Aufgaben:Exercise 5.8: Equalization in Matrix Vector Notation - 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.~~8_Entzerrung_in_Matrix–Vektor–Notation~~]]		[[de:Aufgaben:Aufgabe 5.8: Entzerrung in Matrix–Vektor–Notation]]

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	:$${\rm\bf{R}} = {\rm\bf{D}} \cdot \left( {\begin{array}{*{20}c}{H_0 } & {} & {} & {} \\{} & {H_1 } & {} & {} \\{} & {} & {H_2 } & {} \\{} & {} & {} & {H_3 } \\\end{array}} \right) .$$		:$${\rm\bf{R}} = {\rm\bf{D}} \cdot \left( {\begin{array}{*{20}c}{H_0 } & {} & {} & {} \\{} & {H_1 } & {} & {} \\{} & {} & {H_2 } & {} \\{} & {} & {} & {H_3 } \\\end{array}} \right) .$$
	*For the diagonal elements one obtains:		*For the diagonal elements one obtains:
	:$$H_\mu = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e}}^{ -		:$$H_\mu = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e}}^{ -{\rm{j\hspace{0.04cm}\cdot \hspace{0.04cm}2\pi }}\hspace{0.04cm}\cdot \hspace{0.04cm} l \hspace{0.04cm}\cdot\hspace{0.04cm}{\mu }/{4}} } \hspace{0.3cm}\Rightarrow \hspace{0.3cm} H_0 = 1,\hspace{0.1cm}H_1 = -0.4 - {\rm{j}} \cdot0.6,\hspace{0.1cm}H_2 = -0.2,\hspace{0.1cm}H_3 = -0.4 + {\rm{j}}\cdot 0.6 $$:$$\Rightarrow \hspace{0.3cm}{\rm\bf{R}} = \left( {R_0 ,R_1 ,R_2 ,R_3} \right)= \left( \hspace{0.15cm}0,\hspace{0.15cm}0,\hspace{0.15cm}-0.2, \hspace{0.15cm}0 \right) \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Re}[r_2]\hspace{0.15cm} \underline{=-0.2},\hspace{0.2cm} {\rm Im}[r_2]\hspace{0.15cm} \underline{=0}, \hspace{0.2cm}{\rm Re}[r_3]\hspace{0.15cm} \underline{=0},\hspace{0.2cm} {\rm Im}[r_3]\hspace{0.15cm} \underline{=0}.$$
	{\rm{j\hspace{0.04cm}\cdot \hspace{0.04cm}2\pi }}
	\hspace{0.04cm}\cdot \hspace{0.04cm} l \hspace{0.04cm}\cdot
	\hspace{0.04cm}{\mu }/{4}} } \hspace{0.3cm}
	\Rightarrow \hspace{0.3cm} H_0 = 1,\hspace{0.1cm}H_1 = -0.4 - {\rm{j}} \~~cdot~~
	0.6,\hspace{0.1cm}H_2 = -0.2,\hspace{0.1cm}H_3 = -0.4 + {\rm{j}}
	\cdot 0.6 $$:$$\Rightarrow \hspace{0.3cm}{\rm\bf{R}} = \left( {R_0 ,R_1 ,R_2 ,R_3
	} \right)= \left( \hspace{0.15cm}0,\hspace{0.15cm}0,\hspace{0.15cm}-0.2, \hspace{0.15cm}0 \right) \hspace{0.3cm}
	\Rightarrow \hspace{0.3cm}{\rm Re}[r_2]\hspace{0.15cm} \underline{=-0.2},\hspace{0.2cm} {\rm Im}[r_2]\hspace{0.15cm} \underline{=0}, \hspace{0.2cm}{\rm Re}[r_3]\hspace{0.15cm} \underline{=0},\hspace{0.2cm} {\rm Im}[r_3]\hspace{0.15cm} \underline{=0}.$$

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	'''(1)'''  The discrete time domain values at the receiver are calculated using the cyclic transmission matrix  ${\rm\bf{H}}_{\rm{C}} $  as follows:		'''(1)'''  The discrete time domain values at the receiver are calculated using the cyclic transmission matrix  ${\rm\bf{H}}_{\rm{C}} $  as follows:
	:$${\rm\bf{r}} = {\rm\bf{d}} \cdot {\rm\bf{H}}_{\rm{C}} = \left( {+1 ,-1 ,+1 ,-1 } \right) \cdot \left( {\begin{array}{*{20}c} {0 } & {0.6 } & {0.4 } & {} \\ {} & {0 } & {0.6 } & {0.4 } \\ \hline {0.4 } & {} & {0 } & {0.6 } \\ {0.6 } & {0.4 } & {} & {0 } \\ \end{array}} \right)$$		:$${\rm\bf{r}} = {\rm\bf{d}} \cdot {\rm\bf{H}}_{\rm{C}} = \left( {+1 ,-1 ,+1 ,-1 } \right) \cdot \left( {\begin{array}{*{20}c} {0 } & {0.6 } & {0.4 } & {} \\ {} & {0 } & {0.6 } & {0.4 } \\ \hline {0.4 } & {} & {0 } & {0.6 } \\ {0.6 } & {0.4 } & {} & {0 } \\ \end{array}} \right)$$
	:$$\Rightarrow \hspace{0.3cm}{\rm\bf{r}} = \left( {r_0 ,r_1 ,r_2 ,r_3 } \right) = \left( {-0.2, +0.2,-0.2, +0.2} \right) \hspace{0.3cm}		:$$\Rightarrow \hspace{0.3cm}{\rm\bf{r}} = \left( {r_0 ,r_1 ,r_2 ,r_3 } \right) = \left( {-0.2, +0.2,-0.2, +0.2} \right) \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Re}[r_0]\hspace{0.15cm} \underline{=-0.2},\hspace{0.2cm} {\rm Im}[r_0]\hspace{0.15cm} \underline{=0}, \hspace{0.2cm}{\rm Re}[r_1]\hspace{0.15cm} \underline{=+0.2},\hspace{0.2cm} {\rm Im}[r_1]\hspace{0.15cm} \underline{=0}. $$
	\Rightarrow \hspace{0.3cm}{\rm Re}[r_0]\hspace{0.15cm} \underline{=-0.2},\hspace{0.2cm} {\rm Im}[r_0]\hspace{0.15cm} \underline{=0}, \hspace{0.2cm}{\rm Re}[r_1]\hspace{0.15cm} \underline{=+0.2},\hspace{0.2cm} {\rm Im}[r_1]\hspace{0.15cm} \underline{=0}. $$


	'''(2)'''  The spectral coefficients  ${\rm\bf{D}}$  result directly from the Discrete Fourier Transform  $\rm (DFT)$  of the time domain coefficients  ${\rm\bf{d}}= (+1, -1, +1, -1)$.		'''(2)'''  The spectral coefficients  ${\rm\bf{D}}$  result directly from the Discrete Fourier Transform  $\rm (DFT)$  of the time domain coefficients  ${\rm\bf{d}}= (+1, -1, +1, -1)$.
	*This time domain sequence corresponds to a discrete cosine function with twice the fundamental frequency  $(2 \cdot f_0)$  and amplitude  $1$.  It follows:		*This time domain sequence corresponds to a discrete cosine function with twice the fundamental frequency  $(2 \cdot f_0)$  and amplitude  $1$.  It follows:
	:$${\rm\bf{D}} = \left( {D_0 ,D_1 ,D_2 ,D_3 } \right) =\left( {0, 0,1, 0} \right)\hspace{0.3cm}		:$${\rm\bf{D}} = \left( {D_0 ,D_1 ,D_2 ,D_3 } \right) =\left( {0, 0,1, 0} \right)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Re}[d_2]\hspace{0.15cm} \underline{=1},\hspace{0.2cm} {\rm Im}[d_2]\hspace{0.15cm} \underline{=0}, \hspace{0.2cm}{\rm Re}[d_3]\hspace{0.15cm} \underline{=0},\hspace{0.2cm} {\rm Im}[d_3]\hspace{0.15cm} \underline{=0}.$$
	\Rightarrow \hspace{0.3cm}{\rm Re}[d_2]\hspace{0.15cm} \underline{=1},\hspace{0.2cm} {\rm Im}[d_2]\hspace{0.15cm} \underline{=0}, \hspace{0.2cm}{\rm Re}[d_3]\hspace{0.15cm} \underline{=0},\hspace{0.2cm} {\rm Im}[d_3]\hspace{0.15cm} \underline{=0}.$$


	'''(3)'''  The vector  ${\rm\bf{R}}$  of spectral coefficients after the channel could be calculated by DFT of the vector   ${\rm\bf{r}}$,  analogous to subtask  '''(2)'''.		'''(3)'''  The vector  ${\rm\bf{R}}$  of spectral coefficients after the channel could be calculated by DFT of the vector   ${\rm\bf{r}}$,  analogous to subtask  '''(2)'''.
	*An alternative solution path is:		*An alternative solution path is:
	:$${\rm\bf{R}} = {\rm\bf{D}} \cdot \left( {\begin{array}{*{20}c}		:$${\rm\bf{R}} = {\rm\bf{D}} \cdot \left( {\begin{array}{*{20}c}{H_0 } & {} & {} & {} \\{} & {H_1 } & {} & {} \\{} & {} & {H_2 } & {} \\{} & {} & {} & {H_3 } \\\end{array}} \right) .$$
	{H_0 } & {} & {} & {} \\
	{} & {H_1 } & {} & {} \\
	{} & {} & {H_2 } & {} \\
	{} & {} & {} & {H_3 } \\
	\end{array}} \right) .$$
	*For the diagonal elements one obtains:		*For the diagonal elements one obtains:
	:$$H_\mu = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e}}^{ -		:$$H_\mu = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e}}^{ -
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	\Rightarrow \hspace{0.3cm} H_0 = 1,\hspace{0.1cm}H_1 = -0.4 - {\rm{j}} \cdot		\Rightarrow \hspace{0.3cm} H_0 = 1,\hspace{0.1cm}H_1 = -0.4 - {\rm{j}} \cdot
	0.6,\hspace{0.1cm}H_2 = -0.2,\hspace{0.1cm}H_3 = -0.4 + {\rm{j}}		0.6,\hspace{0.1cm}H_2 = -0.2,\hspace{0.1cm}H_3 = -0.4 + {\rm{j}}
	\cdot 0.6 $$		\cdot 0.6 $$:$$\Rightarrow \hspace{0.3cm}{\rm\bf{R}} = \left( {R_0 ,R_1 ,R_2 ,R_3
	:$$\Rightarrow \hspace{0.3cm}{\rm\bf{R}} = \left( {R_0 ,R_1 ,R_2 ,R_3
	} \right)= \left( \hspace{0.15cm}0,\hspace{0.15cm}0,\hspace{0.15cm}-0.2, \hspace{0.15cm}0 \right) \hspace{0.3cm}		} \right)= \left( \hspace{0.15cm}0,\hspace{0.15cm}0,\hspace{0.15cm}-0.2, \hspace{0.15cm}0 \right) \hspace{0.3cm}
	\Rightarrow \hspace{0.3cm}{\rm Re}[r_2]\hspace{0.15cm} \underline{=-0.2},\hspace{0.2cm} {\rm Im}[r_2]\hspace{0.15cm} \underline{=0}, \hspace{0.2cm}{\rm Re}[r_3]\hspace{0.15cm} \underline{=0},\hspace{0.2cm} {\rm Im}[r_3]\hspace{0.15cm} \underline{=0}.$$		\Rightarrow \hspace{0.3cm}{\rm Re}[r_2]\hspace{0.15cm} \underline{=-0.2},\hspace{0.2cm} {\rm Im}[r_2]\hspace{0.15cm} \underline{=0}, \hspace{0.2cm}{\rm Re}[r_3]\hspace{0.15cm} \underline{=0},\hspace{0.2cm} {\rm Im}[r_3]\hspace{0.15cm} \underline{=0}.$$

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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:5.8_Entzerrung_in_Matrix–Vektor–Notation]]

Guenter at 14:50, 31 January 2022

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	}}		}}

	[[File:~~P_ID1663__A_5_8~~.png\|right\|frame\|Block diagram of OFDM transmission ~~'''KORREKTUR''': cyclic prefix ...~~]]		[[File:EN_A_5_8.png\|right\|frame\|Block diagram of OFDM transmission]]
	We consider the blocks of an OFDM system shown in the diagram,  assuming a system with  $N = 4$  carriers and a channel with  $L = 2$  echoes.		We consider the blocks of an OFDM system shown in the diagram,  assuming a system with  $N = 4$  carriers and a channel with  $L = 2$  echoes.
	*Only a single frame is considered and for the transmission vector  (in the time domain),  we apply:		*Only a single frame is considered and for the transmission vector  (in the time domain),  we apply:

Guenter at 16:49, 24 January 2022

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	'''(3)'''  The vector  ${\rm\bf{R}}$  of spectral coefficients ~~downstream of~~ the channel could be calculated by DFT of the vector   ${\rm\bf{r}}$ , analogous to subtask  '''(2)'''.		'''(3)'''  The vector  ${\rm\bf{R}}$  of spectral coefficients after the channel could be calculated by DFT of the vector   ${\rm\bf{r}}$,  analogous to subtask  '''(2)'''.
	*An alternative solution path is:		*An alternative solution path is:
	:$${\rm\bf{R}} = {\rm\bf{D}} \cdot \left( {\begin{array}{*{20}c}		:$${\rm\bf{R}} = {\rm\bf{D}} \cdot \left( {\begin{array}{*{20}c}
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	'''(5)'''  The equalization calculated in  '''(4)'''  follows the "zero forcing" approach   ⇒   <u>solution 1</u>.		'''(5)'''  The equalization calculated in  '''(4)'''  follows the  "zero forcing"  approach   ⇒   <u>solution 1</u>.

	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 16:37, 24 January 2022

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	[[File:P_ID1663__A_5_8.png\|right\|frame\|Block diagram of OFDM transmission '''KORREKTUR''': cyclic prefix ...]]		[[File:P_ID1663__A_5_8.png\|right\|frame\|Block diagram of OFDM transmission '''KORREKTUR''': cyclic prefix ...]]
	We consider the blocks of an OFDM system shown in the diagram, assuming a system with  $N = 4$  carriers and a channel with  $L = 2$  echoes.		We consider the blocks of an OFDM system shown in the diagram,  assuming a system with  $N = 4$  carriers and a channel with  $L = 2$  echoes.
	*Only a single frame is considered and for the transmission vector (in the time domain), we apply:		*Only a single frame is considered and for the transmission vector  (in the time domain),  we apply:
	:$${\rm\bf{d}} = (d_0, \ d_1,\ d_2,\ d_3 ) = (+1, -1, +1, -1 ).$$		:$${\rm\bf{d}} = (d_0, \ d_1,\ d_2,\ d_3 ) = (+1, -1, +1, -1 ).$$
	*Let the channel impulse response be described by		*Let the channel impulse response be described by
	:$${\rm\bf{h}} = (h_0, \ h_1,\ h_2 ) = (0, \ 0.6, \ 0.4 ).$$		:$${\rm\bf{h}} = (h_0, \ h_1,\ h_2 ) = (0, \ 0.6, \ 0.4 ).$$
	*To represent the cyclic prefix in this task, instead of the extended ~~transmit~~ vector with the associated transmission matrix  ${\rm\bf{H}}_{\rm ext}$,  we use the cyclic transmission matrix		*To represent the cyclic prefix in this task,  instead of the extended transmission vector with the associated transmission matrix  ${\rm\bf{H}}_{\rm ext}$,  we use the cyclic transmission matrix
	:$${\rm\bf{H}}_{\rm{C}} = \left( {\begin{array}{*{20}c} {h_0 } & {h_1 } & {h_2 } & {} \\ {} & {h_0 } & {h_1 } & {h_2 } \\ \hline {h_2 } & {} & {h_0 } & {h_1 } \\ {h_1 } & {h_2 } & {} & {h_0 } \\ \end{array}} \right).$$		:$${\rm\bf{H}}_{\rm{C}} = \left( {\begin{array}{*{20}c} {h_0 } & {h_1 } & {h_2 } & {} \\ {} & {h_0 } & {h_1 } & {h_2 } \\ \hline {h_2 } & {} & {h_0 } & {h_1 } \\ {h_1 } & {h_2 } & {} & {h_0 } \\ \end{array}} \right).$$
	*For the spectral coefficients at the receiver, according to the Discrete Fourier Transform  $\rm (DFT)$:		*For the spectral coefficients at the receiver,  according to the Discrete Fourier Transform  $\rm (DFT)$:
	:$${\rm\bf{R}} = {\rm\bf{D}} \cdot \left( {\begin{array}{*{20}c} {H_0 } & {} & {} & {} \\ {} & {H_1 } & {} & {} \\ {} & {} & {H_2 } & {} \\ {} & {} & {} & {H_3 } \\ \end{array}} \right) ,$$		:$${\rm\bf{R}} = {\rm\bf{D}} \cdot \left( {\begin{array}{*{20}c} {H_0 } & {} & {} & {} \\ {} & {H_1 } & {} & {} \\ {} & {} & {H_2 } & {} \\ {} & {} & {} & {H_3 } \\ \end{array}} \right) ,$$
	:where the diagonal elements are to be calculated as follows:		:where the diagonal elements are to be calculated as follows:
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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".

	*For the Discrete Fourier Transform (DFT) in matrix-vector notation holds:		*For the Discrete Fourier Transform (DFT) in matrix-vector notation holds:
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	<quiz display=simple>		<quiz display=simple>
	{Calculate the discrete reception values  $r = (r_0, r_1, r_2, r_3)$  in the time domain.  Enter  $r_0$  and  $r_1$  as a check:		{Calculate the discrete reception values  $r = (r_0, r_1, r_2, r_3)$  in the time domain.  Enter  $r_0$  and  $r_1$  as a check.
	\|type="{}"}		\|type="{}"}
	${\rm Re}\big[r_0\big] \ = \ $ { -0.202--0.198 }		${\rm Re}\big[r_0\big] \ = \ $ { -0.202--0.198 }
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	${\rm Im}\big[D_3\big] \ = \ $ { 0. }		${\rm Im}\big[D_3\big] \ = \ $ { 0. }

	{Calculate the discrete spectral coefficients  ${\rm\bf{R}}= (R_0, R_1, R_2, R_3)$  according to the channel.  Enter  $R_2$  and  $R_3$  as a check:		{Calculate the discrete spectral coefficients  ${\rm\bf{R}}= (R_0, R_1, R_2, R_3)$  according to the channel.  Enter  $R_2$  and  $R_3$  as a check.
	\|type="{}"}		\|type="{}"}
	${\rm Re}\big[R_2\big] \ = \ $ { -0.202--0.198 }		${\rm Re}\big[R_2\big] \ = \ $ { -0.202--0.198 }
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	${\rm Im}\big[R_3\big] \ = \ $ { 0. }		${\rm Im}\big[R_3\big] \ = \ $ { 0. }

	{ Determine the discrete equalizer coefficients  ${\rm\bf{e}}= (e_0, e_1, e_2, e_3)$:		{ Determine the discrete equalizer coefficients  ${\rm\bf{e}}= (e_0, e_1, e_2, e_3)$.
	\|type="{}"}		\|type="{}"}
	${\rm Re}\big[e_0\big] \ = \ $ { 1 1% }		${\rm Re}\big[e_0\big] \ = \ $ { 1 1% }
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	\|type="()"}		\|type="()"}
	+ As "Zero Forcing" approach,		+ As "Zero Forcing" approach,
	- as "~~matched filter~~" approach,		- as "Matched Filter" approach,
	- as "Minimum Mean Square Error  $\rm (MMSE)$”approach.		- as "Minimum Mean Square Error  $\rm (MMSE)$”approach.
	</quiz>		</quiz>

Hwang at 10:26, 29 December 2021

2021-12-29T10:26:40Z

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	{{quiz-Header\|Buchseite=~~Modulationsverfahren~~/~~Realisierung von OFDM-Systemen~~		{{quiz-Header\|Buchseite=Modulation_Methods/Implementation_of_OFDM_Systems
	}}		}}

Hwang at 10:24, 29 December 2021

2021-12-29T10:24:04Z

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Guenter at 15:59, 14 December 2021

2021-12-14T15:59:19Z

← Older revision		Revision as of 17:59, 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^]]