {D_{\mu ,k} \cdot w^{ - \nu \cdot \mu } } \quad {\rm{mit}} \quad w = {\rm{e}}^{ - {\rm{j}} {\rm{2\pi}}/N}.$$
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<br><br><br><br><br><br><br><br><br><br>
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Für die 16–QAM soll in dieser Aufgabe von folgender Signalraumkonstellation ausgegangen werden:
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Notes:
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[[File:P_ID1666__A_5_7_Signalraum.png]]
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*The exercise belongs to the chapter&nbsp;  [[Modulation_Methods/Realisierung_von_OFDM-Systemen|Implementation of OFDM Systems]].
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===Fragebogen===
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*Reference is also made to the chapter&nbsp;  [[Signal_Representation/Discrete_Fourier_Transform_(DFT)|Discrete Fourier Transform]].  
 +
*The equation of the IDFT is with &nbsp;$ν = 0$, ... , $N–1$:
 +
::$$\quad d_{\nu,\ k} = \sum\limits_{\mu = 0}^{N - 1} {D_{\mu, \ k} \cdot w^{ - \nu \cdot \mu } } \quad {\rm{mit}} \quad w = {\rm{e}}^{ - {\rm{j}} {\rm{2\pi}}/N}.$$
 +
<br clear=all>
 +
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
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{Geben Sie die maximale Datenbitrate des Systems an:
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{Specify the maximum data bit rate of the system.
 
|type="{}"}
 
|type="{}"}
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$R_B$ = { 64 35 } $Kbit/s$
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$R_{\rm B} \ = \ $ { 64 3% } $\ \rm kbit/s$
  
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{Geben Sie für die gegebene 16–QAM–Signalraumzuordnung die komplexen Trägerkoeffizienten für die folgenden Eingangsbitfolgen an:
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{For the given 16-QAM signal space allocation,&nbsp; specify the complex carrier coefficients &nbsp;$D_\mu$&nbsp; for the following input bit sequences.
 
|type="{}"}
 
|type="{}"}
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Bitfolge 1111:  Re{$D_0$} = { -1 3% }
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${\rm Re}\big [D_0 \big ] \ = \ $ { -1.03--0.97 } $\ \ \text{for the bit sequence 1111}$  
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Im{$D_0$} = { -1 3% }  
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${\rm Im}\big [D_0\big ] \ = \ $ { -1.03--0.97 }  
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Bitfolge 0111:  Re{$D_1$} = { -1 3% }  
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${\rm Re}\big [D_1\big ] \ = \ $ { -1.03--0.97 } $\ \ \text{for the bit sequence 0111}$
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Im{$D_1$} = { 1 3% }  
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${\rm Im}\big [D_1\big ] \ = \ $ { 1 }  
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Bitfolge 1000:  Re{$D_2$} = { 3 3% }  
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${\rm Re}\big [D_2\big ] \ = \ $ { 3 3% } $\ \ \text{for the bit sequence 1000}$  
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Im{$D_2$} = { -3 3% }  
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${\rm Im}\big [D_2\big ] \ = \ $ { -3.09--2.91 }  
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Bitfolge 0000:  Re{$D_3$} = { 3 3% }  
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${\rm Re}\big [D_3\big ] \ = \ $ { 3 3% } $\ \ \text{for the bit sequence 0000}$
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Im{$D_3$} = { 3 3% }  
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${\rm Im}\big [D_3\big ] \ = \ $ { 3 3% }  
  
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{Berechnen Sie daraus die diskreten Zeitbereichswerte innerhalb des Rahmens.
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{From this,&nbsp; calculate the discrete time domain values &nbsp;$d_\nu$&nbsp; within the frame.
 
|type="{}"}
 
|type="{}"}
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Re{$d_0$} = { 4 3% }
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${\rm Re}\big [d_0\big ] \ = \ $ { 4 1% }
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Im{$d_0$} = { 0 3% }
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${\rm Im}\big [d_0\big ] \ = \ $ { 0. }
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Re{$d_1$} = { -2 3% }
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${\rm Re}\big [d_1\big ] \ = \ $ { -2.02--1.98 }
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Im{$d_1$} = { 2 3% }
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${\rm Im}\big [d_1\big ] \ = \ $ { 2 1% }
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Re{$d_2$} = { 0 3% }
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${\rm Re}\big [d_2\big ] \ = \ $ { 0. }
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Im{$d_2$} = { -8 3% }
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${\rm Im}\big [d_2\big ] \ = \ $ { -8.08--7.92 }
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Re{$d_3$} = { -6 3% }
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${\rm Re}\big [d_3\big ] \ = \ $ { -6.06--5.94 }
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Im{$d_3$} = { 6 3% }
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${\rm Im}\big [d_3\big ] \ = \ $ { 6 1% }
  
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{Welche Aussagen sind für den Crest–Faktor zutreffend, der das Verhältnis von Spitzenwert zu Effektivwert einer Wechselgröße bezeichnet?
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{Which statements are true for the crest factor,&nbsp; which denotes the ratio of the peak value to the rms value of an alternating quantity?
 
|type="[]"}
 
|type="[]"}
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- Der Crest–Faktor ist bei einem OFDM–System eher gering.
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- The crest factor is rather low for an OFDM system.
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+Der Crest–Faktor kann bei OFDM–Systemen sehr groß werden.
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+ The crest factor can become very large in OFDM systems.
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+ Ein großer Crest–Faktor kann zu Realisierungsproblemen führen.
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+ A large crest factor can lead to implementation problems.
  
 
</quiz>
 
</quiz>
  
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===Musterlösung===
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===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
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'''1.'''  Da hier kein Guard–Intervall berücksichtigt wird, ist die Symboldauer T gleich der Rahmendauer $T_R = 0.25 ms$. Bei N = 4 Trägern und 16–QAM gilt für die Bitrate am Eingang:
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'''(1)'''&nbsp; Since no guard interval is considered here,&nbsp; the symbol duration&nbsp; $T$&nbsp; is equal to the frame duration&nbsp; $T_{\rm{R}} = 0.25 \ \rm ms$.  
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$$R_{\rm{B}} = \frac{1}{T_{\rm{B}}} = \frac{4 \cdot {\rm{log}_2}\hspace{0.08cm}(16)}{T} = \frac{4 \cdot 4}{0.25\,\,{\rm ms}}\hspace{0.15cm}\underline {= 64\,\,{\rm kbit/s}}.$$
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*For&nbsp; $N = 4$&nbsp; carriers and&nbsp; $\rm 16–QAM$,&nbsp; the bit rate at the input is:
 +
:$$R_{\rm{B}} = \frac{1}{T_{\rm{B}}} = \frac{4 \cdot {\rm{log}_2}\hspace{0.08cm}(16)}{T} = \frac{4 \cdot 4}{0.25\,\,{\rm ms}}\hspace{0.15cm}\underline {= 64\,\,{\rm kbit/s}}.$$
  
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'''2.'''  Aus der Signalraumzuordnung folgt für die Trägerkoeffizienten (auf den Index k wird verzichtet):
 
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$${\rm{Bitfolge}}\hspace{0.5cm}1111:\hspace{1cm} D_0  =  -1 - {\rm{j}},$$
 
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$${\rm{Bitfolge}}\hspace{0.5cm}0111:\hspace{1cm} D_1  =  -1 + {\rm{j}},$$
 
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$$ {\rm{Bitfolge}}\hspace{0.5cm}1000:\hspace{1cm} D_2  =  +3 - 3{\rm{j}},$$
 
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$${\rm{Bitfolge}}\hspace{0.5cm}0000:\hspace{1cm} D_3  =  +3 + 3{\rm{j}}.$$
 
  
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'''3.'''  Die angegebene IDFT–Gleichung lautet mit N = 4:
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'''(2)'''&nbsp; From the signal space allocation,&nbsp; it follows for the carrier coefficients&nbsp; $($the index&nbsp; $k$&nbsp; is omitted$)$:
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$$d_{\nu } = \sum\limits_{\mu = 0}^{N - 1} {D_{\mu } \cdot {\rm{e}}^{ \hspace{0.04cm} {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} \pi/2 \hspace{0.04cm}\cdot \hspace{0.04cm}\nu \hspace{0.04cm}\cdot \hspace{0.04cm} \mu } } .$$
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:$${\text{Bit sequence}}\hspace{0.2cm}1111:\hspace{0.5cm} D_0  =  -1 - {\rm{j}}\hspace{0.4cm}\Rightarrow\hspace{0.3cm}{\rm Re}[D_0]\hspace{0.15cm}\underline{=-1},\hspace{0.2cm}{\rm Im}[D_0]\hspace{0.15cm}\underline{=-1},$$
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Daraus erhält man für ν = 0, ... , 3:
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:$${\text{Bit sequence}}\hspace{0.2cm}0111:\hspace{0.5cm} D_1  =  -1 + {\rm{j}}\hspace{0.4cm}\Rightarrow\hspace{0.3cm}{\rm Re}[D_1]\hspace{0.15cm}\underline{=-1},\hspace{0.2cm}{\rm Im}[D_1]\hspace{0.15cm}\underline{=+1},$$  
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$$d_0  = D_0 + D_1 +D_2 +D_3 = 4,$$  
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:$${\text{Bit sequence}}\hspace{0.2cm}1000:\hspace{0.5cm} D_2 =  +3 - 3{\rm{j}},\hspace{0.15cm}\Rightarrow\hspace{0.3cm}{\rm Re}[D_2]\hspace{0.15cm}\underline{=+3},\hspace{0.2cm}{\rm Im}[D_2]\hspace{0.15cm}\underline{=-3},$$  
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$$d_1 =  D_0 + {\rm{j}} \cdot D_1 - D_2 -{\rm{j}} \cdot D_3 = -2 + 2 \cdot {\rm{j}},$$  
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:$${\text{Bit sequence}}\hspace{0.2cm}0000:\hspace{0.5cm} D_3 =  +3 + 3{\rm{j}}\hspace{0.2cm}\Rightarrow\hspace{0.3cm}{\rm Re}[D_3]\hspace{0.15cm}\underline{=+3},\hspace{0.2cm}{\rm Im}[D_3]\hspace{0.15cm}\underline{=+3}.$$  
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$$d_2  =  D_0 - D_1 + D_2 - D_3 = -8 \cdot {\rm{j}},$$
 
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$$d_3 =  D_0 - {\rm{j}} \cdot D_1 - D_2 +{\rm{j}} \cdot D_3 = -6 + 6 \cdot {\rm{j}}.$$
 
  
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'''4.'''  Richtig sind die beiden letzten Lösungsvorschläge. Bei OFDM ist der Crest–Faktor eher groß, was bei den verwendeten Verstärkerschaltungen zu Problemen in Bezug auf Linearitätsanforderungen und Energieeffizienz führen kann.
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 +
'''(3)'''&nbsp;  The given&nbsp; $\rm IDFT$&nbsp; equation is with&nbsp; $N = 4$:
 +
:$$d_{\nu } = \sum\limits_{\mu = 0}^{N - 1} {D_{\mu } \cdot {\rm{e}}^{ \hspace{0.04cm} {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} \pi/2 \hspace{0.04cm}\cdot \hspace{0.04cm}\nu \hspace{0.04cm}\cdot \hspace{0.04cm} \mu } } .$$
 +
*From this we obtain for&nbsp; $ν = 0$, ... , $3$:
 +
:$$d_0  =  D_0 + D_1 +D_2 +D_3 = 4 \hspace{2.9cm}\Rightarrow\hspace{0.4cm}{\rm Re}[d_0]\hspace{0.15cm}\underline{=4},\hspace{0.2cm}{\rm Im}[d_0]\hspace{0.15cm}\underline{=0},$$
 +
:$$d_1  =  D_0 + {\rm{j}} \cdot D_1 - D_2 -{\rm{j}} \cdot D_3 = -2 + 2 \cdot {\rm{j}}\hspace{0.4cm}\Rightarrow\hspace{0.4cm}{\rm Re}[d_1]\hspace{0.15cm}\underline{=-2},\hspace{0.2cm}{\rm Im}[d_1]\hspace{0.15cm}\underline{=+2},$$
 +
:$$d_2  =  D_0 - D_1 + D_2 - D_3 = -8 \cdot {\rm{j}}\hspace{2.1cm}\Rightarrow\hspace{0.4cm}{\rm Re}[d_2]\hspace{0.15cm}\underline{=0},\hspace{0.2cm}{\rm Im}[d_2]\hspace{0.15cm}\underline{=-8},$$
 +
:$$d_3  =  D_0 - {\rm{j}} \cdot D_1 - D_2 +{\rm{j}} \cdot D_3 = -6 + 6 \cdot {\rm{j}}\hspace{0.4cm}\Rightarrow\hspace{0.4cm}{\rm Re}[d_3]\hspace{0.15cm}\underline{=-6},\hspace{0.2cm}{\rm Im}[d_3]\hspace{0.15cm}\underline{=+6}.$$
 +
 
 +
 
 +
'''(4)'''&nbsp; The&nbsp; <u>last two solutions</u>&nbsp; are correct:
 +
* For OFDM,&nbsp; the crest factor is rather large.
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*This can lead to problems in terms of linearity requirements and energy efficiency for the amplifier circuits used.
  
 
{{ML-Fuß}}
 
{{ML-Fuß}}
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[[Category:Aufgaben zu Modulationsverfahren|^5.6 Realisierung von OFDM-Systemen^]]
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[[Category:Modulation Methods: Exercises|^5.6 Realization of OFDM Systems^]]

Latest revision as of 18:03, 11 January 2022

Block diagram of the IDFT

In this exercise,  we take a closer look at an OFDM transmitter implemented using the  "Inverse Discrete Fourier Transform"  $\rm (IDFT)$.    Thereby it is valid:

  • The system has  $N = 4$  carriers.
  • The frame duration is  $T_{\ \rm R} = 0.25 \ \rm ms$.
  • A guard interval is not used.
  • In each frame  $16$  bits are transmitted.
  • The upper right diagram shows the block  "IDFT"  of the OFDM transmitter structure.
  • Here,  four bits each result in a complex symbol according to the  $\rm16–QAM$  signal space allocation sketched below left.


Suggested 16–QAM signal space allocation











Notes:

$$\quad d_{\nu,\ k} = \sum\limits_{\mu = 0}^{N - 1} {D_{\mu, \ k} \cdot w^{ - \nu \cdot \mu } } \quad {\rm{mit}} \quad w = {\rm{e}}^{ - {\rm{j}} {\rm{2\pi}}/N}.$$


Questions

1

Specify the maximum data bit rate of the system.

$R_{\rm B} \ = \ $

$\ \rm kbit/s$

2

For the given 16-QAM signal space allocation,  specify the complex carrier coefficients  $D_\mu$  for the following input bit sequences.

${\rm Re}\big [D_0 \big ] \ = \ $

$\ \ \text{for the bit sequence 1111}$
${\rm Im}\big [D_0\big ] \ = \ $

${\rm Re}\big [D_1\big ] \ = \ $

$\ \ \text{for the bit sequence 0111}$
${\rm Im}\big [D_1\big ] \ = \ $

${\rm Re}\big [D_2\big ] \ = \ $

$\ \ \text{for the bit sequence 1000}$
${\rm Im}\big [D_2\big ] \ = \ $

${\rm Re}\big [D_3\big ] \ = \ $

$\ \ \text{for the bit sequence 0000}$
${\rm Im}\big [D_3\big ] \ = \ $

3

From this,  calculate the discrete time domain values  $d_\nu$  within the frame.

${\rm Re}\big [d_0\big ] \ = \ $

${\rm Im}\big [d_0\big ] \ = \ $

${\rm Re}\big [d_1\big ] \ = \ $

${\rm Im}\big [d_1\big ] \ = \ $

${\rm Re}\big [d_2\big ] \ = \ $

${\rm Im}\big [d_2\big ] \ = \ $

${\rm Re}\big [d_3\big ] \ = \ $

${\rm Im}\big [d_3\big ] \ = \ $

4

Which statements are true for the crest factor,  which denotes the ratio of the peak value to the rms value of an alternating quantity?

The crest factor is rather low for an OFDM system.
The crest factor can become very large in OFDM systems.
A large crest factor can lead to implementation problems.


Solution

Solution

(1)  Since no guard interval is considered here,  the symbol duration  $T$  is equal to the frame duration  $T_{\rm{R}} = 0.25 \ \rm ms$.

  • For  $N = 4$  carriers and  $\rm 16–QAM$,  the bit rate at the input is:
$$R_{\rm{B}} = \frac{1}{T_{\rm{B}}} = \frac{4 \cdot {\rm{log}_2}\hspace{0.08cm}(16)}{T} = \frac{4 \cdot 4}{0.25\,\,{\rm ms}}\hspace{0.15cm}\underline {= 64\,\,{\rm kbit/s}}.$$


(2)  From the signal space allocation,  it follows for the carrier coefficients  $($the index  $k$  is omitted$)$:

$${\text{Bit sequence}}\hspace{0.2cm}1111:\hspace{0.5cm} D_0 = -1 - {\rm{j}}\hspace{0.4cm}\Rightarrow\hspace{0.3cm}{\rm Re}[D_0]\hspace{0.15cm}\underline{=-1},\hspace{0.2cm}{\rm Im}[D_0]\hspace{0.15cm}\underline{=-1},$$
$${\text{Bit sequence}}\hspace{0.2cm}0111:\hspace{0.5cm} D_1 = -1 + {\rm{j}}\hspace{0.4cm}\Rightarrow\hspace{0.3cm}{\rm Re}[D_1]\hspace{0.15cm}\underline{=-1},\hspace{0.2cm}{\rm Im}[D_1]\hspace{0.15cm}\underline{=+1},$$
$${\text{Bit sequence}}\hspace{0.2cm}1000:\hspace{0.5cm} D_2 = +3 - 3{\rm{j}},\hspace{0.15cm}\Rightarrow\hspace{0.3cm}{\rm Re}[D_2]\hspace{0.15cm}\underline{=+3},\hspace{0.2cm}{\rm Im}[D_2]\hspace{0.15cm}\underline{=-3},$$
$${\text{Bit sequence}}\hspace{0.2cm}0000:\hspace{0.5cm} D_3 = +3 + 3{\rm{j}}\hspace{0.2cm}\Rightarrow\hspace{0.3cm}{\rm Re}[D_3]\hspace{0.15cm}\underline{=+3},\hspace{0.2cm}{\rm Im}[D_3]\hspace{0.15cm}\underline{=+3}.$$


(3)  The given  $\rm IDFT$  equation is with  $N = 4$:

$$d_{\nu } = \sum\limits_{\mu = 0}^{N - 1} {D_{\mu } \cdot {\rm{e}}^{ \hspace{0.04cm} {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} \pi/2 \hspace{0.04cm}\cdot \hspace{0.04cm}\nu \hspace{0.04cm}\cdot \hspace{0.04cm} \mu } } .$$
  • From this we obtain for  $ν = 0$, ... , $3$:
$$d_0 = D_0 + D_1 +D_2 +D_3 = 4 \hspace{2.9cm}\Rightarrow\hspace{0.4cm}{\rm Re}[d_0]\hspace{0.15cm}\underline{=4},\hspace{0.2cm}{\rm Im}[d_0]\hspace{0.15cm}\underline{=0},$$
$$d_1 = D_0 + {\rm{j}} \cdot D_1 - D_2 -{\rm{j}} \cdot D_3 = -2 + 2 \cdot {\rm{j}}\hspace{0.4cm}\Rightarrow\hspace{0.4cm}{\rm Re}[d_1]\hspace{0.15cm}\underline{=-2},\hspace{0.2cm}{\rm Im}[d_1]\hspace{0.15cm}\underline{=+2},$$
$$d_2 = D_0 - D_1 + D_2 - D_3 = -8 \cdot {\rm{j}}\hspace{2.1cm}\Rightarrow\hspace{0.4cm}{\rm Re}[d_2]\hspace{0.15cm}\underline{=0},\hspace{0.2cm}{\rm Im}[d_2]\hspace{0.15cm}\underline{=-8},$$
$$d_3 = D_0 - {\rm{j}} \cdot D_1 - D_2 +{\rm{j}} \cdot D_3 = -6 + 6 \cdot {\rm{j}}\hspace{0.4cm}\Rightarrow\hspace{0.4cm}{\rm Re}[d_3]\hspace{0.15cm}\underline{=-6},\hspace{0.2cm}{\rm Im}[d_3]\hspace{0.15cm}\underline{=+6}.$$


(4)  The  last two solutions  are correct:

  • For OFDM,  the crest factor is rather large.
  • This can lead to problems in terms of linearity requirements and energy efficiency for the amplifier circuits used.

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