Aufgaben:Exercise 5.2: Inverse Discrete Fourier Transform: Difference between revisions

Latest revision as of 17:56, 16 March 2026

Five different sets for the spectral coefficients $D(\mu)$

With the Discrete Fourier Transform $\rm (DFT)$,

the $N$ spectral range coefficients $D(\mu)$ are calculated

from the $N$ time coefficients $d(\nu)$ ⇒ samples of the continuous-time signal $x(t)$.

With $\nu = 0$, ... , $N – 1$ and $\mu = 0$, ... , $N – 1$ holds:

$$D(\mu) = \frac{1}{N} \cdot \sum_{\nu = 0 }^{N-1}d(\nu)\cdot {w}^{\hspace{0.05cm}\nu \hspace{0.05cm} \cdot \hspace{0.05cm}\mu} \hspace{0.05cm}.$$

Here $w$ denotes the complex rotation factor:

$$w = {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2 \pi /N}= \cos \left( {2 \pi}/{N}\right)-{\rm j} \cdot \sin \left( {2 \pi}/{N}\right)\hspace{0.05cm}.$$

For the Inverse Discrete Fourier Transform $\rm (DFT)$ ⇒ "inverse function" of the DFT, the following applies accordingly:

$$d(\nu) = \sum_{\mu = 0 }^{N-1}D(\mu) \cdot {w}^{-\nu \hspace{0.05cm} \cdot \hspace{0.05cm}\mu} \hspace{0.05cm}.$$

In this task, the time coefficients $d(\nu)$ are to be determined for various sequences $D(\mu)$ (which are labelled $\rm A$, ... , $\rm E$ in the table above). Thus, $N = 8$ always applies.

Hints:

This task belongs to the chapter Discrete Fourier Transformation (DFT).

The topic dealt with here is also dealt with in the interactive applet Discrete Fourier Transform and Inverse.

Questions

$d(0)\ = \ $

$d(1)\ = \ $

$d(0)\ = \ $

$d(1)\ = \ $

$d(0)\ = \ $

$d(1)\ = \ $

$d(0)\ = \ $

$d(1)\ = \ $

$d(0)\ = \ $

$d(1)\ = \ $

Solution

(1) From the IDFT equation, with $D(\mu) = 0$ for $\mu \ne 0$:

$$d(\nu) = D(0) \cdot w^0 = D(0) =1\hspace{0.5cm}(0 \le \nu \le 7)\ \hspace{0.5cm} \Rightarrow\hspace{0.5cm}\hspace{0.15 cm}\underline{d(0) = d(1) = 1}.$$

This parameter set describes the discrete form of the Fourier correspondence of the DC signal:

$$x(t) = 1 \hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm}X(f) = {\delta}(f) \hspace{0.05cm}.$$

(2) All spectral coefficients are zero except $D_1 = D_7 = 0.5$. It follows for $0 ≤ ν ≤ 7$:

$$d(\nu) = 0.5 \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /4) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} + 0.5 \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /4) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu}\hspace{0.05cm}.$$

However, due to periodicity, also holds:

$$d(\nu) = 0.5 \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /4) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} + 0.5 \cdot {\rm e}^{{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /4) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} = \cos \left({\pi}/{4} \cdot \nu \right) \hspace{0.3cm} \Rightarrow \hspace{0.3cm}\hspace{0.15 cm}\underline{d(0) = 1}, \hspace{0.2cm}\hspace{0.15 cm}\underline{d(1) = {1}/{\sqrt{2}} \approx 0.707}\hspace{0.05cm}.$$

It is therefore the discrete-time equivalent of

$$x(t) = \cos(2 \pi \cdot f_{\rm A} \cdot t) \hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm}X(f) = {1}/{2} \cdot {\delta}(f + f_{\rm A}) + {1}/{2} \cdot {\delta}(f - f_{\rm A}) \hspace{0.05cm},$$

where $f_{\rm A}$ denotes the smallest frequency that can be represented in the DFT.

(3) Compared to subtask (2), the oscillation frequency is now twice as large, namely $2 f_{\rm A}$ instead of $f_{\rm A}$:

$$x(t) = \cos(2 \pi \cdot (2f_{\rm A}) \cdot t) \hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm}X(f) = {1}/{2} \cdot {\delta}(f + 2f_{\rm A}) + {1}/{2} \cdot {\delta}(f - 2f_{\rm A}) \hspace{0.05cm},$$

Thus the sequence $\langle \hspace{0.1cm}d(ν)\hspace{0.1cm}\rangle $ describes two periods of the cosine oscillation, and it holds for $0 ≤ ν ≤ 7$:

$$ d(\nu) = 0.5 \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /2) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} + 0.5 \cdot {\rm e}^{{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /2) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} = \cos \left({\pi}/{2} \cdot \nu \right)\hspace{0.3cm} \Rightarrow \hspace{0.3cm}\hspace{0.15 cm}\underline{d(0) = 1, \hspace{0.2cm}d(1) = 0}\hspace{0.05cm}.$$

(4) By further doubling the cosine frequency to $4 f_{\rm A}$ one finally arrives at the continuous-time Fourier correspondence

$$d(\nu) = 0.5 \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \pi \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} + 0.5 \cdot {\rm e}^{{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \pi \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} = \cos \left(\pi \cdot \nu \right)\hspace{0.05cm}$$

and thus to the time coefficients

$$d(0) =d(2) =d(4) =d(6) \hspace{0.15 cm}\underline{= +1}, \hspace{0.2cm}d(1) =d(3) =d(5) =d(7) \hspace{0.15 cm}\underline{= -1}\hspace{0.05cm}.$$

Note that here the two Dirac functions coincide in the discrete-time representation due to periodicity.
The coefficients $D (+4) = 0.5$ and $D (-4) = 0.5$ together give $D (4) = 1$.

(5) The Discrete Fourier Transform is also linear. Therefore, the superposition principle is still applicable:

The coefficients $D(\mu )$ from column $\rm E$ result as the sums of columns $\rm A$ and $\rm D$.
Therefore, the alternating sequence $\langle \hspace{0.1cm}d(ν) \hspace{0.1cm}\rangle $ becomes the sequence shifted up by $1$ according to subtask (4):

$$ \hspace{0.15 cm}\underline{d(0) =d(2) =d(4) =d(6)= 2}, \hspace{0.2cm}\hspace{0.15 cm}\underline{d(1) =d(3) =d(5) =d(7) = 0}\hspace{0.05cm}.$$

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Signaldarstellung/Diskrete Fouriertransformation (DFT)
+{{quiz-Header|Buchseite=Signal_Representation/Discrete_Fourier_Transform_(DFT)
 }}
-[[File:P_ID1138__Sig_A_5_2.png|250px|right|Verwendete Spektralkoeffizienten]]
+[[File:P_ID1138__Sig_A_5_2.png|250px|right|frame|Five different sets for the spectral coefficients&nbsp; $D(\mu)$]]
-Bei der ''Diskreten Fouriertransformation'' (DFT) werden aus den $N$ Zeitkoeffizienten $d(\nu)$ &nbsp; &rArr;   &nbsp; Abtastwerte des zeitkontinuierlichen Signals $x(t)$ – die $N$ Spektralbereichskoeffizienten $D(\mu)$ berechnet. Mit $\nu = 0, ... , N – 1$ und $\mu = 0, ... , N – 1$ gilt:
+With the&nbsp; '''Discrete Fourier Transform'''&nbsp; $\rm (DFT)$,
+*the&nbsp; $N$&nbsp; spectral range coefficients&nbsp; $D(\mu)$&nbsp;  are calculated
+*from the&nbsp; $N$&nbsp; time coefficients&nbsp; $d(\nu)$ &nbsp; &rArr;   &nbsp; samples of the continuous-time signal&nbsp; $x(t)$.
+With&nbsp; $\nu = 0$, ... , $N – 1$&nbsp; and&nbsp; $\mu = 0$, ... , $N – 1$&nbsp; holds:
-$$D(\mu) = \frac{1}{N} \cdot \sum_{\nu = 0 }^{N-1}
+:$$D(\mu) = \frac{1}{N} \cdot \sum_{\nu = 0 }^{N-1}d(\nu)\cdot  {w}^{\hspace{0.05cm}\nu \hspace{0.05cm} \cdot \hspace{0.05cm}\mu} \hspace{0.05cm}.$$
-  d(\nu)\cdot  {w}^{\hspace{0.05cm}\nu \hspace{0.03cm} \cdot \hspace{0.05cm}\mu} \hspace{0.05cm}.$$
-Hierbei bezeichnet $w$ den komplexen Drehfaktor:
+Here&nbsp; $w$&nbsp; denotes the complex rotation factor:
-$$w  = {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2 \pi /N}
+:$$w  = {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} 2 \pi /N}= \cos \left(  {2 \pi}/{N}\right)-{\rm j} \cdot \sin \left(  {2 \pi}/{N}\right)\hspace{0.05cm}.$$
- = \cos \left(  {2 \pi}/{N}\right)-{\rm j} \cdot \sin \left(  {2 \pi}/{N}\right)
- \hspace{0.05cm}.$$
-Für die ''Inverse Diskrete Fouriertransformation'' (IDFT)  &nbsp; ⇒  &nbsp; „Umkehrfunktion” der DFT  gilt entsprechend:
+For the&nbsp; '''Inverse Discrete Fourier Transform'''&nbsp; $\rm (DFT)$  &nbsp; ⇒  &nbsp; "inverse function" of the DFT, the following applies accordingly:
-$$d(\nu) =  \sum_{\mu = 0 }^{N-1}
+:$$d(\nu) =  \sum_{\mu = 0 }^{N-1}D(\mu) \cdot  {w}^{-\nu \hspace{0.05cm} \cdot \hspace{0.05cm}\mu} \hspace{0.05cm}.$$
- D(\mu) \cdot  {w}^{-\nu \hspace{0.03cm} \cdot \hspace{0.05cm}\mu} \hspace{0.05cm}.$$
-In dieser Aufgabe sollen für verschiedene Beispielfolgen $D(\mu)$ – die in obiger Tabelle mit „A”, ... , „E” bezeichnet sind – die Zeitkoeffizienten $d(\nu)$ ermittelt werden. Es gilt somit stets $N = 8$.
+In this task, the time coefficients&nbsp; $d(\nu)$&nbsp;  are to be determined for various sequences&nbsp; $D(\mu)$&nbsp;  (which are labelled&nbsp; $\rm A$, ... ,&nbsp; $\rm E$&nbsp; in the table above).&nbsp; Thus,&nbsp; $N = 8$ always applies.
-''Hinweise:''
-*Die Aufgabe gehört zum  Kapitel [[Signaldarstellung/Diskrete_Fouriertransformation_(DFT)|Diskrete Fouriertransformation (DFT)]].
-*Sollte die Eingabe des Zahlenwertes &bdquo;0&rdquo; erforderlich sein, so geben Sie bitte &bdquo;0.&rdquo; ein.
-*Zu der hier behandelten Thematik gibt es auch ein Interaktionsmodul:
-:[[Diskrete Fouriertransformation]]
-Hinweis: Die Aufgabe bezieht sich auf die theoretischen Grundlagen von Kapitel 5.2. Diese können Sie sich auch mit folgendem Interaktionsmodul verdeutlichen:
-Diskrete Fouriertransformation
-===Fragebogen===
+''Hints:''
+*This task belongs to the chapter&nbsp; [[Signal_Representation/Discrete_Fourier_Transform_(DFT)|Discrete Fourier Transformation (DFT)]].
+*The topic dealt with here is also dealt with in the interactive applet&nbsp; [[Applets:Discrete_Fouriertransform_and_Inverse|Discrete Fourier Transform and Inverse]].
+===Questions===
 <quiz display=simple>
-{Wie lauten die Zeitkoeffizienten $d(\nu)$ für die $D(\mu)$–Werte von Spalte A?
+{What are the time coefficients&nbsp; $d(\nu)$&nbsp; for the&nbsp; $D(\mu)$&nbsp; values of column&nbsp; $\rm A$?
 |type="{}"}
-$D(\mu )$ gemäß &bdquo;A&rdquo;: &nbsp;&nbsp; $d(0)$ =&nbsp; { 1 3% }
+$d(0)\ = \ $  { 1 3% }
-$d(1)$ =&nbsp; { 1 3% }
+$d(1)\ = \ $ { 1 3% }
-{Wie lauten die Zeitkoeffizienten $d(ν)$ für die $D(\mu)$–Werte von Spalte B?
+{What are the time coefficients&nbsp; $d(ν)$&nbsp; for the&nbsp; $D(\mu)$&nbsp; values of column&nbsp; $\rm B$?
 |type="{}"}
-$D(\mu )$ gemäß B: $d(0) =$ { 1 }
+$d(0)\ = \ $ { 1 3% }
-$D(\mu )$ gemäß B: $d(1) =$ { 0.707 3% }
+$d(1)\ = \ $ { 0.707 3% }
-{Wie lauten die Zeitkoeffizienten $d(ν)$ für die $D(\mu)$ –Werte von Spalte C?
+{What are the time coefficients&nbsp; $d(ν)$&nbsp; for the&nbsp; $D(\mu)$&nbsp; values of column&nbsp; $\rm C$?
 |type="{}"}
-$D(\mu )$ gemäß C: $d(0) =$ { 1 }
+$d(0)\ = \ $ { 1 3% }
-$D(\mu )$ gemäß C: $d(1) =$ { 0 }
+$d(1)\ = \ $ { 0. }
-{Wie lauten die Zeitkoeffizienten $d(ν)$ für die $D(\mu)$–Werte von Spalte D?
+{What are the time coefficients&nbsp; $d(ν)$&nbsp; for the&nbsp; $D(\mu)$&nbsp; values of column&nbsp; $\rm D$?
 |type="{}"}
-$D(\mu )$ gemäß D: $d(0) =$ { 1 }
+$d(0)\ = \ ${ 1 3% }
-$D(\mu )$ gemäß D: $d(1) =$ { -1 }
+$d(1)\ = \ $ { -1.03--0.97 }
-{Wie lauten die Zeitkoeffizienten $d(ν)$ für die $D(\mu)$–Wertevon Spalte E?
+{What are the time coefficients&nbsp; $d(ν)$&nbsp; for the&nbsp; $D(\mu)$&nbsp; values of column&nbsp; $\rm E$?
 |type="{}"}
-$D(\mu )$ gemäß A: $d(0) =$ { 2 }
+$d(0)\ = \ $ { 2 3% }
-$D(\mu )$ gemäß A: $d(1) =$ { 0 }
+$d(1)\ = \ $ { 0. }
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''1.''' Aus der IDFT–Gleichung wird mit $D(\mu)$ = 0 für $\mu \approx$ 0:
+'''(1)'''&nbsp;  From the IDFT equation,&nbsp; with&nbsp; $D(\mu) = 0$&nbsp; for&nbsp; $\mu \ne 0$:
-$$d(\nu) = D(0) \cdot w^0 = D(0) =1\hspace{0.5cm}(0 \le \nu \le 7)$$
+:$$d(\nu) = D(0) \cdot w^0 = D(0) =1\hspace{0.5cm}(0 \le \nu \le 7)\ \hspace{0.5cm} \Rightarrow\hspace{0.5cm}\hspace{0.15 cm}\underline{d(0) = d(1) = 1}.$$
+*This parameter set describes the discrete form of the Fourier correspondence of the DC signal:
+:$$x(t) = 1 \hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm}X(f) = {\delta}(f) \hspace{0.05cm}.$$
-$$\Rightarrow\hspace{0.5cm}\hspace{0.15 cm}\underline{d(0) = d(1) = 1}.$$
-Dieser Parametersatz beschreibt die diskrete Form der Fourierkorrespondenz des Gleichsignals:
+'''(2)'''&nbsp;  All spectral coefficients are zero except&nbsp; $D_1 = D_7 = 0.5$.&nbsp; It follows for&nbsp; $0 ≤ ν ≤ 7$:
-$$x(t) = 1 \hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm}
+:$$d(\nu) = 0.5 \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /4) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} + 0.5 \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /4) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu}\hspace{0.05cm}.$$
- X(f) = {\delta}(f) \hspace{0.05cm}.$$
-'''2.'''  Alle Spektralkoeffizienten sind 0 mit Ausnahme von $D_1$ = $D_7$ = 0.5. Daraus folgt für 0 ≤ $ν$ ≤ 7:
+*However, due to periodicity, also holds:
-$$d(\nu) = 0.5 \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /4) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} + 0.5 \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /4) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu}
+:$$d(\nu)  =  0.5 \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /4) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} + 0.5 \cdot {\rm e}^{{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /4) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} = \cos \left({\pi}/{4} \cdot \nu \right)  \hspace{0.3cm} \Rightarrow  \hspace{0.3cm}\hspace{0.15 cm}\underline{d(0) = 1}, \hspace{0.2cm}\hspace{0.15 cm}\underline{d(1) = {1}/{\sqrt{2}} \approx 0.707}\hspace{0.05cm}.$$
-  \hspace{0.05cm}.$$
-Aufgrund der Periodizität gilt aber auch:
+*It is therefore the discrete-time equivalent of
-$$d(\nu)  =  0.5 \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /4) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} + 0.5 \cdot {\rm e}^{{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /4) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} = \cos \left({\pi}/{4} \cdot \nu \right)$$ $$\Rightarrow  \hspace{0.3cm}\hspace{0.15 cm}\underline{d(0) = 1}, \hspace{0.2cm}\hspace{0.15 cm}\underline{d(1) = {1}/{\sqrt{2}} \approx 0.707}
+:$$x(t) = \cos(2 \pi \cdot f_{\rm A} \cdot t) \hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm}X(f) =  {1}/{2} \cdot {\delta}(f + f_{\rm A}) +  {1}/{2} \cdot {\delta}(f - f_{\rm A}) \hspace{0.05cm},$$
- \hspace{0.05cm}.$$
-Es handelt sich also um das zeitdiskrete Äquivalent zu
+:where&nbsp; $f_{\rm A}$&nbsp; denotes the smallest frequency that can be represented in the DFT.
-$$x(t) = \cos(2 \pi \cdot f_{\rm A} \cdot t) \hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm}
- X(f) = \frac {1}{2} \cdot {\delta}(f + f_{\rm A}) + \frac {1}{2} \cdot {\delta}(f - f_{\rm A}) \hspace{0.05cm},$$
-wobei $f_A$ die kleinste in der DFT darstellbare Frequenz bezeichnet.
-'''3.'''  Gegenüber Aufgabe 2) ist nun die Frequenz doppelt so groß, nämlich 2 · $f_A$ anstelle von $f_A$:
+'''(3)'''&nbsp; Compared to subtask&nbsp; '''(2)''',&nbsp; the oscillation frequency is now twice as large, namely&nbsp; $2 f_{\rm A}$&nbsp; instead of&nbsp; $f_{\rm A}$:
-$$x(t) = \cos(2 \pi \cdot (2f_{\rm A}) \cdot t) \hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm}
+:$$x(t) = \cos(2 \pi \cdot (2f_{\rm A}) \cdot t) \hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm}X(f) = {1}/{2} \cdot {\delta}(f + 2f_{\rm A}) + {1}/{2} \cdot {\delta}(f - 2f_{\rm A}) \hspace{0.05cm},$$
- X(f) = \frac {1}{2} \cdot {\delta}(f + 2f_{\rm A}) + \frac {1}{2} \cdot {\delta}(f - 2f_{\rm A}) \hspace{0.05cm},$$
-Damit beschreibt die Folge 〈 $d(ν)$〉 zwei Perioden der Cosinusschwingung, und es gilt für 0 ≤ $ν$ ≤ 7:
+*Thus the sequence&nbsp;  $\langle \hspace{0.1cm}d(ν)\hspace{0.1cm}\rangle $&nbsp; describes two periods of the cosine oscillation, and it holds for&nbsp; $0 ≤ ν ≤ 7$:
-$$ d(\nu)  =  0.5 \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /2) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} + 0.5 \cdot {\rm e}^{{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /2) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} = \cos \left({\pi}/{2} \cdot \nu \right)$$  $$\Rightarrow  \hspace{0.3cm}\hspace{0.15 cm}\underline{d(0) = 1, \hspace{0.2cm}d(1) = 0}
+:$$ d(\nu)  =  0.5 \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /2) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} + 0.5 \cdot {\rm e}^{{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /2) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} = \cos \left({\pi}/{2} \cdot \nu \right)\hspace{0.3cm}  \Rightarrow  \hspace{0.3cm}\hspace{0.15 cm}\underline{d(0) = 1, \hspace{0.2cm}d(1) = 0}\hspace{0.05cm}.$$
- \hspace{0.05cm}.$$
-'''4.'''   Durch eine weitere Verdoppelung der Cosinusfrequenz auf 4 $f_A$ kommt man schließlich zur zeitkontinuierlichen Fourierkorrespondenz
+'''(4)'''&nbsp; By further doubling the cosine frequency to&nbsp; $4 f_{\rm A}$&nbsp; one finally arrives at the continuous-time Fourier correspondence
-$$d(\nu) = 0.5 \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \pi  \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} + 0.5 \cdot {\rm e}^{{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \pi  \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} = \cos \left(\pi \cdot \nu \right)
+:$$d(\nu) = 0.5 \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \pi  \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} + 0.5 \cdot {\rm e}^{{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} \pi  \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} = \cos \left(\pi \cdot \nu \right)\hspace{0.05cm}$$
- \hspace{0.05cm}$$
-und damit zu den Zeitkoeffizienten
+:and thus to the time coefficients
-$$d(0) =d(2) =d(4) =d(6) \hspace{0.15 cm}\underline{= +1}, \hspace{0.2cm}d(1) =d(3) =d(5) =d(7)  \hspace{0.15 cm}\underline{= -1}
+:$$d(0) =d(2) =d(4) =d(6) \hspace{0.15 cm}\underline{= +1}, \hspace{0.2cm}d(1) =d(3) =d(5) =d(7)  \hspace{0.15 cm}\underline{= -1}\hspace{0.05cm}.$$
- \hspace{0.05cm}.$$
+*Note that here the two Dirac functions coincide in the discrete-time representation due to periodicity.
+*The coefficients&nbsp; $D (+4) = 0.5$&nbsp; and&nbsp; $D (-4) = 0.5$&nbsp; together give&nbsp; $D (4) = 1$.
-Zu beachten ist, dass die beiden Diracfunktionen in der zeitdiskreten Darstellung aufgrund der Periodizität zusammenfallen. Das heißt: Die Koeffizienten $D$ (4) = 0.5 und $D$ (-4) = 0.5 ergeben zusammen $D$ (4) = 1.
-'''5.'''  Die Diskrete Fouriertransformation ist ebenfalls linear. Deshalb ist das Superpositionsprinzip weiterhin anwendbar. Die Koeffizienten $D(\mu )$ aus Spalte E ergeben sich als die Summen der Spalten A und D. Deshalb wird aus der alternierenden Folge 〈 $d(ν)$〉 entsprechend Teilaufgabe 4) die um 1 nach oben verschobene Folge:
+'''(5)'''&nbsp; The Discrete Fourier Transform is also linear. Therefore, the superposition principle is still applicable:
+*The coefficients&nbsp; $D(\mu )$&nbsp; from column&nbsp; $\rm E$&nbsp; result as the sums of columns&nbsp; $\rm A$&nbsp; and&nbsp; $\rm D$.
+*Therefore, the alternating sequence&nbsp;  $\langle \hspace{0.1cm}d(ν) \hspace{0.1cm}\rangle $&nbsp;  becomes the sequence shifted up by&nbsp; $1$&nbsp; according to subtask&nbsp; '''(4)''':
-$$ \hspace{0.15 cm}\underline{d(0) =d(2) =d(4) =d(6)= 2}, \hspace{0.2cm}\hspace{0.15 cm}\underline{d(1) =d(3) =d(5) =d(7)  = 0}
+:$$ \hspace{0.15 cm}\underline{d(0) =d(2) =d(4) =d(6)= 2}, \hspace{0.2cm}\hspace{0.15 cm}\underline{d(1) =d(3) =d(5) =d(7)  = 0}\hspace{0.05cm}.$$
- \hspace{0.05cm}.$$
 {{ML-Fuß}}
 __NOEDITSECTION__
-[[Category:Aufgaben zu Signaldarstellung|^5. Zeit- und frequenzdisktrete Signaldarstellung^]]
+[[Category:Signal Representation: Exercises|^5.2 Discrete Fourier Transform^]]
+[[de:Aufgaben:Aufgabe 5.2: Inverse Diskrete Fouriertransformation]]