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

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{{quiz-Header|Buchseite=Modulationsverfahren/Realisierung von OFDM-Systemen
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{{quiz-Header|Buchseite=Modulation_Methods/Implementation_of_OFDM_Systems
 
}}
 
}}
  
−
[[File:P_ID1663__A_5_8.png|right|]]
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[[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. Es wird nur ein einziger Rahmen betrachtet und für den Sendevektor (im Zeitbereich) gelte:
+
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.
−
$${\rm\bf{d}} = (d_0, d_1,d_2,d_3 ) = (+1, -1, +1, -1 ).$$
+
*Only a single frame is considered and for the transmission vector  (in the time domain),  we apply:
−
Die Kanalimpulsantwort sei beschrieben durch
+
:$${\rm\bf{d}} = (d_0, \ d_1,\ d_2,\ d_3 ) = (+1, -1, +1, -1 ).$$
−
$${\rm\bf{h}} = (h_0, h_1,h_2 ) = (0, 0.6, 0.4 ).$$
+
*Let the channel impulse response be described by
−
Zur Repräsentation des zyklischen Präfixes verwenden wir in dieser Aufgabe statt des erweiterten Sendevektors mit der zugehörigen Übertragungsmatrix $H_{ext}$ die zyklische Übertragungsmatrix
+
:$${\rm\bf{h}} = (h_0, \ h_1,\ h_2 ) = (0, \ 0.6, \ 0.4 ).$$
−
$${\rm\bf{H}}_{\rm{C}} = \left( {\begin{array}{*{20}c} {h_0 } & {h_1 } & {h_2 } & {} \\ {} & {h_0 } & {h_1 } & {h_2 } \\ \hline {h_2 } & {} & {h_0 } & {h_1 } \\ {h_1 } & {h_2 } & {} & {h_0 } \\ \end{array}} \right).$$
+
*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
−
Für die Spektralkoeffizienten am Empfänger gelte nach der Diskreten Fouriertransformation (DFT):
+
:$${\rm\bf{H}}_{\rm{C}} = \left( {\begin{array}{*{20}c} {h_0 } & {h_1 } & {h_2 } & {} \\ {} & {h_0 } & {h_1 } & {h_2 } \\ \hline {h_2 } & {} & {h_0 } & {h_1 } \\ {h_1 } & {h_2 } & {} & {h_0 } \\ \end{array}} \right).$$
−
$${\rm\bf{R}} = {\rm\bf{D}} \cdot \left( {\begin{array}{*{20}c} {H_0 } & {} & {} & {} \\ {} & {H_1 } & {} & {} \\ {} & {} & {H_2 } & {} \\ {} & {} & {} & {H_3 } \\ \end{array}} \right) ,$$
+
*For the spectral coefficients at the receiver,  according to the Discrete Fourier Transform  $\rm (DFT)$:
−
wobei die Diagonalelemente wie folgt zu berechnen sind:
+
:$${\rm\bf{R}} = {\rm\bf{D}} \cdot \left( {\begin{array}{*{20}c} {H_0 } & {} & {} & {} \\ {} & {H_1 } & {} & {} \\ {} & {} & {H_2 } & {} \\ {} & {} & {} & {H_3 } \\ \end{array}} \right) ,$$
−
$$H_\mu = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e}}^{ - {\rm{j\hspace{0.04cm}\cdot \hspace{0.04cm}2\pi }} \hspace{0.04cm}\cdot \hspace{0.04cm} l \hspace{0.04cm}\cdot \hspace{0.04cm} {\mu }/{4}} } .$$
+
:where the diagonal elements are to be calculated as follows:
−
Die Entzerrung am Empfänger erfolgt durch Multiplikation im Frequenzbereich mit den Koeffizienten
+
:$$H_\mu = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e}}^{ - {\rm{j\hspace{0.04cm}\cdot \hspace{0.04cm}2\pi }} \hspace{0.04cm}\cdot \hspace{0.04cm} l \hspace{0.04cm}\cdot \hspace{0.04cm} {\mu }/{4}} } .$$
−
$$ e_\mu = {1}/{H_\mu }.$$
+
*The equalization at the receiver is done by multiplication in the frequency domain by the coefficients  $ e_\mu = {1}/{H_\mu }.$
−
'''Hinweis:''' Diese Aufgabe bezieht sich auf die theoretischen Grundlagen von [http://en.lntwww.de/Modulationsverfahren/Realisierung_von_OFDM-Systemen Kapitel 5.6] dieses Buches sowie auf das [http://en.lntwww.de/Signaldarstellung/Diskrete_Fouriertransformation_(DFT) Kapitel 5.2] im Buch „Signaldarstellung”. 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 Empfangsswerte $r = (r_0, r_1, r_2, r_3)$ 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="{}"}
−
Re{$r_0$} = { -0.2 3% }
+
${\rm Re}\big[r_0\big] \ = \ $ { -0.202--0.198 }
−
Im{$r_0$} = { 0 3% }  
+
${\rm Im}\big[r_0\big] \ = \ $ { 0. }  
−
Re{$r_1$} = { 0.2 3% }
+
${\rm Re}\big[r_1\big] \ = \ $ { 0.2 1% }
−
Im{$r_1$} = { 0 3% }  
+
${\rm Im}\big[r_1\big] \ = \ $ { 0. }  
  
−
{Wie lauten die diskreten Spektralbereichskoeffizienten <b>D</b> = (<i>D</i><sub>0</sub>, <i>D</i><sub>1</sub>, <i>D</i><sub>2</sub>, <i>D</i><sub>3</sub>) am Sender?  Geben Sie zur Kontrolle <i>D</i><sub>2</sub> und <i>D</i><sub>3</sub> 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="{}"}
−
Re{$D_2$}= { 1 3% }
+
${\rm Re}\big[D_2\big] \ = \ $ { 1 3% }
−
Im{$D_2$} = { 0 3% }
+
${\rm Im}\big[D_2\big] \ = \ $ { 0. }
−
Re{$D_3$}= { 0 3% }
+
${\rm Re}\big[D_3\big] \ = \ $ { 0. }
−
Im{$D_3$} = { 0 3% }
+
${\rm Im}\big[D_3\big] \ = \ $ { 0. }
  
−
{Berechnen Sie die diskreten Spektralkoeffizienten <b>R</b> = (<i>R</i><sub>0</sub>, <i>R</i><sub>1</sub>, <i>R</i><sub>2</sub>, <i>R</i><sub>3</sub>) nach dem Kanal. Geben Sie zur Kontrolle <i>R</i><sub>2</sub> und <i>R</i><sub>3</sub> 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="{}"}
−
Re{$R_2$} =  { -0.2 3% }  
+
${\rm Re}\big[R_2\big] \ = \ $   { -0.202--0.198 }  
−
Im{$R_2$} = { 0 3% }  
+
${\rm Im}\big[R_2\big] \ = \ $ { 0. }  
−
Re{$R_3$} = { 0 3% }  
+
${\rm Re}\big[R_3\big] \ = \ $  { 0. }  
−
Im{$R_3$} = { 0 3% }
+
${\rm Im}\big[R_3\big] \ = \ $ { 0. }
  
−
{ Bestimmen Sie die diskreten Entzerrerkoeffizienten <b>e</b> = (<i>e</i><sub>0</sub>, <i>e</i><sub>1</sub>, <i>e</i><sub>2</sub>, <i>e</i><sub>3</sub>):
+
{ Determine the discrete equalizer coefficients &nbsp;${\rm\bf{e}}= (e_0, e_1, e_2, e_3)$.
 
|type="{}"}
 
|type="{}"}
−
Re{$e_0$} = { 1 3% }
+
${\rm Re}\big[e_0\big] \ = \ $ { 1 1% }
−
Im{$e_0$}  = { 0 3% }
+
${\rm Im}\big[e_0\big] \ = \ $ { 0. }
−
Re{$e_1$} = { -0.77 3% }
+
${\rm Re}\big[e_1\big] \ = \ $ { -0.78--0.76 }
−
Im{$e_1$}  = { 1.15 3% }
+
${\rm Im}\big[e_1\big] \ = \ $ { 1.15 1% }
−
Re{$e_2$} = { -5 3% }
+
${\rm Re}\big[e_2\big] \ = \ $ { -5.05--4.95 }
−
Im{$e_2$}  = { 0 3% }
+
${\rm Im}\big[e_2\big] \ = \ $ { 0. }
−
Re{$e_3$} = { -0.77 3% }
+
${\rm Re}\big[e_3\big] \ = \ $ { -0.78--0.76 }
−
Im{$e_3$}  = { -1.15 3% }
+
${\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===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
−
'''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:
−
$${\rm\bf{r}} = {\rm\bf{d}} \cdot {\rm\bf{H_C}} = \left( {+1 ,-1 ,+1 ,-1 } \right) \cdot \left( {\begin{array}{*{20}c} {0 } & {0.6 } & {0.4 } & {} \\ {} & {0 } & {0.6 } & {0.4 } \\ \hline {0.4 } & {} & {0 } & {0.6 } \\ {0.6 } & {0.4 } & {} & {0 } \\ \end{array}} \right)$$
+
:$${\rm\bf{r}} = {\rm\bf{d}} \cdot {\rm\bf{H}}_{\rm{C}} = \left( {+1 ,-1 ,+1 ,-1 } \right) \cdot \left( {\begin{array}{*{20}c} {0 } & {0.6 } & {0.4 } & {} \\ {} & {0 } & {0.6 } & {0.4 } \\ \hline {0.4 } & {} & {0 } & {0.6 } \\ {0.6 } & {0.4 } & {} & {0 } \\ \end{array}} \right)$$
−
$$\Rightarrow \hspace{0.3cm}{\rm\bf{r}} = \left( {r_0 ,r_1 ,r_2 ,r_3 } \right) = \left( {-0.2, +0.2,-0.2, +0.2} \right) .$$
+
:$$\Rightarrow \hspace{0.3cm}{\rm\bf{r}} = \left( {r_0 ,r_1 ,r_2 ,r_3 } \right) = \left( {-0.2, +0.2,-0.2, +0.2} \right) \hspace{0.3cm}
−
Damit sind alle Empfangswerte <i>r</i><sub>0</sub> <u>= &ndash;0.2,</u> <i>r</i><sub>1</sub> <u>= +0.2</u>, <i>r</i><sub>2</sub> = <u>&ndash;0.2</u> und <u><i>r</i><sub>3</sub> = +0.2</u> rein reell.
+
\Rightarrow \hspace{0.3cm}{\rm Re}[r_0]\hspace{0.15cm} \underline{=-0.2},\hspace{0.2cm} {\rm Im}[r_0]\hspace{0.15cm} \underline{=0}, \hspace{0.2cm}{\rm Re}[r_1]\hspace{0.15cm} \underline{=+0.2},\hspace{0.2cm} {\rm Im}[r_1]\hspace{0.15cm} \underline{=0}. $$
 +
 
 +
 
 +
'''(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:
 +
:$${\rm\bf{D}} = \left( {D_0 ,D_1 ,D_2 ,D_3 } \right) =\left( {0, 0,1, 0} \right)\hspace{0.3cm}
 +
\Rightarrow \hspace{0.3cm}{\rm Re}[d_2]\hspace{0.15cm} \underline{=1},\hspace{0.2cm} {\rm Im}[d_2]\hspace{0.15cm} \underline{=0}, \hspace{0.2cm}{\rm Re}[d_3]\hspace{0.15cm} \underline{=0},\hspace{0.2cm} {\rm Im}[d_3]\hspace{0.15cm} \underline{=0}.$$
  
−
'''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:
 
−
$${\rm\bf{D}} = \left( {D_0 ,D_1 ,D_2 ,D_3 } \right) \hspace{0.15cm} \underline{=\left( {0, 0,1, 0} \right)} .$$
 
  
−
'''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:
+
'''(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;
−
$${\rm\bf{R}} = {\rm\bf{D}} \cdot \left( {\begin{array}{*{20}c}
+
*An alternative solution path is:
 +
:$${\rm\bf{R}} = {\rm\bf{D}} \cdot \left( {\begin{array}{*{20}c}
 
   {H_0 } & {} & {} & {}  \\
 
   {H_0 } & {} & {} & {}  \\
 
   {} & {H_1 } & {} & {}  \\
 
   {} & {H_1 } & {} & {}  \\
Line 80: Line 97:
 
   {} & {} & {} & {H_3 }  \\
 
   {} & {} & {} & {H_3 }  \\
 
\end{array}} \right)  .$$
 
\end{array}} \right)  .$$
−
Für die Diagonalelemente erhält man:
+
*For the diagonal elements one obtains:
−
$$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 }}
 
{\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}\cdot \hspace{0.04cm} l \hspace{0.04cm}\cdot
−
\hspace{0.04cm}{\mu }/{4}} }$$
+
\hspace{0.04cm}{\mu }/{4}} } \hspace{0.3cm}
−
$$\Rightarrow \hspace{0.3cm} H_0 = 1,\hspace{0.1cm}H_1 = -0.4 - {\rm{j}} \cdot
+
\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
+
:$$\Rightarrow \hspace{0.3cm}{\rm\bf{R}} = \left( {R_0 ,R_1 ,R_2 ,R_3
−
} \right) \hspace{0.15cm} \underline{= \left( {\hspace{0.15cm}0,\hspace{0.15cm}0,-0.2, \hspace{0.15cm}0} \right)} .$$
+
} \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:
 +
:$$e_1 = \frac {1}{-0.4 - {\rm{j}} \cdot 0.6} \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm{Re}}[e_1] = \frac {-0.4}{0.4^2 + 0.6^2}\hspace{0.15cm} \underline{ \approx -0.77},\hspace{0.3cm} {\rm{Im}}[e_1] = \frac {0.6}{0.4^2 + 0.6^2} \hspace{0.15cm} \underline{\approx +1.15},$$
 +
:$$e_3 = \frac {1}{-0.4 + {\rm{j}} \cdot 0.6} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} {\rm{Re}}[e_3] = \frac {-0.4}{0.4^2 + 0.6^2}\hspace{0.15cm} \underline{ \approx -0.77},\hspace{0.3cm} {\rm{Im}}[e_3] = \frac {-0.6}{0.4^2 + 0.6^2} \hspace{0.15cm} \underline{\approx -1.15}.$$
 +
 
  
−
'''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>.
 
  
 
{{ML-Fuß}}
 
{{ML-Fuß}}
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−
[[Category:Aufgaben zu  Modulationsverfahren|^5.6 Realisierung von OFDM-Systemen^]]
+
[[Category:Modulation Methods: Exercises|^5.6 Realization of OFDM Systems^]]

Latest revision as of 16:50, 31 January 2022

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:
$$H_\mu = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e}}^{ - {\rm{j\hspace{0.04cm}\cdot \hspace{0.04cm}2\pi }} \hspace{0.04cm}\cdot \hspace{0.04cm} l \hspace{0.04cm}\cdot \hspace{0.04cm}{\mu }/{4}} } \hspace{0.3cm} \Rightarrow \hspace{0.3cm} H_0 = 1,\hspace{0.1cm}H_1 = -0.4 - {\rm{j}} \cdot 0.6,\hspace{0.1cm}H_2 = -0.2,\hspace{0.1cm}H_3 = -0.4 + {\rm{j}} \cdot 0.6 $$
$$\Rightarrow \hspace{0.3cm}{\rm\bf{R}} = \left( {R_0 ,R_1 ,R_2 ,R_3 } \right)= \left( \hspace{0.15cm}0,\hspace{0.15cm}0,\hspace{0.15cm}-0.2, \hspace{0.15cm}0 \right) \hspace{0.3cm} \Rightarrow \hspace{0.3cm}{\rm Re}[r_2]\hspace{0.15cm} \underline{=-0.2},\hspace{0.2cm} {\rm Im}[r_2]\hspace{0.15cm} \underline{=0}, \hspace{0.2cm}{\rm Re}[r_3]\hspace{0.15cm} \underline{=0},\hspace{0.2cm} {\rm Im}[r_3]\hspace{0.15cm} \underline{=0}.$$


(4)  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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