Difference between revisions of "Aufgaben:Exercise 5.8: Equalization in Matrix Vector Notation"
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| − | {{quiz-Header|Buchseite= | + | {{quiz-Header|Buchseite=Modulation_Methods/Implementation_of_OFDM_Systems |
}} | }} | ||
| − | [[File: | + | [[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. | |
| − | * | + | *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 |
:$${\rm\bf{h}} = (h_0, \ h_1,\ h_2 ) = (0, \ 0.6, \ 0.4 ).$$ | :$${\rm\bf{h}} = (h_0, \ h_1,\ h_2 ) = (0, \ 0.6, \ 0.4 ).$$ | ||
| − | * | + | *To represent the cyclic prefix in this task, instead of the extended 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 $\rm (DFT)$: |
:$${\rm\bf{R}} = {\rm\bf{D}} \cdot \left( {\begin{array}{*{20}c} {H_0 } & {} & {} & {} \\ {} & {H_1 } & {} & {} \\ {} & {} & {H_2 } & {} \\ {} & {} & {} & {H_3 } \\ \end{array}} \right) ,$$ | :$${\rm\bf{R}} = {\rm\bf{D}} \cdot \left( {\begin{array}{*{20}c} {H_0 } & {} & {} & {} \\ {} & {H_1 } & {} & {} \\ {} & {} & {H_2 } & {} \\ {} & {} & {} & {H_3 } \\ \end{array}} \right) ,$$ | ||
| − | : | + | :where the diagonal elements are to be calculated as follows: |
:$$H_\mu = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e}}^{ - {\rm{j\hspace{0.04cm}\cdot \hspace{0.04cm}2\pi }} \hspace{0.04cm}\cdot \hspace{0.04cm} l \hspace{0.04cm}\cdot \hspace{0.04cm} {\mu }/{4}} } .$$ | :$$H_\mu = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e}}^{ - {\rm{j\hspace{0.04cm}\cdot \hspace{0.04cm}2\pi }} \hspace{0.04cm}\cdot \hspace{0.04cm} l \hspace{0.04cm}\cdot \hspace{0.04cm} {\mu }/{4}} } .$$ | ||
| − | * | + | *The equalization at the receiver is done by multiplication in the frequency domain by the coefficients $ e_\mu = {1}/{H_\mu }.$ |
| Line 24: | Line 24: | ||
| − | + | 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: |
| − | :$${\rm\bf{F}} = \left( {\begin{array}{*{20}c} 1 & 1 & \cdots & 1 \\ 1 & {} & {} & {} \\ \vdots & {} & {{\rm{e}}^{ - {\rm{j2\pi }}{\kern 1pt} \nu {\kern 1pt} \mu /N} } & {} \\ 1 & {} & {} & {} \\ \end{array}} \right), \qquad {\rm{DFT\; | + | :$${\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> | ||
| − | { | + | {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 } | ||
| Line 43: | Line 43: | ||
${\rm Im}\big[r_1\big] \ = \ $ { 0. } | ${\rm Im}\big[r_1\big] \ = \ $ { 0. } | ||
| − | { | + | {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. |
|type="{}"} | |type="{}"} | ||
${\rm Re}\big[D_2\big] \ = \ $ { 1 3% } | ${\rm Re}\big[D_2\big] \ = \ $ { 1 3% } | ||
| Line 50: | Line 50: | ||
${\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. |
|type="{}"} | |type="{}"} | ||
${\rm Re}\big[R_2\big] \ = \ $ { -0.202--0.198 } | ${\rm Re}\big[R_2\big] \ = \ $ { -0.202--0.198 } | ||
| Line 57: | Line 57: | ||
${\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)$. |
|type="{}"} | |type="{}"} | ||
${\rm Re}\big[e_0\big] \ = \ $ { 1 1% } | ${\rm Re}\big[e_0\big] \ = \ $ { 1 1% } | ||
| Line 68: | Line 68: | ||
${\rm Im}\big[e_3\big] \ = \ $ { -1.16--1.14 } | ${\rm Im}\big[e_3\big] \ = \ $ { -1.16--1.14 } | ||
| − | { | + | {What is the name of the equalization approach used? |
|type="()"} | |type="()"} | ||
| − | + | + | + As "Zero Forcing" approach, |
| − | - | + | - as "Matched Filter" approach, |
| − | - | + | - as "Minimum Mean Square Error $\rm (MMSE)$”approach. |
</quiz> | </quiz> | ||
| − | === | + | ===Solution=== |
{{ML-Kopf}} | {{ML-Kopf}} | ||
| − | '''(1)''' | + | '''(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} | :$$\Rightarrow \hspace{0.3cm}{\rm\bf{r}} = \left( {r_0 ,r_1 ,r_2 ,r_3 } \right) = \left( {-0.2, +0.2,-0.2, +0.2} \right) \hspace{0.3cm} | ||
| Line 83: | Line 83: | ||
| − | '''(2)''' | + | '''(2)''' The spectral coefficients ${\rm\bf{D}}$ result directly from the Discrete Fourier Transform $\rm (DFT)$ of the time domain coefficients ${\rm\bf{d}}= (+1, -1, +1, -1)$. |
| − | * | + | *This time domain sequence corresponds to a discrete cosine function with twice the fundamental frequency $(2 \cdot f_0)$ and amplitude $1$. It follows: |
:$${\rm\bf{D}} = \left( {D_0 ,D_1 ,D_2 ,D_3 } \right) =\left( {0, 0,1, 0} \right)\hspace{0.3cm} | :$${\rm\bf{D}} = \left( {D_0 ,D_1 ,D_2 ,D_3 } \right) =\left( {0, 0,1, 0} \right)\hspace{0.3cm} | ||
\Rightarrow \hspace{0.3cm}{\rm Re}[d_2]\hspace{0.15cm} \underline{=1},\hspace{0.2cm} {\rm Im}[d_2]\hspace{0.15cm} \underline{=0}, \hspace{0.2cm}{\rm Re}[d_3]\hspace{0.15cm} \underline{=0},\hspace{0.2cm} {\rm Im}[d_3]\hspace{0.15cm} \underline{=0}.$$ | \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}.$$ | ||
| − | '''(3)''' | + | '''(3)''' The vector ${\rm\bf{R}}$ of spectral coefficients after the channel could be calculated by DFT of the vector ${\rm\bf{r}}$, analogous to subtask '''(2)'''. |
| − | * | + | *An alternative solution path is: |
:$${\rm\bf{R}} = {\rm\bf{D}} \cdot \left( {\begin{array}{*{20}c} | :$${\rm\bf{R}} = {\rm\bf{D}} \cdot \left( {\begin{array}{*{20}c} | ||
{H_0 } & {} & {} & {} \\ | {H_0 } & {} & {} & {} \\ | ||
| Line 97: | Line 97: | ||
{} & {} & {} & {H_3 } \\ | {} & {} & {} & {H_3 } \\ | ||
\end{array}} \right) .$$ | \end{array}} \right) .$$ | ||
| − | * | + | *For the diagonal elements one obtains: |
:$$H_\mu = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e}}^{ - | :$$H_\mu = \sum\limits_{l = 0}^2 {h_l \cdot {\rm{e}}^{ - | ||
{\rm{j\hspace{0.04cm}\cdot \hspace{0.04cm}2\pi }} | {\rm{j\hspace{0.04cm}\cdot \hspace{0.04cm}2\pi }} | ||
| Line 110: | Line 110: | ||
| − | '''(4)''' | + | '''(4)''' The equalizer coefficients result in $e_\mu = 1/H_\mu$. |
| − | * | + | *With the result from subtask '''(3)''', the coefficients $e_0 = 1$, $e_2 = -5$ are real: |
:$${\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}.$$ | :$${\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_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}.$$ | :$$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}.$$ | ||
| − | '''(5)''' | + | '''(5)''' The equalization calculated in '''(4)''' follows the "zero forcing" approach ⇒ <u>solution 1</u>. |
{{ML-Fuß}} | {{ML-Fuß}} | ||
| Line 125: | Line 125: | ||
| − | [[Category:Modulation Methods: Exercises|^5.6 | + | [[Category:Modulation Methods: Exercises|^5.6 Realization of OFDM Systems^]] |
Latest revision as of 16:50, 31 January 2022
Block diagram of OFDM transmission
We consider the blocks of an OFDM system shown in the diagram, assuming a system with $N = 4$ carriers and a channel with $L = 2$ echoes.
- Only a single frame is considered and for the transmission vector (in the time domain), we apply:
- $${\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}} } .$$
- The equalization at the receiver is done by multiplication in the frequency domain by the coefficients $ e_\mu = {1}/{H_\mu }.$
Notes:
- The exercise belongs to the chapter Implementation of OFDM Systems.
- Reference is also made to the chapter Discrete Fourier Transform in the book "Signal Representation".
- For the Discrete Fourier Transform (DFT) in matrix-vector notation holds:
- $${\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}}^*.$$
Questions
Solution
(1) The discrete time domain values at the receiver are calculated using the cyclic transmission matrix ${\rm\bf{H}}_{\rm{C}} $ as follows:
- $${\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.
