Aufgaben:Exercise 5.7Z: Application of the IDFT - Revision history

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	[[Category:Modulation Methods: Exercises\|^5.6 Realization of OFDM Systems^]]		[[Category:Modulation Methods: Exercises\|^5.6 Realization of OFDM Systems^]]
	[[de:Aufgaben:5.~~7Z_Anwendung_der_IDFT~~]]		[[de:Aufgaben:Aufgabe 5.7Z: Anwendung der IDFT]]

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	:$$d(\nu) = (1 - {\rm j}) \cdot {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu} +(1 + {\rm j}) \cdot {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {7\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu}.$$		:$$d(\nu) = (1 - {\rm j}) \cdot {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu} +(1 + {\rm j}) \cdot {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {7\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu}.$$
	*However,  due to periodicity,  also holds:		*However,  due to periodicity,  also holds:
	:$$d(\nu) = (1 - {\rm j}) \cdot {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu} +(1 + {\rm j}) \cdot {\rm{e}}^{ +{\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu}=		:$$d(\nu) = (1 - {\rm j}) \cdot {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu} +(1 + {\rm j}) \cdot {\rm{e}}^{ +{\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu}=\left[ {\rm{e}}^{ + {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu} + {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu}\right]+ {\rm{j}} \cdot\left[ {\rm{e}}^{ + {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu} - {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu}\right].$$
	\left[ {\rm{e}}^{ + {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu} + {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu}\right]+ {\rm{j}} \cdot\left[ {\rm{e}}^{ + {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu} - {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu}\right].$$
	*Using Euler's theorem,  this expression can be transformed as follows:		*Using Euler's theorem,  this expression can be transformed as follows:
	:$$d(\nu) = 2 \cdot \cos \left( {\pi}/{4}\cdot \nu \right)+ 2 \cdot \sin \left( {\pi}/{4}\cdot \nu \right).$$		:$$d(\nu) = 2 \cdot \cos \left( {\pi}/{4}\cdot \nu \right)+ 2 \cdot \sin \left( {\pi}/{4}\cdot \nu \right).$$
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	'''(3)'''  According to the general equation:		'''(3)'''  According to the general equation:
	:$$d(1) = \sum\limits_{\mu = 0}^{7} D(\mu)\cdot {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\mu} = \left[ D(1) + D(7) \right]\cdot \cos \left( {\pi}/{4} \right) + \left[ D(3) + D(5) \right]\cdot \cos \left( {3\pi}/{4} \right)+		:$$d(1) = \sum\limits_{\mu = 0}^{7} D(\mu)\cdot {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\mu} = \left[ D(1) + D(7) \right]\cdot \cos \left( {\pi}/{4} \right) + \left[ D(3) + D(5) \right]\cdot \cos \left( {3\pi}/{4} \right)+{\rm j} \cdot \left[ D(2) - D(6) \right]\cdot \sin \left( {\pi}/{2} \right) + D(4) \cdot {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}}.$$
	{\rm j} \cdot \left[ D(2) - D(6) \right]\cdot \sin \left( {\pi}/{2} \right) + D(4) \cdot {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}}.$$
	*The first three terms give pure real results:		*The first three terms give pure real results:
	:$${\rm Re}[d(1)] = (1+1) \cdot \frac{1}{\sqrt{2}}-(3+3) \cdot \frac{1}{\sqrt{2}}+ {\rm j} \cdot4{\rm j} \cdot 1 = -\frac{4}{\sqrt{2}}-4\hspace{0.15cm}\underline { \approx -6.829}.$$		:$${\rm Re}[d(1)] = (1+1) \cdot \frac{1}{\sqrt{2}}-(3+3) \cdot \frac{1}{\sqrt{2}}+ {\rm j} \cdot4{\rm j} \cdot 1 = -\frac{4}{\sqrt{2}}-4\hspace{0.15cm}\underline { \approx -6.829}.$$

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	[[Category:Modulation Methods: Exercises\|^5.6 Realization of OFDM Systems^]]		[[Category:Modulation Methods: Exercises\|^5.6 Realization of OFDM Systems^]]
			[[de:Aufgaben:5.7Z_Anwendung_der_IDFT]]

Guenter at 08:35, 12 January 2022

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Because of  $D(μ) = 0$  for  $μ ≠ 0$  all time coefficients  $d(ν) = D(0)= 1 - {\rm j}$.  Thus ~~also holds~~:		'''(1)'''  Because of  $D(μ) = 0$  for  $μ ≠ 0$  all time coefficients  $d(ν) = D(0)= 1 - {\rm j}$.  Thus:
	:$${\rm Re}[d(1)] \hspace{0.15cm}\underline {=+ 1}, \hspace{0.3cm}{\rm Im}[d(1)] \hspace{0.15cm}\underline {= -1}.$$		:$${\rm Re}[d(1)] \hspace{0.15cm}\underline {=+ 1}, \hspace{0.3cm}{\rm Im}[d(1)] \hspace{0.15cm}\underline {= -1}.$$


	'''(2)'''  Here, all spectral coefficients are zero except  $D_1 = 1 - {\rm j}$  and  $D_7 = 1 + {\rm j}$.  It follows for all time coefficients  $(0 ≤ ν ≤ 7)$:		'''(2)'''  Here,  all spectral coefficients are zero except  $D_1 = 1 - {\rm j}$  and  $D_7 = 1 + {\rm j}$.  It follows for all time coefficients  $(0 ≤ ν ≤ 7)$:
	:$$d(\nu) = (1 - {\rm j}) \cdot {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu} +(1 + {\rm j}) \cdot {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {7\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu}.$$		:$$d(\nu) = (1 - {\rm j}) \cdot {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu} +(1 + {\rm j}) \cdot {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {7\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu}.$$
	*However, due to periodicity, also holds:		*However,  due to periodicity,  also holds:
	:$$d(\nu) = (1 - {\rm j}) \cdot {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu} +(1 + {\rm j}) \cdot {\rm{e}}^{ +{\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu}=		:$$d(\nu) = (1 - {\rm j}) \cdot {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu} +(1 + {\rm j}) \cdot {\rm{e}}^{ +{\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu}=
	\left[ {\rm{e}}^{ + {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu} + {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu}\right]+ {\rm{j}} \cdot\left[ {\rm{e}}^{ + {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu} - {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu}\right].$$		\left[ {\rm{e}}^{ + {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu} + {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu}\right]+ {\rm{j}} \cdot\left[ {\rm{e}}^{ + {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu} - {\rm{e}}^{ - {\rm{j}}\hspace{0.04cm}\cdot \hspace{0.04cm} {\rm{\pi}}/4\hspace{0.04cm}\cdot \hspace{0.04cm}\nu}\right].$$
	*Using Euler's theorem, this expression can be transformed as follows:		*Using Euler's theorem,  this expression can be transformed as follows:
	:$$d(\nu) = 2 \cdot \cos \left( {\pi}/{4}\cdot \nu \right)+ 2 \cdot \sin \left( {\pi}/{4}\cdot \nu \right).$$		:$$d(\nu) = 2 \cdot \cos \left( {\pi}/{4}\cdot \nu \right)+ 2 \cdot \sin \left( {\pi}/{4}\cdot \nu \right).$$
	*This time function  $d(ν)$  is purely real and characterizes a harmonic oscillation with amplitude  $ 2 \cdot \sqrt{2}$  and phase  $φ = 45^\circ$.		*This time function  $d(ν)$  is purely real and characterizes a harmonic oscillation with amplitude  $ 2 \cdot \sqrt{2}$  and phase  $φ = 45^\circ$.

Guenter at 08:30, 12 January 2022

2022-01-12T08:30:34Z

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	*The exercise belongs to the chapter  [[Modulation_Methods/Realisierung_von_OFDM-Systemen\|Implementation of OFDM Systems]].		*The exercise belongs to the chapter  [[Modulation_Methods/Realisierung_von_OFDM-Systemen\|Implementation of OFDM Systems]].
	*Reference is also made to the chapter  [[Signal_Representation/Discrete_Fourier_Transform_(DFT)\|Discrete Fourier Transform]]  in the book "Signal Representation".		*Reference is also made to the chapter  [[Signal_Representation/Discrete_Fourier_Transform_(DFT)\|Discrete Fourier Transform]]  in the book "Signal Representation".
	*You can check your results with the interactive applet  [[Applets:~~Diskrete_Fouriertransformation_und_Inverse~~\|Discrete Fourier Transform and Inverse]].		*You can check your results with the interactive applet  [[Applets:Discrete_Fouriertransform_and_Inverse\|Discrete Fourier Transform and Inverse]].

Guenter at 08:28, 12 January 2022

2022-01-12T08:28:39Z

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	Here,  $w$  abbreviates the complex rotation factor:		Here,  $w$  abbreviates the complex rotation factor:
	:$$w = {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2 \pi /N} = \cos \left( {2 \pi}/{N}\right)-{\rm j} \cdot \sin \left( {2 \pi}/{N}\right) \hspace{0.05cm}.$$		:$$w = {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2 \pi /N} = \cos \left( {2 \pi}/{N}\right)-{\rm j} \cdot \sin \left( {2 \pi}/{N}\right) \hspace{0.05cm}.$$
	Correspondingly,  for the  [[Signal_Representation/Discrete_Fourier_Transform_(DFT)#~~Inverse_Diskrete_Fouriertransformation~~\|Inverse ~~Discrete~~ Fourier ~~Transform~~]]  $\rm (IDFT)$  practically as an  "inverse function"  of the DFT applies:		Correspondingly,  for the  [[Signal_Representation/Discrete_Fourier_Transform_(DFT)#Inverse_discrete_Fourier_transform\|Inverse discrete Fourier transform]]  $\rm (IDFT)$  practically as an  "inverse function"  of the DFT applies:
	:$$d(\nu) = \sum_{\mu = 0 }^{N-1} D(\mu) \cdot {w}^{-\nu \hspace{0.08cm} \cdot \hspace{0.05cm}\mu} \hspace{0.05cm}.$$		:$$d(\nu) = \sum_{\mu = 0 }^{N-1} D(\mu) \cdot {w}^{-\nu \hspace{0.08cm} \cdot \hspace{0.05cm}\mu} \hspace{0.05cm}.$$
	*In this exercise,  the time coefficients  $d(ν)$  are to be determined for different complex-valued example sequences  $D(μ)$ – which are labeled  $\rm A$,  $\rm B$  and  $\rm C$  in the table.		*In this exercise,  the time coefficients  $d(ν)$  are to be determined for different complex-valued example sequences  $D(μ)$ – which are labeled  $\rm A$,  $\rm B$  and  $\rm C$  in the table.

Guenter at 08:26, 12 January 2022

2022-01-12T08:26:16Z

Hwang at 09:39, 29 December 2021

2021-12-29T09:39:40Z

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Guenter at 15:58, 14 December 2021

2021-12-14T15:58:56Z

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	[[Category:Modulation Methods: Exercises\|^5.6 ~~Realisierung von~~ OFDM~~-Systemen~~^]]		[[Category:Modulation Methods: Exercises\|^5.6 Realization of OFDM Systems^]]

Guenter: Guenter moved page Aufgaben:Aufgabe 5.7Z: Anwendung der IDFT to Aufgaben:Exercise 5.7Z: Application of the IDFT

2021-12-14T15:55:22Z

Guenter moved page Aufgaben:Aufgabe 5.7Z: Anwendung der IDFT to Aufgaben:Exercise 5.7Z: Application of the IDFT

← Older revision	Revision as of 17:55, 14 December 2021
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	*The exercise belongs to the chapter  [[Modulation_Methods/Realisierung_von_OFDM-Systemen\|Implementation of OFDM Systems]].		*The exercise belongs to the chapter  [[Modulation_Methods/Realisierung_von_OFDM-Systemen\|Implementation of OFDM Systems]].
	*Reference is also made to the chapter  [[Signal_Representation/Discrete_Fourier_Transform_(DFT)\|Discrete Fourier Transform]]  in the book "Signal Representation".		*Reference is also made to the chapter  [[Signal_Representation/Discrete_Fourier_Transform_(DFT)\|Discrete Fourier Transform]]  in the book "Signal Representation".
	*You can check your results with the interactive applet  [[Applets:~~Diskrete_Fouriertransformation_und_Inverse~~\|Discrete Fourier Transform and Inverse]].		*You can check your results with the interactive applet  [[Applets:Discrete_Fouriertransform_and_Inverse\|Discrete Fourier Transform and Inverse]].

← Older revision		Revision as of 10:26, 12 January 2022
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	[[File:P_ID1670__Mod_Z_5_7.png\|right\|frame\|Three sets  $\rm A$,  $\rm B$  and  $\rm C$  <br>for the spectral coefficients]]		[[File:P_ID1670__Mod_Z_5_7.png\|right\|frame\|Three sets  $\rm A$,  $\rm B$  and  $\rm C$  <br>for the spectral coefficients]]
	In the  [[Signal_Representation/Discrete_Fourier_Transform_(DFT)\|Discrete Fourier Transform]]  $\rm (DFT)$,  the ~~time sample values~~  $d(ν)$  with ~~the~~ control variable  $ν = 0$, ... , $N – 1$  the ~~discrete spectral coefficients~~  $D(μ)$  with  $μ = 0$, ... , $N – 1$  ~~is calculated~~ as follows:		In the  [[Signal_Representation/Discrete_Fourier_Transform_(DFT)\|Discrete Fourier Transform]]  $\rm (DFT)$,  the discrete spectral coefficients  $D(μ)$  with control variable  $μ = 0$, ... , $N – 1$  are calculated from the time sample values  $d(ν)$  with  $ν = 0$, ... , $N – 1$ as follows:
	:$$D(\mu) = \frac{1}{N} \cdot \sum_{\nu = 0 }^{N-1} d(\nu)\cdot {w}^{\hspace{0.05cm}\nu \hspace{0.08cm} \cdot \hspace{0.05cm}\mu} \hspace{0.05cm}.$$		:$$D(\mu) = \frac{1}{N} \cdot \sum_{\nu = 0 }^{N-1} d(\nu)\cdot {w}^{\hspace{0.05cm}\nu \hspace{0.08cm} \cdot \hspace{0.05cm}\mu} \hspace{0.05cm}.$$
	Here,  $w$  abbreviates the complex rotation factor~~, which is defined as follows~~:		Here,  $w$  abbreviates the complex rotation factor:
	:$$w = {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2 \pi /N} = \cos \left( {2 \pi}/{N}\right)-{\rm j} \cdot \sin \left( {2 \pi}/{N}\right) \hspace{0.05cm}.$$		:$$w = {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2 \pi /N} = \cos \left( {2 \pi}/{N}\right)-{\rm j} \cdot \sin \left( {2 \pi}/{N}\right) \hspace{0.05cm}.$$
	Correspondingly, for the  [[Signal_Representation/Discrete_Fourier_Transform_(DFT)#Inverse_Diskrete_Fouriertransformation\|Inverse Discrete Fourier Transform]]  $\rm (IDFT)$  practically as an "inverse function" of the DFT ~~the following~~ applies:		Correspondingly,  for the  [[Signal_Representation/Discrete_Fourier_Transform_(DFT)#Inverse_Diskrete_Fouriertransformation\|Inverse Discrete Fourier Transform]]  $\rm (IDFT)$  practically as an  "inverse function"  of the DFT applies:
	:$$d(\nu) = \sum_{\mu = 0 }^{N-1} D(\mu) \cdot {w}^{-\nu \hspace{0.08cm} \cdot \hspace{0.05cm}\mu} \hspace{0.05cm}.$$		:$$d(\nu) = \sum_{\mu = 0 }^{N-1} D(\mu) \cdot {w}^{-\nu \hspace{0.08cm} \cdot \hspace{0.05cm}\mu} \hspace{0.05cm}.$$
	In this exercise, the time coefficients  $d(ν)$  are to be determined for different complex-valued example sequences  $D(μ)$ – which are labeled  $\rm A$,  $\rm B$  and  $\rm C$  in the table.  Thus,  $N = 8$ is always valid.		*In this exercise,  the time coefficients  $d(ν)$  are to be determined for different complex-valued example sequences  $D(μ)$ – which are labeled  $\rm A$,  $\rm B$  and  $\rm C$  in the table.
			*Thus,  $N = 8$ is always valid.


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			Notes:


	''Notes:''
	*The exercise belongs to the chapter  [[Modulation_Methods/Realisierung_von_OFDM-Systemen\|Implementation of OFDM Systems]].		*The exercise belongs to the chapter  [[Modulation_Methods/Realisierung_von_OFDM-Systemen\|Implementation of OFDM Systems]].
	*Reference is also made to the chapter  [[Signal_Representation/Discrete_Fourier_Transform_(DFT)\|Discrete Fourier Transform]]  in the book "Signal Representation".		*Reference is also made to the chapter  [[Signal_Representation/Discrete_Fourier_Transform_(DFT)\|Discrete Fourier Transform]]  in the book "Signal Representation".