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

Guenter at 15:59, 16 May 2021

2021-05-16T15:59:51Z

Guenter at 15:26, 16 May 2021

2021-05-16T15:26:46Z

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"

Noah at 10:12, 21 March 2021

2021-03-21T10:12:44Z

Noah at 10:12, 21 March 2021

2021-03-21T10:12:24Z

Noah at 19:14, 20 March 2021

2021-03-20T19:14:22Z

Oezdemir: Oezdemir moved page Aufgaben:Aufgabe 5.2Z: DFT eines Dreieckimpulses to Aufgaben:Exercise 5.2Z: DFT of a Triangular Pulse

2020-11-29T18:51:35Z

Oezdemir moved page Aufgabe 5.2Z: DFT eines Dreieckimpulses to Exercise 5.2Z: DFT of a Triangular Pulse

Oezdemir at 18:50, 29 November 2020

2020-11-29T18:50:00Z

Javier: Text replacement - "Signaldarstellung/Diskrete Fouriertransformation (DFT)" to "Signal_Representation/Discrete_Fourier_Transform_(DFT)"

2020-09-01T13:29:44Z

Text replacement - "Signaldarstellung/Diskrete Fouriertransformation (DFT)" to "Signal_Representation/Discrete_Fourier_Transform_(DFT)"

@@ Line 99: / Line 99: @@
 '''(2)'''&nbsp; According to the diagram&nbsp; $T_{\rm A} = T/4$.
-*With&nbsp; $T = 1 \ \text{ms}$&nbsp; one thus obtains&nbsp; $\underline{T_{\rm A} = 0.25 \ \text{ms}}$.
+*With&nbsp; $T = 1 \ \text{ms}$&nbsp; one obtains&nbsp; $\underline{T_{\rm A} = 0.25 \ \text{ms}}$.
@@ 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}.$$
 This result could have been predicted without calculation:
-*The DFT coefficients&nbsp; $D(\mu)$&nbsp; are at the same time the Fourier coefficients of the function&nbsp; $T_{\rm P} = 2T$&nbsp; periodised at the distance&nbsp; $x_{\rm Per}(t)$. This is shown as a dashed line in the graph on the information page.
+*The DFT coefficients&nbsp; $D(\mu)$&nbsp; are at the same time the Fourier coefficients of the function&nbsp; $x_{\rm Per}(t)$&nbsp; periodised at the distance&nbsp; $T_{\rm P} = 2T$. <br>This is shown as a dashed line in the graph on the information page.
-*Due to symmetry properties, however, all even Fourier coefficients of the function&nbsp; $x_{\rm Per}(t)$&nbsp; are equal to zero: &nbsp; &rArr; &nbsp; $D(4)\hspace{0.15cm}\underline{=0},$ &nbsp;  $D(6)\hspace{0.15cm}\underline{=0}$.
+*Due to symmetry properties, however, all even Fourier coefficients of the function&nbsp; $x_{\rm Per}(t)$&nbsp; are equal to zero &nbsp; &rArr; &nbsp; $D(4)\hspace{0.15cm}\underline{=0},$ &nbsp;  $D(6)\hspace{0.15cm}\underline{=0}$.
-'''(6)'''&nbsp; The coefficient&nbsp; $D(7)$&nbsp; describes the periodised spectral function at the frequency&nbsp; $f = 7 \cdot f_{\rm A}$. Due to periodicity and symmetry property holds:
+'''(6)'''&nbsp; The coefficient&nbsp; $D(7)$&nbsp; describes the periodised spectral function at the frequency&nbsp; $f = 7 \cdot f_{\rm A}$.&nbsp; Due to periodicity and symmetry property holds:
 :$$D(7) = D(-1) = D^{\star}(1) \hspace{0.05cm}.$$
 Preferably, we calculate this DFT coefficient:

← Older revision		Revision as of 15:26, 16 May 2021
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−	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]]:
	:$$X(f) = A \cdot T \cdot {\rm si}^2(\pi f T)\hspace{0.5cm}\text{with}\hspace{0.5cm}{\rm si}(x)=\sin(x)/x\hspace{0.05cm}.$$		:$$X(f) = A \cdot T \cdot {\rm si}^2(\pi f T)\hspace{0.5cm}\text{with}\hspace{0.5cm}{\rm si}(x)=\sin(x)/x\hspace{0.05cm}.$$

@@ Line 4: / Line 4: @@
 [[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
 :$$x(t) = \left\{ \begin{array}{l} A \cdot \left( 1 - {|t|}/{T} \right ) \\
 \hspace{0.25cm} 0 \\  \end{array} \right.\quad
-\begin{array}{*{10}c}   {\rm{f\ddot{u}r}}
+\begin{array}{*{10}c}   {\rm{for}}
-\\   {\rm{f\ddot{u}r}}  \\ \end{array}\begin{array}{*{10}c}
+\\   {\rm{for}}  \\ \end{array}\begin{array}{*{10}c}
 {\left| \hspace{0.005cm} t\hspace{0.05cm} \right| \le T,}  \\
 {\left|\hspace{0.005cm} t \hspace{0.05cm} \right| > T.}  \\
@@ Line 20: / Line 20: @@
 The spectrum&nbsp; $X(f)$&nbsp; is obtained by applying&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse#The_First_Fourier_Integral|the first Fourier Integral]]:
-:$$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}.$$
-The spectral function is now to be approximated by a ''Discrete Fourier Transform''&nbsp; (DFT) with&nbsp; $N = 8$&nbsp;, where the &nbsp; $N$&nbsp; coefficients for the time domain &nbsp; &rArr; &nbsp; $d(0)$, ... , $d(7)$&nbsp; can be taken from the graph.
+The spectral function is now to be approximated by a&nbsp;  Discrete Fourier Transform&nbsp; $\rm (DFT)$&nbsp;  with&nbsp; $N = 8$&nbsp;, where the &nbsp; $N$&nbsp; coefficients for the time domain &nbsp; &rArr; &nbsp; $d(0)$, ... , $d(7)$&nbsp; can be taken from the graph.
-The corresponding spectral coefficients&nbsp; $D(0)$, ... ,&nbsp; $D(7)$&nbsp; are to be determined, whereby for the indices&nbsp; $\mu = 0$, ... , $N–1$&nbsp; applies:
+The corresponding spectral coefficients&nbsp; $D(0)$, ... ,&nbsp; $D(7)$&nbsp; are to be determined.&nbsp;  For the indices&nbsp; $\mu = 0$, ... , $N–1$&nbsp; applies:
 :$$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}.$$
-If we denote the distance between two samples in the time domain by&nbsp; $T_{\rm A}$&nbsp; and the corresponding frequency distance between two lines by &nbsp; $f_{\rm A}$, the following relationship applies:
+If we denote the distance between two samples in the time domain by&nbsp; $T_{\rm A}$&nbsp; and the corresponding frequency distance between two lines by $f_{\rm A}$, following relationship applies:
 :$$N \cdot f_{\rm A} \cdot T_{\rm A} = 1 \hspace{0.05cm}.$$
@@ Line 40: / Line 40: @@
 *This task belongs to the chapter&nbsp; [[Signal_Representation/Discrete_Fourier_Transform_(DFT)|Discrete Fourier Transformation (DFT)]].
-*You can check your solutions with the interactive applet&nbsp; [[Applets:Diskrete_Fouriertransformation_und_Inverse|Discrete Fourier Transform and Inverse]]&nbsp;.
+*The topic dealt with here is also dealt with in the interactive applet&nbsp; [[Applets:Discrete_Fouriertransform_and_Inverse|Discrete Fourier Transform and Inverse]].

← Older revision		Revision as of 12:39, 23 March 2021
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	__NOEDITSECTION__		__NOEDITSECTION__
−	[[Category:~~Exercises for~~ Signal Representation\|^5.2 Discrete Fourier Transform^]]	+	[[Category:Signal Representation: Exercises\|^5.2 Discrete Fourier Transform^]]

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

@@ Line 14: / Line 14: @@
 The signal parameters have the following values:
-* Amplitude&nbsp; $A = 4 \ \text{V}$,
+* amplitude&nbsp; $A = 4 \ \text{V}$,
 * equivalent pulse duration&nbsp; $\Delta t = T = 1 \, \text{ms}$.
@@ Line 51: / Line 51: @@
 <quiz display=simple>
-{Geben Sie die Zeitkoeffizienten an. Wie groß sind&nbsp; $d(0)$,&nbsp; $d(3)$&nbsp; und&nbsp; $d(6)$?
+{Give the time coefficients. What are the values of&nbsp; $d(0)$,&nbsp; $d(3)$&nbsp; and&nbsp; $d(6)$?
 |type="{}"}
 $d(0)\ = \ $ { 4 3% } &nbsp;$\text{V}$
@@ Line 58: / Line 58: @@
-{Wie groß ist der Abstand&nbsp; $T_{\rm A}$&nbsp; zweier Zeitabtastwerte?
+{What is the distance&nbsp; $T_{\rm A}$&nbsp; between two time samples?
 |type="{}"}
 $T_{\rm A}\ = \ ${ 0.25 3% } &nbsp;$\text{ms}$
-{Wie groß ist der Abstand&nbsp; $f_{\rm A}$&nbsp; zweier DFT–Frequenzabtastwerte?
+{What is the distance&nbsp; $f_{\rm A}$&nbsp; between two DFT frequency samples?
 |type="{}"}
 $f_{\rm A}\ = \ ${ 0.5 3% } &nbsp;$\text{kHz}$
-{Berechnen Sie den Koeffizienten&nbsp; $D(0)$&nbsp; und interpretieren Sie das Ergebnis.
+{Calculate the coefficient&nbsp; $D(0)$&nbsp; and interpret the result.
 |type="{}"}
 $D(0)\ = \ $ { 2 3% } &nbsp;$\text{V}$
-{Berechnen Sie den Koeffizienten&nbsp; $D(2)$&nbsp; und interpretieren Sie das Ergebnis, auch im Hinblick auf die Koeffizienten&nbsp; $D(4)$&nbsp; und&nbsp; $D(6)$.
+{Calculate the coefficient&nbsp; $D(2)$&nbsp; and interpret the result, also in terms of the coefficients&nbsp; $D(4)$&nbsp; and&nbsp; $D(6)$.
 |type="{}"}
 $D(2)\ = \ ${ 0. } &nbsp;$\text{V}$
-{Berechnen und interpretieren Sie den DFT–Koeffizienten&nbsp; $D(7)$.
+{Calculate and interpret the DFT coefficient&nbsp; $D(7)$.
 |type="{}"}
 $D(7)\ = \ $ { 0.854 3% } &nbsp;$\text{V}$
@@ Line 88: / Line 88: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp;  Aus der Grafik ergeben sich mit&nbsp; $A = 4 \ {\rm V}$&nbsp; folgende Werte:
+'''(1)'''&nbsp;  From the graph the following values result with&nbsp; $A = 4 \ {\rm V}$&nbsp;:
 :$${d(0) = 4\,{\rm V}, \hspace{0.1cm}d(1) = d(7) = 3\,{\rm V}, \hspace{0.1cm}
   \hspace{0.1cm}d(2) = d(6) = 2\,{\rm V}, \hspace{0.1cm}
@@ Line 98: / Line 98: @@
-'''(2)'''&nbsp; Entsprechend der Grafik gilt&nbsp; $T_{\rm A} = T/4$.
+'''(2)'''&nbsp; According to the diagram&nbsp; $T_{\rm A} = T/4$.
-*Mit&nbsp; $T = 1 \ \text{ms}$&nbsp; erhält man somit&nbsp; $\underline{T_{\rm A} = 0.25 \ \text{ms}}$.
+*With&nbsp; $T = 1 \ \text{ms}$&nbsp; one thus obtains&nbsp; $\underline{T_{\rm A} = 0.25 \ \text{ms}}$.
-'''(3)'''&nbsp; Für die Abstände der Abtastwerte im Zeit– und Frequenzbereich gilt:
+'''(3)'''&nbsp; For the distances of the samples in the time and frequency domain applies:
 :$$N \cdot f_{\rm A} \cdot T_{\rm A} = 1 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}f_{\rm
   A}= \frac{1}{ 8 \cdot 0.25\, {\rm ms}}\hspace{0.15 cm}\underline{= 0.5\, {\rm kHz}}\hspace{0.05cm}.$$
-'''(4)'''&nbsp; Mit&nbsp; $N = 8$&nbsp; und&nbsp; $\mu = 0$&nbsp; folgt aus der DFT–Gleichung:
+'''(4)'''&nbsp; With&nbsp; $N = 8$&nbsp; and&nbsp; $\mu = 0$&nbsp;, it follows from the DFT equation:
 :$$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}.$$
-*Der DFT–Wert $D(0)$ beschreibt den Spektralwert bei&nbsp; $f = 0$, wobei folgender Zusammenhang gilt:
+*The DFT value $D(0)$ describes the spectral value at&nbsp; $f = 0$, where the following relation holds:
 :$$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}.$$
-*Dieser Wert stimmt mit dem theoretischen Wert &nbsp;   $(A \cdot T)$&nbsp; überein.
+*This value agrees with the theoretical value &nbsp;   $(A \cdot T)$&nbsp;.
-'''(5)'''&nbsp; Mit&nbsp; $N = 8$&nbsp; und&nbsp; $\mu = 2$&nbsp; erhält man:
+'''(5)'''&nbsp; With&nbsp; $N = 8$&nbsp; and&nbsp; $\mu = 2$&nbsp; we obtain:
 :$$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} =
@@ Line 123: / Line 123: @@
   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}.$$
-Dieses Ergebnis hätte man auch ohne Rechnung vorhersagen können:
+This result could have been predicted without calculation:
-*Die DFT-Koeffizienten&nbsp; $D(\mu)$&nbsp; sind gleichzeitig die Fourierkoeffizienten der im Abstand&nbsp; $T_{\rm P} = 2T$&nbsp; periodifizierten Funktion&nbsp; $x_{\rm Per}(t)$. Diese ist in der Grafik auf der Angabenseite gestrichelt eingezeichnet.
+*The DFT coefficients&nbsp; $D(\mu)$&nbsp; are at the same time the Fourier coefficients of the function&nbsp; $T_{\rm P} = 2T$&nbsp; periodised at the distance&nbsp; $x_{\rm Per}(t)$. This is shown as a dashed line in the graph on the information page.
-*Aufgrund von Symmetrieeigenschaften sind aber alle geradzahligen Fourierkoeffizienten der Funktion&nbsp; $x_{\rm Per}(t)$&nbsp; gleich Null: &nbsp; &rArr; &nbsp; $D(4)\hspace{0.15cm}\underline{=0},$ &nbsp;  $D(6)\hspace{0.15cm}\underline{=0}$.
+*Due to symmetry properties, however, all even Fourier coefficients of the function&nbsp; $x_{\rm Per}(t)$&nbsp; are equal to zero: &nbsp; &rArr; &nbsp; $D(4)\hspace{0.15cm}\underline{=0},$ &nbsp;  $D(6)\hspace{0.15cm}\underline{=0}$.
-'''(6)'''&nbsp; Der Koeffizient&nbsp; $D(7)$&nbsp; beschreibt die periodifizierte Spektralfunktion bei der Frequenz&nbsp; $f = 7 \cdot f_{\rm A}$. Aufgrund der Periodizität und von Symmetrieeigenschaft gilt:
+'''(6)'''&nbsp; The coefficient&nbsp; $D(7)$&nbsp; describes the periodised spectral function at the frequency&nbsp; $f = 7 \cdot f_{\rm A}$. Due to periodicity and symmetry property holds:
 :$$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(\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)$$
 :$$\Rightarrow \; \; D(1)  =  \frac{2 + \sqrt{2}}{4} \approx 0.854{\rm V}\hspace{0.05cm}.$$
-Da&nbsp; $D(1)$&nbsp; rein reell ist, gilt&nbsp; $D(7) = D(1) \; \underline{= 0.854 \ {\rm V}}$.
+Since&nbsp; $D(1)$&nbsp; is purely real,&nbsp; $D(7) = D(1) \; \underline{= 0.854 \ {\rm V}}$.
-Daraus ergeben sich für die zugehörigen Werte der kontinuierlichen Spektralfunktion:
+This gives for the corresponding values of the continuous spectral function:
 :$$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}.$$
-*Wegen der impliziten periodischen Fortsetzung durch die DFT stimmt der so berechnete Wert mit dem tatsächlichen Wert&nbsp; $(4 \cdot A \cdot T/\pi^2 = 1.621 · 10^{-3}\text{ V/Hz})$ 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&nbsp; $(4 \cdot A \cdot T/\pi^2 = 1.621 · 10^{-3}\text{ V/Hz})$ nicht exakt überein.
-*Der relative Fehler beträgt ca.&nbsp; $5.3\%$.
+*The relative error is approx.&nbsp; $5.3\%$.
 {{ML-Fuß}}

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID1140__Sig_Z_5_2.png|right|frame|Diskretisierung eines Dreieckimpulses]]
+[[File:P_ID1140__Sig_Z_5_2.png|right|frame|Discretisation of a triangular pulse]]
-Betrachtet wird der skizzierte Dreieckimpuls
+Consider the sketched triangular momentum
 :$$x(t) = \left\{ \begin{array}{l} A \cdot \left( 1 - {|t|}/{T} \right ) \\
 \hspace{0.25cm} 0 \\  \end{array} \right.\quad
@@ Line 12: / Line 12: @@
 {\left|\hspace{0.005cm} t \hspace{0.05cm} \right| > T.}  \\
 \end{array}$$
-Die Signalparameter haben folgende Werte:
+The signal parameters have the following values:
 * Amplitude&nbsp; $A = 4 \ \text{V}$,
-* äquivalente Impulsdauer&nbsp; $\Delta t = T = 1 \, \text{ms}$.
+* equivalent pulse duration&nbsp; $\Delta t = T = 1 \, \text{ms}$.
-Das Spektrum&nbsp; $X(f)$&nbsp; erhält man durch Anwendung des&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse#Das_erste_Fourierintegral|ersten Fourierintegrals]]:
+The spectrum&nbsp; $X(f)$&nbsp; is obtained by applying&nbsp; [[Signal_Representation/Fourier_Transform_and_Its_Inverse#The_First_Fourier_Integral|the first Fourier Integral]]:
 :$$X(f) = A \cdot T \cdot {\rm si}^2(\pi f T)\hspace{0.05cm}.$$
-Die Spektralfunktion soll nun durch eine ''Diskrete Fouriertransformation''&nbsp; (DFT) mit&nbsp; $N = 8$&nbsp; angenähert werden, wobei die&nbsp; $N$&nbsp; Koeffizienten für den Zeitbereich &nbsp; &rArr; &nbsp; $d(0)$, ... , $d(7)$&nbsp; der Grafik entnommen werden können.
+The spectral function is now to be approximated by a ''Discrete Fourier Transform''&nbsp; (DFT) with&nbsp; $N = 8$&nbsp;, where the &nbsp; $N$&nbsp; coefficients for the time domain &nbsp; &rArr; &nbsp; $d(0)$, ... , $d(7)$&nbsp; can be taken from the graph.
-Die dazugehörigen Spektralkoeffizienten&nbsp; $D(0)$, ... ,&nbsp; $D(7)$&nbsp; sind zu ermitteln, wobei für die Indizes&nbsp; $\mu = 0$, ... , $N–1$&nbsp; gilt:
+The corresponding spectral coefficients&nbsp; $D(0)$, ... ,&nbsp; $D(7)$&nbsp; are to be determined, whereby for the indices&nbsp; $\mu = 0$, ... , $N–1$&nbsp; applies:
 :$$D(\mu) = \frac{1}{N} \cdot \sum_{\nu = 0 }^{N-1}
    d(\nu)\cdot  {w}^{\hspace{0.05cm}\nu \hspace{0.05cm} \cdot \hspace{0.05cm}\mu} \hspace{0.05cm}.$$
-Bezeichnet man den Abstand zweier Abtastwerte im Zeitbereich mit&nbsp; $T_{\rm A}$&nbsp; und den entsprechenden Frequenzabstand zweier Linien mit&nbsp; $f_{\rm A}$, so gilt folgender Zusammenhang:
+If we denote the distance between two samples in the time domain by&nbsp; $T_{\rm A}$&nbsp; and the corresponding frequency distance between two lines by &nbsp; $f_{\rm A}$, the following relationship applies:
 :$$N \cdot f_{\rm A} \cdot T_{\rm A} = 1 \hspace{0.05cm}.$$
@@ Line 37: / Line 37: @@
-''Hinweise:''
+''Hints:''
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Signal_Representation/Discrete_Fourier_Transform_(DFT)|Diskrete Fouriertransformation (DFT)]].
+*This task belongs to the chapter&nbsp; [[Signal_Representation/Discrete_Fourier_Transform_(DFT)|Discrete Fourier Transformation (DFT)]].
-*Ihre Lösungen können Sie mit dem interaktiven Applet&nbsp; [[Applets:Diskrete_Fouriertransformation_und_Inverse|Diskrete Fouriertransformation und Inverse]]&nbsp; überrprüfen.
+*You can check your solutions with the interactive applet&nbsp; [[Applets:Diskrete_Fouriertransformation_und_Inverse|Discrete Fourier Transform and Inverse]]&nbsp;.
@@ Line 48: / Line 48: @@
-===Fragebogen===
+===Question===
 <quiz display=simple>
@@ Line 86: / Line 86: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp;  Aus der Grafik ergeben sich mit&nbsp; $A = 4 \ {\rm V}$&nbsp; folgende Werte:

← Older revision	Revision as of 18:51, 29 November 2020
(No difference)

← Older revision		Revision as of 18:50, 29 November 2020
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	__NOEDITSECTION__		__NOEDITSECTION__
−	[[Category:Exercises for Signal Representation\|^5. ~~Zeit- und frequenzdiskrete Signaldarstellung~~^]]	+	[[Category:Exercises for Signal Representation\|^5.2 Discrete Fourier Transform^]]

← Older revision		Revision as of 13:29, 1 September 2020
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−	{{quiz-Header\|Buchseite=~~Signaldarstellung~~/~~Diskrete Fouriertransformation~~ (DFT)	+	{{quiz-Header\|Buchseite=Signal_Representation/Discrete_Fourier_Transform_(DFT)
	}}		}}