Modulation Methods/Implementation of OFDM Systems - Revision history

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 '''(1)''' &nbsp; To represent the cyclic prefix,&nbsp; we use the cyclic transmission matrix  &nbsp;${\rm\bf{H} }_{\rm{C} }$,&nbsp; instead of the extended transmitted vector with the corresponding transmission matrix,&nbsp; resulting in the received vector &nbsp;${\rm \bf r}=  {\rm \bf d} \cdot {\rm \bf H}_{\rm{C} }$:&nbsp;
-:$${\rm\bf{H} }_{\rm{C} }  = \left( {\begin{array}{*{20}c}
+:$${\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),  \hspace{1cm}{\rm\bf{r} } = \left( {r_0 ,r_1 ,r_2 ,r_3 } \right) = \left( {d_0 ,d_1 ,d_2 ,d_3 } \right) \cdot \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) $$:$$\Rightarrow \hspace{0.3cm} r_0 = d_0 \cdot h_0 + d_2 \cdot h_2 + d_3 \cdot h_1, \hspace{0.5cm}r_1 = d_0 \cdot h_1 + d_1 \cdot h_0 + d_3 \cdot h_2,$$:$$\Rightarrow \hspace{0.3cm} r_2 = d_0 \cdot h_2 + d_1 \cdot h_1 + d_2 \cdot h_0, \hspace{0.5cm}r_3 = d_1 \cdot h_2 + d_2 \cdot h_1 + d_3 \cdot h_0.$$'''(2)''' &nbsp;  The&nbsp; (discrete)&nbsp; Fourier transform of the received  vector is calculated to be:$${\rm\bf{R} } = \frac{1}{N} \cdot {\rm\bf{r} } \cdot {\rm\bf{F} } = {\rm\bf{D} } \cdot \left( {\begin{array}{*{20}c}{H_0 } & {} & {} & {}  \\{} & {H_1 } & {} & {}  \\{} & {} & {H_2 } & {}  \\{} & {} & {} & {H_3 }  \\\end{array} } \right) ,\hspace{0.25cm} {\rm mit}\hspace{0.25cm} H_\mu = \sum\limits_{l = 0}^2 {h_l \cdot{\rm{e} }^{ - {\rm{j \hspace{0.05cm}\cdot \hspace{0.03cm} 2\pi } }\hspace{0.05cm}\cdot \hspace{0.03cm}l \hspace{0.05cm}\cdot \hspace{0.03cm} \mu /4} }  .$$'''(3)'''  &nbsp; For numerical calculations,&nbsp; we assume a known BPSK-encoded transmitted  sequence &nbsp;$\rm \bf D$&nbsp; (in the frequency domain)&nbsp; and the following channel impulse response &nbsp;$\bf h$:&nbsp;:$${\rm\bf{D} } = \frac{1}{N} \cdot {\rm\bf{d} } \cdot {\rm\bf{F} } =\left( D_0, D_1,D_2,D_3\right) = \left( +1,\ -1,\ +1,\ -1\right),$$:$${\rm\bf{h} }= \left( h_0, h_1,h_2\right) = \left(0.5,\ 0.3,\ 0.2\right).$$'''(4)'''  &nbsp; First,&nbsp; we determine the elements &nbsp;$H_{\mu}$&nbsp; of the diagonal matrix::$$\begin{array}{l}H_0 = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e} }^0 = 0.5 + 0.3 + 0.2 = 1,} \\H_1 = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e} }^{ - {\rm{j\hspace{0.05cm}\cdot \hspace{0.03cm} 2\pi } }\hspace{0.05cm}\cdot \hspace{0.03cm} l \hspace{0.05cm}\cdot \hspace{0.03cm} {1}/{4} } } = 0.5 \cdot {\rm{e} }^0 + 0.3 \cdot {\rm{e} }^{ - {\rm{j\hspace{0.05cm}\cdot \hspace{0.03cm}\pi } } /2 } +0.2 \cdot {\rm{e} }^{ - {\rm{j\hspace{0.05cm}\cdot \hspace{0.03cm}\pi } } } = 0.3 - {\rm{j} } \cdot 0.3, \\H_2 = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e} }^{ - {\rm{j\hspace{0.05cm}\cdot \hspace{0.03cm} 2\pi } }\hspace{0.05cm}\cdot \hspace{0.03cm} l \hspace{0.05cm}\cdot \hspace{0.03cm} {2}/{4} } } = 0.5 \cdot {\rm{e} }^0 + 0.3 \cdot {\rm{e} }^{ - {\rm{j\hspace{0.05cm}\cdot \hspace{0.03cm}\pi } } }+ 0.2 \cdot {\rm{e} }^{ - {\rm{j\hspace{0.05cm}\cdot \hspace{0.03cm}2\pi } } } = 0.4, \\H_3 = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e} }^{ - {\rm{j\hspace{0.05cm}\cdot \hspace{0.03cm} 2\pi } }\hspace{0.05cm}\cdot \hspace{0.03cm} l \hspace{0.05cm}\cdot \hspace{0.03cm} {3}/{4} } } = 0.5 \cdot {\rm{e} }^0 + 0.3 \cdot {\rm{e} }^{ - {\rm{j\hspace{0.05cm}\cdot \hspace{0.03cm} {3}/{2} \pi } } }+ 0.2 \cdot {\rm{e} }^{ - {\rm{j\hspace{0.05cm}\cdot \hspace{0.03cm}3\pi } } } = 0.3 + {\rm{j} } \cdot 0.3. \\\end{array}$$'''(5)'''  &nbsp; Thus,&nbsp; the vector of frequency support points at the receiver is given by:$$\begin{align*}{\rm\bf{R} }  &=  \left( {\rm{1, -1, \; \; 1, -1} } \right) \cdot \left( {\begin{array}{*{20}c}1 & {} & {} & {}  \\{} & {0.3 - {\rm{j} } \cdot 0.3} & {} & {}  \\{} & {} & {0.4} & {}  \\{} & {} & {} & {0.3 + {\rm{j} } \cdot 0.3}  \\\end{array} } \right) \ = \  {\rm{ (1, -0.3 + j \cdot 0.3, \; \; 0.4, -0.3 - j \cdot 0.3) } }.\end{align*}$$
 '''(6)'''  &nbsp; One chooses the equalizer coefficients according to &nbsp;$e_{\mu} = 1/H_{\mu}$,&nbsp; where &nbsp;$\mu = 0$, ... , $3$ &nbsp; holds:

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 Alternatively,&nbsp; one can use the &nbsp;"cyclic matrix" &nbsp;$\rm \bf  H_C$&nbsp; with &nbsp;$N$&nbsp; rows and &nbsp;$N$&nbsp; columns as well as the &nbsp;"Fourier transform &nbsp;$\rm \bf  F$&nbsp; in matrix–vector notation":&nbsp;
-:$${\rm\bf{H}}_{\rm{C}}  = \left( {\begin{array}{*{20}c}
+:$${\rm\bf{H}}_{\rm{C}}  = \left( {\begin{array}{*{20}c}{h_0 } & {h_1 } &  \cdots  &  \cdots  & {h_L } & {} & {} & {}  \\{} & {h_0 } & {h_1 } &  \cdots  &  \cdots  & {h_L } & {} & {}  \\{} & {} &  \ddots  &  \ddots  & {} & {} &  \ddots  & {}  \\{} & {} & {} & {h_0 } & {h_1 } &  \cdots  &  \cdots  & {h_L }  \\\hline{h_L } & {} & {} & {} & {h_0 } & {h_1 } &  \cdots  & {h_{L - 1} }  \\\vdots  &  \ddots  & {} & {} & {} &  \ddots  & {} &  \vdots   \\\vdots  & {} &  \ddots  & {} & {} & {} &  \ddots  &  \vdots   \\{h_1 } &  \cdots  &  \cdots  & {h_L } & {} & {} & {} & {h_0 }  \\\end{array}} \right), \hspace{1cm}  {\rm\bf{F}} = \left( {\begin{array}{*{20}c}1 & 1 &  \cdots  & 1  \\1 & {} & {} & {}  \\\vdots  & {} & {{\rm{e}}^{ - {\rm{j \hspace{0.05cm}\cdot \hspace{0.03cm} 2\pi }}{\kern 1pt}\cdot \hspace{0.02cm}\nu {\kern 1pt} \cdot\mu /N} } & {}  \\1 & {} & {} & {}  \\\end{array}} \right) .$$*The Discrete Fourier Transform&nbsp; $\rm (DFT)$&nbsp; can be represented by &nbsp;$1/N · \bf F$&nbsp; and its inverse&nbsp; $\rm (IDFT)$&nbsp; by &nbsp;$\rm \bf F^{\star}$ such that for the transmitted  vector: &nbsp;$\rm {\bf d} = {\bf D} · {\bf F}^{\star}$.*The &nbsp;$N$&nbsp; spectral coefficients are described by the vector &nbsp;${\bf D} = 1/N · {\bf d} · {\bf F}$&nbsp; and the  received vector is &nbsp;${\bf r} = {\bf d} · {\bf H}_{\rm C} = {\bf D} · {\bf F}^{\star} · {\bf H}_{\rm C}$.*The (discrete) Fourier transform &nbsp;$\rm \bf R$&nbsp; of the received vector &nbsp;$\rm \bf r$&nbsp; can then be written in the following way::$${\rm\bf{R}} = \frac{1}{N} \cdot {\rm\bf{r}} \cdot {\rm\bf{F}} = {\rm\bf{D}} \cdot \left( {\begin{array}{*{20}c}{H_0 } & {} & {} & {}  \\{} & {H_1 } & {} & {}  \\{} & {} &  \ddots  & {}  \\{} & {} & {} & {H_{N - 1} }  \\\end{array}} \right),\hspace{0.25cm} {\rm with}\hspace{0.25cm}  H_\mu = \sum\limits_{l = 0}^L {h_l \cdot{\rm{e}}^{ - {\rm{j \hspace{0.05cm}\cdot \hspace{0.03cm} 2\pi }}\hspace{0.05cm}\cdot \hspace{0.03cm} l \hspace{0.05cm}\cdot \hspace{0.03cm}\mu /N} }.$$
 [[File:EN_Mod_T_5_6_S8b.png|right|frame| Block diagram of the OFDM receiver]]

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	\cdot \hspace{0.02cm}\nu {\kern 1pt} \cdot\mu /N} } & {} \\		\cdot \hspace{0.02cm}\nu {\kern 1pt} \cdot\mu /N} } & {} \\
	1 & {} & {} & {} \\		1 & {} & {} & {} \\
	\end{array}} \right) .$$		\end{array}} \right) .$$The Discrete Fourier Transform  $\rm (DFT)$  can be represented by  $1/N · \bf F$  and its inverse  $\rm (IDFT)$  by  $\rm \bf F^{\star}$ such that for the transmitted vector:  $\rm {\bf d} = {\bf D} · {\bf F}^{\star}$.The  $N$  spectral coefficients are described by the vector  ${\bf D} = 1/N · {\bf d} · {\bf F}$  and the received vector is  ${\bf r} = {\bf d} · {\bf H}_{\rm C} = {\bf D} · {\bf F}^{\star} · {\bf H}_{\rm C}$.The (discrete) Fourier transform  $\rm \bf R$  of the received vector  $\rm \bf r$  can then be written in the following way::$${\rm\bf{R}} = \frac{1}{N} \cdot {\rm\bf{r}} \cdot {\rm\bf{F}} = {\rm\bf{D}} \cdot \left( {\begin{array}{{20}c}

	*The Discrete Fourier Transform  $\rm (DFT)$  can be represented by  $1/N · \bf F$  and its inverse  $\rm (IDFT)$  by  $\rm \bf F^{\star}$ such that for the transmitted vector:  $\rm {\bf d} = {\bf D} · {\bf F}^{\star}$.

	*The  $N$  spectral coefficients are described by the vector  ${\bf D} = 1/N · {\bf d} · {\bf F}$  and the received vector is  ${\bf r} = {\bf d} · {\bf H}_{\rm C} = {\bf D} · {\bf F}^{\star} · {\bf H}_{\rm C}$.

	*The (discrete) Fourier transform  $\rm \bf R$  of the received vector  $\rm \bf r$  can then be written in the following way:
	:$${\rm\bf{R}} = \frac{1}{N} \cdot {\rm\bf{r}} \cdot {\rm\bf{F}} = {\rm\bf{D}} \cdot \left( {\begin{array}{*{20}c}
	{H_0 } & {} & {} & {} \\		{H_0 } & {} & {} & {} \\
	{} & {H_1 } & {} & {} \\		{} & {H_1 } & {} & {} \\

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	{h_2 } & {} & {h_0 } & {h_1 } \\		{h_2 } & {} & {h_0 } & {h_1 } \\
	{h_1 } & {h_2 } & {} & {h_0 } \\		{h_1 } & {h_2 } & {} & {h_0 } \\
	\end{array} } \right) $$		\end{array} } \right) $$:$$\Rightarrow \hspace{0.3cm} r_0 = d_0 \cdot h_0 + d_2 \cdot h_2 + d_3 \cdot h_1, \hspace{0.5cm}
	:$$\Rightarrow \hspace{0.3cm} r_0 = d_0 \cdot h_0 + d_2 \cdot h_2 + d_3 \cdot h_1, \hspace{0.5cm}		r_1 = d_0 \cdot h_1 + d_1 \cdot h_0 + d_3 \cdot h_2,$$:$$\Rightarrow \hspace{0.3cm} r_2 = d_0 \cdot h_2 + d_1 \cdot h_1 + d_2 \cdot h_0, \hspace{0.5cm}
	r_1 = d_0 \cdot h_1 + d_1 \cdot h_0 + d_3 \cdot h_2,$$		r_3 = d_1 \cdot h_2 + d_2 \cdot h_1 + d_3 \cdot h_0.$$'''(2)'''   The  (discrete)  Fourier transform of the received vector is calculated to be:$${\rm\bf{R} } = \frac{1}{N} \cdot {\rm\bf{r} } \cdot {\rm\bf{F} } = {\rm\bf{D} } \cdot \left( {\begin{array}{*{20}c}
	:$$\Rightarrow \hspace{0.3cm} r_2 = d_0 \cdot h_2 + d_1 \cdot h_1 + d_2 \cdot h_0, \hspace{0.5cm}
	r_3 = d_1 \cdot h_2 + d_2 \cdot h_1 + d_3 \cdot h_0.$$

	'''(2)'''   The  (discrete)  Fourier transform of the received vector is calculated to be
	:$${\rm\bf{R} } = \frac{1}{N} \cdot {\rm\bf{r} } \cdot {\rm\bf{F} } = {\rm\bf{D} } \cdot \left( {\begin{array}{*{20}c}
	{H_0 } & {} & {} & {} \\		{H_0 } & {} & {} & {} \\
	{} & {H_1 } & {} & {} \\		{} & {H_1 } & {} & {} \\
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	{} & {} & {} & {H_3 } \\		{} & {} & {} & {H_3 } \\
	\end{array} } \right) ,\hspace{0.25cm} {\rm mit}\hspace{0.25cm} H_\mu = \sum\limits_{l = 0}^2 {h_l \cdot		\end{array} } \right) ,\hspace{0.25cm} {\rm mit}\hspace{0.25cm} H_\mu = \sum\limits_{l = 0}^2 {h_l \cdot
	{\rm{e} }^{ - {\rm{j \hspace{0.05cm}\cdot \hspace{0.03cm} 2\pi } }\hspace{0.05cm}\cdot \hspace{0.03cm}l \hspace{0.05cm}\cdot \hspace{0.03cm} \mu /4} } .$$		{\rm{e} }^{ - {\rm{j \hspace{0.05cm}\cdot \hspace{0.03cm} 2\pi } }\hspace{0.05cm}\cdot \hspace{0.03cm}l \hspace{0.05cm}\cdot \hspace{0.03cm} \mu /4} } .$$'''(3)'''   For numerical calculations,  we assume a known BPSK-encoded transmitted sequence  $\rm \bf D$  (in the frequency domain)  and the following channel impulse response  $\bf h$: :$${\rm\bf{D} } = \frac{1}{N} \cdot {\rm\bf{d} } \cdot {\rm\bf{F} } =
			\left( D_0, D_1,D_2,D_3\right) = \left( +1,\ -1,\ +1,\ -1\right),$$:$$
	'''(3)'''   For numerical calculations,  we assume a known BPSK-encoded transmitted sequence  $\rm \bf D$  (in the frequency domain)  and the following channel impulse response  $\bf h$:
	:$${\rm\bf{D} } = \frac{1}{N} \cdot {\rm\bf{d} } \cdot {\rm\bf{F} } =
	\left( D_0, D_1,D_2,D_3\right) = \left( +1,\ -1,\ +1,\ -1\right),$$
	:$$
	{\rm\bf{h} }= \left( h_0, h_1,h_2\right) = \left(		{\rm\bf{h} }= \left( h_0, h_1,h_2\right) = \left(
	0.5,\ 0.3,\ 0.2\right).$$		0.5,\ 0.3,\ 0.2\right).$$'''(4)'''   First,  we determine the elements  $H_{\mu}$  of the diagonal matrix::$$\begin{array}{l}

	'''(4)'''   First,  we determine the elements  $H_{\mu}$  of the diagonal matrix:
	:$$\begin{array}{l}
	H_0 = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e} }^0 = 0.5 + 0.3 + 0.2 = 1,} \\		H_0 = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e} }^0 = 0.5 + 0.3 + 0.2 = 1,} \\
	H_1 = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e} }^{ - {\rm{j\hspace{0.05cm}\cdot \hspace{0.03cm} 2\pi } }\hspace{0.05cm}\cdot \hspace{0.03cm} l \hspace{0.05cm}\cdot \hspace{0.03cm} {1}/{4} } } = 0.5 \cdot {\rm{e} }^0 + 0.3 \cdot {\rm{e} }^{ - {\rm{j\hspace{0.05cm}\cdot \hspace{0.03cm}\pi } } /2 } +		H_1 = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e} }^{ - {\rm{j\hspace{0.05cm}\cdot \hspace{0.03cm} 2\pi } }\hspace{0.05cm}\cdot \hspace{0.03cm} l \hspace{0.05cm}\cdot \hspace{0.03cm} {1}/{4} } } = 0.5 \cdot {\rm{e} }^0 + 0.3 \cdot {\rm{e} }^{ - {\rm{j\hspace{0.05cm}\cdot \hspace{0.03cm}\pi } } /2 } +
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	H_3 = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e} }^{ - {\rm{j\hspace{0.05cm}\cdot \hspace{0.03cm} 2\pi } }\hspace{0.05cm}\cdot \hspace{0.03cm} l \hspace{0.05cm}\cdot \hspace{0.03cm} {3}/{4} } } = 0.5 \cdot {\rm{e} }^0 + 0.3 \cdot {\rm{e} }^{ - {\rm{j\hspace{0.05cm}\cdot \hspace{0.03cm} {3}/{2} \pi } } }		H_3 = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e} }^{ - {\rm{j\hspace{0.05cm}\cdot \hspace{0.03cm} 2\pi } }\hspace{0.05cm}\cdot \hspace{0.03cm} l \hspace{0.05cm}\cdot \hspace{0.03cm} {3}/{4} } } = 0.5 \cdot {\rm{e} }^0 + 0.3 \cdot {\rm{e} }^{ - {\rm{j\hspace{0.05cm}\cdot \hspace{0.03cm} {3}/{2} \pi } } }
	+ 0.2 \cdot {\rm{e} }^{ - {\rm{j\hspace{0.05cm}\cdot \hspace{0.03cm}3\pi } } } = 0.3 + {\rm{j} } \cdot 0.3. \\		+ 0.2 \cdot {\rm{e} }^{ - {\rm{j\hspace{0.05cm}\cdot \hspace{0.03cm}3\pi } } } = 0.3 + {\rm{j} } \cdot 0.3. \\
	\end{array}$$		\end{array}$$'''(5)'''   Thus,  the vector of frequency support points at the receiver is given by:$$\begin{align}{\rm\bf{R} } &= \left( {\rm{1, -1, \; \; 1, -1} } \right) \cdot \left( {\begin{array}{{20}c}
	'''(5)'''   Thus,  the vector of frequency support points at the receiver is given by
	:$$\begin{align}{\rm\bf{R} } &= \left( {\rm{1, -1, \; \; 1, -1} } \right) \cdot \left( {\begin{array}{{20}c}
	1 & {} & {} & {} \\		1 & {} & {} & {} \\
	{} & {0.3 - {\rm{j} } \cdot 0.3} & {} & {} \\		{} & {0.3 - {\rm{j} } \cdot 0.3} & {} & {} \\
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	'''(7)'''   The rectified symbol sequence with  ${\bf e} = (e_0, e_1, e_2, e_3)$  finally results in		'''(7)'''   The rectified symbol sequence with  ${\bf e} = (e_0, e_1, e_2, e_3)$  finally results in
	:$$\hat {\rm\bf{D} } = {\rm\bf{R} } \cdot {\rm\bf{e} }^{\rm{T} } = (R_0 ,R_1 ,R_2 ,R_3) \cdot \left( {\begin{array}{*{20}c}		:$$\hat {\rm\bf{D} } = {\rm\bf{R} } \cdot {\rm\bf{e} }^{\rm{T} } = (R_0 ,R_1 ,R_2 ,R_3) \cdot \left( {\begin{array}{*{20}c}{e_0 } \\{e_1 } \\{e_2 } \\{e_3 } \\\end{array}} \right) = \left( +1, -1, \; +1, -1 \right).$$
	{e_0 } \\
	{e_1 } \\
	{e_2 } \\
	{e_3 } \\
	\end{array}} \right) = \left( +1, -1, \; +1, -1 \right).$$
	⇒   This corresponds exactly to the transmitted symbol sequence  $\bf D$.  That is: <br>		⇒   This corresponds exactly to the transmitted symbol sequence  $\bf D$.  That is: <br>

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	{{Display}}		{{Display}}
			[[de:Modulationsverfahren/Realisierung_von_OFDM-Systemen]]

Noah at 16:05, 11 March 2023

2023-03-11T16:05:26Z

Guenter at 16:33, 23 January 2023

2023-01-23T16:33:27Z

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	$\text{Example 2:}$		$\text{Example 2:}$
	This diagram again shows the real part of the received OFDM signal,  but now with  "guard gap".  The assumptions of  [[Modulation_Methods/Implementation_of_OFDM_Systems#Intercarrier_interference_and_intersymbol_interference\|$\text{Example 1}$]]  have been kept.		This diagram again shows the real part of the received OFDM signal,  but now with  "guard gap".  The assumptions of  [[Modulation_Methods/Implementation_of_OFDM_Systems#Intercarrier_interference_and_intersymbol_interference\|$\text{Example 1}$]]  have been kept.
	[[File:~~EN_Mod_T_5_6_S5a~~.png \|right\|frame\| OFDM reception signal over multipath channel with guard gap ~~'''KORREKTUR''': duration~~]]		[[File:EN_Mod_T_5_6_S4b_v2.png \|right\|frame\| OFDM reception signal over multipath channel with guard gap]]

	The diagram shows:		The diagram shows:

Guenter at 15:27, 19 January 2023

2023-01-19T15:27:37Z

Hwang at 11:33, 19 January 2023

2023-01-19T11:33:42Z

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	{e_3 } \\		{e_3 } \\
	\end{array}} \right) = \left( +1, -1, \; +1, -1 \right).$$		\end{array}} \right) = \left( +1, -1, \; +1, -1 \right).$$
	⇒   This corresponds exactly to the transmit symbol sequence  $\bf D$.  That is: <br>		⇒   This corresponds exactly to the transmit '''KORREKTUR: transmitted?''' symbol sequence  $\bf D$.  That is: <br>

	:'''Knowing the channel,  the signal can be completely equalized,  requiring only a single multiplication per symbol  (carrier)'''. }}		:'''Knowing the channel,  the signal can be completely equalized,  requiring only a single multiplication per symbol  (carrier)'''. }}

Hwang at 11:33, 19 January 2023

2023-01-19T11:33:05Z

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	{\rm{e} }^{ - {\rm{j \hspace{0.05cm}\cdot \hspace{0.03cm} 2\pi } }\hspace{0.05cm}\cdot \hspace{0.03cm}l \hspace{0.05cm}\cdot \hspace{0.03cm} \mu /4} } .$$		{\rm{e} }^{ - {\rm{j \hspace{0.05cm}\cdot \hspace{0.03cm} 2\pi } }\hspace{0.05cm}\cdot \hspace{0.03cm}l \hspace{0.05cm}\cdot \hspace{0.03cm} \mu /4} } .$$

	'''(3)'''   For numerical calculations,  we assume a known BPSK-encoded transmit sequence  $\rm \bf D$  (in the frequency domain)  and the following channel impulse response  $\bf h$:		'''(3)'''   For numerical calculations,  we assume a known BPSK-encoded transmit '''KORREKTUR: transmitted?''' sequence  $\rm \bf D$  (in the frequency domain)  and the following channel impulse response  $\bf h$:
	:$${\rm\bf{D} } = \frac{1}{N} \cdot {\rm\bf{d} } \cdot {\rm\bf{F} } =		:$${\rm\bf{D} } = \frac{1}{N} \cdot {\rm\bf{d} } \cdot {\rm\bf{F} } =
	\left( D_0, D_1,D_2,D_3\right) = \left( +1,\ -1,\ +1,\ -1\right),$$		\left( D_0, D_1,D_2,D_3\right) = \left( +1,\ -1,\ +1,\ -1\right),$$

← Older revision		Revision as of 18:05, 11 March 2023
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	#Further one recognizes  "intersymbol interference"  $\rm (ISI)$  in the green framed time interval  $0 ≤ t < τ_1$:   <br>       The red predecessor symbol  $k-\hspace{-0.08cm}1$   $($frequency  $3 · f_0)$  interferes with the black symbol  $k$   $($frequency $1 · f_0)$. }}		#Further one recognizes  "intersymbol interference"  $\rm (ISI)$  in the green framed time interval  $0 ≤ t < τ_1$:   <br>       The red predecessor symbol  $k-\hspace{-0.08cm}1$   $($frequency  $3 · f_0)$  interferes with the black symbol  $k$   $($frequency $1 · f_0)$. }}

	==Guard ~~gap~~ to reduce intersymbol interference==		==Guard interval to reduce intersymbol interference==
	<br>		<br>
	A first possible solution for the second problem  $\rm (ISI)$  is the introduction of a guard ~~gap~~ of length  $T_{\rm G}$:		A first possible solution for the second problem  $\rm (ISI)$  is the introduction of a guard interval of length  $T_{\rm G}$:
	[[File: P_ID1643__Mod_T_5_6_S4b_1_neu.png\|right\|frame\|Principle of the  "guard ~~gap~~"]]		[[File: P_ID1643__Mod_T_5_6_S4b_1_neu.png\|right\|frame\|Principle of the  "guard interval"]]
	*Here,  the signal between two symbols is set to zero for the duration of the protection time  $T_{\rm G}$.		*Here,  the signal between two symbols is set to zero for the duration of the protection time  $T_{\rm G}$.

	*As a result,  possible pulse trailers of symbol  $k-\hspace{-0.08cm}1$  no longer extend into the following symbol  $(k)$,  provided that the guard ~~gap~~ is selected  "wider"  than the maximum channel delay.		*As a result,  possible pulse trailers of symbol  $k-\hspace{-0.08cm}1$  no longer extend into the following symbol  $(k)$,  provided that the guard interval is selected  "wider"  than the maximum channel delay.

	*The new frame duration  $T_{\rm R}$ – i.e. the distance between successive transmitted symbols – is thus given by		*The new frame duration  $T_{\rm R}$ – i.e. the distance between successive transmitted symbols – is thus given by
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	{{GraueBox\|TEXT=		{{GraueBox\|TEXT=
	$\text{Example 2:}$		$\text{Example 2:}$
	This diagram again shows the real part of the received OFDM signal,  but now with  "guard ~~gap~~".  The assumptions of  [[Modulation_Methods/Implementation_of_OFDM_Systems#Intercarrier_interference_and_intersymbol_interference\|$\text{Example 1}$]]  have been kept.		This diagram again shows the real part of the received OFDM signal,  but now with  "guard interval".  The assumptions of  [[Modulation_Methods/Implementation_of_OFDM_Systems#Intercarrier_interference_and_intersymbol_interference\|$\text{Example 1}$]]  have been kept.
	[[File:EN_Mod_T_5_6_S4b_v2.png \|right\|frame\| OFDM reception signal over multipath channel with guard ~~gap~~]]		[[File:EN_Mod_T_5_6_S4b_v2.png \|right\|frame\| OFDM reception signal over multipath channel with guard interval]]

	The diagram shows:		The diagram shows:
	# In addition,  $T_{\rm G} = T/4$  is set,  which corresponds to the limiting case  $T_{\rm G} = τ_{\rm max}$  for the present channel.		# In addition,  $T_{\rm G} = T/4$  is set,  which corresponds to the limiting case  $T_{\rm G} = τ_{\rm max}$  for the present channel.
	# By using a guard ~~gap~~ of corresponding width,   intersymbol interference  $\rm (ISI)$  can be avoided   ⇒   in interval  $k$  only one frequency occurs.		# By using a guard interval of corresponding width,   intersymbol interference  $\rm (ISI)$  can be avoided   ⇒   in interval  $k$  only one frequency occurs.
	# '''But''':  Intercarrier interference  $\rm (ICI)$  cannot be prevented by this,  because the symbols still have a transient phase and thus jumps.		# '''But''':  Intercarrier interference  $\rm (ICI)$  cannot be prevented by this,  because the symbols still have a transient phase and thus jumps.

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	The  "guard ~~gap~~"  approach will not be considered further.  Rather,  a better alternative is presented in the next section.}}		The  "guard interval"  approach will not be considered further.  Rather,  a better alternative is presented in the next section.}}

	==Cyclic Prefix==		==Cyclic Prefix==
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	*For this,  the end of a symbol in the time interval  $T \ – \ T_{\rm G} ≤ t < T$  is prefixed again to the actual symbol.		*For this,  the end of a symbol in the time interval  $T \ – \ T_{\rm G} ≤ t < T$  is prefixed again to the actual symbol.
	*This procedure thus generates a  »'''cyclic prefix'''«.		*This procedure thus generates a  »'''cyclic prefix'''«.
	*As with the  "guard ~~gap~~",  the interval duration increases from symbol duration  $T$  to the new frame duration  $T_{\rm R} = T + T_{\rm G}$.		*As with the  "guard interval",  the interval duration increases from symbol duration  $T$  to the new frame duration  $T_{\rm R} = T + T_{\rm G}$.
	*The new number of samples of the extended discrete-time signal in the  $k$–th interval is then:		*The new number of samples of the extended discrete-time signal in the  $k$–th interval is then:
	:$$N_{\rm{total}} = N + N_{\rm{G}} = N \cdot (1 + T_{\rm{G}} /T) .$$		:$$N_{\rm{total}} = N + N_{\rm{G}} = N \cdot (1 + T_{\rm{G}} /T) .$$
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	{{GraueBox\|TEXT=		{{GraueBox\|TEXT=
	$\text{Example 3:}$  The diagram shows the operation of the guard interval in the continuous-time case.		$\text{Example 3:}$  The diagram shows the operation of the guard interval in the continuous-time case.
	The parameters from the consideration of the guard ~~gap~~ in  [[Modulation_Methods/Implementation_of_OFDM_Systems#Intercarrier_interference_and_intersymbol_interference\|$\text{Example 1}$]]  still apply,  although only one symbol  $($with frequency  $f_0)$ is now considered.		The parameters from the consideration of the guard interval in  [[Modulation_Methods/Implementation_of_OFDM_Systems#Intercarrier_interference_and_intersymbol_interference\|$\text{Example 1}$]]  still apply,  although only one symbol  $($with frequency  $f_0)$ is now considered.
	[[File:EN_Mod_T_5_6_S5b_neu.png \|right\|frame\| Received OFDM signal over multipath channel with cyclic prefix]]		[[File:EN_Mod_T_5_6_S5b_neu.png \|right\|frame\| Received OFDM signal over multipath channel with cyclic prefix]]

← Older revision		Revision as of 17:27, 19 January 2023
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	*Under the simplifying assumption of a transmission spectrum  $S(f)$  hard limited to $1/T$,  the bandwidth efficiency – see [Kam04]<ref>Kammeyer, K.D.:  Nachrichtenübertragung.  Stuttgart: B.G. Teubner, 4. Auflage, 2004.</ref>:		*Under the simplifying assumption of a transmission spectrum  $S(f)$  hard limited to $1/T$,  the bandwidth efficiency – see [Kam04]<ref>Kammeyer, K.D.:  Nachrichtenübertragung.  Stuttgart: B.G. Teubner, 4. Auflage, 2004.</ref>:
	:$$\beta = \frac{ \text{symbol rate} }{ {\rm bandwidth} } = \frac{1/(T + T_{\rm G})}{1/T} = \frac{1}{{1 + T_{\rm{G}} /T}}.$$		:$$\beta = \frac{ \text{symbol rate} }{ {\rm bandwidth} } = \frac{1/(T + T_{\rm G})}{1/T} = \frac{1}{{1 + T_{\rm{G}} /T}}.$$
	*However,  in a system using the so-called  "matched filter approach",  increasing the frame duration from  $T$  to  $T_{\rm G} + T$  leads to a decrease in the signal-to-noise ratio if the impulse responses  $g_{\rm S}(t)$  and  $g_{\rm E}(t)$  of the transmission and ~~reception '''KORREKTUR:~~ receiver~~'''~~ filters are matched to the symbol duration  $T$.		*However,  in a system using the so-called  "matched filter approach",  increasing the frame duration from  $T$  to  $T_{\rm G} + T$  leads to a decrease in the signal-to-noise ratio if the impulse responses  $g_{\rm S}(t)$  and  $g_{\rm E}(t)$  of the transmission and receiver filters are matched to the symbol duration  $T$.
	*The resulting  signal–to–noise ratio  $\rm (SNR)$  of the overall system  (in dB)  can be calculated as follows,  taking into account the guard interval:		*The resulting  signal–to–noise ratio  $\rm (SNR)$  of the overall system  (in dB)  can be calculated as follows,  taking into account the guard interval:
	:$${\rm{SNR}}_{\hspace{0.08cm}{\rm{ {\rm{with} }\hspace{0.08cm} GI} } } = {\rm{SNR}}_{\hspace{0.08cm}{\rm{{\rm{without}}\hspace{0.08cm} GI}}} + 10 \cdot \lg (\beta ), \quad {\rm{where}}$$		:$${\rm{SNR}}_{\hspace{0.08cm}{\rm{ {\rm{with} }\hspace{0.08cm} GI} } } = {\rm{SNR}}_{\hspace{0.08cm}{\rm{{\rm{without}}\hspace{0.08cm} GI}}} + 10 \cdot \lg (\beta ), \quad {\rm{where}}$$
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	*Using the convolution theorem for linear time-invariant systems,  one can then also write the spectrum as a product of two discrete Fourier transforms:		*Using the convolution theorem for linear time-invariant systems,  one can then also write the spectrum as a product of two discrete Fourier transforms:
	:$$R_\mu = {\rm{DFT} }\{ d_\nu \} \cdot {\rm{DFT} }\{ h_\nu \} = D_\mu \cdot H_\mu.$$		:$$R_\mu = {\rm{DFT} }\{ d_\nu \} \cdot {\rm{DFT} }\{ h_\nu \} = D_\mu \cdot H_\mu.$$
	*To compensate for the influence of the channel on the ~~reception '''KORREKTUR:~~ received~~?'''~~ sequence,  it is convenient to multiply the spectrum by the inverse transfer function  $1/H_{\mu}$.		*To compensate for the influence of the channel on the received sequence,  it is convenient to multiply the spectrum by the inverse transfer function  $1/H_{\mu}$.
	*This  "zero forcing"  approach leads to the ideal signal reconstruction in the noise-free case.  The equalization can be done point by point:		*This  "zero forcing"  approach leads to the ideal signal reconstruction in the noise-free case.  The equalization can be done point by point:
	:$$\hat {D}_\mu = \frac{1}{ {H_\mu } } \cdot R_\mu.$$}}		:$$\hat {D}_\mu = \frac{1}{ {H_\mu } } \cdot R_\mu.$$}}
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	*The vector of a channel with  $L$  echoes is  $\mathbf h = (h_0$, ... , $h_L)$.  The transmission matrix with  $N$  rows and  $N + L$  columns is:		*The vector of a channel with  $L$  echoes is  $\mathbf h = (h_0$, ... , $h_L)$.  The transmission matrix with  $N$  rows and  $N + L$  columns is:
	:$${\rm\bf{H}} = \left( {\begin{array}{*{20}c} {h_0 } & {h_1 } & \cdots & {h_L } & {} & {} & {} \\ {} & {h_0 } & {h_1 } & \cdots & {h_L } & {} & {} \\ {} & {} & \ddots & \ddots & {} & \ddots & {} \\ {} & {} & {} & {h_0 } & {h_1 } & \cdots & {h_L } \\ \end{array}} \right).$$		:$${\rm\bf{H}} = \left( {\begin{array}{*{20}c} {h_0 } & {h_1 } & \cdots & {h_L } & {} & {} & {} \\ {} & {h_0 } & {h_1 } & \cdots & {h_L } & {} & {} \\ {} & {} & \ddots & \ddots & {} & \ddots & {} \\ {} & {} & {} & {h_0 } & {h_1 } & \cdots & {h_L } \\ \end{array}} \right).$$
	*Here,  $N$  indicates the number of carriers and hence the number of time samples of the IDFT.  With the ~~transmit '''KORREKTUR:~~ transmitted~~'''~~ vector  ${\bf d} = (d_0$,  ...  , $d_{N–1})$  the ~~receive '''KORREKTUR:~~ received~~'''~~ vector is:		*Here,  $N$  indicates the number of carriers and hence the number of time samples of the IDFT.  With the transmitted vector  ${\bf d} = (d_0$,  ...  , $d_{N–1})$  the received vector is:
	:$$\bf r = d · H.$$		:$$\bf r = d · H.$$

	*Considering the cyclic prefix,  the extended ~~transmit '''KORREKTUR:~~ transmitted~~'''~~ vector is obtained:		*Considering the cyclic prefix,  the extended transmitted vector is obtained:
	:$${\rm\bf{d}}_{{\rm{ext}}} = (d_{N - N_G } , \ \ldots \ ,d_{N - 1} ,d_0 , \ \ldots \ ,d_{N - 1} ).$$		:$${\rm\bf{d}}_{{\rm{ext}}} = (d_{N - N_G } , \ \ldots \ ,d_{N - 1} ,d_0 , \ \ldots \ ,d_{N - 1} ).$$
	*Now,  one could extend the above transmission matrix  $\bf H$  likewise accordingly on   $(N + N_{\rm G})$  rows   and   $(N + L + N_{\rm G})$  columns   as well as remove the prefix at the receiver again,  which is not to be pursued here further.		*Now,  one could extend the above transmission matrix  $\bf H$  likewise accordingly on   $(N + N_{\rm G})$  rows   and   $(N + L + N_{\rm G})$  columns   as well as remove the prefix at the receiver again,  which is not to be pursued here further.
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	\end{array}} \right) .$$		\end{array}} \right) .$$

	*The Discrete Fourier Transform  $\rm (DFT)$  can be represented by  $1/N · \bf F$  and its inverse  $\rm (IDFT)$  by  $\rm \bf F^{\star}$ such that for the ~~transmit '''KORREKTUR:~~ transmitted~~'''~~ vector:  $\rm {\bf d} = {\bf D} · {\bf F}^{\star}$.		*The Discrete Fourier Transform  $\rm (DFT)$  can be represented by  $1/N · \bf F$  and its inverse  $\rm (IDFT)$  by  $\rm \bf F^{\star}$ such that for the transmitted vector:  $\rm {\bf d} = {\bf D} · {\bf F}^{\star}$.

	*The  $N$  spectral coefficients are described by the vector  ${\bf D} = 1/N · {\bf d} · {\bf F}$  and the ~~reception '''KORREKTUR:~~ received~~'''~~ vector is  ${\bf r} = {\bf d} · {\bf H}_{\rm C} = {\bf D} · {\bf F}^{\star} · {\bf H}_{\rm C}$.		*The  $N$  spectral coefficients are described by the vector  ${\bf D} = 1/N · {\bf d} · {\bf F}$  and the received vector is  ${\bf r} = {\bf d} · {\bf H}_{\rm C} = {\bf D} · {\bf F}^{\star} · {\bf H}_{\rm C}$.

	*The (discrete) Fourier transform  $\rm \bf R$  of the ~~reception '''KORREKTUR:~~ received~~'''~~ vector  $\rm \bf r$  can then be written in the following way:		*The (discrete) Fourier transform  $\rm \bf R$  of the received vector  $\rm \bf r$  can then be written in the following way:
	:$${\rm\bf{R}} = \frac{1}{N} \cdot {\rm\bf{r}} \cdot {\rm\bf{F}} = {\rm\bf{D}} \cdot \left( {\begin{array}{*{20}c}		:$${\rm\bf{R}} = \frac{1}{N} \cdot {\rm\bf{r}} \cdot {\rm\bf{F}} = {\rm\bf{D}} \cdot \left( {\begin{array}{*{20}c}
	{H_0 } & {} & {} & {} \\		{H_0 } & {} & {} & {} \\
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	{{BlaueBox\|TEXT=		{{BlaueBox\|TEXT=
	$\text{Conclusion:}$		$\text{Conclusion:}$
	*The ~~reception '''KORREKTUR:~~ received~~'''~~ symbol on the  $\mu$–th carrier is:		*The received symbol on the  $\mu$–th carrier is:
	:$$R_{\mu} = D_{\mu} · H_{\mu} \ \ (\mu = 0, \text{...}\ ,\ N–1).$$		:$$R_{\mu} = D_{\mu} · H_{\mu} \ \ (\mu = 0, \text{...}\ ,\ N–1).$$
	*This can thus be equalized using the  "Zero Forcing"  approach:		*This can thus be equalized using the  "Zero Forcing"  approach:
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	$\text{Example 5:}$		$\text{Example 5:}$
	We assume a system with  $N = 4$  carriers and a channel with  $L = 2$  echoes,		We assume a system with  $N = 4$  carriers and a channel with  $L = 2$  echoes,
	*so that for the ~~transmit '''KORREKTUR:~~ transmitted~~'''~~ vector  ${\bf d} = (d_0, d_1, d_2, d_3)$,  and		*so that for the transmitted vector  ${\bf d} = (d_0, d_1, d_2, d_3)$,  and
	*for the channel impulse response  ${\bf h} = (h_0, h_1, h_2)$.		*for the channel impulse response  ${\bf h} = (h_0, h_1, h_2)$.


	'''(1)'''   To represent the cyclic prefix,  we use the cyclic transmission matrix  ${\rm\bf{H} }_{\rm{C} }$,  instead of the extended ~~transmission '''KORREKTUR:~~ transmitted~~?'''~~ vector with the corresponding transmission matrix,  resulting in the ~~reception '''KORREKTUR:~~ received~~'''~~ vector  ${\rm \bf r}= {\rm \bf d} \cdot {\rm \bf H}_{\rm{C} }$:		'''(1)'''   To represent the cyclic prefix,  we use the cyclic transmission matrix  ${\rm\bf{H} }_{\rm{C} }$,  instead of the extended transmitted vector with the corresponding transmission matrix,  resulting in the received vector  ${\rm \bf r}= {\rm \bf d} \cdot {\rm \bf H}_{\rm{C} }$:
	:$${\rm\bf{H} }_{\rm{C} } = \left( {\begin{array}{*{20}c}		:$${\rm\bf{H} }_{\rm{C} } = \left( {\begin{array}{*{20}c}
	{h_0 } & {h_1 } & {h_2 } & { } \\		{h_0 } & {h_1 } & {h_2 } & { } \\
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	r_3 = d_1 \cdot h_2 + d_2 \cdot h_1 + d_3 \cdot h_0.$$		r_3 = d_1 \cdot h_2 + d_2 \cdot h_1 + d_3 \cdot h_0.$$

	'''(2)'''   The  (discrete)  Fourier transform of the ~~receive '''KORREKTUR:~~ received~~'''~~ vector is calculated to be		'''(2)'''   The  (discrete)  Fourier transform of the received vector is calculated to be
	:$${\rm\bf{R} } = \frac{1}{N} \cdot {\rm\bf{r} } \cdot {\rm\bf{F} } = {\rm\bf{D} } \cdot \left( {\begin{array}{*{20}c}		:$${\rm\bf{R} } = \frac{1}{N} \cdot {\rm\bf{r} } \cdot {\rm\bf{F} } = {\rm\bf{D} } \cdot \left( {\begin{array}{*{20}c}
	{H_0 } & {} & {} & {} \\		{H_0 } & {} & {} & {} \\
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	{\rm{e} }^{ - {\rm{j \hspace{0.05cm}\cdot \hspace{0.03cm} 2\pi } }\hspace{0.05cm}\cdot \hspace{0.03cm}l \hspace{0.05cm}\cdot \hspace{0.03cm} \mu /4} } .$$		{\rm{e} }^{ - {\rm{j \hspace{0.05cm}\cdot \hspace{0.03cm} 2\pi } }\hspace{0.05cm}\cdot \hspace{0.03cm}l \hspace{0.05cm}\cdot \hspace{0.03cm} \mu /4} } .$$

	'''(3)'''   For numerical calculations,  we assume a known BPSK-encoded ~~transmit '''KORREKTUR:~~ transmitted~~?'''~~ sequence  $\rm \bf D$  (in the frequency domain)  and the following channel impulse response  $\bf h$:		'''(3)'''   For numerical calculations,  we assume a known BPSK-encoded transmitted sequence  $\rm \bf D$  (in the frequency domain)  and the following channel impulse response  $\bf h$:
	:$${\rm\bf{D} } = \frac{1}{N} \cdot {\rm\bf{d} } \cdot {\rm\bf{F} } =		:$${\rm\bf{D} } = \frac{1}{N} \cdot {\rm\bf{d} } \cdot {\rm\bf{F} } =
	\left( D_0, D_1,D_2,D_3\right) = \left( +1,\ -1,\ +1,\ -1\right),$$		\left( D_0, D_1,D_2,D_3\right) = \left( +1,\ -1,\ +1,\ -1\right),$$
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	{e_3 } \\		{e_3 } \\
	\end{array}} \right) = \left( +1, -1, \; +1, -1 \right).$$		\end{array}} \right) = \left( +1, -1, \; +1, -1 \right).$$
	⇒   This corresponds exactly to the ~~transmit '''KORREKTUR:~~ transmitted~~?'''~~ symbol sequence  $\bf D$.  That is: <br>		⇒   This corresponds exactly to the transmitted symbol sequence  $\bf D$.  That is: <br>

	:'''Knowing the channel,  the signal can be completely equalized,  requiring only a single multiplication per symbol  (carrier)'''. }}		:'''Knowing the channel,  the signal can be completely equalized,  requiring only a single multiplication per symbol  (carrier)'''. }}