Aufgaben:Exercise 5.8: Equalization in Matrix Vector Notation: Difference between revisions

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{{quiz-Header|Buchseite=Modulationsverfahren/Realisierung von OFDM-Systemen
{{quiz-Header|Buchseite=Modulation_Methods/Implementation_of_OFDM_Systems
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[[File:P_ID1663__A_5_8.png|right|frame|Blockschaltbild der OFDM-Übertragung]]
[[File:EN_A_5_8.png|right|frame|Block diagram of OFDM transmission]]
Wir betrachten die in der Grafik hinterlegten Blöcke eines OFDM–Systems, wobei wir von einem System mit $N = 4$ Trägern und einem Kanal mit $L = 2$ Echos ausgehen.  
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.
*Es wird nur ein einziger Rahmen betrachtet und für den Sendevektor (im Zeitbereich) gelte:
*Only a single frame is considered and for the transmission vector  (in the time domain),  we apply:
:$${\rm\bf{d}} = (d_0, d_1,d_2,d_3 ) = (+1, -1, +1, -1 ).$$
:$${\rm\bf{d}} = (d_0, \ d_1,\ d_2,\ d_3 ) = (+1, -1, +1, -1 ).$$
*Die Kanalimpulsantwort sei beschrieben durch
*Let the channel impulse response be described by
:$${\rm\bf{h}} = (h_0, h_1,h_2 ) = (0, 0.6, 0.4 ).$$
:$${\rm\bf{h}} = (h_0, \ h_1,\ h_2 ) = (0, \ 0.6, \ 0.4 ).$$
*Zur Repräsentation des zyklischen Präfixes verwenden wir in dieser Aufgabe statt des erweiterten Sendevektors mit der zugehörigen Übertragungsmatrix $H_{ext}$ die zyklische Übertragungsmatrix
*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
:$${\rm\bf{H}}_{\rm{C}} = \left( {\begin{array}{*{20}c} {h_0 } & {h_1 } & {h_2 } & {} \\ {} & {h_0 } & {h_1 } & {h_2 } \\ \hline {h_2 } & {} & {h_0 } & {h_1 } \\ {h_1 } & {h_2 } & {} & {h_0 } \\ \end{array}} \right).$$
:$${\rm\bf{H}}_{\rm{C}} = \left( {\begin{array}{*{20}c} {h_0 } & {h_1 } & {h_2 } & {} \\ {} & {h_0 } & {h_1 } & {h_2 } \\ \hline {h_2 } & {} & {h_0 } & {h_1 } \\ {h_1 } & {h_2 } & {} & {h_0 } \\ \end{array}} \right).$$
*Für die Spektralkoeffizienten am Empfänger gelte nach der Diskreten Fouriertransformation (DFT):
*For the spectral coefficients at the receiver,  according to the Discrete Fourier Transform  $\rm (DFT)$:
:$${\rm\bf{R}} = {\rm\bf{D}} \cdot \left( {\begin{array}{*{20}c} {H_0 } & {} & {} & {} \\ {} & {H_1 } & {} & {} \\ {} & {} & {H_2 } & {} \\ {} & {} & {} & {H_3 } \\ \end{array}} \right) ,$$
:$${\rm\bf{R}} = {\rm\bf{D}} \cdot \left( {\begin{array}{*{20}c} {H_0 } & {} & {} & {} \\ {} & {H_1 } & {} & {} \\ {} & {} & {H_2 } & {} \\ {} & {} & {} & {H_3 } \\ \end{array}} \right) ,$$
:wobei die Diagonalelemente wie folgt zu berechnen sind:
:where the diagonal elements are to be calculated as follows:
:$$H_\mu = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e}}^{ - {\rm{j\hspace{0.04cm}\cdot \hspace{0.04cm}2\pi }} \hspace{0.04cm}\cdot \hspace{0.04cm} l \hspace{0.04cm}\cdot \hspace{0.04cm} {\mu }/{4}} } .$$
:$$H_\mu = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e}}^{ - {\rm{j\hspace{0.04cm}\cdot \hspace{0.04cm}2\pi }} \hspace{0.04cm}\cdot \hspace{0.04cm} l \hspace{0.04cm}\cdot \hspace{0.04cm} {\mu }/{4}} } .$$
*Die Entzerrung am Empfänger erfolgt durch Multiplikation im Frequenzbereich mit den Koeffizienten $ e_\mu = {1}/{H_\mu }.$
*The equalization at the receiver is done by multiplication in the frequency domain by the coefficients  $ e_\mu = {1}/{H_\mu }.$




''Hinweise:''
*Die Aufgabe gehört zum  Kapitel [[Modulationsverfahren/Realisierung_von_OFDM-Systemen|Realisierung von OFDM-Systemen]].
*Bezug genommen wird auch  auf das Kapitel  [[Signaldarstellung/Diskrete_Fouriertransformation_(DFT)|Diskrete Fouriertransformation]] im Buch „Signaldarstellung”.
*Sollte die Eingabe des Zahlenwertes „0” erforderlich sein, so geben Sie bitte „0.” ein.
*Für die Diskrete Fouriertransformation (DFT) gilt in Matrix–Vektor–Notation:
:$${\rm\bf{F}} = \left( {\begin{array}{*{20}c} 1 & 1 & \cdots & 1 \\ 1 & {} & {} & {} \\ \vdots & {} & {{\rm{e}}^{ - {\rm{j2\pi }}{\kern 1pt} \nu {\kern 1pt} \mu /N} } & {} \\ 1 & {} & {} & {} \\ \end{array}} \right), \qquad {\rm{DFT\; mit}} \; {1}/{N} \cdot {\rm\bf{F}}; \qquad {\rm{IDFT \; mit}} \; {\rm\bf{F}}^*.$$






===Fragebogen===
 
 
 
Notes:
*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".
*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>
<quiz display=simple>
{Berechnen Sie die diskreten Empfangswerte $r = (r_0, r_1, r_2, r_3)$ im Zeitbereich. Geben Sie zur Kontrolle $r_0$ und $r_1$ ein:
{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="{}"}
|type="{}"}
${\rm Re}[r_0] \ = \ $ { -0.202--0.198 }
${\rm Re}\big[r_0\big] \ = \ $ { -0.202--0.198 }
${\rm Im}[r_0] \ = \ $ { 0. }  
${\rm Im}\big[r_0\big] \ = \ $ { 0. }  
${\rm Re}[r_1] \ = \ $ { 0.2 1% }
${\rm Re}\big[r_1\big] \ = \ $ { 0.2 1% }
${\rm Im}[r_0] \ = \ $ { 0. }  
${\rm Im}\big[r_1\big] \ = \ $ { 0. }  


{Wie lauten die diskreten Spektralbereichskoeffizienten ${\rm\bf{D}}= (D_0, D_1, D_2, D_3)$ am Sender?  Geben Sie zur Kontrolle $D_2$ und $D_3$ ein.
{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="{}"}
|type="{}"}
${\rm Re}[D_2] \ = \ $ { 1 3% }
${\rm Re}\big[D_2\big] \ = \ $ { 1 3% }
${\rm Im}[D_2] \ = \ $ { 0. }
${\rm Im}\big[D_2\big] \ = \ $ { 0. }
${\rm Re}[D_3] \ = \ $ { 0. }
${\rm Re}\big[D_3\big] \ = \ $ { 0. }
${\rm Im}[D_3] \ = \ $ { 0. }
${\rm Im}\big[D_3\big] \ = \ $ { 0. }


{Berechnen Sie die diskreten Spektralkoeffizienten ${\rm\bf{R}}= (R_0, R_1, R_2, R_3)$ nach dem Kanal. Geben Sie zur Kontrolle $R_2$ und $R_3$ ein:
{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="{}"}
|type="{}"}
${\rm Re}[R_2] \ = \ $  { -0.202--0.198 }  
${\rm Re}\big[R_2\big] \ = \ $  { -0.202--0.198 }  
${\rm Im}[R_2] \ = \ $ { 0. }  
${\rm Im}\big[R_2\big] \ = \ $ { 0. }  
${\rm Re}[R_3] \ = \ $  { 0. }  
${\rm Re}\big[R_3\big] \ = \ $  { 0. }  
${\rm Im}[R_3] \ = \ $ { 0. }
${\rm Im}\big[R_3\big] \ = \ $ { 0. }


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


{Wie bezeichnet man den verwendeten Entzerrungsansatz?
{What is the name of the equalization approach used?
|type="[]"}
|type="()"}
+ Als „Zero Forcing”–Ansatz,
+ As "Zero Forcing" approach,
- als „Matched Filter”–Ansatz,
- as "Matched Filter" approach,
- als „Minimum Mean Square Error (MMSE)”–Ansatz.
- as "Minimum Mean Square Error&nbsp; $\rm (MMSE)$”approach.
</quiz>
</quiz>


===Musterlösung===
===Solution===
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'''1.'''  Die diskreten Zeitbereichswerte am Empfänger berechnen sich mit der zyklischen Übertragungsmatrix $H_C$ wie folgt:
'''(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:
$${\rm\bf{r}} = {\rm\bf{d}} \cdot {\rm\bf{H_C}} = \left( {+1 ,-1 ,+1 ,-1 } \right) \cdot \left( {\begin{array}{*{20}c} {0 } & {0.6 } & {0.4 } & {} \\ {} & {0 } & {0.6 } & {0.4 } \\ \hline {0.4 } & {} & {0 } & {0.6 } \\ {0.6 } & {0.4 } & {} & {0 } \\ \end{array}} \right)$$
:$${\rm\bf{r}} = {\rm\bf{d}} \cdot {\rm\bf{H}}_{\rm{C}} = \left( {+1 ,-1 ,+1 ,-1 } \right) \cdot \left( {\begin{array}{*{20}c} {0 } & {0.6 } & {0.4 } & {} \\ {} & {0 } & {0.6 } & {0.4 } \\ \hline {0.4 } & {} & {0 } & {0.6 } \\ {0.6 } & {0.4 } & {} & {0 } \\ \end{array}} \right)$$
$$\Rightarrow \hspace{0.3cm}{\rm\bf{r}} = \left( {r_0 ,r_1 ,r_2 ,r_3 } \right) = \left( {-0.2, +0.2,-0.2, +0.2} \right) .$$
:$$\Rightarrow \hspace{0.3cm}{\rm\bf{r}} = \left( {r_0 ,r_1 ,r_2 ,r_3 } \right) = \left( {-0.2, +0.2,-0.2, +0.2} \right) \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Re}[r_0]\hspace{0.15cm} \underline{=-0.2},\hspace{0.2cm} {\rm Im}[r_0]\hspace{0.15cm} \underline{=0}, \hspace{0.2cm}{\rm Re}[r_1]\hspace{0.15cm} \underline{=+0.2},\hspace{0.2cm} {\rm Im}[r_1]\hspace{0.15cm} \underline{=0}. $$
Damit sind alle Empfangswerte <i>r</i><sub>0</sub> <u>= &ndash;0.2,</u> <i>r</i><sub>1</sub> <u>= +0.2</u>, <i>r</i><sub>2</sub> = <u>&ndash;0.2</u> und <u><i>r</i><sub>3</sub> = +0.2</u> rein reell.
 
 
'''(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:
:$$H_\mu = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e}}^{ -{\rm{j\hspace{0.04cm}\cdot \hspace{0.04cm}2\pi }}\hspace{0.04cm}\cdot \hspace{0.04cm} l \hspace{0.04cm}\cdot\hspace{0.04cm}{\mu }/{4}} } \hspace{0.3cm}\Rightarrow \hspace{0.3cm} H_0 = 1,\hspace{0.1cm}H_1 = -0.4 - {\rm{j}} \cdot0.6,\hspace{0.1cm}H_2 = -0.2,\hspace{0.1cm}H_3 = -0.4 + {\rm{j}}\cdot 0.6 $$:$$\Rightarrow \hspace{0.3cm}{\rm\bf{R}} = \left( {R_0 ,R_1 ,R_2 ,R_3} \right)= \left( \hspace{0.15cm}0,\hspace{0.15cm}0,\hspace{0.15cm}-0.2, \hspace{0.15cm}0 \right)  \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Re}[r_2]\hspace{0.15cm} \underline{=-0.2},\hspace{0.2cm} {\rm Im}[r_2]\hspace{0.15cm} \underline{=0}, \hspace{0.2cm}{\rm Re}[r_3]\hspace{0.15cm} \underline{=0},\hspace{0.2cm} {\rm Im}[r_3]\hspace{0.15cm} \underline{=0}.$$
 
 
'''(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:
:$${\rm Re}[e_0]\hspace{0.15cm} \underline{=1},\hspace{0.2cm} {\rm Im}[e_0]\hspace{0.15cm} \underline{=0}, \hspace{0.2cm}{\rm Re}[e_2]\hspace{0.15cm} \underline{=-5},\hspace{0.2cm} {\rm Im}[e_2]\hspace{0.15cm} \underline{=0}.$$


'''2.''' Die Spektralkoeffizienten <b>D</b> ergeben sich direkt aus der Diskreten Fouriertransformation (DFT) der Zeitbereichskoeffizienten <b>d</b> = (+1, &ndash;1, +1, &ndash;1). Diese Zeitbereichsfolge entspricht einer diskreten Cosinusfunktion mit der doppelten Grundfrequnz (2 &middot; <i>f</i><sub>0</sub>) und der Amplitude 1. Daraus folgt:
*For the other two coefficients:
$${\rm\bf{D}} = \left( {D_0 ,D_1 ,D_2 ,D_3 } \right) \hspace{0.15cm} \underline{=\left( {0, 0,1, 0} \right)} .$$
:$$e_1 = \frac {1}{-0.4 - {\rm{j}} \cdot 0.6} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm{Re}}[e_1] = \frac {-0.4}{0.4^2 + 0.6^2}\hspace{0.15cm} \underline{ \approx -0.77},\hspace{0.3cm} {\rm{Im}}[e_1] = \frac {0.6}{0.4^2 + 0.6^2} \hspace{0.15cm} \underline{\approx +1.15},$$
:$$e_3 = \frac {1}{-0.4 + {\rm{j}} \cdot 0.6} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} {\rm{Re}}[e_3] = \frac {-0.4}{0.4^2 + 0.6^2}\hspace{0.15cm} \underline{ \approx -0.77},\hspace{0.3cm} {\rm{Im}}[e_3] = \frac {-0.6}{0.4^2 + 0.6^2} \hspace{0.15cm} \underline{\approx -1.15}.$$


'''3.''' Der Vektor <b>R</b> der Spektralkoeffizienten nach dem Kanal könnte analog zur Teilaufgabe b) durch die DFT des Vektors <b>r</b> berechnet werden. Ein alternativer Lösungsweg lautet:
$${\rm\bf{R}} = {\rm\bf{D}} \cdot \left( {\begin{array}{*{20}c}
  {H_0 } & {} & {} & {}  \\
  {} & {H_1 } & {} & {}  \\
  {} & {} & {H_2 } & {}  \\
  {} & {} & {} & {H_3 }  \\
\end{array}} \right)  .$$
Für die Diagonalelemente erhält man:
$$H_\mu = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e}}^{ -
{\rm{j\hspace{0.04cm}\cdot \hspace{0.04cm}2\pi }}
\hspace{0.04cm}\cdot \hspace{0.04cm} l \hspace{0.04cm}\cdot
\hspace{0.04cm}{\mu }/{4}} }$$
$$\Rightarrow \hspace{0.3cm} H_0 = 1,\hspace{0.1cm}H_1 = -0.4 - {\rm{j}} \cdot
0.6,\hspace{0.1cm}H_2 = -0.2,\hspace{0.1cm}H_3 = -0.4 + {\rm{j}}
\cdot 0.6 $$
$$\Rightarrow \hspace{0.3cm}{\rm\bf{R}} = \left( {R_0 ,R_1 ,R_2 ,R_3
} \right) \hspace{0.15cm} \underline{= \left( {\hspace{0.15cm}0,\hspace{0.15cm}0,-0.2, \hspace{0.15cm}0} \right)}  .$$


'''4.''' Die Entzerrerkoeffizienten ergeben sich zu <i>e<sub>&mu;</sub></i> = 1/<i>H<sub>&mu;</sub></i>. Mit dem Ergebnis zu Teilaufgabe c) sind die Koeffizienten <i>e</i><sub>0</sub> <u>= 1</u> und <i>e</i><sub>2</sub> <u>= &ndash;5</u> reell, während für <i>&mu;</i> = 1, <i>&mu;</i> = 3 gilt:
'''(5)'''&nbsp; The equalization calculated in&nbsp; '''(4)'''&nbsp; follows the&nbsp; "zero forcing"&nbsp; approach &nbsp; &rArr; &nbsp; <u>solution 1</u>.
$$e_1 = \frac {1}{-0.4 - {\rm{j}} \cdot 0.6}$$
$$\Rightarrow \hspace{0.3cm} {\rm{Re}}[e_1] = \frac {-0.4}{0.4^2 + 0.6^2}\hspace{0.15cm} \underline{ \approx -0.77},\hspace{0.3cm} {\rm{Im}}[e_1] = \frac {0.6}{0.4^2 + 0.6^2} \hspace{0.15cm} \underline{\approx 1.15},$$
$$e_3 = \frac {1}{-0.4 + {\rm{j}} \cdot 0.6}$$
$$\Rightarrow \hspace{0.3cm} {\rm{Re}}[e_3] = \frac {-0.4}{0.4^2 + 0.6^2}\hspace{0.15cm} \underline{ \approx -0.77},\hspace{0.3cm} {\rm{Im}}[e_3] = \frac {-0.6}{0.4^2 + 0.6^2} \hspace{0.15cm} \underline{\approx -1.15}.$$
'''5.''' Die unter d) berechnete Entzerrung folgt dem <u>&bdquo;Zero Forcing&rdquo;&ndash;Ansatz</u>.


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[[Category:Aufgaben zu  Modulationsverfahren|^5.6 Realisierung von OFDM-Systemen^]]
[[Category:Modulation Methods: Exercises|^5.6 Realization of OFDM Systems^]]
[[de:Aufgaben:Aufgabe 5.8: Entzerrung in Matrix–Vektor–Notation]]

Latest revision as of 17:57, 16 March 2026

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.

  • Only a single frame is considered and for the transmission vector  (in the time domain),  we apply:
$${\rm\bf{d}} = (d_0, \ d_1,\ d_2,\ d_3 ) = (+1, -1, +1, -1 ).$$
  • 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,  instead of the extended transmission vector with the associated transmission matrix  ${\rm\bf{H}}_{\rm ext}$,  we use the cyclic transmission matrix
$${\rm\bf{H}}_{\rm{C}} = \left( {\begin{array}{*{20}c} {h_0 } & {h_1 } & {h_2 } & {} \\ {} & {h_0 } & {h_1 } & {h_2 } \\ \hline {h_2 } & {} & {h_0 } & {h_1 } \\ {h_1 } & {h_2 } & {} & {h_0 } \\ \end{array}} \right).$$
  • For the spectral coefficients at the receiver,  according to the Discrete Fourier Transform  $\rm (DFT)$:
$${\rm\bf{R}} = {\rm\bf{D}} \cdot \left( {\begin{array}{*{20}c} {H_0 } & {} & {} & {} \\ {} & {H_1 } & {} & {} \\ {} & {} & {H_2 } & {} \\ {} & {} & {} & {H_3 } \\ \end{array}} \right) ,$$
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  $ e_\mu = {1}/{H_\mu }.$





Notes:

  • 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

1 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.

${\rm Re}\big[r_0\big] \ = \ $
${\rm Im}\big[r_0\big] \ = \ $
${\rm Re}\big[r_1\big] \ = \ $
${\rm Im}\big[r_1\big] \ = \ $

2 What are the discrete spectral domain coefficients  ${\rm\bf{D}}= (D_0, D_1, D_2, D_3)$  at the transmitter?  Enter  $D_2$  and  $D_3$  as a check.

${\rm Re}\big[D_2\big] \ = \ $
${\rm Im}\big[D_2\big] \ = \ $
${\rm Re}\big[D_3\big] \ = \ $
${\rm Im}\big[D_3\big] \ = \ $

3 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.

${\rm Re}\big[R_2\big] \ = \ $
${\rm Im}\big[R_2\big] \ = \ $
${\rm Re}\big[R_3\big] \ = \ $
${\rm Im}\big[R_3\big] \ = \ $

4 Determine the discrete equalizer coefficients  ${\rm\bf{e}}= (e_0, e_1, e_2, e_3)$.

${\rm Re}\big[e_0\big] \ = \ $
${\rm Im}\big[e_0\big] \ = \ $
${\rm Re}\big[e_1\big] \ = \ $
${\rm Im}\big[e_1\big] \ = \ $
${\rm Re}\big[e_2\big] \ = \ $
${\rm Im}\big[e_2\big] \ = \ $
${\rm Re}\big[e_3\big] \ = \ $
${\rm Im}\big[e_3\big] \ = \ $

5 What is the name of the equalization approach used?

As "Zero Forcing" approach,
as "Matched Filter" approach,
as "Minimum Mean Square Error  $\rm (MMSE)$”approach.


Solution

Solution

(1)  The discrete time domain values at the receiver are calculated using the cyclic transmission matrix  ${\rm\bf{H}}_{\rm{C}} $  as follows:

$${\rm\bf{r}} = {\rm\bf{d}} \cdot {\rm\bf{H}}_{\rm{C}} = \left( {+1 ,-1 ,+1 ,-1 } \right) \cdot \left( {\begin{array}{*{20}c} {0 } & {0.6 } & {0.4 } & {} \\ {} & {0 } & {0.6 } & {0.4 } \\ \hline {0.4 } & {} & {0 } & {0.6 } \\ {0.6 } & {0.4 } & {} & {0 } \\ \end{array}} \right)$$
$$\Rightarrow \hspace{0.3cm}{\rm\bf{r}} = \left( {r_0 ,r_1 ,r_2 ,r_3 } \right) = \left( {-0.2, +0.2,-0.2, +0.2} \right) \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Re}[r_0]\hspace{0.15cm} \underline{=-0.2},\hspace{0.2cm} {\rm Im}[r_0]\hspace{0.15cm} \underline{=0}, \hspace{0.2cm}{\rm Re}[r_1]\hspace{0.15cm} \underline{=+0.2},\hspace{0.2cm} {\rm Im}[r_1]\hspace{0.15cm} \underline{=0}. $$


(2)  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:
$${\rm\bf{D}} = \left( {D_0 ,D_1 ,D_2 ,D_3 } \right) =\left( {0, 0,1, 0} \right)\hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Re}[d_2]\hspace{0.15cm} \underline{=1},\hspace{0.2cm} {\rm Im}[d_2]\hspace{0.15cm} \underline{=0}, \hspace{0.2cm}{\rm Re}[d_3]\hspace{0.15cm} \underline{=0},\hspace{0.2cm} {\rm Im}[d_3]\hspace{0.15cm} \underline{=0}.$$


(3)  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:
$${\rm\bf{R}} = {\rm\bf{D}} \cdot \left( {\begin{array}{*{20}c}{H_0 } & {} & {} & {} \\{} & {H_1 } & {} & {} \\{} & {} & {H_2 } & {} \\{} & {} & {} & {H_3 } \\\end{array}} \right) .$$
  • For the diagonal elements one obtains:
$$H_\mu = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e}}^{ -{\rm{j\hspace{0.04cm}\cdot \hspace{0.04cm}2\pi }}\hspace{0.04cm}\cdot \hspace{0.04cm} l \hspace{0.04cm}\cdot\hspace{0.04cm}{\mu }/{4}} } \hspace{0.3cm}\Rightarrow \hspace{0.3cm} H_0 = 1,\hspace{0.1cm}H_1 = -0.4 - {\rm{j}} \cdot0.6,\hspace{0.1cm}H_2 = -0.2,\hspace{0.1cm}H_3 = -0.4 + {\rm{j}}\cdot 0.6 $$:$$\Rightarrow \hspace{0.3cm}{\rm\bf{R}} = \left( {R_0 ,R_1 ,R_2 ,R_3} \right)= \left( \hspace{0.15cm}0,\hspace{0.15cm}0,\hspace{0.15cm}-0.2, \hspace{0.15cm}0 \right) \hspace{0.3cm}\Rightarrow \hspace{0.3cm}{\rm Re}[r_2]\hspace{0.15cm} \underline{=-0.2},\hspace{0.2cm} {\rm Im}[r_2]\hspace{0.15cm} \underline{=0}, \hspace{0.2cm}{\rm Re}[r_3]\hspace{0.15cm} \underline{=0},\hspace{0.2cm} {\rm Im}[r_3]\hspace{0.15cm} \underline{=0}.$$


(4)  The equalizer coefficients result in  $e_\mu = 1/H_\mu$.

  • With the result from subtask  (3),  the coefficients  $e_0 = 1$,  $e_2 = -5$ 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)  The equalization calculated in  (4)  follows the  "zero forcing"  approach   ⇒   solution 1.

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