Aufgaben:Exercise 5.2: Inverse Discrete Fourier Transform - Revision history

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	[[Category:Signal Representation: Exercises\|^5.2 Discrete Fourier Transform^]]		[[Category:Signal Representation: Exercises\|^5.2 Discrete Fourier Transform^]]
	[[de:Aufgaben:5.~~2_Inverse_DFT~~]]		[[de:Aufgaben:Aufgabe 5.2: Inverse Diskrete Fouriertransformation]]

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	With  $\nu = 0$, ... , $N – 1$  and  $\mu = 0$, ... , $N – 1$  holds:		With  $\nu = 0$, ... , $N – 1$  and  $\mu = 0$, ... , $N – 1$  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}.$$
	d(\nu)\cdot {w}^{\hspace{0.05cm}\nu \hspace{0.05cm} \cdot \hspace{0.05cm}\mu} \hspace{0.05cm}.$$

	Here  $w$  denotes the complex rotation factor:		Here  $w$  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}.$$
	= \cos \left( {2 \pi}/{N}\right)-{\rm j} \cdot \sin \left( {2 \pi}/{N}\right)
	\hspace{0.05cm}.$$

	For the  '''Inverse Discrete Fourier Transform'''  $\rm (DFT)$   ⇒   "inverse function" of the DFT, the following applies accordingly:		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(\nu) = \sum_{\mu = 0 }^{N-1}D(\mu) \cdot {w}^{-\nu \hspace{0.05cm} \cdot \hspace{0.05cm}\mu} \hspace{0.05cm}.$$
	D(\mu) \cdot {w}^{-\nu \hspace{0.05cm} \cdot \hspace{0.05cm}\mu} \hspace{0.05cm}.$$

	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.		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.
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	*This parameter set describes the discrete form of the Fourier correspondence of the DC signal:		*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(t) = 1 \hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm}X(f) = {\delta}(f) \hspace{0.05cm}.$$
	X(f) = {\delta}(f) \hspace{0.05cm}.$$


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

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

	*However, due to periodicity, also holds:		*However, due to periodicity, also holds:

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

	*It is therefore the discrete-time equivalent of		*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(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},$$
	X(f) = {1}/{2} \cdot {\delta}(f + f_{\rm A}) + {1}/{2} \cdot {\delta}(f - f_{\rm A}) \hspace{0.05cm},$$

	:where  $f_{\rm A}$  denotes the smallest frequency that can be represented in the DFT.		:where  $f_{\rm A}$  denotes the smallest frequency that can be represented in the DFT.
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	'''(3)'''  Compared to subtask  '''(2)''',  the oscillation frequency is now twice as large, namely  $2 f_{\rm A}$  instead of  $f_{\rm A}$:		'''(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(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},$$
	X(f) = {1}/{2} \cdot {\delta}(f + 2f_{\rm A}) + {1}/{2} \cdot {\delta}(f - 2f_{\rm A}) \hspace{0.05cm},$$

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

	:$$ d(\nu) = 0.5 \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /2) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} + 0.5 \cdot {\rm e}^{{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /2) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} = \cos \left({\pi}/{2} \cdot \nu \right)\hspace{0.3cm} \Rightarrow \hspace{0.3cm}\hspace{0.15 cm}\underline{d(0) = 1, \hspace{0.2cm}d(1) = 0}		:$$ 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)'''  By further doubling the cosine frequency to  $4 f_{\rm A}$  one finally arrives at the continuous-time Fourier correspondence		'''(4)'''  By further doubling the cosine frequency to  $4 f_{\rm A}$  one finally arrives at the continuous-time Fourier correspondence

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

	:and thus to the time coefficients		: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.		*Note that here the two Dirac functions coincide in the discrete-time representation due to periodicity.
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	*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)''':		*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{0.15 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}}		{{ML-Fuß}}

Maintenance script: Add German interlanguage link

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Add German interlanguage link

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	[[Category:Signal Representation: Exercises\|^5.2 Discrete Fourier Transform^]]		[[Category:Signal Representation: Exercises\|^5.2 Discrete Fourier Transform^]]
			[[de:Aufgaben:5.2_Inverse_DFT]]

Guenter at 14:38, 16 May 2021

2021-05-16T14:38:34Z

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  From the IDFT equation, with  $D(\mu) = 0$  for  $\mu \ne 0$:		'''(1)'''  From the IDFT equation,  with  $D(\mu) = 0$  for  $\mu \ne 0$:

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

	*This set ~~of parameters~~ describes the discrete form of the Fourier correspondence of the DC signal:		*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(t) = 1 \hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm}
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	'''(2)'''  All spectral coefficients are zero except  $D_1 = D_7 = 0.5$. It follows for  $0 ≤ ν ≤ 7$:		'''(2)'''  All spectral coefficients are zero except  $D_1 = D_7 = 0.5$.  It follows for  $0 ≤ ν ≤ 7$:

	:$$d(\nu) = 0.5 \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /4) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} + 0.5 \cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /4) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu}		:$$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}
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	'''(3)'''  Compared to subtask  '''(2)''' , the oscillation frequency is now twice as large, namely  $2 f_{\rm A}$  instead of  $f_{\rm A}$:		'''(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(t) = \cos(2 \pi \cdot (2f_{\rm A}) \cdot t) \hspace{0.2cm}\circ\!\!-\!\!\!-\!\!\!-\!\!\bullet\, \hspace{0.2cm}
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	'''(5)'''  The Discrete Fourier Transform is also linear. Therefore, the superposition principle is still applicable:		'''(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  $\rm D$.		*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)'''~~ ~~:		*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{0.15 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}

Guenter at 14:17, 16 May 2021

2021-05-16T14:17:10Z

← Older revision		Revision as of 16:17, 16 May 2021
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	*This task belongs to the chapter  [[Signal_Representation/Discrete_Fourier_Transform_(DFT)\|Discrete Fourier Transformation (DFT)]].		*This task belongs to the chapter  [[Signal_Representation/Discrete_Fourier_Transform_(DFT)\|Discrete Fourier Transformation (DFT)]].

	*The topic dealt with here is also dealt with in the interactive applet  [[Applets:~~Diskrete_Fouriertransformation_und_Inverse~~\|Discrete Fourier Transform and Inverse]].		*The topic dealt with here is also dealt with in the interactive applet  [[Applets:Discrete_Fouriertransform_and_Inverse\|Discrete Fourier Transform and Inverse]].

Guenter at 14:15, 16 May 2021

2021-05-16T14:15:56Z

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	[[File:P_ID1138__Sig_A_5_2.png\|250px\|right\|frame\|Five different sets for the spectral coefficients  $D(\mu)$]]		[[File:P_ID1138__Sig_A_5_2.png\|250px\|right\|frame\|Five different sets for the spectral coefficients  $D(\mu)$]]

	With the ''Discrete Fourier Transform''  ~~(DFT), the following are obtained~~		With the  '''Discrete Fourier Transform'''  $\rm (DFT)$,
	*from the  $N$~~  time coefficients  $d(~~\~~nu)$   ⇒   samples of the continuous-time signal  $x~~(t)$ –		*the  $N$  spectral range coefficients  $D(\mu)$  are calculated
	*the  $N$  spectral range coefficients  $D(\mu)$

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

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

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

	For the ''Inverse Discrete Fourier Transform''  (~~IDFT~~)   ⇒   ~~„inverse function”~~ of the DFT, the following applies accordingly:		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(\nu) = \sum_{\mu = 0 }^{N-1}
	D(\mu) \cdot {w}^{-\nu \hspace{0.05cm} \cdot \hspace{0.05cm}\mu} \hspace{0.05cm}.$$		D(\mu) \cdot {w}^{-\nu \hspace{0.05cm} \cdot \hspace{0.05cm}\mu} \hspace{0.05cm}.$$

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


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	<quiz display=simple>		<quiz display=simple>
	{What are the time coefficients  $d(\nu)$  for the  $D(\mu)$~~–values~~ of column  $\rm A$?		{What are the time coefficients  $d(\nu)$  for the  $D(\mu)$  values of column  $\rm A$?
	\|type="{}"}		\|type="{}"}
	$d(0)\ = \ $ { 1 3% }		$d(0)\ = \ $ { 1 3% }
	$d(1)\ = \ $ { 1 3% }		$d(1)\ = \ $ { 1 3% }

	{What are the time coefficients  $d(ν)$  for the  $D(\mu)$~~–values~~ of column  $\rm B$?		{What are the time coefficients  $d(ν)$  for the  $D(\mu)$  values of column  $\rm B$?
	\|type="{}"}		\|type="{}"}
	$d(0)\ = \ $ { 1 3% }		$d(0)\ = \ $ { 1 3% }
	$d(1)\ = \ $ { 0.707 3% }		$d(1)\ = \ $ { 0.707 3% }

	{What are the time coefficients  $d(ν)$  for the  $D(\mu)$~~–values~~ of column  $\rm C$?		{What are the time coefficients  $d(ν)$  for the  $D(\mu)$  values of column  $\rm C$?
	\|type="{}"}		\|type="{}"}
	$d(0)\ = \ $ { 1 3% }		$d(0)\ = \ $ { 1 3% }
	$d(1)\ = \ $ { 0. }		$d(1)\ = \ $ { 0. }

	{What are the time coefficients  $d(ν)$  for the  $D(\mu)$~~–values~~ of column  $\rm D$?		{What are the time coefficients  $d(ν)$  for the  $D(\mu)$  values of column  $\rm D$?
	\|type="{}"}		\|type="{}"}
	$d(0)\ = \ ${ 1 3% }		$d(0)\ = \ ${ 1 3% }
	$d(1)\ = \ $ { -1.03--0.97 }		$d(1)\ = \ $ { -1.03--0.97 }

	{What are the time coefficients  $d(ν)$  for the  $D(\mu)$~~–values~~ of column  $\rm E$?		{What are the time coefficients  $d(ν)$  for the  $D(\mu)$  values of column  $\rm E$?
	\|type="{}"}		\|type="{}"}
	$d(0)\ = \ $ { 2 3% }		$d(0)\ = \ $ { 2 3% }

Javier: Text replacement - "Category:Exercises for Signal Representation" to "Category:Signal Representation: Exercises"

2021-03-23T12:39:51Z

Text replacement - "Category:Exercises for Signal Representation" to "Category:Signal Representation: Exercises"

← Older revision		Revision as of 14:39, 23 March 2021
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	[[Category:~~Exercises for~~ Signal Representation\|^5.2 Discrete Fourier Transform^]]		[[Category:Signal Representation: Exercises\|^5.2 Discrete Fourier Transform^]]

Noah at 19:01, 20 March 2021

2021-03-20T19:01:43Z

Show changes

Oezdemir: Oezdemir moved page Aufgaben:Aufgabe 5.2: Inverse Diskrete Fouriertransformation to Aufgaben:Exercise 5.2: Inverse Discrete Fourier Transform

2020-11-29T18:50:47Z

Oezdemir moved page Aufgaben:Aufgabe 5.2: Inverse Diskrete Fouriertransformation to Aufgaben:Exercise 5.2: Inverse Discrete Fourier Transform

← Older revision	Revision as of 20:50, 29 November 2020
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Oezdemir at 18:49, 29 November 2020

2020-11-29T18:49:54Z

← Older revision		Revision as of 20:49, 29 November 2020
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	[[Category:Exercises for Signal Representation\|^5. ~~Zeit- und frequenzdiskrete Signaldarstellung~~^]]		[[Category:Exercises for Signal Representation\|^5.2 Discrete Fourier Transform^]]