View source for Aufgaben:Exercise 5.8: Equalization in Matrix Vector Notation


{{quiz-Header|Buchseite=Modulation_Methods/Implementation_of_OFDM_Systems
}}

[[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,&nbsp; assuming a system with &nbsp;$N = 4$&nbsp; carriers and a channel with &nbsp;$L = 2$&nbsp; echoes.
*Only a single frame is considered and for the transmission vector&nbsp; (in the time domain),&nbsp; we apply:
:$${\rm\bf{d}} = (d_0, \ d_1,\ d_2,\ d_3 ) = (+1,  -1,  +1,  -1 ).$$
*Let the channel impulse response be described by
:$${\rm\bf{h}} = (h_0, \ h_1,\ h_2 ) = (0, \ 0.6, \ 0.4 ).$$
*To represent the cyclic prefix in this task,&nbsp; instead of the extended transmission vector with the associated transmission matrix &nbsp;${\rm\bf{H}}_{\rm ext}$,&nbsp; we use the cyclic transmission matrix
:$${\rm\bf{H}}_{\rm{C}} = \left( {\begin{array}{*{20}c} {h_0 } & {h_1 } & {h_2 } & {} \\ {} & {h_0 } & {h_1 } & {h_2 } \\ \hline {h_2 } & {} & {h_0 } & {h_1 } \\ {h_1 } & {h_2 } & {} & {h_0 } \\ \end{array}} \right).$$
*For the spectral coefficients at the receiver,&nbsp; according to the Discrete Fourier Transform&nbsp; $\rm (DFT)$:
:$${\rm\bf{R}} = {\rm\bf{D}} \cdot \left( {\begin{array}{*{20}c} {H_0 } & {} & {} & {} \\ {} & {H_1 } & {} & {} \\ {} & {} & {H_2 } & {} \\ {} & {} & {} & {H_3 } \\ \end{array}} \right) ,$$
:where the diagonal elements are to be calculated as follows:
:$$H_\mu = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e}}^{ - {\rm{j\hspace{0.04cm}\cdot \hspace{0.04cm}2\pi }} \hspace{0.04cm}\cdot \hspace{0.04cm} l \hspace{0.04cm}\cdot \hspace{0.04cm} {\mu }/{4}} } .$$
*The equalization at the receiver is done by multiplication in the frequency domain by the coefficients &nbsp;$ e_\mu = {1}/{H_\mu }.$








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)|Discrete Fourier Transform]]&nbsp; in the book&nbsp; "Signal Representation".
 
*For the Discrete Fourier Transform (DFT) in matrix-vector notation holds:
:$${\rm\bf{F}} = \left( {\begin{array}{*{20}c} 1 & 1 & \cdots & 1 \\ 1 & {} & {} & {} \\ \vdots & {} & {{\rm{e}}^{ - {\rm{j2\pi }}{\kern 1pt} \nu {\kern 1pt} \mu /N} } & {} \\ 1 & {} & {} & {} \\ \end{array}} \right), \qquad {\rm{DFT\; with}} \; {1}/{N} \cdot {\rm\bf{F}}; \qquad {\rm{IDFT \; with}} \; {\rm\bf{F}}^*.$$



===Questions===

<quiz display=simple>
{Calculate the discrete reception values &nbsp;$r = (r_0, r_1, r_2, r_3)$&nbsp; in the time domain.&nbsp; Enter &nbsp;$r_0$&nbsp; and &nbsp;$r_1$&nbsp; as a check.
|type="{}"}
${\rm Re}\big[r_0\big] \ = \ $ { -0.202--0.198 }
${\rm Im}\big[r_0\big] \ = \ $ { 0. } 
${\rm Re}\big[r_1\big] \ = \ $ { 0.2 1% }
${\rm Im}\big[r_1\big] \ = \ $ { 0. } 

{What are the discrete spectral domain coefficients &nbsp;${\rm\bf{D}}= (D_0, D_1, D_2, D_3)$&nbsp; at the transmitter?&nbsp;  Enter &nbsp;$D_2$&nbsp; and &nbsp;$D_3$&nbsp; as a check.
|type="{}"}
${\rm Re}\big[D_2\big] \ = \ $ { 1 3% }
${\rm Im}\big[D_2\big] \ = \ $ { 0. }
${\rm Re}\big[D_3\big] \ = \ $ { 0. }
${\rm Im}\big[D_3\big] \ = \ $ { 0. }

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

{ Determine the discrete equalizer coefficients &nbsp;${\rm\bf{e}}= (e_0, e_1, e_2, e_3)$.
|type="{}"}
${\rm Re}\big[e_0\big] \ = \ $ { 1 1% }
${\rm Im}\big[e_0\big] \ = \ $ { 0. }
${\rm Re}\big[e_1\big] \ = \ $ { -0.78--0.76 }
${\rm Im}\big[e_1\big] \ = \ $ { 1.15 1% }
${\rm Re}\big[e_2\big] \ = \ $ { -5.05--4.95 }
${\rm Im}\big[e_2\big] \ = \ $ { 0. }
${\rm Re}\big[e_3\big] \ = \ $ { -0.78--0.76 }
${\rm Im}\big[e_3\big] \ = \ $ { -1.16--1.14 }

{What is the name of the equalization approach used?
|type="()"}
+ As "Zero Forcing" approach,
- as "Matched Filter" approach,
- as "Minimum Mean Square Error&nbsp; $\rm (MMSE)$”approach.
</quiz>

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


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


'''(3)'''&nbsp;  The vector&nbsp; ${\rm\bf{R}}$&nbsp; of spectral coefficients after the channel could be calculated by DFT of the vector &nbsp; ${\rm\bf{r}}$,&nbsp; analogous to subtask&nbsp; '''(2)'''.&nbsp; 
*An alternative solution path is:
:$${\rm\bf{R}} = {\rm\bf{D}} \cdot \left( {\begin{array}{*{20}c}
   {H_0 } & {} & {} & {}  \\
   {} & {H_1 } & {} & {}  \\
   {} & {} & {H_2 } & {}  \\
   {} & {} & {} & {H_3 }  \\
\end{array}} \right)  .$$
*For the diagonal elements one obtains:
:$$H_\mu = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e}}^{ -
{\rm{j\hspace{0.04cm}\cdot \hspace{0.04cm}2\pi }}
\hspace{0.04cm}\cdot \hspace{0.04cm} l \hspace{0.04cm}\cdot
\hspace{0.04cm}{\mu }/{4}} } \hspace{0.3cm} 
\Rightarrow \hspace{0.3cm} H_0 = 1,\hspace{0.1cm}H_1 = -0.4 - {\rm{j}} \cdot
0.6,\hspace{0.1cm}H_2 = -0.2,\hspace{0.1cm}H_3 = -0.4 + {\rm{j}}
\cdot 0.6 $$
:$$\Rightarrow \hspace{0.3cm}{\rm\bf{R}} = \left( {R_0 ,R_1 ,R_2 ,R_3
} \right)= \left( \hspace{0.15cm}0,\hspace{0.15cm}0,\hspace{0.15cm}-0.2, \hspace{0.15cm}0 \right)  \hspace{0.3cm}
\Rightarrow \hspace{0.3cm}{\rm Re}[r_2]\hspace{0.15cm} \underline{=-0.2},\hspace{0.2cm} {\rm Im}[r_2]\hspace{0.15cm} \underline{=0}, \hspace{0.2cm}{\rm Re}[r_3]\hspace{0.15cm} \underline{=0},\hspace{0.2cm} {\rm Im}[r_3]\hspace{0.15cm} \underline{=0}.$$


'''(4)'''&nbsp;  The equalizer coefficients result in&nbsp; $e_\mu = 1/H_\mu$. 
*With the result from subtask&nbsp; '''(3)''',&nbsp; the coefficients&nbsp; $e_0 = 1$,&nbsp; $e_2 = -5$ are real:
:$${\rm Re}[e_0]\hspace{0.15cm} \underline{=1},\hspace{0.2cm} {\rm Im}[e_0]\hspace{0.15cm} \underline{=0}, \hspace{0.2cm}{\rm Re}[e_2]\hspace{0.15cm} \underline{=-5},\hspace{0.2cm} {\rm Im}[e_2]\hspace{0.15cm} \underline{=0}.$$

*For the other two coefficients:
:$$e_1 = \frac {1}{-0.4 - {\rm{j}} \cdot 0.6} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm{Re}}[e_1] = \frac {-0.4}{0.4^2 + 0.6^2}\hspace{0.15cm} \underline{ \approx -0.77},\hspace{0.3cm} {\rm{Im}}[e_1] = \frac {0.6}{0.4^2 + 0.6^2} \hspace{0.15cm} \underline{\approx +1.15},$$
:$$e_3 = \frac {1}{-0.4 + {\rm{j}} \cdot 0.6} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} {\rm{Re}}[e_3] = \frac {-0.4}{0.4^2 + 0.6^2}\hspace{0.15cm} \underline{ \approx -0.77},\hspace{0.3cm} {\rm{Im}}[e_3] = \frac {-0.6}{0.4^2 + 0.6^2} \hspace{0.15cm} \underline{\approx -1.15}.$$


'''(5)'''&nbsp;  The equalization calculated in&nbsp; '''(4)'''&nbsp; follows the&nbsp; "zero forcing"&nbsp; approach &nbsp; &rArr; &nbsp; <u>solution 1</u>.

{{ML-Fuß}}



[[Category:Modulation Methods: Exercises|^5.6 Realization of OFDM Systems^]]