Aufgaben:Exercise 5.7: OFDM Transmitter using IDFT - Revision history

Guenter at 17:03, 11 January 2022

2022-01-11T17:03:37Z

Guenter at 16:56, 11 January 2022

2022-01-11T16:56:00Z

Guenter at 16:54, 11 January 2022

2022-01-11T16:54:08Z

Hwang at 08:58, 29 December 2021

2021-12-29T08:58:30Z

Hwang at 15:00, 20 December 2021

2021-12-20T15:00:33Z

Guenter at 15:58, 14 December 2021

2021-12-14T15:58:25Z

Guenter: Guenter moved page Aufgaben:Aufgabe 5.7: OFDM–Sender mittels IDFT to Aufgaben:Exercise 5.7: OFDM Transmitter using IDFT

2021-12-14T15:54:31Z

Guenter moved page Aufgabe 5.7: OFDM–Sender mittels IDFT to Exercise 5.7: OFDM Transmitter using IDFT

Javier: Text replacement - "„" to """

2021-05-28T14:39:15Z

Text replacement - "„" to """

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2021-05-28T14:33:35Z

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Javier: Text replacement - "Category:Aufgaben zu Modulationsverfahren" to "Category:Modulation Methods: Exercises"

2021-03-23T12:52:13Z

Text replacement - "Category:Aufgaben zu Modulationsverfahren" to "Category:Modulation Methods: Exercises"

@@ Line 60: / Line 60: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp;  Since no guard interval is considered here, the symbol duration&nbsp; $T$&nbsp; is equal to the frame duration&nbsp; $T_{\rm{R}} = 0.25 \ \rm ms$.
+'''(1)'''&nbsp;  Since no guard interval is considered here,&nbsp; the symbol duration&nbsp; $T$&nbsp; is equal to the frame duration&nbsp; $T_{\rm{R}} = 0.25 \ \rm ms$.
 *For&nbsp; $N = 4$&nbsp; carriers and&nbsp; $\rm 16–QAM$,&nbsp; the bit rate at the input is:
 :$$R_{\rm{B}} = \frac{1}{T_{\rm{B}}} = \frac{4 \cdot {\rm{log}_2}\hspace{0.08cm}(16)}{T} = \frac{4 \cdot 4}{0.25\,\,{\rm ms}}\hspace{0.15cm}\underline {= 64\,\,{\rm kbit/s}}.$$
-'''(2)'''&nbsp;  From the signal space allocation, it follows for the carrier coefficients&nbsp; $($the index&nbsp; $k$&nbsp; is omitted$)$:
+'''(2)'''&nbsp;  From the signal space allocation,&nbsp; it follows for the carrier coefficients&nbsp; $($the index&nbsp; $k$&nbsp; is omitted$)$:
-:$${\rm{Bit \quad sequence}}\hspace{0.2cm}1111:\hspace{0.5cm} D_0  =  -1 - {\rm{j}}\hspace{0.4cm}\Rightarrow\hspace{0.3cm}{\rm Re}[D_0]\hspace{0.15cm}\underline{=-1},\hspace{0.2cm}{\rm Im}[D_0]\hspace{0.15cm}\underline{=-1},$$
+:$${\text{Bit sequence}}\hspace{0.2cm}1111:\hspace{0.5cm} D_0  =  -1 - {\rm{j}}\hspace{0.4cm}\Rightarrow\hspace{0.3cm}{\rm Re}[D_0]\hspace{0.15cm}\underline{=-1},\hspace{0.2cm}{\rm Im}[D_0]\hspace{0.15cm}\underline{=-1},$$
-:$${\rm{Bit \quad sequence}}\hspace{0.2cm}0111:\hspace{0.5cm} D_1  =  -1 + {\rm{j}}\hspace{0.4cm}\Rightarrow\hspace{0.3cm}{\rm Re}[D_1]\hspace{0.15cm}\underline{=-1},\hspace{0.2cm}{\rm Im}[D_1]\hspace{0.15cm}\underline{=+1},$$
+:$${\text{Bit sequence}}\hspace{0.2cm}0111:\hspace{0.5cm} D_1  =  -1 + {\rm{j}}\hspace{0.4cm}\Rightarrow\hspace{0.3cm}{\rm Re}[D_1]\hspace{0.15cm}\underline{=-1},\hspace{0.2cm}{\rm Im}[D_1]\hspace{0.15cm}\underline{=+1},$$
-:$${\rm{Bit \quad sequence}}\hspace{0.2cm}1000:\hspace{0.5cm} D_2  =  +3 - 3{\rm{j}},\hspace{0.15cm}\Rightarrow\hspace{0.3cm}{\rm Re}[D_2]\hspace{0.15cm}\underline{=+3},\hspace{0.2cm}{\rm Im}[D_2]\hspace{0.15cm}\underline{=-3},$$
+:$${\text{Bit sequence}}\hspace{0.2cm}1000:\hspace{0.5cm} D_2  =  +3 - 3{\rm{j}},\hspace{0.15cm}\Rightarrow\hspace{0.3cm}{\rm Re}[D_2]\hspace{0.15cm}\underline{=+3},\hspace{0.2cm}{\rm Im}[D_2]\hspace{0.15cm}\underline{=-3},$$
-:$${\rm{Bit \quad sequence}}\hspace{0.2cm}0000:\hspace{0.5cm} D_3  =  +3 + 3{\rm{j}}\hspace{0.2cm}\Rightarrow\hspace{0.3cm}{\rm Re}[D_3]\hspace{0.15cm}\underline{=+3},\hspace{0.2cm}{\rm Im}[D_3]\hspace{0.15cm}\underline{=+3}.$$
+:$${\text{Bit sequence}}\hspace{0.2cm}0000:\hspace{0.5cm} D_3  =  +3 + 3{\rm{j}}\hspace{0.2cm}\Rightarrow\hspace{0.3cm}{\rm Re}[D_3]\hspace{0.15cm}\underline{=+3},\hspace{0.2cm}{\rm Im}[D_3]\hspace{0.15cm}\underline{=+3}.$$
-'''(3)'''&nbsp;  The given IDFT equation is with&nbsp; $N = 4$:
+'''(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; $ν = 0$, ... , $3$:
@@ Line 81: / Line 81: @@
-'''(4)'''&nbsp;  The <u>last two solutions</u> are correct:
+'''(4)'''&nbsp;  The&nbsp; <u>last two solutions</u>&nbsp; are correct:
-* For OFDM, the crest factor is rather large.
+* For OFDM,&nbsp; the crest factor is rather large.
 *This can lead to problems in terms of linearity requirements and energy efficiency for the amplifier circuits used.

@@ Line 17: / Line 17: @@
 Notes:
 *The exercise belongs to the chapter&nbsp;  [[Modulation_Methods/Realisierung_von_OFDM-Systemen|Implementation of OFDM Systems]].
-*Reference is also made to the chapter&nbsp;  [[Signal_Representation/Discrete_Fourier_Transform_(DFT)|Diskrete Fouriertransformation]].
+*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;$ν = 0$, ... , $N–1$:
 ::$$\quad d_{\nu,\ k} = \sum\limits_{\mu = 0}^{N - 1} {D_{\mu, \ k} \cdot w^{ - \nu \cdot \mu } } \quad {\rm{mit}} \quad w = {\rm{e}}^{ - {\rm{j}} {\rm{2\pi}}/N}.$$

@@ Line 4: / Line 4: @@
 [[File:P_ID1662__A_5_7.png|right|frame|Block diagram of the IDFT ]]
-In this exercise, we take a closer look at an OFDM transmitter implemented using the &nbsp;''Inverse Discrete Fourier Transform''&nbsp; $\rm (IDFT)$.&nbsp; &nbsp; Thereby it is valid:
+In this exercise,&nbsp; we take a closer look at an OFDM transmitter implemented using the &nbsp;"Inverse Discrete Fourier Transform"&nbsp; $\rm (IDFT)$.&nbsp; &nbsp; Thereby it is valid:
 * The system has &nbsp;$N = 4$&nbsp; carriers.
 * The frame duration is &nbsp;$T_{\ \rm R} = 0.25 \ \rm ms$.
 * A guard interval is not used.
 * In each frame &nbsp;$16$&nbsp; bits are transmitted.
-*The upper right diagram shows the block "IDFT" of the OFDM transmitter structure.
+*The upper right diagram shows the block&nbsp; "IDFT"&nbsp; of the OFDM transmitter structure.
-*Here, four bits each result in a complex symbol according to the&nbsp; $\rm16–QAM$ signal space allocation sketched below left.
+*Here,&nbsp; four bits each result in a complex symbol according to the&nbsp; $\rm16–QAM$&nbsp; signal space allocation sketched below left.
-[[File:P_ID1666__A_5_7_Signalraum.png|links|frame|Suggested 16–QAM signal space allocation]]
+[[File:P_ID1666__A_5_7_Signalraum.png|left|frame|Suggested 16–QAM signal space allocation]]
+<br><br><br><br><br><br><br><br><br><br>
 Notes:
+*The exercise belongs to the chapter&nbsp;  [[Modulation_Methods/Realisierung_von_OFDM-Systemen|Implementation of OFDM Systems]].
 *Reference is also made to the chapter&nbsp;  [[Signal_Representation/Discrete_Fourier_Transform_(DFT)|Diskrete Fouriertransformation]].
 *The equation of the IDFT is with &nbsp;$ν = 0$, ... , $N–1$:
-::$$\quad d_{\nu ,k} = \sum\limits_{\mu = 0}^{N - 1} {D_{\mu ,k} \cdot w^{ - \nu \cdot \mu } } \quad {\rm{mit}} \quad w = {\rm{e}}^{ - {\rm{j}} {\rm{2\pi}}/N}.$$
+::$$\quad d_{\nu,\ k} = \sum\limits_{\mu = 0}^{N - 1} {D_{\mu, \ k} \cdot w^{ - \nu \cdot \mu } } \quad {\rm{mit}} \quad w = {\rm{e}}^{ - {\rm{j}} {\rm{2\pi}}/N}.$$
 <br clear=all>
 ===Questions===
@@ Line 41: / Line 29: @@
 $R_{\rm B} \ = \ $ { 64 3% } $\ \rm kbit/s$
-{For the given 16-QAM signal space allocation, specify the complex carrier coefficients &nbsp;$D_\mu$&nbsp; for the following input bit sequences.
+{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="{}"}
 ${\rm Re}\big [D_0 \big ] \ = \ $ { -1.03--0.97 }  $\ \ \text{for the bit sequence 1111}$
@@ Line 52: / Line 40: @@
 ${\rm Im}\big [D_3\big ] \ = \ $ { 3 3% }
-{From this, calculate the discrete time domain values &nbsp;$d_\nu$&nbsp; within the frame.
+{From this,&nbsp; calculate the discrete time domain values &nbsp;$d_\nu$&nbsp; within the frame.
 |type="{}"}
 ${\rm Re}\big [d_0\big ] \ = \ $ { 4 1% }
@@ Line 63: / Line 51: @@
 ${\rm Im}\big [d_3\big ] \ = \ $ { 6 1% }
-{Which statements are true for the crest factor, which denotes the ratio of the peak value to the rms value of an alternating quantity?
+{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="[]"}
 - The crest factor is rather low for an OFDM system.

@@ Line 37: / Line 37: @@
 <quiz display=simple>
-{Geben Sie die maximale Datenbitrate des Systems an.
+{Specify the maximum data bit rate of the system.
 |type="{}"}
 $R_{\rm B} \ = \ $ { 64 3% } $\ \rm kbit/s$
-{Geben Sie für die gegebene 16–QAM–Signalraumzuordnung die komplexen Trägerkoeffizienten &nbsp;$D_\mu$&nbsp; für die folgenden Eingangsbitfolgen an.
+{For the given 16-QAM signal space allocation, specify the complex carrier coefficients &nbsp;$D_\mu$&nbsp; for the following input bit sequences.
 |type="{}"}
-${\rm Re}\big [D_0 \big ] \ = \ $ { -1.03--0.97 }  $\ \ \text{für die Bitfolge 1111}$
+${\rm Re}\big [D_0 \big ] \ = \ $ { -1.03--0.97 }  $\ \ \text{for the bit sequence 1111}$
 ${\rm Im}\big [D_0\big ] \ = \ $ { -1.03--0.97 }
-${\rm Re}\big [D_1\big ] \ = \ $ { -1.03--0.97 }  $\ \ \text{für die Bitfolge 0111}$
+${\rm Re}\big [D_1\big ] \ = \ $ { -1.03--0.97 }  $\ \ \text{for the bit sequence 0111}$
 ${\rm Im}\big [D_1\big ] \ = \ $ { 1 }
-${\rm Re}\big [D_2\big ] \ = \ $ { 3 3% } $\ \ \text{für die Bitfolge 1000}$
+${\rm Re}\big [D_2\big ] \ = \ $ { 3 3% } $\ \ \text{for the bit sequence 1000}$
 ${\rm Im}\big [D_2\big ] \ = \ $ { -3.09--2.91 }
-${\rm Re}\big [D_3\big ] \ = \ $ { 3 3% } $\ \ \text{für die Bitfolge 0000}$
+${\rm Re}\big [D_3\big ] \ = \ $ { 3 3% } $\ \ \text{for the bit sequence 0000}$
 ${\rm Im}\big [D_3\big ] \ = \ $ { 3 3% }
-{Berechnen Sie daraus die diskreten Zeitbereichswerte &nbsp;$d_\nu$&nbsp; innerhalb des Rahmens.
+{From this, calculate the discrete time domain values &nbsp;$d_\nu$&nbsp; within the frame.
 |type="{}"}
 ${\rm Re}\big [d_0\big ] \ = \ $ { 4 1% }
@@ Line 63: / Line 63: @@
 ${\rm Im}\big [d_3\big ] \ = \ $ { 6 1% }
-{Welche Aussagen sind für den Crest–Faktor zutreffend, der das Verhältnis von Spitzenwert zu Effektivwert einer Wechselgröße bezeichnet?
+{Which statements are true for the crest factor, which denotes the ratio of the peak value to the rms value of an alternating quantity?
 |type="[]"}
-- Der Crest–Faktor ist bei einem OFDM–System eher gering.
+- The crest factor is rather low for an OFDM system.
-+ Der Crest–Faktor kann bei OFDM–Systemen sehr groß werden.
++ The crest factor can become very large in OFDM systems.
-+ Ein großer Crest–Faktor kann zu Realisierungsproblemen führen.
++ A large crest factor can lead to implementation problems.
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp;  Da hier kein Guard–Intervall berücksichtigt wird, ist die Symboldauer&nbsp; $T$&nbsp; gleich der Rahmendauer&nbsp; $T_{\rm{R}} = 0.25 \ \rm ms$.
+'''(1)'''&nbsp;  Since no guard interval is considered here, the symbol duration&nbsp; $T$&nbsp; is equal to the frame duration&nbsp; $T_{\rm{R}} = 0.25 \ \rm ms$.
-*Bei&nbsp; $N = 4$&nbsp; Trägern und&nbsp; $\rm 16–QAM$&nbsp; gilt für die Bitrate am Eingang:
+*For&nbsp; $N = 4$&nbsp; carriers and&nbsp; $\rm 16–QAM$,&nbsp; the bit rate at the input is:
 :$$R_{\rm{B}} = \frac{1}{T_{\rm{B}}} = \frac{4 \cdot {\rm{log}_2}\hspace{0.08cm}(16)}{T} = \frac{4 \cdot 4}{0.25\,\,{\rm ms}}\hspace{0.15cm}\underline {= 64\,\,{\rm kbit/s}}.$$
-'''(2)'''&nbsp;  Aus der Signalraumzuordnung folgt für die Trägerkoeffizienten&nbsp; $($auf den Index&nbsp; $k$&nbsp; wird verzichtet$)$:
+'''(2)'''&nbsp;  From the signal space allocation, it follows for the carrier coefficients&nbsp; $($the index&nbsp; $k$&nbsp; is omitted$)$:
-:$${\rm{Bitfolge}}\hspace{0.2cm}1111:\hspace{0.5cm} D_0  =  -1 - {\rm{j}}\hspace{0.4cm}\Rightarrow\hspace{0.3cm}{\rm Re}[D_0]\hspace{0.15cm}\underline{=-1},\hspace{0.2cm}{\rm Im}[D_0]\hspace{0.15cm}\underline{=-1},$$
+:$${\rm{Bit \quad sequence}}\hspace{0.2cm}1111:\hspace{0.5cm} D_0  =  -1 - {\rm{j}}\hspace{0.4cm}\Rightarrow\hspace{0.3cm}{\rm Re}[D_0]\hspace{0.15cm}\underline{=-1},\hspace{0.2cm}{\rm Im}[D_0]\hspace{0.15cm}\underline{=-1},$$
-:$${\rm{Bitfolge}}\hspace{0.2cm}0111:\hspace{0.5cm} D_1  =  -1 + {\rm{j}}\hspace{0.4cm}\Rightarrow\hspace{0.3cm}{\rm Re}[D_1]\hspace{0.15cm}\underline{=-1},\hspace{0.2cm}{\rm Im}[D_1]\hspace{0.15cm}\underline{=+1},$$
+:$${\rm{Bit \quad sequence}}\hspace{0.2cm}0111:\hspace{0.5cm} D_1  =  -1 + {\rm{j}}\hspace{0.4cm}\Rightarrow\hspace{0.3cm}{\rm Re}[D_1]\hspace{0.15cm}\underline{=-1},\hspace{0.2cm}{\rm Im}[D_1]\hspace{0.15cm}\underline{=+1},$$
-:$${\rm{Bitfolge}}\hspace{0.2cm}1000:\hspace{0.5cm} D_2  =  +3 - 3{\rm{j}},\hspace{0.15cm}\Rightarrow\hspace{0.3cm}{\rm Re}[D_2]\hspace{0.15cm}\underline{=+3},\hspace{0.2cm}{\rm Im}[D_2]\hspace{0.15cm}\underline{=-3},$$
+:$${\rm{Bit \quad sequence}}\hspace{0.2cm}1000:\hspace{0.5cm} D_2  =  +3 - 3{\rm{j}},\hspace{0.15cm}\Rightarrow\hspace{0.3cm}{\rm Re}[D_2]\hspace{0.15cm}\underline{=+3},\hspace{0.2cm}{\rm Im}[D_2]\hspace{0.15cm}\underline{=-3},$$
-:$${\rm{Bitfolge}}\hspace{0.2cm}0000:\hspace{0.5cm} D_3  =  +3 + 3{\rm{j}}\hspace{0.2cm}\Rightarrow\hspace{0.3cm}{\rm Re}[D_3]\hspace{0.15cm}\underline{=+3},\hspace{0.2cm}{\rm Im}[D_3]\hspace{0.15cm}\underline{=+3}.$$
+:$${\rm{Bit \quad sequence}}\hspace{0.2cm}0000:\hspace{0.5cm} D_3  =  +3 + 3{\rm{j}}\hspace{0.2cm}\Rightarrow\hspace{0.3cm}{\rm Re}[D_3]\hspace{0.15cm}\underline{=+3},\hspace{0.2cm}{\rm Im}[D_3]\hspace{0.15cm}\underline{=+3}.$$
-'''(3)'''&nbsp;  Die angegebene IDFT–Gleichung lautet mit&nbsp; $N = 4$:
+'''(3)'''&nbsp;  The given IDFT equation is with&nbsp; $N = 4$:
 :$$d_{\nu } = \sum\limits_{\mu = 0}^{N - 1} {D_{\mu } \cdot {\rm{e}}^{ \hspace{0.04cm} {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} \pi/2 \hspace{0.04cm}\cdot \hspace{0.04cm}\nu \hspace{0.04cm}\cdot \hspace{0.04cm} \mu } } .$$
-*Daraus erhält man für&nbsp; $ν = 0$, ... , $3$:
+*From this we obtain for&nbsp; $ν = 0$, ... , $3$:
 :$$d_0  =  D_0 + D_1 +D_2 +D_3 = 4 \hspace{2.9cm}\Rightarrow\hspace{0.4cm}{\rm Re}[d_0]\hspace{0.15cm}\underline{=4},\hspace{0.2cm}{\rm Im}[d_0]\hspace{0.15cm}\underline{=0},$$
 :$$d_1  =  D_0 + {\rm{j}} \cdot D_1 - D_2 -{\rm{j}} \cdot D_3 = -2 + 2 \cdot {\rm{j}}\hspace{0.4cm}\Rightarrow\hspace{0.4cm}{\rm Re}[d_1]\hspace{0.15cm}\underline{=-2},\hspace{0.2cm}{\rm Im}[d_1]\hspace{0.15cm}\underline{=+2},$$
@@ Line 94: / Line 94: @@
-'''(4)'''&nbsp;  Richtig sind die <u>beiden letzten Lösungsvorschläge</u>:
+'''(4)'''&nbsp;  The <u>last two solutions</u> are correct:
-* Bei OFDM ist der Crest–Faktor eher groß.
+* For OFDM, the crest factor is rather large.
-*Dies kann bei den verwendeten Verstärkerschaltungen zu Problemen in Bezug auf Linearitätsanforderungen und Energieeffizienz führen.
+*This can lead to problems in terms of linearity requirements and energy efficiency for the amplifier circuits used.
 {{ML-Fuß}}

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Modulationsverfahren/Realisierung von OFDM-Systemen
+{{quiz-Header|Buchseite=Modulation_Methods/Implementation_of_OFDM_Systems
 }}
-[[File:P_ID1662__A_5_7.png|right|frame|Blockschaltbild der IDFT ]]
+[[File:P_ID1662__A_5_7.png|right|frame|Block diagram of the IDFT ]]
-In dieser Aufgabe wird ein OFDM–Sender genauer betrachtet, der mit Hilfe der &nbsp;''Inversen Diskreten Fouriertransformation''&nbsp; $\rm (IDFT)$&nbsp; realisiert ist.&nbsp; Dabei gelte:
+In this exercise, we take a closer look at an OFDM transmitter implemented using the &nbsp;''Inverse Discrete Fourier Transform''&nbsp; $\rm (IDFT)$.&nbsp; &nbsp; Thereby it is valid:
-* Das System habe &nbsp;$N = 4$&nbsp; Träger.
+* The system has &nbsp;$N = 4$&nbsp; carriers.
-* Die Rahmendauer sei &nbsp;$T_{\ \rm R} = 0.25 \ \rm ms$.
+* The frame duration is &nbsp;$T_{\ \rm R} = 0.25 \ \rm ms$.
-* Ein Guard–Intervall wird nicht verwendet.
+* A guard interval is not used.
-* In jedem Rahmen werden &nbsp;$16$&nbsp; Bit übertragen.
+* In each frame &nbsp;$16$&nbsp; bits are transmitted.
-*Die rechte obere Grafik zeigt den Block "IDFT" der OFDM–Senderstruktur.
+*The upper right diagram shows the block "IDFT" of the OFDM transmitter structure.
-*Jeweils vier Bit ergeben hierbei ein komplexes Symbol gemäß der unten links skizzierten&nbsp; $\rm16–QAM$–Signalraumzuordung.
+*Here, four bits each result in a complex symbol according to the&nbsp; $\rm16–QAM$ signal space allocation sketched below left.
-[[File:P_ID1666__A_5_7_Signalraum.png|links|frame|Vorgeschlagene 16–QAM-Signalraumzuordnung]]
+[[File:P_ID1666__A_5_7_Signalraum.png|links|frame|Suggested 16–QAM signal space allocation]]
@@ Line 27: / Line 27: @@
-''Hinweise:''
+''Notes:''
-*Die Aufgabe gehört zum  Kapitel&nbsp;  [[Modulation_Methods/Realisierung_von_OFDM-Systemen|Realisierung von OFDM-Systemen]].
+*The task belongs to the chapter&nbsp;  [[Modulation_Methods/Realisierung_von_OFDM-Systemen|Implementation of OFDM Systems]].
-*Bezug genommen wird auch  auf das Kapitel&nbsp;  [[Signal_Representation/Discrete_Fourier_Transform_(DFT)|Diskrete Fouriertransformation]].
+*Reference is also made to the chapter&nbsp;  [[Signal_Representation/Discrete_Fourier_Transform_(DFT)|Diskrete Fouriertransformation]].
-*Die Gleichung der IDFT lautet mit &nbsp;$ν = 0$, ... , $N–1$:
+*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>
-===Fragebogen===
+===Questions===
 <quiz display=simple>

← Older revision		Revision as of 15:58, 14 December 2021
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−	[[Category:Modulation Methods: Exercises\|^5.6 ~~Realisierung von~~ OFDM~~-Systemen~~^]]	+	[[Category:Modulation Methods: Exercises\|^5.6 Realization of OFDM Systems^]]

← Older revision		Revision as of 12:52, 23 March 2021
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−	[[Category:~~Aufgaben zu Modulationsverfahren~~\|^5.6 Realisierung von OFDM-Systemen^]]	+	[[Category:Modulation Methods: Exercises\|^5.6 Realisierung von OFDM-Systemen^]]