Aufgaben:Exercise 5.2Z: DFT of a Triangular Pulse - Revision history

Maintenance script: Fix interlanguage link: resolve redirect chain

2026-03-16T15:56:25Z

Fix interlanguage link: resolve redirect chain

← Older revision		Revision as of 17:56, 16 March 2026
Line 13:		Line 13:


	The spectrum  $X(f)$  is obtained by applying  [[Signal_Representation/~~Fourier_Transform_and_its_Inverse~~#The_first_Fourier_integral\|the first Fourier Integral]]:		The spectrum  $X(f)$  is obtained by applying  [[Signal_Representation/The_Fourier_Transform_and_its_Inverse#The_first_Fourier_integral\|the first Fourier Integral]]:
	:$$X(f) = A \cdot T \cdot {\rm si}^2(\pi f T)\hspace{0.5cm}\text{with}\hspace{0.5cm}{\rm si}(x)=\sin(x)/x\hspace{0.05cm}.$$		:$$X(f) = A \cdot T \cdot {\rm si}^2(\pi f T)\hspace{0.5cm}\text{with}\hspace{0.5cm}{\rm si}(x)=\sin(x)/x\hspace{0.05cm}.$$

Line 92:		Line 92:

	'''(3)'''  For the distances of the samples in the time and frequency domain applies:		'''(3)'''  For the distances of the samples in the time and frequency domain applies:
	:$$N \cdot f_{\rm A} \cdot T_{\rm A} = 1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}f_{\~~rmA~~}= \frac{1}{ 8 \cdot 0.25\, {\rm ms}}\hspace{0.15 cm}\underline{= 0.5\, {\rm kHz}}\hspace{0.05cm}.$$		:$$N \cdot f_{\rm A} \cdot T_{\rm A} = 1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}f_{\rm A}= \frac{1}{ 8 \cdot 0.25\, {\rm ms}}\hspace{0.15 cm}\underline{= 0.5\, {\rm kHz}}\hspace{0.05cm}.$$


Line 127:		Line 127:
	__NOEDITSECTION__		__NOEDITSECTION__
	[[Category:Signal Representation: Exercises\|^5.2 Discrete Fourier Transform^]]		[[Category:Signal Representation: Exercises\|^5.2 Discrete Fourier Transform^]]
	[[de:Aufgaben:5.~~2Z_DFT_eines_Dreieckimpulses~~]]		[[de:Aufgaben:Aufgabe 5.2Z: DFT eines Dreieckimpulses]]

Maintenance script: Join 1 split formula blocks

2026-03-16T14:17:03Z

Join 1 split formula blocks

← Older revision		Revision as of 16:17, 16 March 2026
Line 5:		Line 5:
	[[File:P_ID1140__Sig_Z_5_2.png\|right\|frame\|Discretisation of a triangular pulse]]		[[File:P_ID1140__Sig_Z_5_2.png\|right\|frame\|Discretisation of a triangular pulse]]
	Consider the sketched triangular pulse		Consider the sketched triangular pulse
	:$$x(t) = \left\{ \begin{array}{l} A \cdot \left( 1 - {\|t\|}/{T} \right ) \\		:$$x(t) = \left\{ \begin{array}{l} A \cdot \left( 1 - {\|t\|}/{T} \right ) \\\hspace{0.25cm} 0 \\ \end{array} \right.\quad\begin{array}{{10}c} {\rm{for}}\\ {\rm{for}} \\ \end{array}\begin{array}{{10}c}{\left\| \hspace{0.005cm} t\hspace{0.05cm} \right\| \le T,} \\{\left\|\hspace{0.005cm} t \hspace{0.05cm} \right\| > T.} \\\end{array}$$
	\hspace{0.25cm} 0 \\ \end{array} \right.\quad
	\begin{array}{*{10}c} {\rm{for}}
	\\ {\rm{for}} \\ \end{array}\begin{array}{*{10}c}
	{\left\| \hspace{0.005cm} t\hspace{0.05cm} \right\| \le T,} \\
	{\left\|\hspace{0.005cm} t \hspace{0.05cm} \right\| > T.} \\
	\end{array}$$
	The signal parameters have the following values:		The signal parameters have the following values:

Maintenance script: Join split $$ formula lines for correct rendering

2026-03-16T13:48:49Z

Join split $$ formula lines for correct rendering

← Older revision		Revision as of 15:48, 16 March 2026
Line 25:		Line 25:

	The corresponding spectral coefficients  $D(0)$, ... ,  $D(7)$  are to be determined.  For the indices  $\mu = 0$, ... , $N–1$  applies:		The corresponding spectral coefficients  $D(0)$, ... ,  $D(7)$  are to be determined.  For the indices  $\mu = 0$, ... , $N–1$  applies:
	:$$D(\mu) = \frac{1}{N} \cdot \sum_{\nu = 0 }^{N-1}		:$$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}.$$
	d(\nu)\cdot {w}^{\hspace{0.05cm}\nu \hspace{0.05cm} \cdot \hspace{0.05cm}\mu} \hspace{0.05cm}.$$

	If we denote the distance between two samples in the time domain by  $T_{\rm A}$  and the corresponding frequency distance between two lines by $f_{\rm A}$, following relationship applies:		If we denote the distance between two samples in the time domain by  $T_{\rm A}$  and the corresponding frequency distance between two lines by $f_{\rm A}$, following relationship applies:
Line 89:		Line 88:
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  From the graph the following values result with  $A = 4 \ {\rm V}$ :		'''(1)'''  From the graph the following values result with  $A = 4 \ {\rm V}$ :
	:$${d(0) = 4\,{\rm V}, \hspace{0.1cm}d(1) = d(7) = 3\,{\rm V}, \hspace{0.1cm}		:$${d(0) = 4\,{\rm V}, \hspace{0.1cm}d(1) = d(7) = 3\,{\rm V}, \hspace{0.1cm}\hspace{0.1cm}d(2) = d(6) = 2\,{\rm V}, \hspace{0.1cm}\hspace{0.1cm}d(3) = d(5) = 1\,{\rm V}, \hspace{0.1cm}\hspace{0.1cm}d(4) = 0}\hspace{0.05cm}. $$
	\hspace{0.1cm}d(2) = d(6) = 2\,{\rm V}, \hspace{0.1cm}		:$$\Rightarrow \hspace{0.15 cm}\underline{d(0) = 4\,{\rm V}, \hspace{0.1cm}d(3) = 1\,{\rm V},\hspace{0.1cm}d(6) = 2\,{\rm V}. \hspace{0.1cm}}\hspace{0.05cm} $$
	\hspace{0.1cm}d(3) = d(5) = 1\,{\rm V}, \hspace{0.1cm}
	\hspace{0.1cm}d(4) = 0}\hspace{0.05cm}. $$
	:$$\Rightarrow \hspace{0.15 cm}\underline{d(0) = 4\,{\rm V}, \hspace{0.1cm}d(3) = 1\,{\rm V},
	\hspace{0.1cm}d(6) = 2\,{\rm V}. \hspace{0.1cm}}
	\hspace{0.05cm} $$


Line 104:		Line 98:

	'''(3)'''  For the distances of the samples in the time and frequency domain applies:		'''(3)'''  For the distances of the samples in the time and frequency domain applies:
	:$$N \cdot f_{\rm A} \cdot T_{\rm A} = 1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}f_{\rm		:$$N \cdot f_{\rm A} \cdot T_{\rm A} = 1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}f_{\rmA}= \frac{1}{ 8 \cdot 0.25\, {\rm ms}}\hspace{0.15 cm}\underline{= 0.5\, {\rm kHz}}\hspace{0.05cm}.$$
	A}= \frac{1}{ 8 \cdot 0.25\, {\rm ms}}\hspace{0.15 cm}\underline{= 0.5\, {\rm kHz}}\hspace{0.05cm}.$$


	'''(4)'''  With  $N = 8$  and  $\mu = 0$ , it follows from the DFT equation:		'''(4)'''  With  $N = 8$  and  $\mu = 0$ , it follows from the DFT equation:
	:$$D(0) = \frac{1}{8}\cdot \sum_{\nu = 0 }^{7}		:$$D(0) = \frac{1}{8}\cdot \sum_{\nu = 0 }^{7}d(\nu) = \frac{1\,{\rm V}}{8}\cdot (4+3+2+1+0+1+2+3)\hspace{0.15 cm}\underline{= 2 \,{\rm V}}\hspace{0.05cm}.$$
	d(\nu) = \frac{1\,{\rm V}}{8}\cdot (4+3+2+1+0+1+2+3)\hspace{0.15 cm}\underline{= 2 \,{\rm V}}\hspace{0.05cm}.$$
	*The DFT value $D(0)$ describes the spectral value at  $f = 0$, where the following relation holds:		*The DFT value $D(0)$ describes the spectral value at  $f = 0$, where the following relation holds:
	:$$X(f=0) = \frac{D(0)}{f_{\rm A}}= \frac{ 2\,{\rm V}}{0.5\,{\rm kHz}}= 4 \cdot 10^{-3}\,{\rm V/Hz}\hspace{0.05cm}.$$		:$$X(f=0) = \frac{D(0)}{f_{\rm A}}= \frac{ 2\,{\rm V}}{0.5\,{\rm kHz}}= 4 \cdot 10^{-3}\,{\rm V/Hz}\hspace{0.05cm}.$$
Line 118:		Line 110:

	'''(5)'''  With  $N = 8$  and  $\mu = 2$  we obtain:		'''(5)'''  With  $N = 8$  and  $\mu = 2$  we obtain:
	:$$D(2) = \frac{1}{8}\cdot \sum_{\nu = 0 }^{7}		:$$D(2) = \frac{1}{8}\cdot \sum_{\nu = 0 }^{7}d(\nu)\cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /2) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} =\frac{1}{8}\cdot \sum_{\nu = 0 }^{7}d(\nu)\cdot (-{\rm j})^{\nu} \hspace{0.3cm}\Rightarrow \hspace{0.3cm} = \frac{1\,{\rm V}}{8}\cdot (4-3\cdot{\rm j}-2+{\rm j}-{\rm j}-2+3\cdot{\rm j})\hspace{0.15 cm}\underline{= 0}\hspace{0.05cm}.$$
	d(\nu)\cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /2) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} =
	\frac{1}{8}\cdot \sum_{\nu = 0 }^{7}
	d(\nu)\cdot (-{\rm j})^{\nu} \hspace{0.3cm}
	\Rightarrow \hspace{0.3cm} = \frac{1\,{\rm V}}{8}\cdot (4-3\cdot{\rm j}-2+{\rm j}-{\rm j}-2+3\cdot{\rm j})\hspace{0.15 cm}\underline{= 0}\hspace{0.05cm}.$$
	This result could have been predicted without calculation:		This result could have been predicted without calculation:
	*The DFT coefficients  $D(\mu)$  are at the same time the Fourier coefficients of the function  $x_{\rm Per}(t)$  periodised at the distance  $T_{\rm P} = 2T$. <br>This is shown as a dashed line in the graph on the information page.		*The DFT coefficients  $D(\mu)$  are at the same time the Fourier coefficients of the function  $x_{\rm Per}(t)$  periodised at the distance  $T_{\rm P} = 2T$. <br>This is shown as a dashed line in the graph on the information page.
Line 132:		Line 120:
	:$$D(7) = D(-1) = D^{\star}(1) \hspace{0.05cm}.$$		:$$D(7) = D(-1) = D^{\star}(1) \hspace{0.05cm}.$$
	Preferably, we calculate this DFT coefficient:		Preferably, we calculate this DFT coefficient:
	:$$D(1) = \frac{1}{8}\cdot \sum_{\nu = 0 }^{7}		:$$D(1) = \frac{1}{8}\cdot \sum_{\nu = 0 }^{7}d(\nu)\cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /4) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} =\frac{1\,{\rm V}}{8}\cdot \left(4 +3\cdot \frac{1 - {\rm j}}{\sqrt{2}}-2\cdot {\rm j}+ \frac{-1 - {\rm j}}{\sqrt{2}}-{\rm j}+\frac{-1 + {\rm j}}{\sqrt{2}}-{\rm j}+ 2\cdot {\rm j}+3\cdot \frac{1 - {\rm j}}{\sqrt{2}}\right)$$
	d(\nu)\cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /4) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} =
	\frac{1\,{\rm V}}{8}\cdot \left(4 +3\cdot \frac{1 - {\rm j}}{\sqrt{2}}-2\cdot {\rm j}+ \frac{-1 - {\rm j}}{\sqrt{2}}-{\rm j}+
	\frac{-1 + {\rm j}}{\sqrt{2}}-{\rm j}+ 2\cdot {\rm j}+3\cdot \frac{1 - {\rm j}}{\sqrt{2}}\right)$$
	:$$\Rightarrow \; \; D(1) = \frac{2 + \sqrt{2}}{4} \approx 0.854{\rm V}\hspace{0.05cm}.$$		:$$\Rightarrow \; \; D(1) = \frac{2 + \sqrt{2}}{4} \approx 0.854{\rm V}\hspace{0.05cm}.$$
	Since  $D(1)$  is purely real,  $D(7) = D(1) \; \underline{= 0.854 \ {\rm V}}$.		Since  $D(1)$  is purely real,  $D(7) = D(1) \; \underline{= 0.854 \ {\rm V}}$.

Maintenance script: Add German interlanguage link

2026-03-16T12:34:01Z

Add German interlanguage link

← Older revision		Revision as of 14:34, 16 March 2026
Line 148:		Line 148:
	__NOEDITSECTION__		__NOEDITSECTION__
	[[Category:Signal Representation: Exercises\|^5.2 Discrete Fourier Transform^]]		[[Category:Signal Representation: Exercises\|^5.2 Discrete Fourier Transform^]]
			[[de:Aufgaben:5.2Z_DFT_eines_Dreieckimpulses]]

Guenter at 15:59, 16 May 2021

2021-05-16T15:59:51Z

← Older revision		Revision as of 17:59, 16 May 2021
Line 99:		Line 99:

	'''(2)'''  According to the diagram  $T_{\rm A} = T/4$.		'''(2)'''  According to the diagram  $T_{\rm A} = T/4$.
	*With  $T = 1 \ \text{ms}$  one ~~thus~~ obtains  $\underline{T_{\rm A} = 0.25 \ \text{ms}}$.		*With  $T = 1 \ \text{ms}$  one obtains  $\underline{T_{\rm A} = 0.25 \ \text{ms}}$.


Line 124:		Line 124:
	\Rightarrow \hspace{0.3cm} = \frac{1\,{\rm V}}{8}\cdot (4-3\cdot{\rm j}-2+{\rm j}-{\rm j}-2+3\cdot{\rm j})\hspace{0.15 cm}\underline{= 0}\hspace{0.05cm}.$$		\Rightarrow \hspace{0.3cm} = \frac{1\,{\rm V}}{8}\cdot (4-3\cdot{\rm j}-2+{\rm j}-{\rm j}-2+3\cdot{\rm j})\hspace{0.15 cm}\underline{= 0}\hspace{0.05cm}.$$
	This result could have been predicted without calculation:		This result could have been predicted without calculation:
	*The DFT coefficients  $D(\mu)$  are at the same time the Fourier coefficients of the function  $T_{\rm P} ~~= 2T~~$  periodised at the distance  $x_{\rm ~~Per~~}~~(t)~~$. This is shown as a dashed line in the graph on the information page.		*The DFT coefficients  $D(\mu)$  are at the same time the Fourier coefficients of the function  $x_{\rm Per}(t)$  periodised at the distance  $T_{\rm P} = 2T$. <br>This is shown as a dashed line in the graph on the information page.
	*Due to symmetry properties, however, all even Fourier coefficients of the function  $x_{\rm Per}(t)$  are equal to zero:   ⇒   $D(4)\hspace{0.15cm}\underline{=0},$   $D(6)\hspace{0.15cm}\underline{=0}$.		*Due to symmetry properties, however, all even Fourier coefficients of the function  $x_{\rm Per}(t)$  are equal to zero   ⇒   $D(4)\hspace{0.15cm}\underline{=0},$   $D(6)\hspace{0.15cm}\underline{=0}$.



	'''(6)'''  The coefficient  $D(7)$  describes the periodised spectral function at the frequency  $f = 7 \cdot f_{\rm A}$. Due to periodicity and symmetry property holds:		'''(6)'''  The coefficient  $D(7)$  describes the periodised spectral function at the frequency  $f = 7 \cdot f_{\rm A}$.  Due to periodicity and symmetry property holds:
	:$$D(7) = D(-1) = D^{\star}(1) \hspace{0.05cm}.$$		:$$D(7) = D(-1) = D^{\star}(1) \hspace{0.05cm}.$$
	Preferably, we calculate this DFT coefficient:		Preferably, we calculate this DFT coefficient:

Guenter at 15:26, 16 May 2021

2021-05-16T15:26:46Z

← Older revision		Revision as of 17:26, 16 May 2021
Line 19:		Line 19:


	The spectrum  $X(f)$  is obtained by applying  [[Signal_Representation/~~Fourier_Transform_and_Its_Inverse~~#~~The_First_Fourier_Integral~~\|the first Fourier Integral]]:		The spectrum  $X(f)$  is obtained by applying  [[Signal_Representation/Fourier_Transform_and_its_Inverse#The_first_Fourier_integral\|the first Fourier Integral]]:
	:$$X(f) = A \cdot T \cdot {\rm si}^2(\pi f T)\hspace{0.5cm}\text{with}\hspace{0.5cm}{\rm si}(x)=\sin(x)/x\hspace{0.05cm}.$$		:$$X(f) = A \cdot T \cdot {\rm si}^2(\pi f T)\hspace{0.5cm}\text{with}\hspace{0.5cm}{\rm si}(x)=\sin(x)/x\hspace{0.05cm}.$$

Guenter at 15:26, 16 May 2021

2021-05-16T15:26:07Z

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

2021-03-23T12:39:19Z

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

← Older revision		Revision as of 14:39, 23 March 2021
Line 147:		Line 147:

	__NOEDITSECTION__		__NOEDITSECTION__
	[[Category:~~Exercises for~~ Signal Representation\|^5.2 Discrete Fourier Transform^]]		[[Category:Signal Representation: Exercises\|^5.2 Discrete Fourier Transform^]]

Noah at 10:12, 21 March 2021

2021-03-21T10:12:44Z

← Older revision		Revision as of 12:12, 21 March 2021
Line 141:		Line 141:
	This gives for the corresponding values of the continuous spectral function:		This gives for the corresponding values of the continuous spectral function:
	:$$X(f=-f_{\rm A}) = X(f=+f_{\rm A}) =\frac{D(1)}{f_{\rm A}}= 1.708 \cdot 10^{-3}\,{\rm V/Hz}\hspace{0.05cm}.$$		:$$X(f=-f_{\rm A}) = X(f=+f_{\rm A}) =\frac{D(1)}{f_{\rm A}}= 1.708 \cdot 10^{-3}\,{\rm V/Hz}\hspace{0.05cm}.$$
	*Because of the implicit periodic continuation by the DFT, the value calculated in this way does not exactly match the actual value  $(4 \cdot A \cdot T/\pi^2 = 1.621 · 10^{-3}\text{ V/Hz})$ ~~nicht exakt überein~~.		*Because of the implicit periodic continuation by the DFT, the value calculated in this way does not exactly match the actual value  $(4 \cdot A \cdot T/\pi^2 = 1.621 · 10^{-3}\text{ V/Hz})$.
	*The relative error is approx.  $5.3\%$.		*The relative error is approx.  $5.3\%$.
	{{ML-Fuß}}		{{ML-Fuß}}

Noah at 10:12, 21 March 2021

2021-03-21T10:12:24Z

← Older revision		Revision as of 12:12, 21 March 2021
Line 14:		Line 14:
	The signal parameters have the following values:		The signal parameters have the following values:

	* ~~Amplitude~~  $A = 4 \ \text{V}$,		* amplitude  $A = 4 \ \text{V}$,

	* equivalent pulse duration  $\Delta t = T = 1 \, \text{ms}$.		* equivalent pulse duration  $\Delta t = T = 1 \, \text{ms}$.
Line 51:		Line 51:

	<quiz display=simple>		<quiz display=simple>
	{~~Geben Sie die Zeitkoeffizienten an~~. ~~Wie groß sind~~  $d(0)$,  $d(3)$  ~~und~~  $d(6)$?		{Give the time coefficients. What are the values of  $d(0)$,  $d(3)$  and  $d(6)$?
	\|type="{}"}		\|type="{}"}
	$d(0)\ = \ $ { 4 3% }  $\text{V}$		$d(0)\ = \ $ { 4 3% }  $\text{V}$
Line 58:		Line 58:


	{~~Wie groß ist der Abstand~~  $T_{\rm A}$  ~~zweier Zeitabtastwerte~~?		{What is the distance  $T_{\rm A}$  between two time samples?
	\|type="{}"}		\|type="{}"}
	$T_{\rm A}\ = \ ${ 0.25 3% }  $\text{ms}$		$T_{\rm A}\ = \ ${ 0.25 3% }  $\text{ms}$


	{~~Wie groß ist der Abstand~~  $f_{\rm A}$  ~~zweier DFT–Frequenzabtastwerte~~?		{What is the distance  $f_{\rm A}$  between two DFT frequency samples?
	\|type="{}"}		\|type="{}"}
	$f_{\rm A}\ = \ ${ 0.5 3% }  $\text{kHz}$		$f_{\rm A}\ = \ ${ 0.5 3% }  $\text{kHz}$


	{~~Berechnen Sie den Koeffizienten~~  $D(0)$  ~~und interpretieren Sie das Ergebnis~~.		{Calculate the coefficient  $D(0)$  and interpret the result.
	\|type="{}"}		\|type="{}"}
	$D(0)\ = \ $ { 2 3% }  $\text{V}$		$D(0)\ = \ $ { 2 3% }  $\text{V}$


	{~~Berechnen Sie den Koeffizienten~~  $D(2)$  ~~und interpretieren Sie das Ergebnis~~, ~~auch im Hinblick auf die Koeffizienten~~  $D(4)$  ~~und~~  $D(6)$.		{Calculate the coefficient  $D(2)$  and interpret the result, also in terms of the coefficients  $D(4)$  and  $D(6)$.
	\|type="{}"}		\|type="{}"}
	$D(2)\ = \ ${ 0. }  $\text{V}$		$D(2)\ = \ ${ 0. }  $\text{V}$


	{~~Berechnen und interpretieren Sie den DFT–Koeffizienten~~  $D(7)$.		{Calculate and interpret the DFT coefficient  $D(7)$.
	\|type="{}"}		\|type="{}"}
	$D(7)\ = \ $ { 0.854 3% }  $\text{V}$		$D(7)\ = \ $ { 0.854 3% }  $\text{V}$
Line 88:		Line 88:
	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  ~~Aus der Grafik ergeben sich mit~~  $A = 4 \ {\rm V}$  ~~folgende Werte~~:		'''(1)'''  From the graph the following values result with  $A = 4 \ {\rm V}$ :
	:$${d(0) = 4\,{\rm V}, \hspace{0.1cm}d(1) = d(7) = 3\,{\rm V}, \hspace{0.1cm}		:$${d(0) = 4\,{\rm V}, \hspace{0.1cm}d(1) = d(7) = 3\,{\rm V}, \hspace{0.1cm}
	\hspace{0.1cm}d(2) = d(6) = 2\,{\rm V}, \hspace{0.1cm}		\hspace{0.1cm}d(2) = d(6) = 2\,{\rm V}, \hspace{0.1cm}
Line 98:		Line 98:


	'''(2)'''  ~~Entsprechend der Grafik gilt~~  $T_{\rm A} = T/4$.		'''(2)'''  According to the diagram  $T_{\rm A} = T/4$.
	*~~Mit~~  $T = 1 \ \text{ms}$  ~~erhält man somit~~  $\underline{T_{\rm A} = 0.25 \ \text{ms}}$.		*With  $T = 1 \ \text{ms}$  one thus obtains  $\underline{T_{\rm A} = 0.25 \ \text{ms}}$.



	'''(3)'''  ~~Für die Abstände der Abtastwerte im Zeit– und Frequenzbereich gilt~~:		'''(3)'''  For the distances of the samples in the time and frequency domain applies:
	:$$N \cdot f_{\rm A} \cdot T_{\rm A} = 1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}f_{\rm		:$$N \cdot f_{\rm A} \cdot T_{\rm A} = 1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}f_{\rm
	A}= \frac{1}{ 8 \cdot 0.25\, {\rm ms}}\hspace{0.15 cm}\underline{= 0.5\, {\rm kHz}}\hspace{0.05cm}.$$		A}= \frac{1}{ 8 \cdot 0.25\, {\rm ms}}\hspace{0.15 cm}\underline{= 0.5\, {\rm kHz}}\hspace{0.05cm}.$$


	'''(4)'''  ~~Mit~~  $N = 8$  ~~und~~  $\mu = 0$  ~~folgt aus der DFT–Gleichung~~:		'''(4)'''  With  $N = 8$  and  $\mu = 0$ , it follows from the DFT equation:
	:$$D(0) = \frac{1}{8}\cdot \sum_{\nu = 0 }^{7}		:$$D(0) = \frac{1}{8}\cdot \sum_{\nu = 0 }^{7}
	d(\nu) = \frac{1\,{\rm V}}{8}\cdot (4+3+2+1+0+1+2+3)\hspace{0.15 cm}\underline{= 2 \,{\rm V}}\hspace{0.05cm}.$$		d(\nu) = \frac{1\,{\rm V}}{8}\cdot (4+3+2+1+0+1+2+3)\hspace{0.15 cm}\underline{= 2 \,{\rm V}}\hspace{0.05cm}.$$
	*~~Der DFT–Wert~~ $D(0)$ ~~beschreibt den Spektralwert bei~~  $f = 0$, ~~wobei folgender Zusammenhang gilt~~:		*The DFT value $D(0)$ describes the spectral value at  $f = 0$, where the following relation holds:
	:$$X(f=0) = \frac{D(0)}{f_{\rm A}}= \frac{ 2\,{\rm V}}{0.5\,{\rm kHz}}= 4 \cdot 10^{-3}\,{\rm V/Hz}\hspace{0.05cm}.$$		:$$X(f=0) = \frac{D(0)}{f_{\rm A}}= \frac{ 2\,{\rm V}}{0.5\,{\rm kHz}}= 4 \cdot 10^{-3}\,{\rm V/Hz}\hspace{0.05cm}.$$
	*~~Dieser Wert stimmt mit dem theoretischen Wert~~   $(A \cdot T)$  ~~überein~~.		*This value agrees with the theoretical value   $(A \cdot T)$ .



	'''(5)'''  ~~Mit~~  $N = 8$  ~~und~~  $\mu = 2$  ~~erhält man~~:		'''(5)'''  With  $N = 8$  and  $\mu = 2$  we obtain:
	:$$D(2) = \frac{1}{8}\cdot \sum_{\nu = 0 }^{7}		:$$D(2) = \frac{1}{8}\cdot \sum_{\nu = 0 }^{7}
	d(\nu)\cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /2) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} =		d(\nu)\cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /2) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} =
Line 123:		Line 123:
	d(\nu)\cdot (-{\rm j})^{\nu} \hspace{0.3cm}		d(\nu)\cdot (-{\rm j})^{\nu} \hspace{0.3cm}
	\Rightarrow \hspace{0.3cm} = \frac{1\,{\rm V}}{8}\cdot (4-3\cdot{\rm j}-2+{\rm j}-{\rm j}-2+3\cdot{\rm j})\hspace{0.15 cm}\underline{= 0}\hspace{0.05cm}.$$		\Rightarrow \hspace{0.3cm} = \frac{1\,{\rm V}}{8}\cdot (4-3\cdot{\rm j}-2+{\rm j}-{\rm j}-2+3\cdot{\rm j})\hspace{0.15 cm}\underline{= 0}\hspace{0.05cm}.$$
	~~Dieses Ergebnis hätte man auch ohne Rechnung vorhersagen können~~:		This result could have been predicted without calculation:
	*~~Die~~ DFT~~-Koeffizienten~~  $D(\mu)$  ~~sind gleichzeitig die Fourierkoeffizienten der im Abstand~~  $T_{\rm P} = 2T$  ~~periodifizierten Funktion~~  $x_{\rm Per}(t)$. ~~Diese ist~~ in ~~der Grafik auf der Angabenseite gestrichelt eingezeichnet~~.		*The DFT coefficients  $D(\mu)$  are at the same time the Fourier coefficients of the function  $T_{\rm P} = 2T$  periodised at the distance  $x_{\rm Per}(t)$. This is shown as a dashed line in the graph on the information page.
	*~~Aufgrund von Symmetrieeigenschaften sind aber alle geradzahligen Fourierkoeffizienten der Funktion~~  $x_{\rm Per}(t)$  ~~gleich Null~~:   ⇒   $D(4)\hspace{0.15cm}\underline{=0},$   $D(6)\hspace{0.15cm}\underline{=0}$.		*Due to symmetry properties, however, all even Fourier coefficients of the function  $x_{\rm Per}(t)$  are equal to zero:   ⇒   $D(4)\hspace{0.15cm}\underline{=0},$   $D(6)\hspace{0.15cm}\underline{=0}$.



	'''(6)'''  ~~Der Koeffizient~~  $D(7)$  ~~beschreibt die periodifizierte Spektralfunktion bei der Frequenz~~  $f = 7 \cdot f_{\rm A}$. ~~Aufgrund der Periodizität und von Symmetrieeigenschaft gilt~~:		'''(6)'''  The coefficient  $D(7)$  describes the periodised spectral function at the frequency  $f = 7 \cdot f_{\rm A}$. Due to periodicity and symmetry property holds:
	:$$D(7) = D(-1) = D^{\star}(1) \hspace{0.05cm}.$$		:$$D(7) = D(-1) = D^{\star}(1) \hspace{0.05cm}.$$
	~~Vorzugsweise berechnen wir diesen DFT–Koeffizienten~~:		Preferably, we calculate this DFT coefficient:
	:$$D(1) = \frac{1}{8}\cdot \sum_{\nu = 0 }^{7}		:$$D(1) = \frac{1}{8}\cdot \sum_{\nu = 0 }^{7}
	d(\nu)\cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /4) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} =		d(\nu)\cdot {\rm e}^{-{\rm j} \hspace{0.05cm}\cdot \hspace{0.05cm} (\pi /4) \hspace{0.05cm}\cdot \hspace{0.05cm} \nu} =
Line 137:		Line 137:
	\frac{-1 + {\rm j}}{\sqrt{2}}-{\rm j}+ 2\cdot {\rm j}+3\cdot \frac{1 - {\rm j}}{\sqrt{2}}\right)$$		\frac{-1 + {\rm j}}{\sqrt{2}}-{\rm j}+ 2\cdot {\rm j}+3\cdot \frac{1 - {\rm j}}{\sqrt{2}}\right)$$
	:$$\Rightarrow \; \; D(1) = \frac{2 + \sqrt{2}}{4} \approx 0.854{\rm V}\hspace{0.05cm}.$$		:$$\Rightarrow \; \; D(1) = \frac{2 + \sqrt{2}}{4} \approx 0.854{\rm V}\hspace{0.05cm}.$$
	Da  $D(1)$  ~~rein reell ist~~, ~~gilt~~  $D(7) = D(1) \; \underline{= 0.854 \ {\rm V}}$.		Since  $D(1)$  is purely real,  $D(7) = D(1) \; \underline{= 0.854 \ {\rm V}}$.

	~~Daraus ergeben sich für die zugehörigen Werte der kontinuierlichen Spektralfunktion~~:		This gives for the corresponding values of the continuous spectral function:
	:$$X(f=-f_{\rm A}) = X(f=+f_{\rm A}) =\frac{D(1)}{f_{\rm A}}= 1.708 \cdot 10^{-3}\,{\rm V/Hz}\hspace{0.05cm}.$$		:$$X(f=-f_{\rm A}) = X(f=+f_{\rm A}) =\frac{D(1)}{f_{\rm A}}= 1.708 \cdot 10^{-3}\,{\rm V/Hz}\hspace{0.05cm}.$$
	*~~Wegen der impliziten periodischen Fortsetzung durch die~~ DFT ~~stimmt der so berechnete Wert mit dem tatsächlichen Wert~~  $(4 \cdot A \cdot T/\pi^2 = 1.621 · 10^{-3}\text{ V/Hz})$ nicht exakt überein.		*Because of the implicit periodic continuation by the DFT, the value calculated in this way does not exactly match the actual value  $(4 \cdot A \cdot T/\pi^2 = 1.621 · 10^{-3}\text{ V/Hz})$ nicht exakt überein.
	*~~Der~~ relative ~~Fehler beträgt ca~~.  $5.3\%$.		*The relative error is approx.  $5.3\%$.
	{{ML-Fuß}}		{{ML-Fuß}}

← Older revision		Revision as of 17:26, 16 May 2021
Line 4:		Line 4:

	[[File:P_ID1140__Sig_Z_5_2.png\|right\|frame\|Discretisation of a triangular pulse]]		[[File:P_ID1140__Sig_Z_5_2.png\|right\|frame\|Discretisation of a triangular pulse]]
	Consider the sketched triangular ~~momentum~~		Consider the sketched triangular pulse
	:$$x(t) = \left\{ \begin{array}{l} A \cdot \left( 1 - {\|t\|}/{T} \right ) \\		:$$x(t) = \left\{ \begin{array}{l} A \cdot \left( 1 - {\|t\|}/{T} \right ) \\
	\hspace{0.25cm} 0 \\ \end{array} \right.\quad		\hspace{0.25cm} 0 \\ \end{array} \right.\quad
	\begin{array}{*{10}c} {\rm{~~f\ddot{u}r~~}}		\begin{array}{*{10}c} {\rm{for}}
	\\ {\rm{~~f\ddot{u}r~~}} \\ \end{array}\begin{array}{*{10}c}		\\ {\rm{for}} \\ \end{array}\begin{array}{*{10}c}
	{\left\| \hspace{0.005cm} t\hspace{0.05cm} \right\| \le T,} \\		{\left\| \hspace{0.005cm} t\hspace{0.05cm} \right\| \le T,} \\
	{\left\|\hspace{0.005cm} t \hspace{0.05cm} \right\| > T.} \\		{\left\|\hspace{0.005cm} t \hspace{0.05cm} \right\| > T.} \\
Line 20:		Line 20:

	The spectrum  $X(f)$  is obtained by applying  [[Signal_Representation/Fourier_Transform_and_Its_Inverse#The_First_Fourier_Integral\|the first Fourier Integral]]:		The spectrum  $X(f)$  is obtained by applying  [[Signal_Representation/Fourier_Transform_and_Its_Inverse#The_First_Fourier_Integral\|the first Fourier Integral]]:
	:$$X(f) = A \cdot T \cdot {\rm si}^2(\pi f T)\hspace{0.05cm}.$$		:$$X(f) = A \cdot T \cdot {\rm si}^2(\pi f T)\hspace{0.5cm}\text{with}\hspace{0.5cm}{\rm si}(x)=\sin(x)/x\hspace{0.05cm}.$$

	The spectral function is now to be approximated by a ''Discrete Fourier Transform''  (DFT) with  $N = 8$ , where the   $N$  coefficients for the time domain   ⇒   $d(0)$, ... , $d(7)$  can be taken from the graph.		The spectral function is now to be approximated by a  Discrete Fourier Transform  $\rm (DFT)$  with  $N = 8$ , where the   $N$  coefficients for the time domain   ⇒   $d(0)$, ... , $d(7)$  can be taken from the graph.

	The corresponding spectral coefficients  $D(0)$, ... ,  $D(7)$  are to be determined~~, whereby for~~ the indices  $\mu = 0$, ... , $N–1$  applies:		The corresponding spectral coefficients  $D(0)$, ... ,  $D(7)$  are to be determined.  For the indices  $\mu = 0$, ... , $N–1$  applies:
	:$$D(\mu) = \frac{1}{N} \cdot \sum_{\nu = 0 }^{N-1}		:$$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}.$$		d(\nu)\cdot {w}^{\hspace{0.05cm}\nu \hspace{0.05cm} \cdot \hspace{0.05cm}\mu} \hspace{0.05cm}.$$

	If we denote the distance between two samples in the time domain by  $T_{\rm A}$  and the corresponding frequency distance between two lines by ~~ ~~ $f_{\rm A}$, ~~the~~ following relationship applies:		If we denote the distance between two samples in the time domain by  $T_{\rm A}$  and the corresponding frequency distance between two lines by $f_{\rm A}$, following relationship applies:
	:$$N \cdot f_{\rm A} \cdot T_{\rm A} = 1 \hspace{0.05cm}.$$		:$$N \cdot f_{\rm A} \cdot T_{\rm A} = 1 \hspace{0.05cm}.$$

Line 40:		Line 40:
	*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)]].

	*~~You can check your solutions~~ with 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]].